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Foundations of Physics
Skill and Practice Worksheets
Credits
CPO Science Curriculum Development Team
Author: Thomas Hsu, Ph.D
Vice Presidents: Thomas Narro and Lynda Pennell
Writers: Mary Beth Abel Hughes, Jim Sammons, Manos Chaniotakis, David Bliss, Stacy
Kissel, Mary Ann Erickson, Scott Eddleman, Lainie Ives, Irene Baker, Erik Benton, and
Patsy DeCoster
Graphic Artists: Bruce Holloway and Polly Crisman
Technical Consultants
Tracy Morrow and Julie Dalton
Foundations of Physics
Skill and Practice Worksheets
Copyright
2004 CPO Science
ISBN 1-58892-102-6
All rights reserved. No part of this work may be reproduced or transmitted in any form or by an
means, electronic or mechanical, including photocopying and recording, or by any information
store or retrieval system, without permission in writing. For permission and other rights under
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Printed and Bound in the United States of America
Skill and Practice Worksheets
Skill Sheets
Skill Sheets are numbered according to textbook section. If more than one skill sheet goes with a particular section, the section number is followed by a letter. The unit icon is shown on each Skill Sheet.
1.1
1.2
2.1
2.2
2.3
3.1
3.2
3.3A
3.3B
4.1
4.2
4.3
5.1
5.2
6.1
6.2
6.3
7.1A
7.1B
7.1C
7.2
7.3A
7.3B
8.1
8.3
9.1
10.1
10.2
10.3
11.2
12.1
12.2
13.1
14.1
Solving Equations
Galileo Galilei
International System of Units
Converting Units
Scientific Notation
Speed Problems
Making Line Graphs
Analyzing Graphs of Motion Without
Numbers
Analyzing Graphs of Motion With Numbers
Acceleration Problems
Acceleration and Speed-Time Graphs
Acceleration Due to Gravity
Isaac Newton
Newton's Second Law
Mass and Weight
Friction
Equilibrium
Adding Displacement Vectors
Vector Components
Pythagorean Theorem
Projectile Motion
Equilibrium in 2-D
Inclined Planes
Circular Motion
Universal Gravitation
Torque
Mechanical Advantage
Work
Potential and Kinetic Energy
Power
Momentum
Rate of Change of Momentum
Harmonic Motion
Waves
15.1
16.1
17.1A
17.1B
17.2
17.3
18.1
18.3
19.2
19.3A
19.3B
20.1
20.2
20.3
21.2
22.3
23.1
23.3
24.3
25.2
26.3
27.1
27.2
27.3
28.1A
28.1B
28.1C
28.2
29.2
29.3
30.1A
30.1B
30.3
Decibel Scale Problems
Light Intensity Problems16.1 Light Intensity
Problems
The Law of Reflection
Refraction
Ray Diagrams
Thin Lens Formula
The Speed of Light
Albert Einstein
Using an Electric Meter
Ohm’s Law
Ben Franklin
Parallel and Series Circuits
Network Circuits
Electrical Power
Coulomb’s Law
Magnetic Earth
Magnetic Fields and Forces
Michael Faraday
Binary Number Problems
Temperature Scales
Heat Transfer
Stress and Strain
Archimedes
Gas Laws
The Structure of the Atom
The Periodic Table
Lise Meitner
Niels Bohr
Dot Diagrams
Chemical Equations
Radioactivity
Marie and Pierre Curie
Ernest Rutherford
Skill Builders
Skill Builders are organized alphabetically and are meant to be used when students need to practice basic skills.
Internet Research Skills
Lab Report Format
Making Graphs
Safety Skills
Significant Digits
Student Evaluation Sheet
Name:
Skill Sheet 1.1
Solving Equations
Concepts in physics can often be expressed using formulas written as mathematical equations. To make
calculations using these formulas, the equations must be rearranged and solved for the variables that are used
in the relationship. As you solve equations, your goal is to isolate the unknown variable on one side of the
equation. As you proceed, you must use transformations that produce equations that are equivalent to the
original. Several inverse operations can be used.
1. Simple examples of solving equations
Sample
To solve it . . .
Solution
Addition:
n – 5 = 12
Add 5 to each side.
n = 17
Subtraction:
q + 4 = 16
Subtract 4 from each side.
q = 12
Multiplication:
t/
Multiply both sides by 2.
t = 36
Division:
4r = 24
Divide both sides by 4.
r=6
Multiplying by a reciprocal:
12 = 3/4 y
Multiply both sides by 4/3.
y = 16
2
= 18
2. Solving equations that require transformations
Solving some equations may require two or more transformations. To do this, follow the steps below:
•
Simplify one or both sides of the equation, if necessary.
•
Use inverse operations to isolate the variable.
Sample
To solve it . . .
1/
3a
+ 8 = 24
Write the original equation.
1/
3a
+ (8 – 8) = 24 – 8
Subtract 8 from both sides.
1/
3a
= 16
Simplify.
3(1/3)a = 3(16)
Multiply both sides by 3.
a = 48
1
Skill Sheet 1.1 Solving Equations
3. Solving equations that have variables on both sides
Some equations have variables on both sides of the equal sign. To solve these equations, collect the variable
terms on the side with the greater variable coefficient.
Sample
To solve it . . .
1/
Write the original equation.
4 (12x
+ 16) = 10 - 3(x – 2)
4(1/4)(12x + 16) = 4[10 - 3(x-2)]
Multiply both sides by 4.
12x + 16 = 40 – 12x + 24
Simplify.
12x + 16 – 16 = 64 – 16 – 12x
Subtract 16 from both sides.
12x + 12x = 48 – 12x + 12x
Add 12x to both sides.
24x = 48
Divide each side by 24.
x=2
4. Solving equations that contain radicals
Some equations contain radicals. Squaring both sides of an equation often simplifies the solution. When squaring
both sides of an equation, often an answer is introduced that is not a solution. These are referred to as extraneous
solutions. You should always substitute your answers into the original equation to check for extraneous solutions.
Sample
To solve it . . .
x–7 = 0
Write the original equation.
x–7+7 = 0+7
Add 7 to both sides.
2
( x) = 7
2
Square both sides.
x = 49
Sample
To solve it . . .
Write the original equation.
x + 13 = 0
x + 13 – 13 = 0 – 13 Subtract 13 from both sides.
2
( x ) = ( –13 )
x = 169
169 + 13 ≠ 0
2
Square both sides.
Simplify.
Checking the solutions in the original equation, you see that x = –1 is not a valid
solution.
2
Skill Sheet 1.1 Solving Equations
Sample
To solve it . . .
(x + 2) = x
2
( x + 2) = x
x+2 = x
Write the original equation.
2
2
2
Square both sides.
Simplify.
0 = x –x–2
Write in standard form.
0 = ( x – 2 )( x + 1 )
Factor.
x = 2 and x = -1
Checking by substitution, you see that the equation has no solution.
(2 + 2) = 2
( – 1 + 2 ) ≠ –1
Checking the solutions in the original equation, you see that x = 2 is a valid solution, but
x = -1 is not a valid solution.
5. Solving equations with quadratic equations
Some common relationships in physics are expressed using quadratic equations. The general
expression for a quadratic equation is: ax2 + bx + c = 0. To find the solutions for these
equations it is often necessary to use the quadratic formula. The formula is:
2
– b ± b – 4ac
x = -------------------------------------2a
Use of this equation may be time-consuming, but it gives an exact solution.
Example: A path of a certain waterfall as it tumbles over a vertical cliff can be given by the equation:
2
h = – 6.03 x + 901
where h = the height of the waterfall and x is the horizontal distance (in feet) from the base of the cliff to where
the water contacts the ground. How far from the base of the cliff does the water contact the ground?
Since h = 0 when the water strikes, the quadratic equation for this problem is written: 0 = -6.03x2 + 901
2
b ± b – 4ac , the solution is written as:
Using the quadratic formula, x = –-------------------------------------2a
– 0 ± 0 + ( – 4 ) ( – 6.03 )901x = ----------------------------------------------------------------2 ( – 6.03 )
x ≈ 12.3 or –12.3
The distance from the base of the cliff to where the water makes contact with the ground is 12.3 feet. The answer,
x = –12.3, is a meaningless value since the distance is in front of the cliff and therefore a positive number.
In addition to the techniques listed above, both quadratic equations and linear equations may be solved and
checked by graphing.
3
Skill Sheet 1.1 Solving Equations
6. Problems
For each of the following problems provide a solution by:
•
Writing an equation or equations that can be used to determine the unknown.
•
Solving the equation(s) for the unknown
1.
Jackie Joyner-Kersee won the gold medal in the Olympic heptathlon in both 1988 and 1992. Her score in
1992 was 7,044 points. This was 247 points higher than her score in 1988. What was her 1988 score?
2.
You decide to ride an elevator until it takes you to the ground floor without pushing any buttons. The
elevator takes you up 4 floors, down 6 floors, up 1 floor, down 8 floors, down 3 floors, up 1 floor, and then
down 6 floors to ground level. On what floor did you begin? (HINT: Note that the first floor is “1” not “0.”)
3.
A peregrine falcon can dive at speeds of 130 kilometers per hour. If at a height of 1.5 kilometers above the
ground the falcon spotted a grouse sitting in the top of a tree 12 meters tall, what is the least amount of time,
measured in seconds, it would take the falcon to reach the grouse?
4.
The distance a fire hose will project its water is given by the formula: d = n/2 + 26 where d is the maximum
distance in feet the hose will spray and n is the nozzle pressure measured in pounds per square inch. How
much pressure would be required to reach a flame 100 feet from the hose?
5.
Tim has fallen through the ice while ice skating. If his body temperature reaches 35° Celsius, he will
experience hypothermia, a possibly life-threatening condition. An EMT records his temperature as
94° Fahrenheit (°F = 9/5°C + 32). Will he experience hypothermia?
6.
Beth and Bob live 100 kilometers apart. They decide to ride their bicycles to meet one another. Beth rides
west toward Bob at 23 kilometers per hour. Bob starts at the same time and rides east toward Beth at
14 kilometers per hour. How long does it take for them to meet?
4
Skill Sheet 1.1 Solving Equations
7.
The period of a mass suspended on a vibrating spring can be approximated by the equation:
T = 2π m
---k
where T is the period, m is the mass of the object, and k is the spring constant of the spring. What is the
spring constant of a spring whose period of vibration is 0.1922 seconds when a mass of 0.010 kilogram is
suspended from the spring?
8.
The path of a diver diving from a 10-foot high diving board is given by the equation:
2
h = – 0.44x + 2.61x + 10
where h is the height of the diver above the water and x is the horizontal distance of the diver from the end of
the diving board. Calculate the horizontal distance from the end of the board at which the diver enters the
water. Assume that the height of the diver above the water is zero.
5
Name:
Skill Sheet 1.2
Galileo Galilei
Galileo Galilei was a mathematician, scientist, inventor, and astronomer. His observations led to significant
advances in our understanding of pendulum motion and free fall. He invented a thermometer, water pump,
military compass, and microscope. He refined a Dutch invention, the telescope, and used it to revolutionize
our understanding of the solar system.
Galileo Galilei was born in Pisa, Italy, in 1564. His father, a musician and wool trader,
hoped his son would find a more profitable career. He sent Galileo to a monastery
school at age 11 to prepare for medical school. After four years, Galileo had decided to
become a monk. His family had daughters who would need dowries in order to marry,
and his father planned on Galileo’s financial help. Galileo was hastily withdrawn from
the monastery school.
Two years later, he enrolled as a medical student at the University of Pisa, though his
real interest was in mathematics and natural philosophy. It soon became apparent that
Galileo did not intend to apply himself to medical studies and it was finally agreed that
he could study mathematics instead.
Galileo was insatiably curious. At age 20, he watched a lamp swinging from a cathedral ceiling. He used his
pulse as a makeshift stopwatch and discovered that the lamp’s long and short swings took the same amount of
time. He wrote about this in an early paper titled “On Motion.” Years later, he drew up plans for a new invention,
a pendulum clock based on his discovery.
Galileo began teaching at the University of Padua in 1592, and stayed for 18 years. Here he invented a simple
thermometer, a water pump, and a compass for accurately aiming cannonballs. He also performed experiments
with falling objects, using an inclined plane to slow the object’s motion so it could be more accurately timed.
Through these experiments, he realized that all objects fall at the same rate unless acted on by another force.
In 1609, Galileo heard that a Dutch eyeglass maker had invented an instrument that made things appear larger.
Soon he had crafted his own ten-powered telescope. The Senate in Venice was impressed with its potential
military uses and Galileo’s finances improved. In a year, he had refined his invention to a 30-powered telescope.
Galileo’s curiosity now turned toward the skies. Using his telescope, he discovered craters on the moon,
sunspots, Jupiter’s four largest moons, and the phases of Venus. His observations led him to conclude that Earth
could not possibly be the center of the universe, as had been commonly accepted since the time of the GrecoEgyptian astronomer Ptolemy in the second century. Instead, Galileo was convinced that fifteenth-century
mathematician Nicolaus Copernicus must have been right: The sun is in the center of the universe and the planets
revolve around it.
Galileo published his conclusion in February 1632. He chose to present his ideas in the form of a conversation
between two characters, the one representing Ptolemy’s view seeming foolish and bullheaded. This was
provocative because the Roman Catholic Church subscribed to Ptolemy’s ideas. That same year, Galileo was
brought before the Inquisition in Rome and convicted of heresy for promoting Copernicus’ ideas. He was
sentenced to house arrest, and lived until his death in 1642 watched over by Inquisition guards.
Questions
1. In your opinion, which of Galileo’s ideas or inventions had the biggest impact on history? Why?
2. Research one of Galileo’s inventions and draw a diagram showing how it worked.
1
Name:
Skill Sheet 2.1
International System of Units
In ancient times, as trade developed between cities and nations, units of measurement were developed to
measure the size of purchases and transactions. Greeks and Egyptians based their measurements of length on
the human foot. Usually, it was based on the king’s foot size. The volume of baskets was measured by how
much goatskin they could hold. Can you see how this could lead to disputes among merchants and their
customers? The International System of Units resolves this problem and others by providing a standard,
interrelated, and reproducible system of measurement.
1. A short history of measurement
The eighteenth century was a time of great beginnings in science. However, by century’s end, scientists found
that their system of measures was increasingly burdensome. Measurements such as the foot were not well
standardized and made it hard to communicate observations. A system that allowed scientists to reproduce and
verify each other’s data was needed.
The metric system was developed to fulfill this need. The system’s basic unit
for measuring length was called the meter, after the Greek word metron
meaning “measure.” But the metric system was not created in a single
development. For example, there were two ideas for the meter—one that used
the length of a pendulum and another that used a fraction of the distance
between Earth’s equator and the north pole. The north pole-to-equator line was
chosen in one of a series of decisions that shaped the metric system. Today, the
General Conference on Weights and Measures, or CGPM (Conférence
Générale des Poids et Mesures), has responsibility for these decisions.
The United States began to incorporate the use of the metric system the late
1800’s. However, most Americans still use the US Customary System
(inherited from the British Imperial System) of feet, inches, and pounds. Not only scientists but most countries—
even England—use what was named the International System of Units. In all these countries, your car’s speed is
measured in kilometers per hour, its gasoline in liters, the cheese you buy in grams, and the temperature in
degrees Celsius.
2. Today's International System of Units
The 11th General Conference on Weights and Measures in 1960 made some of the most important recent
revisions to the universal measurement system. A meter was defined as the distance light travels in a small
fraction of a second. A kilogram was reaffirmed as the mass of platinum-iridium kept in Paris. The International
System of Units was renamed Système International d’Unités and the new “modernized” metric system given the
official symbol SI.
Most students regard “metrics” as a set of memorized prefixes that increase a measurement by tens. This is
certainly true, but it overlooks one of the most important characteristics of SI units. The SI unit for volume, the
liter, is derived from the meter. A liter is that volume contained in a cube that measures 10 centimeters on each
side. The SI unit for weight, the kilogram, was originally the weight of one liter of pure water at standard
temperature and pressure. Although today a kilogram is defined by the prototype platinum-iridium kilogram kept
in Paris, both definitions are close enough to be interchangeable except in the most precise work. This
interrelated characteristic makes SI measurements very easy for scientists and non-scientists.
1
Skill Sheet 2.1 International System of Units
Consider the carpenter who is installing a hot tub. He needs to know the weight of the tub filled with water to
determine whether to strengthen the supports:
Using English units
Using SI units
The carpenter measures a tub interior and finds that
it is 6 feet long, 2 feet deep, and 3 feet wide. He
calculates the tub’s volume to be 36 cubic feet.
The carpenter measures the interior of a tub and
finds that it is 2 meters long, 60 centimeters (0.6 m)
deep, and 1 meter wide. He calculates the tub’s
volume to be 1.2 cubic meters.
There are 7.48 gallons per cubic foot. Therefore, he
multiplies the volume by 7.48 gallons/cubic foot
and finds that the tub holds 269.3 gallons of water
when filled.
He knows that a cubic meter is equivalent to 1,000
liters, so he shifts the decimal and the volume
becomes 1,200 liters.
He multiplies gallons by 8.36 pounds/gallon and
finds the water weight will be 2251.3 pounds.
He also knows that a liter of water weighs
1 kilogram, so in shifting the decimal he arrived at
the weight directly–1,200 kilograms.
He did all of the above in his head.
3. SI prefixes
Prefixes in the SI system indicate the multiplication factor to be used with the measurement unit. For example,
the prefix kilo multiplies the unit by 1,000. A kilometer is equal to 1,000 meters. A kilogram equals 1,000 grams.
Prefix
Symbol
Multiplication factor
pico–
p
0.000000000001
= 10-12
nano–
n
0.000000001
= 10-9
micro–
µ
0.000001
= 10-6
milli–
m
0.001
= 10-3
centi–
c
0.01
= 10-2
deci–
d
0.1
= 10-1
1
= 100
No prefix
deka–
da
10
= 101
hecto–
h
100
= 102
kilo–
k
1,000
= 103
mega–
M
1,000,000
= 106
giga–
G
1,000,000,000
= 109
tera–
T
1, 000, 000, 000, 000
= 1012
2
Skill Sheet 2.1 International System of Units
4. Practical units of SI measurement
In practice, many of the possible prefix and unit measurements are seldom used. In the table below, the most
commonly used SI units of length, volume, and weight are provided.
The most commonly used SI units
Prefix
Length
Volume
Weight
milli-
millimeter
milliliter
milligram
centi-
centimeter
meter
liter
gram
kilometer
(meter3)
kilogram
deci(unit)
decahectokilo-
Consider this progression:
millimeter −−>
centimeter −−>
meter −−>
kilometer −−>
Now the prefix multipliers:
0.001
0.01
1
1,000
Finally, consider the magnitude of change between each of these steps:
millimeter to centimeter = 10
centimeter to meter
= 100
meter to kilometer
= 1,000
This progression explains why there are gaps in the table and illustrates the practical side of measurement. As the
quantity to be measured increases, the size of the most practical unit increases geometrically. If a meter is too
small for a measurement, the next largest prefix-unit, hectometers, will probably not be large enough either.
Here is another example of SI units in practice. The prefix kilo can be joined with the unit liter to form kiloliter,
but you will never see it written and here’s why. Small laboratory ware and kitchenware are calibrated in
milliliters and liters. Remember that a liter of water weighs one kilogram (2.2 pounds), which is easy enough to
pick up. But a kiloliter is 1,000 times heavier! How would you measure and handle that amount of water? Easy—
you would measure its container in meters as the carpenter did in the hot tub example above.
3
Skill Sheet 2.1 International System of Units
5. Comparing SI units
1.
2.
3.
Is a cubic meter equivalent to another measure?
Large volumes are determined by measuring the length, width, and height of their containers. The result is
expressed as the cube of the unit of length used. Because large containers are measured in meters, the result
is expressed in cubic meters. Can you determine another SI unit that is equal to a cubic meter?
•
Remember that a liter is a cube that measures 10 centimeters on each side.
•
Now visualize a cubic meter. How many liter-cubes would line one side? (10 liter-cubes)
•
How many liter-cubes would fit into your virtual cubic meter? (1,000 liter-cubes)
•
How many liters are in a kiloliter? Use the prefix table to answer this question. (1,000 liters)
•
What is the relationship between a cubic meter and a kiloliter? (They are equivalent.)
Now you realize that in the practical world of measurement, large volumes are always measured in cubic
meters. That is why the table in Part 4 shows meters3 in parentheses where kiloliters would have appeared.
Lake volumes are measured from maps in cubic meters, natural gas is delivered to homes in cubic meters,
and topsoil lost to erosion is measured in cubic meters
What is the relationship between a cubic centimeter and a milliliter?
•
What is the length of one side of a liter-cube in centimeters? Use the information found in Practical units
of SI measurement, above. (10 cm)
•
What is the volume of a liter in cubic centimeters? (Use the volume formula, l × w × h. (1,000 cm3)
•
How many milliliters are in a liter? Use the prefix table to answer this question. (1,000 mL)
•
What is the relationship between a cubic centimeter and a milliliter? (They are equivalent.)
What is the relationship between a milliliter of water and a gram?
A milliliter of salt water weights more than a milliliter of fresh water. Therefore, to discover the relationship
between a milliliter of water and a gram, we must define the nature of the water. We will use pure water at
standard temperature and pressure.
•
How much does a liter of pure water at standard temperature and pressure weigh? (1 kg; see Part 2)
•
How many grams are in a kilogram? Use the prefix table to answer this question. (1,000 g)
•
How many milliliters are in a liter? Use the prefix table to answer this question. (1,000 mL)
•
What is the relationship between a milliliter of pure water at standard temperature and pressure and a
gram? (One milliliter weighs one gram.)
It’s important to note that the results of the first two SI challenges were equivalent; you can exchange one for the
other. Although the result of this challenge is numerically equivalent, volume and weight are completely different
concepts. That is why you must state that a milliliter of water WEIGHS one gram, not that a milliliter of water
equals one gram.
4
Skill Sheet 2.1 International System of Units
6. Applying the results of the SI challenges
European kitchens always include a tare scale. A tare scale weighs just like any other scale except that it has a
button that returns the weight to zero even if there is something on it. Cooking with a tare scale and SI units is
easy. Here is how it’s done:
Many recipes like brownies and flavored noodles require an amount of water and milk to be added to a mix. How
might a tare scale allow you to prepare flavored noodles using only a tare balance, a pot, and a mixing spoon?
•
Place the pot with the dry noodles on the balance. Press tare to zero the scale.
•
Run water into the pot until the balance reads in grams the amount of water needed in milliliters.
•
Press tare again to zero the balance. Pour milk from the container until the scale reads in grams the amount
of milk needed in milliliters. Then, heat and stir the noodle, water, and milk mixture. Bon appétit!
7. Converting between two SI Units
Here is how you convert one SI unit to another.
1.
2.
3.
4.
5.
Ask yourself whether the new unit is larger or smaller than the old unit. For example, in converting from
meters to centimeters, the new unit is smaller than the old unit.
If the new unit is smaller, the new quantity must be larger to maintain equality. If the new unit is larger, the
new quantity must be smaller. To make this idea clear, think about a mother and a small child walking across
the room. The mother takes a few large steps, but the child must take many small steps.
If the quantity of the new unit must be larger, the decimal is moved to the right. If the quantity of the new
unit must be smaller, the decimal is moved to the left.
Find the multiplication factor for the old and new units in the SI prefixes table in Part 3. Subtract the
exponents algebraically. Disregard the sign of the result.
Move the decimal in the old quantity in the direction found in Step 2. Move the number of places found in
Step 4. Add zeros if necessary. Write this number as the new quantity.
Example:
How many centimeters equal 2.35 meters?
1.
2.
3.
4.
The new unit (centimeters) is smaller than meters. Therefore, the new quantity must be larger.
The decimal must move right.
The multiplication factor exponent for meter = 0, multiplication factor exponent for centimeter = -2.
Difference without regarding sign = 2.
Therefore, the decimal moves right two places. The new quantity = 235 centimeters.
Practice:
1.
2.
3.
4.
5.
6.
3.45 milligrams = _______________ grams
3.004 meters = _______________ centimeters
112.3 grams = _______________ kilograms
6567.09 millimeters = _______________ centimeters
5.2 liters = _______________ milliliters
How many centimeters are in a kilometer? _______________ centimeters
5
Name:
Skill Sheet 2.2
Converting Units
All measurements have two parts, an amount shown as a number and a unit shown as a word. For example,
your height might be 64 (the amount) inches (the unit). This measurement is equivalent to 5 feet 4 inches.
Why might you use feet and inches to describe your height versus inches only? The type of unit you use
depends on how large or how small a measurement is. For example, the distance to your school might be
158,400 inches or 2.5 miles. Do you see why you might use miles to describe this distance? You will practice
converting between units in this skill sheet.
1. Canceling units or ‘crossing out’
Canceling units is the key to converting units. Here are the four concepts involved. Take it a step at a time to see
what is happening.
1.
One factor multiplied by a fraction is equal to a single fraction:
1m
cm × 1 m23 cm × ------------------ = 23
-----------------------------100 cm
100 cm
2.
Units are just like algebraic variables. Think of them as multiplied by their amounts:
23
cm × 1 m- 23 × cm × 1 × m
-----------------------------= ----------------------------------------100 cm
100 × cm
3.
Anything divided by itself is equal to 1:
1--- = 1
1
4.
23
------ = 1
23
cm
------- = 1
cm
One times anything is equal to that thing, so multiplying by 1 does not change the value:
1×3 = 3
1 × cup = cup
1 × gram = gram
See how each of these concepts works together. The first two concepts are applied to three separate fractions:
1,000 m 1 mile
100 × km × 1,000 × m × 1 × mile100 km × -------------------- × -------------------- = ------------------------------------------------------------------------------1 km
1,609 m
1 × km × 1,609 × m
Scanning this combined fraction above, we see two cases of like terms in the numerator and the denominator.
Kilometers (km) and meters (m) appear in the numerator and the denominator. The third concept above says that
anything divided by itself is equal to 1. Therefore, we can cancel out these terms.
1
1
1,000 m 1 mile
100
×
km
×
1,000
×
m
× 1 × mile
100 km × -------------------- × -------------------- = -------------------------------------------------------------------------------1 km
1,609 m
1 × km × 1,609 × m
The fourth concept says that the two 1’s that resulted from canceling will not change the final value. Note that the
single remaining unit (miles) is carried over as the unit in the result. To provide a final check, determine whether
the resulting unit—in this case, miles—makes sense. Does it make sense to say that 100 kilometers is equal to
62.1 miles?
100 × 1 × 1,000 × 1 × 1 × mile- = 62.1 miles
------------------------------------------------------------------------1 × 1 × 1,609 × 1
1
Skill Sheet 2.2 Converting Units
2. Choosing conversion factors
The first step of converting units is to select a conversion factor that matches the units of the problem. Sometimes
you need more than one conversion factor to solve the problem. For each problem below, circle one or more
conversion factors that you would use to solve each problem. You do not need to solve these problems.
Problem
Conversion Factors
Example:
3, 043 meters equals how many kilometers?
10
mm---------------1 cm
10
cm-------------1m
1 km -------------------1, 000 m
The problem involves meters and kilometers. The circled conversion factor shows the relationship between
meters and kilometers (1 km = 1,000 m) so this is the conversion factor to use.
1.
183 cm equals how many meters?
10
mm---------------1 cm
1m----------------100 cm
1,000
m------------------1 km
1.
53 mm equals how many centimeters?
10
mm---------------1 cm
10
cm-------------1m
1,000
m------------------1 km
1.
73,680 cm equals how many kilometers?
10
mm---------------1 cm
1m----------------100 cm
1 km ------------------1,000 m
3. Applying the conversion factor correctly
Conversion problems are solved with one or more conversion factors. In the conversion process, the starting unit
is canceled, leaving only the ending unit in the answer. To do this, the unit to be canceled must appear in the
denominator of the conversion factor. Sometimes this requires you to invert the conversion factor. Solve each
problem with the conversion factor as is or inverted.
Problem
Conversion factors
Example:
1.5 miles is equal to how many kilometers?.
1 kilometer
1.5 miles × ---------------------------- = 2.4 kilometers
0.624 miles
1.
0.624
miles--------------------------1 kilometer
1 kilometer--------------------------0.624 miles
In this problem, the conversion factor
needs to be inverted.
4.3 centimeters is equal to how many millimeters?
10
millimeters---------------------------------1 centimeter
M
1.
8,700 milligrams is equal to how many grams?
1.
4.3 Astronomical Units is equal to how many kilometers?
10
milligrams
--------------------------------1 gram
2
149,597,870.7
kilometers-----------------------------------------------------------1 Astronomical Unit
Skill Sheet 2.2 Converting Units
4. Practice problems
Solve the following problems using the conversion factors found on the back cover of your text, Foundations of
Physics. Additional conversion factors are included with each problem. Round final answers to the nearest tenth.
In problems with more than one conversion factor, work each conversion factor one at a time, working from left
to right. Treat the result of the first conversion as though it were the beginning of a new problem and ensure that
the new unit to be canceled is in the denominator of the next conversion factor.
Use the following rules to check your work:
•
If the ending unit is larger than the starting unit, the ending amount must be
smaller that the starting amount. Example: 12 eggs = 1 dozen
Dozen is a larger unit than egg; 1 is smaller than 12.
•
If the ending unit is smaller, the ending amount must be larger than the
starting amount.
Example: 1 meter = 100 centimeters
Centimeter is smaller than meter; 100 is larger than 1.
•
The final unit after conversion must answer the original question.
1.
Fill in the following table:
Starting amount and unit
Ending amount and unit
3.0 inches
_____ meters
3.7 gallons
_____ liters
47.0 pounds
_____ kilograms
3.0 pints
_____ liters
230 grams
_____ kilograms
42 millimeters
_____ centimeters
1,000 milliliters
_____ liters
24.3 meters
_____ kilometers
Conversion factors: 0.4536 kilograms = 1 pound or
2.
0.4536
kg----------------------;
1 pound
8 pints = 1 gallon or
8 pints-----------------1 gallon
The volume of a European hot tub is 2,800 liters, but the building code for floor joist size to support the tub
is in gallons. How many gallons should the builder use to calculate the weight of the filled tub?
3
Skill Sheet 2.2 Converting Units
3.
A bullet fired from a .22-caliber rifle leaves the barrel at 1,200 feet per second. How fast is that in meters per
second?
meters--------------------------------Conversion factor: 0.3048 meters = 1 foot or 0.3048
1 foot
4.
One reason that SI units are not popular in the United States is that converting English units directly into SI
units results in numbers with decimals. What would the weight be of a 2-pound can of coffee in grams?
Conversion factor:
grams
----------------------------453.6 grams = 1 pound or 453.6
1 pound
5.
The beverage industry in the United States has been eager to use SI units. One liter of a beverage has 1,000
milliliters. Calculate how many milliliters are in one quart. Why do you think it would be a good marketing
move to sell beverages by the liter rather than by the quart?
Conversion factor:
4 quarts
4 quarts = 1 gallon or -----------------1 gallon
6.
A young French girl went to the market and bought 200 grams of cheese for her mother. About how many
ounces of cheese did she buy?
Conversion factors:
grams
----------------------------453.6 grams = 1 pound or 453.6
1 pound
16 ounces = 1 pound or
7.
16 ounces
-----------------------1 pound
Challenge problem: Here is a good multipart problem that gives you an eye-opening idea of the immense
distances of space. It is also completely imaginary. Although we know sound cannot travel through a
vacuum, imagine that sound can travel in space at the same speed it travels through air under standard
conditions. Our sun has just erupted in an enormous solar flare. We will see it in about 8 1/2 minutes because
of the speed of light. But how long after the flare would we hear it under these imaginary conditions? Round
to the nearest whole number after each step.
Conversion factors:
Distance to sun = 93,000,000 miles
Speed of sound under standard conditions = 343 meters per second.
One kilometer = 0.62 miles
One kilometer = 1,000 meters
One hour = 3,600 seconds.
One year = 8,766 hours.
HINT: Convert the sun's distance to meters and calculate the number of seconds, then convert the number of
seconds to years.
4
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Skill Sheet 2.3
Scientific Notation
Visualizing and working with very large and very small numbers is easier when you use scientific notation. In
this skill sheet, you will practice using scientific notation.
1. When is scientific notation useful?
Scientific notation is useful for large and small numbers, but not when the numbers need to be highly precise.
This is because at high precision, each number must be displayed. The two examples below illustrate when
scientific notation is useful and when it is not.
Scientific notation is useful when the number is not very precise:
Speed of light = 300,000,000 cm/sec = 3.00 × 108 cm/sec
Here, the speed of light is not a precise number. We do not know if the speed of
light is 299, 999, 999 cm/sec or 300,000,000.321 cm/sec.
Scientific notation is not useful when the number is precise:
A precise length of 1 meter = 1,650,763.73 wavelengths of orange-red light
from a krypton-86 lamp.
We cannot easily write this number using scientific notation.
A
B
1. Which of the following numbers in these
sentences would be useful to write in
scientific notation?
The amount of money he had
in his bank account was
$30,892.23.
The net worth of that
businessman is
$53,000,000.000.
2. Which of the following numbers should
not be written in scientific notation?
503, 099, 111, 834.45
0.00000000672
2. Why is scientific notation useful?
Multiplying and dividing in scientific notation is so easy that you can often make estimates in your head. To
multiply in scientific notation, multiply the two bases and add the two exponents. For example, you can easily
calculate the square of the speed of light (3.00 × 108 cm/sec) using scientific notation:
Multiply bases:
Add exponents:
Assemble the number:
3.00 × 3.00 = 9.00
108 × 108 = 1016
9.00 × 1016 cm/sec
Dividing two numbers in scientific notation is similar to multiplying them. To divide, the first base is divided by
the second base. Then the second exponent is subtracted from the first.
4.00 × 108 ÷ 2.00 × 102
=
Divide bases:
Subtract exponents:
Assemble the number:
4.00 ÷ 2.00 = 2.00
108 ÷ 102 = 10(8 - 2) = 106
2.00 × 106
1
Skill Sheet 2.3 Scientific Notation
3. Converting from standard notation to scientific notation
•
Note the position of the decimal in the standard notation number. If there is no decimal, add one at the end of
the number.
•
Count the number of places as you move the decimal until the base number is equal or greater than 1, but
less than 10.
•
Usually the base is shown to two decimal places. Round the base number to two decimal places if necessary.
•
Write the exponent, equal to the number of places moved, as a power of ten. If you moved the decimal to the
left, the exponent is positive. If you moved it to the right, the exponent is negative.
Convert these standard notation numbers to scientific notation:
Standard notation
1.
6,700,000
2.
300,000,000
3.
3,600,000
4.
0.000645
5.
150,000,000
6.
0.001
7.
186,000
8.
0.001341
Scientific notation
4. Converting from scientific notation to standard notation
•
If the exponent is negative, count places left from the decimal in the base number equal to the exponent.
•
If the exponent is positive, count places right from the decimal in the base number equal to the exponent.
4.00 × 108 = 400 × 106 Now add six more zeros to 400 to make 400,000,000
Convert these scientific notation numbers to standard notation:
Scientific notation
1.
3.47 × 108
2.
7.94 × 103
3.
1.96 × 10-3
4.
4.50 × 109
5.
9.61 × 10-6
6.
2.02 × 105
7.
7.01 × 10-10
8.
4.44 × 101
Standard notation
2
Skill Sheet 2.3 Scientific Notation
5. Scientific notation using a calculator
You can convert between standard and scientific notation with a scientific calculator. Calculators vary in
keystrokes used, but most are similar to the keystroke steps listed here. Compare these steps to the instructions
for your scientific calculator.
To enter a number in scientific notation:
•
Type the base number using the number keys and the decimal key.
•
Press the exponent key. On some calculators this key is labeled <EE>. Other common exponent keys are
<Ex >, <Exp>, and <x10>. This automatically sets up the power of ten exponent entry.
•
If the exponent is negative, press the <(-)> and then type exponent. Doing this will make the exponent a
negative number. On some calculators, you type the negative sign after typing the exponent.
Check your work in Part 3 by using these steps with a calculator.
To convert a number in scientific notation to standard notation:
•
If the calculator has different display modes, set it for standard notation display.
•
Enter a number in scientific notation using the steps above.
•
Press the <=> key.
Check your work in Part 4 by using these steps with a calculator.
6. Scientific notation operations using a calculator
Calculators can do math operations (+, -, ·, ÷) in scientific notation and mixed standard and scientific notation.
The procedure is no different from what you have used with standard notation. Simply enter the first number, the
operator, the second number, and then type <=>. Some calculators allow the display mode to be changed between
standard and scientific notation. A calculator set to standard notation display mode allows you to divide two
numbers in scientific notation, but the result appears in standard notation. To display the result in scientific
notation, change the display mode. Solve these operations with your calculator in scientific notation mode.
1.
2.34 × 103 × 6 =
2.
2.34 × 103 × 1.10 × 104 =
3.
7.02 × 103 ÷ 2.34 × 102 =
4.
4.32 × 109 + 7.87 × 109 =
5.
5.21 × 10-4 – 3.01 × 10-4 =
6.
5.06 × 102 + 410 =
7.
7.22 × 105 × 3.33 × 103 =
8.
8.64 × 104 – 5.02 × 103 =
9.
9.87 × 10-7 ÷ 2.10 × 10-3 =
10.
6.88 × 10-3 + 2.45 × 10-3 =
3
Name:
Skill Sheet 3.1
Speed Problems
This skill sheet will allow you to practice solving speed problems. To determine the speed of an object, you
need to know the distance traveled and the time taken to travel that distance. However, by rearranging the
formula for speed, v = d/t, you can also determine the distance traveled or the time it took for the object to
travel that distance, if you know the speed. For example,
Equation…
Gives you…
If you know…
v = d/t
speed
time and distance
d=v×t
distance
speed and time
t = d/v
time
distance and speed
1. Solving problems
Solve the following problems using the speed equation. The first problem is done for you.
1.
What is the speed of a cheetah that travels 112.0 meters in 4.0 seconds?
m- = 28
mspeed = d--- = 112.0
---------------------------t
4.0 sec
sec
2.
A bicyclist travels 60.0 kilometers in 3.5 hours. What is the cyclist’s average speed?
3.
What is the average speed of a car that traveled 300.0 miles in 5.5 hours?
4.
How much time would it take for the sound of thunder to travel 1,500 meters if sound travels at a speed of
330 m/sec?
5.
How much time would it take for an airplane to reach its destination if it traveled at an average speed of
790 kilometers/hour for a distance of 4,700 kilometers?
1
Skill Sheet 3.1 Speed Problems
6.
How far can a person run in 15 minutes if he or she runs at an average speed of 16 km/hr?
(HINT: Remember to convert minutes to hours)
7.
A snail can move approximately 0.30 meters per minute. How many meters can the snail cover in
15 minutes?
2. Unit conversion
So far we have been mostly using the metric system for our problems. Now we will convert to the English system
of measurement. Remember that there are 1,609 meters in one mile. Do not forget to include all units and cancel
appropriately. These questions refer to problems in Part 1.
1.
In problem 1.1, what is the cheetah’s speed in miles/hour?
28
m- -----------------1 mile - 3---------------------, 600 sec
----------×
×
sec 1,609 m
1 hour
=
63 miles
-------------------hour
2.
In problem 1.5, what is the airplane’s speed in miles/ hour?
3.
In problem 1.6, what is the runner’s distance traveled in miles?
4.
You know that there are 1,609 meters in a mile. The number of feet in a mile is 5,280. Use these equalities to
answer the following problems:
a. How many centimeters equals one inch?
b. What is the speed of the snail in problem 1.7 in inches per minute?
2
Name:
Skill Sheet 3.2
Making Line Graphs
Graphs allow you to present data in a form that is easily and quickly understood. Graphs are especially good
for describing changing data. Here is how a line graph is made.
1. Examine your data
Graph data consist of data pairs. Each data pair represents two
variables. The independent variable will be plotted on the x(horizontal) axis and the dependent variable will be plotted on the
y-(vertical) axis. It’s much easier to find the dependent than the
independent variable. Test for the dependent variable by asking
yourself which one depends on the other. For example, in a data
set of money earned for hours worked, the dependent variable is
money earned. That is because the money depends on the hours
worked. Use this test to determine the dependent variable in any
data set.
Next, determine the numerical range between the smallest and
largest value for each variable. If you started working with zero
hours and worked 60 hours to earn this money, the range for hours
would be 60. During this time, if you started with $50 and finished
working with $320, the range for dollars would be 270. Be sure to
calculate the range for the independent and depend variables
separately.
2. Examine your graph
Check the space that you will use for your graph. If you are using a piece of graph paper, allow some space on the
side and bottom for labels and other information. Now count the number of lines right from the y-axis to nearly
the edge of the space. This is the maximum graph space that you have for the independent variable. Repeat this
process for the dependent variable by counting the number of lines up from the x-axis to nearly the top of the
space. This is the maximum graph space that you have for the dependent variable.
3. Determine the graph scale
You are now going to set the scale for the independent and dependent variables. It is important that you calculate
the scales separately. The independent and dependent variables usually have different scales. We know that the
dependent variable, money, has a range of $270. Now imagine that the graph space has a maximum of 20 lines on
the y-axis before it nearly runs off the page. The first line (the x-axis) will be labeled $50. What will the next
higher line be labeled? The point here is that if the value is too small, some of the money data will run out of the
graph space. But if the value is too large, the plotted line will be small and hard to read. Divide the number of
lines into the range to find a starting value. Increase the scale to an easier-to-use scale if necessary.
$270 ÷ 20 lines = $13.50/line
Increase to an easier scale: $15/line
1
Skill Sheet 3.2 Making Line Graphs
Now label the y-axis at $15 per line. It is not necessary to label each line; perhaps each fourth line as a multiple
of $60. Repeat this process for the independent variable.
1.
2.
Use the above information and the graphic to help you make a graph of the following data: (0, $50), (10,
$95), (20, $140), (30, $185), (40, $230), (50, $275), (60, $320). Use your own graph paper or the graph
paper on the last page of this skill sheet.
Use the data and the graph to determine the amount earned per hour during the 60 hours of work time.
4. Determine the independent and dependent variables
Two variables are listed in each row of the first two columns of the table below. Identify the independent and
dependent variable in each data pair. Rewrite the data pair under the correct heading in the next two columns of
the table. The first data pair is done for you.
Data pair not necessarily in order
Independent
Temperature
Hours of heating
Hours of heating
Reaction time
Alcohol consumed
Number of people in a family
Cost per week for groceries
Stream flow rate
Amount of rainfall
Tree age
Average height
5. Find the data range
Calculate the data range for each variable:
Lowest value
Highest value
0
28
10
87
0
4.2
-5
23
0
113
100
1250
2
Range
Dependent
Temperature
Skill Sheet 3.2 Making Line Graphs
6. Set the graph scale
Using the variable range and the number of lines, calculate the scale for an axis and then determine an easy-touse scale. Write the easy-to-use scale in the column with the heading “Adjusted scale.”
Range
Number of lines
13
24
83
43
31
35
4.2
33
12
33
900
15
Range ÷ Number of lines
3
Calculated scale
Adjusted scale
Skill Sheet 3.2 Making Line Graphs
Graph paper
4
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Skill Sheet 3.3A Analyzing Graphs of Motion Without Numbers
Graphs change columns of figures into images that are easy to interpret. Position-time and speed-time graphs
describe the movement of objects. Here are stories for you to tell as graphs and a graph for you to tell as a
story. Both will sharpen your graph interpreting skills.
1. Position-time graphs
Data
Remember the “The Three Little Pigs”?
•
The wolf started from his house.
•
Traveled to the straw house.
•
Stayed to blow it down and eat dinner.
•
Traveled to the stick house.
•
Again stayed, blew it down, and ate seconds.
•
Traveled to the brick house.
•
Died in the stew pot at the brick house.
The wolf started at his house, and the graph starts at the origin. Each time the wolf moves farther from his house,
the line moves upward with passing time. At each pig’s house, the line continues to the right but neither rises nor
falls, indicating that the wolf has stopped moving relative to his starting point. We can deduce that the pigs’
houses are generally in a line away from the wolf’s house and that the brick house was the farthest away. How
would the line look if the brick house were on the way back to the wolf’s house? Remember that position refers
to the starting point—in this case, the wolf’s house.
2. Speed-time graphs
A speed-time graph displays the speed of an object over time and is based on position-time data. You know that
speed is the relationship between distance and time, R = D/T. Look at the first part of the wolf’s trip. The line
rises steadily to a new position and a new time. It would be easy to calculate a speed for this first leg. What if the
wolf traveled this first leg faster? The new line would rise to the same position, but it would take less time. That
would make the new line steeper. Here is the speed-time graph for the wolf:
The wolf moved at the same speed toward his first two “visits.” His
third trip was slightly slower. Except for this slight difference, the wolf
was either at one speed or stopped. That is why this graph is so
angular.
1
Skill Sheet 3.3A Analyzing Graphs of Motion Without Numbers
3. Stories for you to tell as graphs
Read each of the following stories. Then sketch in the line for a position-time graph and a speed-time graph.
1.
2.
“Little Red Riding Hood.” Graph Red Riding Hood's movements.
Data:
•
Little Red Riding Hood set out for Grandmother’s cottage at a good walking pace.
•
She stopped briefly to talk to the wolf.
•
She walked a bit slower because they were talking as they walked to the wild flowers.
•
She stopped to pick flowers for quite a while.
•
Realizing she was late, Red Riding Hood ran the rest of the way to Grandmother’s cottage.
The Tortoise and the Hare. Use two lines to graph both the tortoise and the hare.
Data:
•
The tortoise and the hare began their race from the combined start-finish line.
•
Quickly outdistancing the tortoise, the hare ran off at a moderate speed.
•
The tortoise took off at a slow but steady speed.
•
The hare, with an enormous lead, stopped for a short nap.
•
With a start, the hare awoke and realized that he had been sleeping for a long time.
•
The hare raced off toward the finish at top speed.
•
Before the hare could catch up, the tortoise’s steady pace won the race with an hour to spare.
2
Skill Sheet 3.3A Analyzing Graphs of Motion Without Numbers
3.
The Skyrocket. Graph the altitude of the rocket.
Data:
•
The skyrocket was placed on the launcher.
•
As the rocket motor burned, the rocket flew faster and faster into the sky.
•
The motor burned out; although the rocket began to slow, it continued to coast ever higher.
•
Eventually, the rocket stopped for a split second before it began to fall back to Earth.
•
Gravity pulled the rocket faster and faster toward Earth until a parachute popped out, slowing its descent.
•
The descent ended as the rocket landed gently on the ground.
4. A story to be told from a graph
Tim, a student at Cumberland Junior High, was determined to ask Caroline for a movie date. Here are the graphs
of his movements from his house to Caroline’s. You write the story.
3
Name:
Skill Sheet 3.3B
Analyzing Graphs of Motion With Numbers
Speed can be calculated from position-time graphs and distance can be calculated from speed-time graphs.
Both calculations rely on the familiar speed equation: R = D/T.
1. Calculating speed from a position-time graph
This graph shows position and time for a sailboat starting from its
home port as it sailed to a distant island. By studying the line, you can
see that the sailboat traveled 10 miles in 2 hours.
The speed equation allows us to calculate that the vessel speed during
this time was 5 miles per hour.
R = D⁄T
R = 10 miles ⁄ 2 hours
R = 5 miles/hour, read as 5 miles per hour
This result can now be transferred to a speed-time graph. Remember
that this speed was measured during the first two hours.
The line showing vessel speed is horizontal because the speed was
constant during the two-hour period.
2. Calculating distance from a speed-time graph
Here is the speed-time graph of the same sailboat later in the voyage. Between the second and third hours, the
wind freshened and the sailboat increased its speed to 7 miles per hour. The speed remained 7 miles per hour to
the end of the voyage.
How far did the sailboat go during this time? We will first calculate the distance traveled between the third and
sixth hours.
1
Skill Sheet 3.3B Analyzing Graphs of Motion With Numbers
On a speed-time graph, distance is equal to the area between the baseline and the plotted line. You know that the
area of a rectangle is found with the equation: A = L × W. Similarly, multiplying the speed from the y-axis by the
time on the x-axis produces distance. Notice how the labels cancel to produce miles:
speed × time = distance
7 miles/hour × ( 6 hours – 3 hours ) = distance
7 miles/hour × 3 hours = distance = 21 miles
Now that we have seen how distance is calculated, we can consider the distance
covered between hours 2 and 3.
The easiest way to visualize this problem is to think in geometric terms. Find the
area of the rectangle labeled “1st problem,” then find the area of the triangle above,
and add the two areas.
Area of triangle A
Geometry formula
The area of a triangle is one-half the area of a rectangle.
time
speed × ---------- = distance
2
( 3 hours – 2 hours )
( 7 miles/hour – 5 miles/hour ) × ----------------------------------------------- = distance = 1 mile
2
Area of rectangle B
Geometry formula
speed × time = distance
5 miles/hour × ( 3 hours – 2 hours ) = distance = 5 miles
Add the two areas
Area A + Area B = distance
1 miles + 5 mile = distance = 6 miles
We can now take the distances found for both sections of the speed graph to
complete our position-time graph:
2
Skill Sheet 3.3B Analyzing Graphs of Motion With Numbers
3. Practice: Finding speed from position-time graphs
For each position-time graph, calculate and plot speed on the speed-time graph to the right.
1.
The bicycle trip through hilly country.
2.
A walk in the park.
3.
Strolling up and down the supermarket aisles.
3
Skill Sheet 3.3B Analyzing Graphs of Motion With Numbers
4. Practice: Finding distance from speed-time graphs
For each speed-time graph, calculate and plot the distance on the position-time graph to the right. For this
practice, assume that movement is always away from the starting position.
1.
The honey bee among the flowers.
2.
Rover runs the street.
3.
The amoeba.
4
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Skill Sheet 4.1
Acceleration Problems
This skill sheet will allow you to practice solving acceleration problems. Remember that acceleration is the
rate of change in the speed of an object. In other words, at what rate does an object speed up or slow down? A
positive value for acceleration refers to the rate of speeding up, and negative value for acceleration refers to
the rate of slowing down. The rate of slowing down is also called deceleration. To determine the rate of
acceleration, you use the formula:
speed – Beginning speedAcceleration = Final
-----------------------------------------------------------------------Change in Time
1. Solving acceleration problems
Solve the following problems using the equation for acceleration. Remember the units for acceleration are meters
per second per second or m/sec2. The first problem is done for you.
1.
A biker goes from a speed of 0.0 m/sec to a final speed of 25.0 m/sec in 10 seconds. What is the acceleration
of the bicycle?
25.0
m- – 0.0
m25.0
m---------------------------------------sec
2.5 m
sec
sec
acceleration = ------------------------------------ = ---------------- = ------------2
10 sec
10 sec
sec
2.
A skater increases her velocity from 2.0 m/sec to 10.0 m/sec in 3.0 seconds. What is the acceleration of the
skater?
3.
While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the
automobile’s acceleration? (Remember that a negative value indicates a slowing down or deceleration.)
4.
A parachute on a racing dragster opens and changes the speed of the car from 85 m/sec to 45 m/sec in a
period of 4.5 seconds. What is the acceleration of the dragster?
5.
The cheetah, which is the fastest land mammal, can accelerate from 0.0 mi/hr to 70.0 mi/hr in 3.0 seconds.
What is the acceleration of the cheetah? Give your answer in units of mph/sec.
1
Skill Sheet 4.1 Acceleration Problems
6.
The Lamborghini Diablo sports car can accelerate from 0.0 km/hr to 99.2 km/hr in 4.0 seconds. What is the
acceleration of this car? Give your answer in units of kilometers per hour/sec.
7.
Which has greater acceleration, the cheetah or the Lamborghini Diablo? (To figure this out, you must
remember that there are 1.6 kilometers in 1 mile.) Be sure to show your calculations.
2. Solving for other variables
Now that you have practiced a few acceleration problems, you can rearrange the acceleration formula so that you
can solve for other variables such as time and final speed.
Final speed = Beginning speed + ( acceleration × time )
speed – Beginning speedTime = Final
-----------------------------------------------------------------------Acceleration
1.
A cart rolling down an incline for 5.0 seconds has an acceleration of 4.0 m/sec2. If the cart has a beginning
speed of 2.0 m/sec, what is its final speed?
2.
A car accelerates at a rate of 3.0 m/sec2. If its original speed is 8.0 m/sec, how many seconds will it take the
car to reach a final speed of 25.0 m/sec?
3.
A car traveling at a speed of 30.0 m/sec encounters an emergency and comes to a complete stop. How much
time will it take for the car to stop if its rate of deceleration is -4.0 m/sec2?
4.
If a car can go from 0.0 to 60.0 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its
starting speed were 50.0 mi/hr?
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Skill Sheet 4.2
Acceleration and Speed-Time Graphs
Acceleration and distance can be calculated from speed-time graphs.
1. Calculating acceleration from a speed-time graph
Acceleration is the rate of change in the speed of an object. The graph below shows that object A accelerated
from rest to 10 miles per hour in two hours. The graph also shows that object B took four hours to accelerate
from rest to the same speed. Therefore, object A accelerated twice as fast as object B.
Clearly, the steepness of the line is related to acceleration. This
angle is the slope of the line and is found by dividing the
change in the y-axis value by the change in the x-axis value.
Acceleration = ∆y
-----∆x
In everyday terms, we can say that the speed of object A
“increased 10 miles per hour in two hours.” Using the slope
formula:
∆y
10 mph – 0 mph
5 mph
Acceleration = ------ = --------------------------------------- = --------------∆x
2 hours – 0 hour
hour
•
Acceleration = ∆y/∆x (the symbol ∆ means “change in”)
•
Acceleration = (10 mph – 0 mph)/(2 hours – 0 hours)
•
Acceleration = 5 mph/hour (read as 5 miles per hour per hour)
Beginning physics students are often thrown by the double per time label attached to all accelerations. It is not so
alien a concept if you break it down into its parts:
The speed changes . . .
. . . during this amount of time:
5 miles per hour
per hour
Accelerations can be negative. If the line slopes downward, ∆y will be a negative number because a larger value
of y will be subtracted from a smaller value of y.
2. Calculating distance from a speed-time graph
The area between the line on a speed-time graph and the baseline is equal to the distance that an object travels.
This follows from the rate formula:
Rate or Speed = Distance
--------------------Time
R = D
---T
Or, rewritten:.
RT = D
miles/hour × 3 hours = 3 miles
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Skill Sheet 4.2 Acceleration and Speed-Time Graphs
Notice how the labels cancel to produce a new label that fits the result.
Here is a speed-time graph of a boat starting from one place and
sailing to another:
The graph shows that the sailboat accelerated between the second and
third hour. We can find the total distance by finding the area between
the line and the baseline. The easiest way to do that is to break the
area into sections that are easy to solve and then add them together.
A + B + C + D = distance
•
Use the formula for the area of a rectangle, A = L × W, to find
areas A, B, and D.
•
Use the formula for finding the area of a triangle, A = l × w/2, to find area C.
A + B + C + D = distance
10 miles + 5 miles + 1 mile + 21 miles = 37 miles
3. Acceleration from speed-time graph practice
Calculate acceleration from each of these graphs.
1.
2.
3.
2
Skill Sheet 4.2 Acceleration and Speed-Time Graphs
4.
Find acceleration for segment 1 and segment 2.
4. Distance from speed-time graph practice
Calculate total distance from each of these graphs.
1.
2.
3.
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Skill Sheet 4.3
Acceleration Due to Gravity
One formal description of gravity is “The acceleration due to the force of gravity.” The relationships among
gravity, speed, and time are identical to those among acceleration, speed, and time. This skill sheet will allow
you to practice solving acceleration problems that involve objects that are in free fall.
1. Gravity, velocity, distance, and time
When solving for velocity, distance, or time with an object accelerated by the
force of gravity, we start with an advantage. The acceleration is known to be
9.8 meters/second/second or 9.8 m/sec2. However, three conditions must be
met before we can use this acceleration:
•
The object must be in free fall.
•
The object must have negligible air resistance.
•
The object must be close to the surface of the Earth.
In all of the examples and problems, we will assume that these conditions have
been met and therefore acceleration due to the force of gravity shall be equal to
9.8 m/sec2 and shall be indicated by g Because the y-axis of a graph is vertical,
change in height shall be indicated by y.
Remember that speed refers to “how fast” in any direction, but velocity refers to “how fast” in a specific
direction. The sign of numbers in these calculations is important. Velocities upward shall be positive, and
velocities downward shall be negative.
2. Solving for velocity
Here is the equation for solving for velocity:
final velocity = initial velocity + ( the acceleration due to the force of gravity × time )
OR
v = v 0 + gt
Example:
How fast will a pebble be traveling 3 seconds after being dropped?
v = v 0 + gt
2
v = 0 + ( – 9.8 meters/sec × 3 sec )
v = – 29.4 meters/sec
(Note that gt is negative because the direction is downward.)
1
Skill Sheet 4.3 Acceleration Due to Gravity
3. Problems
1.
A penny dropped into a wishing well reaches the bottom in 1.50 seconds. What was the velocity at impact?
2.
A pitcher threw a baseball straight up at 35.8 meters per second. What was the ball’s velocity after
2.5 seconds? (Note that, although the baseball is still climbing, gravity is accelerating it downward.)
3.
In a bizarre but harmless accident, Superman fell from the top of the Eiffel Tower. How fast was Superman
traveling when he hit the ground 7.8 seconds after falling?
4.
A water balloon was dropped from a high window and struck its target 1.1 seconds later. If the balloon left
the person’s hand at –5 meters/sec, what was its velocity on impact?
4. Solving for distance
Imagine that an object falls for one second. We know that at the end of the
second it will be traveling at 9.8 meters/second. However, it began its fall at
zero meters/second. Therefore, its average velocity is half of
9.8 meters/second. We can find distance by multiplying this average velocity
by time. Here is the equation for solving for distance. Look to find these
concepts in the equation:
the acceleration due to the force of gravity × time
distance = ---------------------------------------------------------------------------------------------------------------------- × time
2
OR
1 2
y = --- gt
2
Example: A pebble dropped from a bridge strikes the water in exactly 4 seconds. How high is the bridge?
1 2
y = --- gt
2
1
y = --- × 9.8 meters/sec × 4 sec × 4 sec
2
1
2
y = --- × 9.8 meters/sec × 4 sec × 4 sec
2
y = 78.4 meters
Note that the terms cancel. The answer written with the correct number of significant figures is 78 meters. The
bridge is 78 meters high.
2
Skill Sheet 4.3 Acceleration Due to Gravity
5. Problems
1.
A stone tumbles into a mineshaft and strikes bottom after falling for 4.2 seconds. How deep is the
mineshaft?
2.
A boy threw a small bundle toward his girlfriend on a balcony 10.0 meters above him. The bundle stopped
rising in 1.5 seconds. How high did the bundle travel? Was that high enough for her to catch it?
3.
A volleyball serve was in the air for 2.2 seconds before it landed untouched in the far corner of the
opponent’s court. What was the maximum height of the serve?
6. Solving for time
The equations demonstrated in Sections 2 and 3 can be used to find time of flight from speed or distance,
respectively. Remember that an object thrown into the air represents two mirror-image flights, one up and the
other down.
Original equation
Time from velocity
Time from distance
Rearranged equation to solve for time
v–v
t = ------------0g
v = v 0 + gt
1 2
y = --- gt
2
t =
2y
-----g
Try these:
1.
At about 55 meters/sec, a falling parachuter (before the parachute opens) no longer accelerates. Air friction
opposes acceleration. Although the effect of air friction begins gradually, imagine that the parachuter is free
falling until terminal speed (the constant falling speed) is reached. How long would that take?
2.
The climber dropped her compass at the end of her 240-meter climb. How long did it take to strike bottom?
3.
For practice and to check your understanding, use these equations to check your work in Sections 2 and 3.
3
Skill Sheets
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Skill Sheet 5.1
Isaac Newton
Isaac Newton is one of the most brilliant figures in scientific history. His three laws of motion are probably
the most important natural laws in all of science. He also made vital contributions to the fields of optics,
calculus, and astronomy.
Isaac Newton was born in 1642 in Lincolnshire, England. His childhood years were
difficult. His father died just before he was born, and when he was 3, his mother
remarried and left her son to live with his grandparents. Newton bitterly resented his
stepfather throughout his life.
An uncle helped Newton remain in school and in 1661, he entered Trinity College at
Cambridge University. He earned his bachelor’s degree in 1665.
Ironically, it was the closing of the university due to the bubonic plague in 1665 that
ushered in the blossoming of Newton’s genius. He returned to Lincolnshire and spent
the next two years in solitary academic pursuit. During this period, he made significant
advances in calculus, worked on a revolutionary theory of the nature of light and color,
developed early versions of his three laws of motion, and gained new insights into the nature of planetary motion.
When Cambridge reopened in 1667, Newton was given a minor position at Trinity and began his academic
career. His studies in optics led to his invention of the reflecting telescope in the early 1670s. In 1672, his first
public paper was presented, on the nature of light and color. Newton longed for public recognition of his work
but dreaded criticism. When another bright young scientist, Robert Hooke, challenged some of his points,
Newton was incensed. An angry exchange of words left Newton reluctant to make public more of his work.
In the 1680s, Newton turned his attention to forces and motion. He worked on applying his three laws of motion
to orbiting bodies, projectiles, pendulums, and free-fall situations. This work led him to formulate his famous
universal law of gravitation.
This concept was truly revolutionary. Less than 50 years earlier, it was commonly believed that some sort of
invisible shield held the planets in orbit. Newton’s law explained that it was the gravitational force between the
sun and the planets that is responsible.
In 1687, Newton published his ideas in a famous work known as the Principia. He jealously guarded the work as
entirely his. He bitterly resented the suggestion that he should acknowledge the exchange of ideas with other
scientists (especially Hooke) as he worked on his treatise.
Newton left Cambridge to take a government position in London in 1696. His years of active scientific research
were over. However, almost three centuries after his death in 1727, Newton remains one of the most important
contributors to our understanding of how the universe works.
Questions
1.
2.
Research the legend of Newton’s apple. Which of Newton’s laws does it help explain?
Newton was outraged when, in 1684, German mathematician Wilhelm Leibniz published a calculus book.
Find out why, and describe how the issue is generally resolved today.
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Skill Sheet 5.2
Newton's Second Law
As you work through the problems here, you will develop your understanding of Newton’s second law of
motion and how it relates to Newton’s first law of motion. The second law states that the acceleration of an
object is directly proportional to the force acting on the object and indirectly proportional to the mass of the
object.
1. Newton’s first law of motion
Newton's first law of motion (the law of inertia) states that the motion of an object will continue until an outside
force changes it. The amount of force needed to change the motion of an object depends on the amount of inertia
the object has. The inertia of an object is related to its mass. You need more force to move or stop an object with
a lot of mass or inertia than you need for an object with less mass or inertia.
In Newton's second law, the acceleration of an object is directly related to the force on it, and inversely related to
the mass of the object. This is shown in the formula below:
acceleration = force
-----------mass
Units for acceleration are m/sec2. Units for force are newtons (N). One newton is equivalent to 1 kg-m/sec2.
Units for mass are kilograms (kg). The equation for acceleration illustrates that acceleration is equal to the ratio
of force to mass. This means that the force on an object causes it to accelerate, but the object’s mass is a measure
of how much it will resist acceleration.
2. Three ways to write Newton’s second law of motion
In the formula for the second law of motion, acceleration equals force divided by mass. What does mass equal?
What does force equal? Rearrange the equation to solve for mass. Then rearrange the equation to solve for force.
What do you want to know?
What do you know?
acceleration (a)
force (F) and mass (m)
mass (m)
acceleration (a) and force (F)
force (F)
acceleration (a) and mass (m)
1
The formula you will use
acceleration = force
-----------mass
Skill Sheet 5.2 Newton's Second Law
3. Solving problems using Newton’s second law
Solve the following problems using Newton’s second law. The first two problems are done for you.
1.
How much force is needed to accelerate a truck with a mass of 2,000 kilograms at a rate of 3m/sec2?
3m
m
F = m × a = 2,000 kg × ---------2- = 6,000 kg ⋅ ---------2- = 6,000 N
sec
sec
2.
What is the mass of an object that requires 15 N to accelerate it at a rate of 1.5 m/sec2?
15 kg-m
-------------------2
sec - = 10 kg
15 N = ------------------m = F
--- = ------------a
1.5 m1.5
m----------------------2
2
sec
sec
3.
What is the rate of acceleration of a 2,000.-kilogram truck if a force of 4,200 N is used to make it start
moving forward?
4.
What is the acceleration of a 0.30 kilogram ball that is hit with a force of 25 N?
5.
How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec2?
2
Skill Sheet 5.2 Newton's Second Law
6.
What is the mass of an object that requires a force of 30 N to accelerate at a rate of 5 m/sec2?
7.
What is the force on a 1,000 kilogram-elevator that is falling freely under the acceleration of gravity only?
8.
What is the mass of an object that needs a force of 4,500 N to accelerate it at a rate of 5 m/sec2?
9.
What is the acceleration of a 6.4 kilogram bowling ball if a force of 12 N is applied to it?
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Skill Sheet 6.1
Mass and Weight
The mass of an object is a characteristic property and doesn't change unless the object changes. Weight is the
result of gravitational attraction and depends on the strength of that attraction. You can calculate changes in
weight based on changes in the force of gravity.
1. Mass, weight, and gravity
The key to these problems is to remember that gravity is a force that acts on mass. The result of that interaction is
a force that we call weight. Mass is always measured in kilograms, but weight is measured in force units, or
newtons.
This is yet another area of confusion within the English system of measurement because pounds are a force unit.
Here is the conversion for pounds force to newtons.
1 pound = 4.45 newtons
The problem is that we also convert pounds into kilograms, the unit of mass. Here is the conversion for pounds
weight to kilograms. You can see that the conversion constants are very similar numerically.
1 pound = 0.454 kilograms
It’s essential to keep these units clear. For that reason, these skill problems will be done in International System
units, newtons and kilograms. Weight in newtons is calculated by multiplying mass (in kilograms) by the
acceleration of gravity. On Earth, the acceleration of gravity is 9.8 meters/second/second. On other planets, the
acceleration of gravity is different.
Example:
What is the weight of a girl with a mass of 50 kilograms in a space station with an artificial gravity of 7 N/kg?
F w = mg
F w = 50 kg × 7 N/kg
F w = 350 N = 78.6 pounds (force units)
2. Mass and weight practice
1.
Science fiction movies depict astronauts landing on large asteroids. The gravitational force of such bodies is
typically less that 0.1 N/kg. Given an astronaut with a mass of 77.3 kilograms:
a. What would his weight be on Earth (g = 9.8 N/kg) in newtons and pounds (force)?
b. What would his weight be on an asteroid (g = 0.08 N/kg) in newtons and pounds (force)?
1
Skill Sheet 6.1 Mass and Weight
2.
If you could embark on Jules Verne’s Journey to the Center of the Earth, you would discover that the force
of gravity would at first increase, as you got closer to the main mass of Earth. But then it would decrease as
your progress left more mass behind you and less in front of you. Imagine now that you are very close to the
center of Earth and the force of gravity has decreased to 0.001 N/kg. What would you weigh in newtons and
pounds (force) if your mass were 52.0 kilograms?
3.
You are a typical young male with a mass of 68.4 kilograms. On Earth, you weigh 670 newtons, or
150.5 pounds (force). If you could survive, what would you weigh on the other planets and the sun? Express
your answer in newtons and pounds (force).
Force of gravity
Sun
newtons
274.4 N/kg
Mercury
3.7 N/kg
Venus
8.9 N/kg
Mars
3.7 N/kg
Saturn
10.4 N/kg
Uranus
8.8 N/kg
Neptune
10.7 N/kg
Pluto
0.7 N/kg
2
pounds
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Skill Sheet 6.2
Friction
Friction is a force that resists motion in all real-world mechanics. This skill sheet will provide you with the
opportunity to practice solving problems that involve static friction and sliding friction. Static friction is the
friction between two surfaces that are not moving. Sliding friction is the friction between two moving
surfaces.
1. Calculating friction
In the illustration below, a force (F) is applied to an object. This force is resisted by
friction (Ff), also a force.
Note that the normal force, (Fn), includes all forces pressing the moving surfaces
together.
Here is the equation for finding friction:
In this equation:
•
Ff = force of friction.
•
µ = coefficient of friction.
•
Fn = normal force (force pressing together).
Example: A cinder block sitting on a sidewalk weighs 90 newtons. The coefficient of friction is 0.4. How much
force is required to start the block sliding?
Ff = µ × F n
Ff = 0.4 × 90 N
Ff = 36 N
To slide the block, F must be greater than 36 N.
2. Static friction
Solve these problems and state your answers in the following way: “More than _____ newtons of force is
needed.” Fill in the blank with the correct number of newtons.
1.
A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh
580 newtons. The coefficient of friction for the lawn is 0.55. How much force is required to start sliding the
wrapped leaves?
1
Skill Sheet 6.2 Friction
2.
Although the collie Lassie could easily pull the 110-newton sled empty, she could not even budge it with
380-newton Timmy aboard. How much force would Lassie have to apply to slide the sled with Timmy
aboard? Assume a coefficient of friction of 0.15.
3.
The plastic wading pool weighs 50 newtons. But the water in it weighs 4,000 newtons. How much force is
required to slide the filled pool if the coefficient of friction is 0.22?
4.
The boys pushed and pushed. At an applied force of 804 newtons, they were just able to move the
1,340-newton car. What was the coefficient of friction?
5.
At an applied force of 530 newtons, the boulder just began to slide. Assuming a coefficient of friction of
0.65, how much did the boulder weigh?
3. Sliding friction
Sliding friction is usually less than static friction. Each of the problems below is related to those in part 2.
1.
Although the pile of leaves in the tarp still weighed 580 newtons, once the family that had raked and
wrapped them up got the huge bundle moving, it was easy to keep it moving. The family estimated that they
were applying 93 newtons of force. What was the coefficient of sliding friction?
2.
Aboard the sled, Timmy kicked off with his feet, and although it wasn't easy, Lassie was able to pull the
combined 490-newton weight of sled and Timmy. Assuming that the sliding coefficient of friction was 0.05,
how much force was Lassie applying to the sled?
3.
The children had been splashing in the pool all day. Their father was surprised that he could drag the plastic
wading pool over the wet grass if he gave it a jerk to get it started. Assume the coefficient of sliding friction
is now 0.08. How much force must he apply to the 4,050-newton pool (and water) once it is moving?
4.
The boys were relieved to find that the 1,340-newton car was easier to push once it got going. Just how hard
did they have to push, assuming a coefficient of friction of 0.4?
5.
Try to do this one logically in your head. The boulder in problem 5 in Part 2 required 530 newtons of force to
just begin to slide at a coefficient of static friction of 0.65. What is the coefficient of sliding friction if the
force required to keep it moving is 265 newtons, half of the 530 newtons? Check your work using the
boulder’s weight you calculated in the earlier problem.
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Skill Sheet 6.3
Equilibrium
When all forces acting on a body are balanced, the forces are in equilibrium. Here are free-body diagrams for
you to use for practice working with equilibrium.
1. Equilibrium
Remember that an unopposed force results in acceleration. Therefore, the forces acting
on a body that is at rest or moving at a constant velocity must be at equilibrium.
Conceptually, free-body equilibrium problems are identical to solving equations.
600 N = 400 N + ? N
200 N = ? N
You can see that the free-body diagram is conceptually interchangeable with the
equation. However, it is important to keep in mind that equilibrium means that forces
are in balance; the condition tells us nothing about the magnitude of those forces.
2. Achieving equilibrium
For each free-body diagram, supply the force or forces necessary to achieve equilibrium.
1.
Draw a force arrow, and write in the force.
2.
Supply the missing force.
1
Skill Sheet 6.3 Equilibrium
3.
Distribute the unknown forces evenly to prevent rotation.
4.
Supply the missing force.
3. Think about these
1.
Here is the classic “asteroid destroys Earth” scenario. The momentum of the
asteroid is beyond the forces that even thermonuclear bombs might apply to
stop its approach. Assuming that you can apply only modest forces, where
might they best be applied to result in a new acceleration that will, as they
say, “save the world”? Draw an arrow to show the best location on the
asteroid to apply force so that it avoids hitting Earth.
2.
Helium balloons stay the same size as you hold them, but swell and burst
as they rise to high altitudes when you let them go. Draw and label force
arrows inside and/or outside the balloons on the graphic at right to show
why the near Earth balloon does not burst, but the high altitude balloon
does eventually burst. Hint: What are the forces on the inside of the
balloon? What are the forces on the outside of the balloons?
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Skill Sheet 7.1A
Adding Displacement Vectors
A displacement vector is a quantity that contains two separate pieces of information: (1) magnitude or size,
and (2) direction. When you add displacement vectors, you end up at a certain position. This new position is
the total displacement from the original position. A vector that connects the starting position with the final
position is called the resultant vector (x).
1. Example vector problem
Andreas walked 5 meters east away from a tree. Then, he walked 3 meters
north. Finally, he walked 1 meter west. Each of these three pathways is a
displacement vector. Use these displacement vectors to find Andreas’s total
displacement from the tree.
Displacement
vector
Direction
Magnitude
(meters)
Total magnitude (total
meters walked)
1
east
5
5
2
north
3
5+3=8
3
west
1
8+1=9
Andreas’s motion can be represented on a graph. To determine his total
displacement from the tree, do the following:
1.
2.
Add the east and west displacement vectors. These are in the x-axis
direction on a graph.
Andreas’s walk = 5 m east + (− 1)m west = 4 m east
Add the north and south displacement vectors. These are in the y-axis
direction on a graph.
Andreas’s walk = 3 m north
Solution: Andreas walked a total of 9 meters. The total displacement is 4 meters east and 3 meters north. The
resultant vector (x) goes from the starting position to the final position of total displacement.
2. Adding displacement vectors
1.
What is the total displacement of a bee that flies 2 meters east, 5 meters north, and 3 meters east?
2.
What is the total displacement of an ant that walks 2 meters west, 3 meters south, 4 meters east, and 1 meter
north?
3.
A ball is kicked 10 meters north, 5 meters west, 15 meters south, 5 meters east, and 5 meters north. Find the
total displacement and the total distance it traveled.
1
Skill Sheet 7.1A Adding Displacement Vectors
3. Adding displacement vectors using x-y coordinates
A resultant vector can be written using x-y coordinates on a graph. The original
position is the origin of a graph where the axes represent east-west and northsouth positions. For example, (2,3)m is a resultant vector with the following
components: 2 meters east and 3 meters north. A resultant vector, (-3,-1)m, has
components 3 meters west and 1 meter south. Use this information to solve the
following problems. Write your answers using x-y coordinates. The first one is
done for you.
1.
Add the following four vectors to find the resultant vector, xR:
x1 = (5,0)m, x2 = (0,–5)m, x3 = (3,0)m, x4 = (–7,0) m
Add the east-west components: 5 m east + 0 m + 3 m east + (− 7) m west =
1 m east
Add the north-south components: 0 m + (−5) m south + 0 m + 0 m = (−5) m south xR = (1,−5)m.
2.
Add the following three vectors to find the resultant vector, xR:
x1 = (–2,0)m, x2= (0,–5,)m, x3 = (3,0)m
3.
Add the following vectors to find the resultant vector. Plot the resultant
vector (xR) on the grid to the left:
x1 = (4,0)m, x2= (–1,2)m, x3 = (0,1)m
4.
Add the following three vectors to find the resultant vector, xR.
x1 = (5,3)m, x2 = (–5,0)m, x3 = (5,2)m
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Skill Sheet 7.1B
Vector Components
There are two ways to represent displacement vectors. One way uses x-y coordinates. In this skill sheet, you
will practice representing vectors that use polar coordinates. Polar coordinates indicate the magnitude of the
vector and its angle from the x-axis on a graph. For example, a displacement vector (2 m, 10°) is 2 meters in
magnitude and at a 10° angle from the x-axis on a graph. Using trigonometry, you can convert polar
coordinates into x-y coordinates.
1. Two ways to write displacement vectors
A displacement vector, x, of magnitude 10 meters that has a component in the
x-direction equal to 7.07 meters and a component in the y-direction equal to
7.07 meters can be represented as x = (7.07,7.07)m.
This same vector can be represented in polar coordinates as x = (10.0,45°)m.
Why is this way of representing the vector like x = (7.07,7.07)m?
If the displacement vector is the hypotenuse of a right triangle, then:
x-component: x = 10cos(45°)m = 10(0.707)m = 7.07 m
y-component: y = 10sin(45°)m = 10(0.707)m = 7.07 m
Note that the angle is measured counterclockwise from the x-axis.
2. Problems
Answer the following problems and show your work. The first problem is done for you.
2.
Find the x-y components of the displacement vector x = (7.0, 30°)m.
x-component = (7.0 meters)cos 30° = 6.1 meters
y-component = (7.0 meters)sin 30° = 3.5 meters
x = (6.1, 3.5)m
What are the x-y coordinates for the displacement vector x = (10, 60°)m? Graph this displacement vector.
3.
Find the x-y components of the velocity vector Z =(17, 90°)m/sec.
1.
1
Skill Sheet 7.1B Vector Components
4.
What are the x-y components of the displacement vector x = (100, 360°)cm.
5.
Graph the vector Z =(– 7.07,7.07)m.
6.
Graph the vector x =(– 2,– 4)cm.
7.
Find the components of the velocity vector Z =(80, 180°)km/h.
8.
Find the x-y components of the acceleration vector x = (5, 225°)m/sec2.
9.
Find the x-y components of the velocity vector Z =(80, 300°)km/h.
10. Find the x-y components of the force vector T = (80, 135°)N.
11. Challenge problem: A robot starts from a certain point and moves west for a distance of 5 meters; it then
goes north for 5 meters, and then east for 10 meters. Assume that the positive x direction is east and the
positive y direction is north. The negative x direction is west and the negative y direction is south.
a. Find the x-y components of the final displacement vector for the robot.
b. Express the final displacement vector in polar coordinates. Use the formula below to find the angle of the
resultant vector. Use the Pythagorean theorem to find r, the length of the resultant vector.
y-- = sin θ
r
2
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Skill Sheet 7.1C
Pythagorean Theorem
When you know the x- and y- components of a vector, you can find its magnitude using the Pythagorean
theorem. This useful theorem states that a2 + b2 = c2, where a, b, and c are the lengths of the sides of any right
triangle. For example, suppose you need to know the distance represented by the displacement vector (4,3)m.
If you walked east 4 meters then north 3 meters, you would walk a total of 7 meters. This is a distance, but it
is not the distance specified by the vector, or the shortest way to go. The vector (4,3)m describes a single
straight line. The length of the line is 5 meters because 42 + 32 = 52.
The Pythagorean theorem can be used to help us calculate the magnitude of a vector once we know its
components along the x- and y- directions. Also, we can find one of the components of the vector if we know the
other component and the magnitude of the vector.
1. Example problem
A displacement vector x = (2,3)m has
these components:
2 meters in the x direction.
3 meters in the y direction.
What is the magnitude of the vector?
Using the Pythagorean theorem, a is the
component along the x direction and b is
the component along the y direction. The
magnitude of the vector is c. We can find
the magnitude by taking the square root of a2 + b2:
2
2
a +b =
2
c
2
2
(2 m) + (3 m) =
2
2
4m +9m =
2
c
c
2
13 m = 3.6 m = c
2. Solving problems
1.
Find the magnitude of the vector u = (3, 4).
2.
Find the magnitude of the vector Z = (–3, –4).
1
2
Skill Sheet 7.1C Pythagorean Theorem
3.
Find the magnitude of the vector Z = (5, 0).
4.
Find the magnitude of the vector x = (12.00, 6.00)cm.
5.
A robot starts from a certain point and moves east for a distance of 5.0 meters, then goes north for
3.0 meters, and then turns west for 2.0 meters.
a. What are the x-y coordinates for the resultant vector?
b. What is the magnitude of the resultant vector for the robot?
6.
Add the vectors Z1 = (5,0), Z2 = (0,–3), and Z3 = (1,0), and find the magnitude of the resultant vector.
7.
Add the vectors Z1 = (–5,0), Z2 = (0,–2), and Z3 = (7,0), and find the magnitude of the resultant vector.
8.
A resultant vector has a magnitude of 25 meters. Its y component is –12 meters. What is its x component?
3. Challenge problems
1.
Express the resultant vector in problem 5, Part 2 in polar coordinates. Assume that the positive x direction is
from west to east and the positive y direction is from south to north.
2.
Add the vectors Z1 = (5,0), Z2 = (0,–5), and Z3 = (5,180°), and find the magnitude of the resultant vector.
3.
Add the vectors Z1 = (5,45°), Z2 = (0,–10), and Z3 = (1,180°), and find the magnitude of the resultant
vector.
2
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Skill Sheet 7.2
Projectile Motion
This is a review of projectile motion. The problems in the skill sheet will give you practice in solving
problems that involve projectile motion.
1. Projectile motion has vertical and horizontal components
Projectile motion has vertical and horizontal components. Gravity affects the
vertical motion of an object. When we drop a ball from a height, we know that its
speed increases as it falls. The increase in vertical speed is due to the acceleration
gravity, g = 9.8 m/sec2. So the vertical speed of the ball will increase by 9.8 m/sec
after each second. After the first second has passed, the speed will be 9.8 m/sec.
After the next second has passed, the speed will be 19.6 m/sec and so on.
The acceleration of gravity affects only the vertical component of the motion.
Horizontal motion is not affected by gravity. So if we neglect the friction from air,
when we throw an object horizontally, its initial horizontal speed will not change.
For example, if we throw a marble horizontally at a speed of 5 m/sec, the marble
will be 5 meters horizontally from our hand after one second, 10 meters after
2 seconds, and so forth.
2. Solving projectile motion problems
Solving projectile motion problems requires using equations. To solve these problems, follow the steps:
•
Read the problem carefully. You may want to diagram the problem to help you understand it.
•
List what you know from the problem and what you need to solve for.
•
Determine which equations for vertical motion or horizontal motion will help you solve the problem. You
may need more than one equation to solve the problem. Some important equations are listed below.
•
Solve the problem and check your work.
2
2
2
Pythagorean theorem
a +b = c
Horizontal motion
v ox = v 0 cos θ
Use this equation to calculate initial horizontal velocity when you know
an angle and magnitude of the initial velocity vector.
Vertical motion
v oy = v 0 sin θ
Use this equation to calculate initial vertical velocity when you know an
angle and magnitude of the initial velocity vector.
Horizontal distance
x = v ox t
This equation is a rearranged version of the speed equation: v = d/t.
Here, x represents d, distance.
Vertical velocity
v y = v oy – gt
1 2
y = v oy t – --- gt
2
v oy
t = -----g
Gravity (g) is included in these equations because vertical speed
accelerates due to gravity when an object is falling.
Vertical distance
The time to reach
maximum height
Use this equation to find the magnitude of a velocity vector, (a, b).
This equation is a rearranged version of acceleration = speed/time.
1
Skill Sheet 7.2 Projectile Motion
3. Sample projectile motion problem
A ball is kicked with an initial total velocity (v0) of 10 m/sec at an angle of 60 degrees off the ground. The time
that it takes for the ball to reach the ground again is twice the time it takes for the ball to reach its maximum
height. Using this information, estimate how far the ball will go horizontally and the maximum height it will
reach.
The horizontal (or x) component of the ball's velocity is:
vox = v0cos(60°) = 10 m/sec × 0.5 = 5 m/sec
The vertical (or y) component of the ball’s velocity is:
v oy = v 0 sin ( 60° ) = 10 m/sec × 0.87 = 8.7 m/sec
The time it takes for the ball to reach its maximum height (t) is written below. The initial vertical velocity is
voy and g is the acceleration due to gravity.
v oy
t = -----g
The total time it takes for the ball to travel horizontally is twice this long:
v oy
8.7 m/sec
t = 2 ×  ------- = 2 ×  -------------------------------- = 1.8 sec
g
9.8 m/sec/sec
With this information, we are now able to answer the questions:
• What is the horizontal range of the ball?
•
What is the maximum height reached by the ball?
The horizontal range equals the speed times the time in the horizontal direction:
x = vox × t = 5 m/sec × 1.8 sec = 9 m
The maximum height—or vertical distance (y)—can be calculated using the formula below.
1 2
y = [ v oy t ] – --- gt
2
In this problem, the ball reaches its maximum height in half the time that the ball travels before reaching the
ground. This time has been calculated to be 1.8 seconds. Therefore, 1/2 times 1.8 seconds is used in the
equation below for vertical distance.
2
1
1
1
y = v oy × --- ( time to travel horizontally ) – --- g × --- ( time to travel horizontally )
2
2
2
1.8
1
2 1.8
y = 8.7 m/sec × ------- sec – --- 9.8 m/sec ------- sec
2
2
2
2
2
= 3.82 m
Skill Sheet 7.2 Projectile Motion
4. Solving problems
Solve the following problems. Show your work.
1.
A cat runs and jumps from one roof top to another which is 5 meters away and 3 meters below. Calculate the
minimum horizontal speed with which the cat must jump off the first roof in order to make it to the other.
a. What do you know?
b. What do you need to solve for?
c. What equation (s) will you use?
d. What is the solution to this problem?
2.
An object is thrown off a cliff with a horizontal speed of 10 m/sec and some unknown initial vertical
velocity. After 3 seconds the object hits the ground which is 30 meters below the cliff. Find the initial
vertical velocity and the total horizontal distance traveled by the object.
a. What do you know?
b. What do you need to solve for?
c. What equation (s) will you use?
d. What is the solution to this problem?
3.
If a marble is released from a height of 10 meters how long would it take for it to hit the ground?
4.
A ball is thrown vertically upwards with a speed of 5 m/sec. How long before the ball hits the ground?
(HINT: Consider that there will be time for the ball to go up and then fall back down.)
5.
A ski jumper competing for an Olympic gold medal wants to jump a horizontal distance of 135 meters. The
takeoff point of the ski jump is at a height of 25 meters. With what horizontal speed must he leave the jump?
6.
An object is launched at an angle of 45 degrees with a speed of 20 m/sec. Calculate the initial velocity
components, the time it takes to hit the ground, the range, and the maximum height it reaches.
3
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Skill Sheet 7.3A
Equilibrium in 2-D
Here you will solve problems that require you to determine the unknown forced needed for an object to be in
equilibrium.
1. Forces and equilibrium
Forces are represented by vectors. They have magnitude (the strength of the
force) and direction. When forces are applied to an object, it will move unless
all forces acting on the object add up to zero. In this case the object is in
equilibrium.
2. Equilibrium problems
Do the following equilibrium problems. The first one is done for you.
1.
2.
A force T = (10, 30°)N is applied on an object. Find the x and y components of the force required for the
object to be in equilibrium.
x component = 10cos(30° + 180°) = –10cos30° N = –8.66 N
y component: = 10sin(30° + 180°) = –10sin30° N = –5.0 N
Find the force required to counteract the force T = (–5,10) kN.
3.
A box weighing 10 kilograms is pulled along the floor with a force of 100 N. The coefficient of friction µ
between the floor and the box is 0.5. Calculate the force required for the box to be in equilibrium.
4.
The force T = (100,–45°)N is applied to an object. Find the x and y components of the force required for
equilibrium.
5.
Find the x and y components of a force needed to counteract the forces T1 = (100,315°)N,
T2 = (100,225°)N, T3 = (60,150°)N.
1
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Skill Sheet 7.3B
Inclined Planes
When an object is placed on an inclined plane, the weight force has a component parallel to the plane and a
component perpendicular to the plane. The parallel component is pulling the object down the plane. Another
force acting on the interface between the object and the place surface is due to friction. The friction force is
acting parallel to the plane and opposite to the direction of motion. In this skill sheet, you will solve problems
that involve objects moving on inclined planes.
1. Solving inclined plane problems
The angle of an inclined plane, θ, is measured from the horizontal.
The horizontal component of the force is mgsinθ where m equals the mass of
an object and g equals the acceleration of gravity (9.8 m/sec/sec).
The vertical component of the force is mgcosθ.
The acceleration of an object on the plane is equal to gsinθ.
The friction force (Ffriction) equals mgcosθ multiplied by the coefficient of
friction (µ).
Ffriction = µmgcosθ
1
Skill Sheet 7.3B Inclined Planes
2. Example problems
1.
Calculate the components of weight of a 10-kilogram box on an inclined plane making an angle 30 degrees
with the horizontal.
2.
What is the acceleration of the box in problem 1 along the plane?
3.
A mass of 30 kilograms is placed on an inclined plane making an angle of 25 degrees with the horizontal.
Find the force parallel and perpendicular to the plane and the acceleration along the plane.
4.
A box weighing 10 kilograms is placed on an inclined plane whose coefficient of friction is 0.30. Calculate
the maximum inclination angle before the box begins to move down the plane.
5.
What is the horizontal component of the force acting on the box in problem 4 at the maximum angle?
6.
What is the vertical component of the force acting on the box in problem 4 at the maximum angle?
7.
What is the total force along the plane acting on the box in problem 4 at the maximum angle?
8.
When the angle of the plane of problem 4 is increased to 45 degrees, what is the acceleration parallel to the
plane? Use the equation below to help you solve this problem:
acceleration = g ( sin θ – µ cos θ )
2
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Skill Sheet 8.1
Circular Motion
You used degrees when you first learned how to measure angles. However, the degree is not the most
convenient unit for using angles to calculate angular speed. For the purpose of angular speed, the radian is a
better unit of angle. One radian equals 57.3 degrees (approximately). Radians are better for angular speed
because a radian is a ratio of two lengths, and it does not have any units in the sense that meters or seconds are
units. This skill sheet provides you with practice in using degrees, radians, and in calculating angular speed.
1. Working with degrees, radians, and angular speed
A full circle has 360 degrees, or 2π radians (π = 3.14).
Convert degrees to radians by multiplying
by π/180°.
Example: How many radians is 45°?
π
45° × ----------- = 0.79 radians
180°
Convert radians to degrees by multiplying
by 180°/π.
Example: How many degrees are represented by 3.5 radians?
180°
3.5 radians × ----------- = 200°
π
Angular speed (ω) is given in radians/second. The formula for angular speed
is shown in the graphic at right.
Linear speed is found by multiplying angular speed by the radius of the
object being considered. The formula for linear speed is: v = ωr.
2. Example problems
1.
Convert to radians:
a. 0 degrees
b. 10 degrees
c. 30 degrees
1
Skill Sheet 8.1 Circular Motion
d. 45 degrees
e. 90 degrees
f. 180 degrees
g. 270 degrees
h. 360 degrees
2.
Convert to degrees: 1.047 radians
3.
A wheel is spinning with an angular speed of 15 radians/sec. What is the angular speed in revolutions per
minute?
4.
A bicycle with a front wheel that is 50 centimeters in diameter and a back wheel that is 74 cm in diameter is
moving along with a linear speed of 16 km/hour. Find the angular speed of the wheels in radians/sec and in
rpm (revolutions per minute).
5.
A wheel that has a radius of 1 meter makes four turns in 3 seconds. Find the angular speed and the linear
speed of this wheel.
6.
A bicycle wheel with a radius of 0.5 meters is rolling with an angular speed of 1.75 rad/sec.What is the linear
speed of the wheel?
7.
A wheel with a radius of 0.25 meters is rolling with an angular speed of 0.75 rad/sec. How far will the wheel
go in one minute?
8.
A ball with a radius of 1 centimeter starts rolling down a ramp. The acceleration of the ball is 2 meters/sec2.
a. What is the angular speed of the ball after 1 second? After 3 seconds?
b. Convert each answer in 8(a) to revolutions per second.
2
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Skill Sheet 8.3
Universal Gravitation
The law of universal gravitation allows you to calculate the gravitational force between two objects from their
masses and the distance between them. The law includes a value called the gravitational constant, or “G.” This
value is the same everywhere in the universe. Calculating the force between small objects like apples or huge
objects like planets, moons, and stars is possible using this law as you will see as you solve the problems in
this skill sheet.
1. What is the law of universal gravitation?
The force between two masses m1 and m2 that are separated by a distance r is given by:
So, when the masses m1 and m2 are given in kilograms and the distance r is given in meters, the force has the
unit of newtons. Remember that the distance r corresponds to the distance between the center of gravity of
the two objects. For example, the gravitational force between two spheres that are touching each other, each
with a radius of 0.3 meter that are touching each other and a mass of 1,000 kilograms, is given by:
F = 6.67 × 10
– 11
2
N-m ⁄ kg
2
1,000
kg × 1,000 kg------------------------------------------------= 0.000185 N
2
( 0.3 m + 0.3 m )
This corresponds to a weight of a mass equal to 18.9 milligrams.
2. Example problems
Answer the following problems. Write your answers using scientific notation.
1.
Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by
a distance of 1 meter.
2.
A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 × 106
meters. Find the mass of the moon.
1
Skill Sheet 8.3 Universal Gravitation
3.
m
For m = 5.9742 × 1024 kilograms and r = 6.378 × 106 meters, what is the value given by this equation: G ----2 ?
r
a. Write the answer in the blank below. Simplify the units of the answer.
b. What does this number remind you of?
c. What real-life values do m and r correspond to?
4.
The distance between Earth and its moon is 3.84 × 108 meters. Earth’s mass is m = 5.9742 × 1024 kg and the
mass of the moon is 7.36 × 1022 kg. What is the force between Earth and the moon?
5.
A satellite is orbiting Earth at a distance of 35 kilometers. The satellite has a mass of 500 kilograms. What is
the force between the planet and the satellite?
6.
The mass of the sun is 1.99 × 1030 kilograms and its distance from Earth is 150 million kilometers
(150 × 109 meters). What is the gravitational force between the sun and Earth?
2
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Skill Sheet 9.1
Torque
In this skill sheet, you will practice solving problems that involve torque. Torque is an action that is created by
an applied force and causes an object to rotate. Any object that rotates has a torque associated with it.
1. What is torque?
Torque, τ, can be calculated by multiplying the applied force, F, by r. The value,
r, is the perpendicular distance between the point of rotation and the line of
action of the force (the line along which the force is applied.
τ = F×r
The unit of torque is newton-meter (N-m).
For many situations the distance r is also called the lever arm.
2. Example torque problem
When you use a wrench to release a rusted bolt, you apply a torque around the axis of the bolt. You might have
noticed that the longer the wrench, the easier it is to perform the task. The length of the wrench is related to the
lever arm. However if you just pull at the end of the wrench you know that there is no way to release the rusted
bolt. The reason is that by pulling you have made the lever arm equal to zero. Zero lever arm, zero torque and the
bolt will keep on rusting.
Since the force has magnitude and direction so does the torque. We talk about torque in the counterclockwise
(CCW) direction, which we call positive and torque in the clockwise (CW) direction which we call negative.
Let’s do some numbers. If you have a wrench of length r = 30 centimeters and you apply a force of 1,000 N at the
end and perpendicular (90°) to the wrench. Because the force is applied downward so the wrench rotates
clockwise around the bolt, the force is negative (-1,000 N). The resulting torque is:
τ =F × r = -1,000 N × 0.3 m = -300 N-m
If you now apply the same force at an angle of 45 degrees from vertical the resulting torque is:
τ =F cos45° × r = -1,000 N cos 45° × 0.3 m = -212 N-m
Although you applied the same force, you get less torque at 45°. If you wanted to create a torque of -300 N-m
while applying the force at a 45 degree angle, you would need to apply -1,414 N of force!
– 300 N-m - = – 1,414 N
F = --------------------------------------cos 45° × 0.3 m
A see-saw works based on the ideas of torque. As you know, the lighter person
(or a cat!) has to sit further out for the saw to be level. You know now that this
is so because the only way to make the torque of the heavy person equal to the
torque of the light person is to increase the lever arm of the light person.
1
Skill Sheet 9.1 Torque
3. Solving problems
Solve the following problems. Show your work. The first
problem is done for you as an example.
1.
For an object to be in rotational equilibrium about a certain
point, the total torque about this point must be zero. For the
example shown in the figure calculate the magnitude and
direction of a force that must be applied at point B for
rotational equilibrium about point P.
Solution: First note that the 35 N force does not create any
torque about point P because this force passes through that point (lever arm = 0).
Let, the force at point B equal FB.
For rotational equilibrium, the following must be true: Torque on to the left of P = Torque to the right of P
FB × 0.25 m = (50 N × 0.1 m) – (75 N × 0.4 m)
25 N-m- = – 100 N
F B = –--------------------0.25 m
For rotational equilibrium, 100 N must be applied downward at point B.
2.
A 10-kilogram mass is suspended from the end of a beam that is 1.2 meters long.
The beam is attached to a wall. Find the magnitude and direction (clockwise or
counterclockwise) of the resulting torque at point B.
3.
Two masses m1and m2 are suspended on an ornament. The ornament is hung
from the ceiling at a point which is 10 centimeters from mass m1 and
30 centimeters from mass m2.
a. If m1 = 6 kg, what does m2 have to be for the ornament
to be in rotational equilibrium?
m
b. Calculate the ratio of -----1- so that the ornament will be
m2
horizontal.
c. Suppose m1 = 10 kg and m2 = 2 kg. You wish to place a third mass, m3 = 5 kg, on the ornament to make it
balance. Should m3 be placed to the right or to the left of the ornament’s suspension point? Explain your
answer.
d. Calculate the exact location where m3 should be placed.
2
Skill Sheet 9.1 Torque
4.
Forces are applied on the beam as shown on the figure at right.
a. Find the torque about point P produced by each of the three forces.
b. Find the net torque about point P.
c. A fourth force is applied to the beam at a distance of 0.30 m to the right of point P. What must the
magnitude and direction of this force be to make the beam in rotational equilibrium?
5.
A flag pole 2.5 meters long is attached to a wall at a 40° angle from vertical. A 50-kilogram mass is
suspended at the end. Calculate the resulting torque at the point of attachment to the wall.
6.
Three 10-newton forces act on a beam as shown to the right.
a. Calculate the torque produced by force 1 about point P.
b. Calculate the torque produced by force 2 about point P.
c. Calculate the torque produced by force 3 about point P.
d. Calculate the net torque about point P.
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Skill Sheet 10.1
Mechanical Advantage
Mechanical advantage (MA) can be defined as the ratio of output force to input force for a machine In other
words, MA tells you how many times a machine multiplies the force put into it. Some machines provide us
with more output force than we applied to the machine—this means MA is greater than one. Some machines
produce an output force smaller than our effort force, and MA is less than one. We choose the type of machine
that will give us the appropriate MA for the work that needs to be performed.
1. What is mechanical advantage?
Mathematically, mechanical advantage may be expressed using either of the following equations:
Fo
force- = ----MA = output
---------------------------input force
Fi
or
Li
input lever arm- = ----MA = -------------------------------------Lo
output lever arm
If we look at the force unit involved in the calculation, the newton (N), we see that it is present in both the
numerator and the denominator of the fraction. Units behave like numbers in mathematical calculations. They
can cancel each other, leaving the value for mechanical advantage as a unit-less quantity.
newtons
-------------------- = N
---- = 1
newtons
N
2. Calculating mechanical advantage
The following set of problems is designed to provide you with practice using the mechanical advantage formula.
The first one is done for you.
1.
A force of 200 newtons is applied to a machine in order to lift a 1,000-newton load. What is the mechanical
advantage of the machine?
force- = 1,000
N- = 5
MA = output
---------------------------------------------input force
200 N
2.
A machine is required to produce an output force of 600 newtons. If the machine has a mechanical
advantage of 6, what input force must be applied to the machine?
3.
An input force of 35 newtons is applied to a machine with a mechanical advantage of 0.75. What is the size
of the load this machine could lift, or how large is the output force?
1
Skill Sheet 10.1 Mechanical Advantage
4.
A machine is designed to push an object with a weight of 12 newtons. If the input force for the machine is
set at 4 N, what is the machine’s mechanical advantage?
5.
A lever has an input arm of 1.5 meters, an output arm of 0.5 meters. What is the mechanical advantage?
6.
A person with a mass of 80 kilograms wants to lift a stone with a mass of 600 kilograms. The person wants
to use a steel bar 2 meters long. Calculate the minimum ratio of the input to the output arm and the maximum
output lever arm.
3. Looking ahead
Machines make work easier. Remember that work is force times distance (W = F × d). The unit for work is the
newton-meter, or often called the joule. Remembering that a joule is the same as a newton-meter will help you
cancel units as you work through the problems in this section.
We put work into a machine (work input) and the machine produces work for us in return (work output). The
work output is never greater than the work input. In fact, work output is always less than work input because of
friction. Friction reduces the amount of energy available to the machine. Less energy for the machine means less
work done by the machine.
In spite of the loss of work due to friction, the machine still makes work easier because machines can provide
mechanical advantage (MA).
Machines can multiply your input force (when MA is greater than 1) so that you can lift a very heavy object.
Machines can also diminish your input force (when MA is less than 1) so that you can handle a very delicate
object that the force of your fingers could damage. Therefore, knowing a machine’s mechanical advantage helps
us choose a machine to perform a specific task.
Use the equations for work and mechanical advantage to solve the following problems. The first one is done for
you.
1.
A force of 30 newtons is applied to a machine through a distance of 10 meters. The machine is designed to
lift an object to a height of 2 meters. If the total work output for the machine is 18 joules, what is the
mechanical advantage of the machine?
input force = 30 N
output force = ( work ÷ distance ) = ( 18 J ÷ 2 m ) = 9 N
output force 9 N
MA = ----------------------------- = ----------- = 0.3
input force 30 N
2.
An input force of 50 newtons is applied through a distance of 10 meters to a machine with a mechanical
advantage of 3. If the work output for the machine is 450 joules and this work is applied through a distance
of 3 meters, what is the output force of the machine?
3.
Two hundred joules of work is put into a machine over a distance of 20 meters. The machine does 150 joules
of work as it lifts a load 10 meters high. What is the machine’s mechanical advantage (MA)?
2
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Skill Sheet 10.2
Work
In science, “work” is defined with an equation. Work is defined as force times distance. By measuring how
much force you have used to move something over a certain distance, you can calculate how much work you
have accomplished. This skill sheet reviews the work equation and provides problems for you to practice
using this equation.
1. What is work?
As you recall, in science work is defined as force acting over a distance. That
is, a force acts upon an object to move it a certain distance. However, to do
work according to this definition, the force must be applied in the same
direction as the movement. For example, if you lift a box off a table, the
force applied is upward and the distance is also upward. This means that you
have done work. However, if you lift the box off the table and then carry it to
a shelf, only the lifting is work. Carrying the box is not work because the
force on the box is upward but the distance is horizontal. However, you
would be doing work if you pushed the box across the floor. Why?
Remember, the only time that work is done is when the force and the
distance are in the same direction. So, in scientific terms, work is the force
that is applied to an object in the same direction as the motion. The formula
for work is:
Work (joules) = Force (newtons) × distance (meters)
W = F×d
You should note and remember that a joule of work is a newton-meter; both units represent the same thing:
WORK. In fact, one joule of work is defined as a force of one newton exerted on an object to move it a distance
of one meter.
1.0 joule = 1.0 N × 1.0 meter = 1.0 newton-meter
2. Applying your knowledge
1.
In your own words, define work in scientific terms. Be complete in your definition.
2.
How are work, force, and distance related?
3.
What are two different units that represent work?
1
Skill Sheet 10.2 Work
3. Solving work problems
Solve the following problems using the formula for work. The first problem is done for you. Write your answers
in joules.
1.
How much work is done on a 10-newton block that is lifted 5 meters off the ground by a pulley?
work = F × d = 10 N × 5 meters = 50 newton-meters = 50 joules
2.
A woman lifts her 100-newton daughter up 1 meter and carries her a distance of 50 meters to her bedroom.
How much work does the woman do?
3.
You pulled your sled through the snow a distance of 500 meters with a force of 200 newtons. How much
work did you do?
4.
An ant sits on the back of a mouse. The mouse carries the ant a distance of 10 meters across the floor. Was
there any work done by the mouse? Explain.
5.
You did 170 joules of work lifting your140-newton backpack. How high did you lift your backpack?
6.
In problem 5, how much did the backpack weigh in pounds? (HINT: There are 4.448 newtons in one pound)
2
Skill Sheet 10.2 Work
7.
A crane does 62,500 joules of work to lift a boulder a distance of 25.0 meters. How much did the boulder
weigh? (HINT: The weight of an object is considered to be a force.)
8.
You lift a 45-newton bag of mulch 1.2 meters and carry it a distance of 10 meters to your garden. How much
work was done?
9.
A 455-newton gymnast jumps upward a distance of 1.50 meters to reach the uneven parallel bars. How
much work did she do before she even began her routine?
10. It took a 500-newton ballerina a force of 250 joules to lift herself upward through the air. She landed a total
of 2.5 meters to the left after completing her jump. How high did she jump?
11. A people-moving conveyor belt moves a 600-newton person a distance of 100 meters through the airport.
How much work was done?
12. A 600-newton passenger at the airport lifts his 100-newton carry-on bag upward a distance of 1 meter. They
ride for 100 meters on the “people mover.” How much work was done in this situation?
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Skill Sheet 10.2 Work
4. Solving work problems that involve angles
Sometimes work problems involve angles. When the applied force is at an
angle to the direction of motion, the only component of the force that
contributes to the work is that which is along the direction of motion. As
shown on the schematic below, when a force (F) is applied at an angle θ to the
direction of motion, the work done by the force (F) as the object moves a
distance (d) as indicated is:
W = Fd cos θ
Note that as the angle θ increases, the work done by the force (F) along (d) is decreasing. When θ becomes 90°,
the work done by the force (F) along d is zero. Use this information to answer the following questions.
A 10-kilogram box is pulled along the floor by a force of 10 newtons that is at an angle
of 30 degrees from horizontal. The box is pulled over a horizontal distance of 2 meters.
1.
How much work is done by the force on the box?
2.
What is the work done by the force when the angle increases to 80 degrees?
5. Solving work problems using the potential energy equation
When objects of mass (m) are moved vertically, the work done by the force of
gravity (mg) is W = mgh, where h is the vertical distance in meters that the mass
has been moved.
For example, the work done against the force of gravity by moving the mass
(m) from position A to position B is W = mgh. Note that this work is now stored
in the system as energy. If the energy at position A is zero, when the object
moves to position B it has potential energy (E) so that E = mgh. Use this
information to answer the following questions.
An 10-kilogram object is pulled up a vertical wall by a force of 500 newton which acts at an
angle 45 degrees.
m = 10 kilograms; θ = 45°; g = 9.8 m/sec2.
1.
What is the work done by the force F on the object as it moves up a distance of 5
meters?
2.
What is the work done against the force of gravity as the object is pulled up 5 meters?
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Skill Sheet 10.3
Potential and Kinetic Energy
In this skill sheet, you will review the forms of energy and formulas for two kinds of energy: potential and
kinetic. After having worked through this skill sheet, calculating the amount of kinetic or potential energy for
an object will be easy!
1. Forms of energy
Energy can be used or stored. When talking about motion, energy that is stored is called potential energy. Energy
that is used when an object is moving is called kinetic energy. Other forms of energy include radiant energy from
the sun, chemical energy from the food you eat, and electrical energy from the outlets in your home. Energy is
measured in joules or newton-meters.
2
m
1 joule = 1 kg ⋅ ---------2- = 1 N ⋅ m = 1 joule
sec
m
1 N = 1 kg ⋅ ---------2sec
2. Potential energy
The word potential means that something is capable of becoming active. Potential energy sometimes is referred
to as stored energy. This type of energy often comes from the position of an object relative to Earth. A diver on
the high board has more potential energy than someone who dives into the pool from the low board.
The formula to calculate the potential energy of an object is the mass of the object times the acceleration of
gravity times the height of the object.
E p = mgh
The mass of the object times the acceleration of gravity (g) is the same as the weight of the object in newtons.
The acceleration of gravity is equal to 9.8 m/sec2.
9.8 m= weight of the object (newtons)
mass of the object (kilograms) × -----------2
sec
3. Kinetic energy
The second category of energy is kinetic energy, the energy of motion. Kinetic energy depends on the mass of the
object as well as the speed of that object. Just imagine a large object moving at a very high speed. You would say
that the object has a lot of energy. Since the object is moving, it has kinetic energy. The formula for kinetic
energy is:
1 2
E k = --- mv
2
To do this calculation, you need to square the velocity value. Next, multiply by the mass, and then divide by 2.
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Skill Sheet 10.3 Potential and Kinetic Energy
4. Solving problems
Now you can practice calculating potential and kinetic energy. Make sure to show your work with all units
present in your calculations as well as in your answer. Write your answers in joules. The first two problems have
been done for you.
1.
A 50-kilogram boy and his 100-kilogram father went jogging. Both ran at a rate of 5 m/sec. Who had more
kinetic energy? Show your work and explain.
Although the boy and his father were running at the same speed, the father has more kinetic energy because
he has more mass.
The kinetic energy of the boy:
2
1--5m 2
m
( 50 kg )  --------- = 625 kg ⋅ ---------2- = 625 joules
2
sec
sec
The kinetic energy of the father:
2
1--5m 2
m
( 100 kg )  --------- = 1, 250 kg ⋅ ---------2- = 1,250 joules
2
sec
sec
2.
What is the potential energy of a 10-newton book sitting on a shelf 2.5 meters high?
The book’s weight (10 newtons) is equal to its mass times the acceleration of gravity. Therefore, you can
easily use this value in the potential energy formula:
potential energy = mgh = ( 10 N ) ( 2.5 m ) = 25 N ⋅ m = 25 joules
3.
Determine the amount of potential energy of a 5-newton book that is moved to three different shelves on a
bookcase. The height of the shelves is 1.0 meter, 1.5 meters, and 2.0 meters.
4.
Two objects were lifted by a machine. One object had a mass of 2 kilograms, and was lifted at a speed of
2 m/sec. The other had a mass of 4 kilograms and was lifted at 3 m/sec. Which object had more kinetic
energy while it was being lifted? Show all calculations.
5.
In problem 4, which object had more potential energy when it was lifted a distance of 10 meters? Show your
calculation. (Remember that gravity = 9.8 m/sec2)
2
Skill Sheet 10.3 Potential and Kinetic Energy
6.
You are standing in your in-line skates at the top of a large hill. Your potential energy is equal to
1,000 joules. The last time you checked, your mass was 60 kilograms.
a. What is your weight in newtons?
b. What is the height of the hill?
c. If you start skating downhill, your potential energy will be converted to kinetic energy. At the bottom of
the hill, your kinetic energy will be equal to your potential energy at the top. What will be your speed at
the bottom of the hill?
7.
Answer the following:
a. A 1-kilogram ball is thrown into the air with an initial velocity of 30 m/sec. How much kinetic energy
does the ball have?
b. How much potential energy does the ball have when it reaches the top of its ascent?
c. How high into the air did the ball travel?
8.
What is the potential energy of a 3-kilogram ball lying on the ground?
3
Skill Sheet 10.3 Potential and Kinetic Energy
9.
What is the kinetic energy of a 2,000-kilogram boat moving at 5 m/sec?
10. What is the velocity of an 500-kilogram elevator that has 4,000 joules of energy?
11. What is the mass of an object that creates 33,750 joules of energy by moving at 30 m/sec?
12. Challenge problem: In the diagram at right, the potential energy of the
ball at position A equals its kinetic energy at position C. At position A, the
ball has zero velocity so its kinetic energy equals zero. At position C, the
ball does not have potential energy because its height is zero. The mass of
the ball is 1 kilogram. Use this information to find the velocity of the ball at
position B.
a. Write an equation that shows how the energy of the ball at position B
relates to its potential energy at position A.
b. What is the velocity of the ball at position B?
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Skill Sheet 11.2
Power
In science, work is defined as the force needed to move an object a certain distance. Suppose that you and a
friend needed to move two 500-newton piles of potting soil to a garden that was 100 meters away. You
accomplished this task in 10 minutes while your friend took 30 minutes. Both of you did the same amount of
work (force × distance), but you did the work in a shorter amount of time. The amount of work done per unit
of time is called power. In this example, you had more power than your friend. This skill sheet will give you
practice with how to calculate power.
1. What is power?
Both you and your friend did the same amount of work.
W = F×d
W = 500 N × 100 m = 50,000 N-m = 50,000 joules
However, you had more power than your friend.
Work (joules)Power (watts) = -----------------------------------Time (seconds)
Let’s do the math to see how this is possible.
Step one: Convert minutes to seconds.
60 seconds
10 minutes × -------------------------- = 600 seconds (You)
minute
60 seconds
30 minutes × -------------------------- = 1, 800 seconds (Friend)
minute
Step two: Find power.
50,000
joules- = 83.3 watts (You)
------------------------------600 seconds
50,000 joules = 27.8 watts (Friend)
----------------------------------1, 800 seconds
As you can see, the same amount of work that is done in less time produces more power. You are familiar with
the word watt from a light bulb. Now you see why a 100-watt bulb is more powerful than a 40-watt bulb. Time
for you to practice solving some problems involving work and power.
1
Skill Sheet 11.2 Power
2. Solving problems
Solve the following problems using the power and work equations. The first problem is done for you.
1.
A motor does 5,000 joules of work in 20 seconds. What is the power of the motor?
joules = 250
joules = 250 watts
power = work
------------ = 5000
-------------------------------------------------time
20 sec
sec
2.
A machine does 1,500 joules of work in 30 seconds. What is the power of this machine?
3.
A sleigh weighing 2,000 newtons is pulled by a horse a distance of 1.0 kilometer (or 1,000 meters) in
45 minutes. What is the power of the horse? (HINT: Convert time to seconds.)
4.
A wagon weighing 1,800 newtons is pulled by a horse at a speed of 0.40 m/sec. What is the power of this
horse?
5.
Suppose a force of 100 newtons is used to push an object a distance of 5 meters in 15 seconds. Find the work
done and the power for this situation.
6.
A force of 100 newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the
work done and the time it takes to do the work.
7.
If a small machine does 2,500 joules of work on an object to move it a distance of 100 meters in 10 seconds,
what is the force needed to do the work? What is the power of the machine doing the work?
8.
A machine uses a force of 200 newtons to do 20,000 joules of work in 20 seconds. Find the distance the
object moved and the power of the machine.(HINT: A joule is the same as a newton-meter.)
9.
A machine that uses 200 watts of power moves an object a distance of 15 meters in 25 seconds. Find the
force needed and the work done by this machine.
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Skill Sheet 12.1
Momentum
This exercise will help you solve problems that involve momentum. The momentum of an object is equal to
its mass times its velocity. When two objects collide, their momentum before the collision is equal to their
momentum after the collision. This statement is called the law of conservation of momentum.
1. What is momentum?
A baseball bat and a ball are a pair of objects that collide with each other. Because of Newton’s third law of
motion, we know that the force the bat has on the baseball is equal, but opposite in direction, to the force of the
ball on the bat. The bat and the baseball illustrate that action and reaction forces come in pairs.
Similarly, the momentum of the bat before it hits the ball will affect how much momentum the ball has after it
and the bat collide. Likewise, the momentum of the ball coming toward the bat determines how much force you
must use when swinging the bat if you are to hit a home run. What is momentum?
The momentum (kg-m/sec) of an object is its mass (in kilograms) multiplied by its velocity (m/sec). The equation
for momentum where p equals momentum, m equals mass, and v equals velocity, is:
p = mv
p = mass in kilograms × velocity in m/sec
2. The law of conservation of momentum
The law of conservation of momentum states that momentum is conserved. This means that the momentum of the
bat and ball before their collision is equal to the momentum of the bat and ball after their collision.
Two colliding objects represent a system. The formula below can be used to find the new velocities of objects if
both keep moving after the collision.
the momentum of a system before a collision = the momentum of a system after a collision
m 1 v 1 ( initial ) + m 2 v 2 (initial) = m 1 v 3 (final) + m 2 v 4 (final)
If two objects are initially at rest, the total momentum of the system is zero. For the final momentum to be zero,
the objects must have equal momenta in opposite directions.
For example, if you are standing on ice skates and throw a bowling ball, the ball’s forward momentum will be
equal to your backward momentum.
the momentum of a system before a collision = 0
0 = the momentum of a system after a collision
0 = m1 v3 + m2 v4
m1 v3 = –( m2 v4 )
If a collision is inelastic and the objects stick together, both have the same final velocity.
m 1 v 1 ( inital ) + m 2 v 2 ( intial ) = ( m 1 + m 2 )v 3 ( final )
1
Skill Sheet 12.1 Momentum
3. Solving momentum problems
Find the momentum of the following objects. The first two problems have been done for you.
1.
A 0.2-kilogram steel ball that is rolling at a velocity of 3.0 m/sec.
3m
m
momentum = m × v = 0.2 kg × --------- = 0.6 kg ⋅ ------sec
sec
2.
A 0.005-kilogram bullet with a velocity of 500 m/sec.
500 m
m
momentum = m × v = 0.005 kg × --------------- = 2.5 kg ⋅ ------sec
sec
3.
A 100-kilogram fullback carrying a football on a play at a velocity of 3.5 m/sec.
4.
A 75-kilogram defensive back chasing the fullback at a velocity of 5 m/sec.
5.
In questions 3 and 4 above, if the fullback collided with the defensive back, who would get thrown
backward? Explain your answer.
6.
If a ball is rolling at a velocity of 1.5 m/sec and has a momentum of 10.0 kg-m/sec, what is the mass of the
ball?
7.
What is the velocity of an object that has a mass of 2.5 kilogram and a momentum of 1,000 kg-m/sec?
2
Skill Sheet 12.1 Momentum
4. Problems involving the law of conservation of momentum
Use the law of conservation of momentum formulas to answer the following problems. The first problem has
been done for you.
1.
You and a friend stand facing each other on ice skates. Your mass is 50 kilograms and your friend’s mass is
60 kilograms. As the two of you push off each other, you move with a velocity of 4 m/sec to the right. What
is your friend’s velocity?
m1 v3 = –( m2 v4 )
( 50 kg ) ( 4 m/sec ) = – ( 60 kg ) ( v 4 )
The answer is: v4 = - 3.33 m/sec or 3.33 m/sec to the left.
2.
A 400-kilogram cannon fires a 10-kilogram cannonball at 20 m/sec. If the cannon is on wheels, at what
velocity does it move backward? (This backward motion is called recoil velocity.)
3.
You stand on a skateboard at rest and throw a rock at 5 m/sec. You move back at 0.5 m/sec. What is the
combined mass of you and the skateboard?
4.
A 2,000-kilogram railroad car moving at 5 m/sec collides with a 6,000-kilogram railroad car at rest. If the
cars coupled together, what is their velocity after the collision?
5.
A 2,000-kilogram railroad car moving at 5 m/sec to the east collides with a 6,000-kilogram railroad car
moving at 3 m/sec to the west. If the cars couple together, what is their velocity after the collision?
6.
A 4-kilogram ball moving at 8 m/sec to the right collides with a 1-kilogram ball at rest. After the collision,
the 4-kilogram ball moves at 4.8 m/sec to the right. What is the velocity of the 1-kilogram ball?
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Skill Sheet 12.2
Rate of Change of Momentum
Momentum is given by the expression p = mv where p is the momentum of an object of mass m moving with
velocity v. The units of momentum are kg-m/sec. Change of momentum (represented ∆p) over a time interval
(represented ∆t) is also called the rate of change of momentum. In this skill sheet, you will practice solving
problems that involve rate of change of momentum.
1. Rate of change of momentum equations
Since, momentum is p = mv, if the mass remains constant during the time ∆t, then:
∆v
∆p
------- = m -----∆t
∆t
∆v
The expression, ------ , represents change in velocity over change in time also known as acceleration. From
∆t
Newton's second law, we know that acceleration equals force divided by mass (a = F/m). Rearranging the
equation, we see that force equals mass times acceleration (F = ma). Similarly, force (F) equals change in
momentum over change in time.
∆v
F = ma = m ------ = ∆p
------∆t
∆t
A mass, m, moving with velocity, v, has momentum mv. If this momentum becomes zero over some change in
time (∆t), then there is a force, F = (mv – 0)/∆t.
•
mv is the initial momentum.
•
0 is the momentum after a change in time ∆t.
When a car accelerates or decelerates, we feel a force that pushes back during acceleration and pushes us forward
during deceleration. When the car brakes slowly, the force is small. However, when the car brakes quickly, the
force increases considerably.
2. Example problems
Example: A person having a weight of 80 kilograms is a passenger in a car going 90 km/hour. The driver puts on
the brakes and the car comes to a stop in 2 seconds. What is the average force felt by the passenger?
First, convert the velocity to a value that is in meter per second: 90 km/hour = 25 m/second. Next, use the
equation that relates force and momentum:
25 – 0 ) m/sec∆p- = m ∆v
------ = 80 kg (---------------------------------Force = -----= 1,000 N
∆t
2 sec
∆t
This is a large force, and for the person to stay in his seat, he must be strapped in with a seat belt.
When the stopping time decreases from 2 seconds to 1 second, the force increases to 2,000 newtons. When the
car is involved in a crash, the change in momentum happens over a much shorter period of time, thereby creating
1
Skill Sheet 12.2 Rate of Change of Momentum
very large forces on the passenger. Air bags and seat belts help by slowing down the person’s momentum change,
resulting in smaller forces and a reduced chance for injury. Let’s look at some numbers.
The car travels at 90 km/hour, crashes, and comes to a stop in 0.1 sec. The air bag inflates and cushions the
person for 1.5 seconds. Let’s calculate the force experienced by the passenger in an automobile without air bags
and in one case with air bags.
•
Without the air bag, the momentum change happens over 0.1 seconds. This results in a force:
25 m/sec
Force = 80 kg --------------------- = 2,000 N
0.1 sec
•
Chances are the passenger is killed.
With the air bag, the force created is:
25 m/sec
Force = 80 kg --------------------- = 1,333 N
1.5 sec
The chances for survival are much higher.
Example: A pile is driven into the ground by hitting it repeatedly. If the pile is hit by the driver mass at a rate of
100 kg/sec and with a speed of 10 m/sec, calculate the resulting average force on the pile.
We are told that the driver mass hits the pile at a rate of 100 kg/sec. What does this mean exactly? We can have a
100-kilogram mass hitting the pile every second, or a 50-kilogram mass hitting the pile every half-second, or a
200-kilogram mass hitting the pile every 2 seconds. You get the idea.
The speed (v) with which the mass hits the pile is 10 m/sec. The mass (m) is 100 kilograms. Time changes occur
at 1 second intervals. The force on the pile is:
∆v
kg
m
kg m
- = 1,000 N
Force = m ------ = 100 ------- 10 ------- = 1,000 ----------2
∆t
sec
sec
sec
3. Problems
1.
A wrecking ball weighing 1,000 kilograms hits a wall with a speed of 2 m/sec and comes to a stop in
1/100 second. Calculate the force experienced by the wall.
2.
A soccer ball weighing 0.15 kilogram is rolling with a speed of 10 m/sec and is stopped by the frictional
force between it and the grass. If the average friction force is 0.5 N, how long would this take?
3.
Water comes out of a fire hose at a rate of 5 kg/sec and with a speed of 50 m/sec. Calculate the force on the
hose. (This is the force that the firefighter has to provide in order to hold the hose.)
2
Skill Sheet 12.2 Rate of Change of Momentum
4.
Water from a fire hose is hitting a wall straight on. The water comes out with a flow rate of 25 kg/sec and
hits the wall with a speed of 30 m/sec. What is the resulting force exerted on the wall by the water?
5.
The water at Niagara Falls flows at a rate of 3 million kilograms per second. The water hits the bottom of the
falls at a speed of 25 m/sec. What is the force generated by the change in momentum of the falling water?
6.
A 50-g (0.05 kilogram) egg that is dropped from a height of 5 meters will hit the floor with a speed of about
10 m/sec. The hard floor forces the egg to stop very quickly. Let’s say that it will stop in 0.001 second.
a. What is the force created on the egg?
b. The egg will break at the force you calculated for 6(a). Imagine that a 50-kilogram person fell down on
the egg falling under the influence of gravity. What would the force of the person on the egg be?
c. Do you think the egg will break if the person fell on it? Why or why not?
d. If we now drop the egg onto a pillow, it will allow the egg to stop over a much longer time compared with
the time it takes for it to stop on the hard surface. The weight and the velocity of the egg is still the same,
but now the time it takes for the egg to come to rest is much longer, about 0.5 second or about 500 times
longer than the time it took to stop on the floor. What would the force on the egg be under these
circumstances?
e. Do you think the egg will break when it drops on the pillow? Why or why not?
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Skill Sheet 13.1
Harmonic Motion
A number of common objects exhibit harmonic motion. A swing, a string on a guitar, sound, and light all
move in a harmonic or wave pattern. We can describe the motion of these objects with special terms like
period, frequency, amplitude, and hertz. In this skill sheet, you will practice using these terms as you work
through the activities, questions, and problems.
1. Reviewing terms
The diagram to the right shows the period of a pendulum. As the ball
on the string is pulled to one side and then let go, the ball moves to the
side opposite the starting place and then returns to the start. This entire
motion equals one cycle. The time it take to move through one cycle is
equal to one period of the pendulum.
As you can see in the diagram, the ball and string always pass through a center point. The distance to which the
ball and string move away from this center point is call the amplitude. For pendulums, amplitude is measured in
degrees. For other kinds of waves, amplitude is measured in units of length like centimeters or meters.
Frequency is a term that refers to how many cycles can occur in one second. For example, the frequency of the
sound wave that corresponds to the musical note “A” is 440 cycles per second or 440 hertz. The unit hertz (Hz) is
defined as the number of cycles per second.
The terms period and frequency are related by the following equation.
1
Skill Sheet 13.1 Harmonic Motion
2. Questions and practice problems
1.
You want to describe the harmonic motion of a swing. You find out that it take 2 seconds for the swing to
complete one cycle. The swing passes through 48 degrees as it goes from high point-to-high point in its
motion (passing through a center).
a. What is the period of the swing?
b. What is the frequency of the swing in hertz?
c. What is the amplitude of the swing?
2.
If you let the swing’s motion continue on its own, what would happen to its amplitude? Why?
3.
Use the graphic to answer the following questions.
a. What is the amplitude of the wave?
b. How many wavelengths are featured in the graphic? In
your response, demonstrate that you understand how
to identify one wavelength.
2
Skill Sheet 13.1 Harmonic Motion
4.
The table below lists data from a pendulum experiment. Use the table to help you answer the questions that
follow.
Trial number
Length of string
Mass of pendulum
Amplitude of pendulum
(cm)
(g)
(degrees)
1
10
5
30
2
10
10
40
3
20
5
30
4
20
10
40
5
30
5
30
6
30
10
40
a. Which of the three variables (length of string, mass of the pendulum, and amplitude) affects the period of
the pendulum the most?
b. For which of the six trials would the pendulum be the slowest? Explain your answer.
c. For which of the six trials would the pendulum be the fastest? Explain your answer.
d. Does the relationship between the mass and period of a pendulum support Newton’s second law of
motion? Explain your answer.
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Skill Sheet 14.1
Waves
What is a wave? How do you calculate the speed of a wave? In this skill sheet you will review how to answer
these questions as you review wave properties.
1. The parts of a wave
1.
On the graphic below label the following parts of a wave: one wavelength, half of a wavelength, the
amplitude, the crest, and the trough.
2.
In the graphic above, how many wavelengths are represented?
3.
Define amplitude of a wave in your own words. What is the amplitude of the wave in the graphic?
4.
How do you calculate the frequency of a wave?
5.
If took 0.05 seconds for the number of wavelengths in the graphic to pass a certain point, what is the
frequency of this wave?
2. The speed of a wave
Below is the formula for the speed of a wave. Use this formula to answer the questions on the next page. Recall
that the frequency of a wave equals 1/period. Be sure to show your work.
1
Skill Sheet 14.1 Waves
1.
The speed of a wave can be calculated by multiplying the frequency by the wavelength. You can also
calculate wave speed by dividing wavelength by the period of the wave. Why does this make sense?
2.
The frequency of a wave is 40 Hz. The speed of the wave is 100 meters per second. What is the wavelength
of this wave?
3.
The wavelength of a wave is 50 centimeters. The frequency is 100 Hz. What is the speed of this wave?
4.
The frequency of wave A is 250 hertz and the wavelength is 30 centimeters. The frequency of wave B is 260
hertz and the wavelength is 25 centimeters. Which is the faster wave?
3. Identifying harmonics
Let’s say you have a machine that supports a 3 meter piece of string. Using this machine you can measure the
frequency at which the string vibrates at each harmonic. Table 1 is partially filled with data. Use your
understanding of harmonics to fill in the rest of the table.
Harmonic #
Frequency
Wavelength
(Hz)
(m)
Speed of the Wave
Frequency times
wavelength
(m/sec)
1
(fundamental)
3
18
2
6
18
3
2
4
12.0
5
15.0
1.5
18
18
6
1.0
1.
When you are looking at a vibrating string, what is the easiest way to determine its harmonic?
2.
What is the wavelength of the fundamental harmonic of a string that is 5 meters long?
2
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Skill Sheet 15.1
Decibel Scale Problems
This skill sheet provides practice problems in using the decibel scale.
The decibel scale relates the amplitude of sound waves to their loudness. Since the range of amplitudes our ears
can hear is very large, a logarithmic scale is used. By compressing the high end of the scale and expanding the
low end of the scale, the logarithmic scale adjusts the very large range of amplitudes so that we can relate the
amplitudes to the way they sound. Increasing a sound’s amplitude by a factor of 10 raises its level on the
logarithmic scale by 20 decibels (dB). Increasing a sound’s amplitude by a factor of 100 raises its level by 40 dB;
by 1,000, 60 dB and so on.
The loudness of the sound is proportional to the amplitude of the sound wave. Each increase of 20 dB will sound
about twice as loud to your ears. Use the following table to help you answer the questions in Parts 1 and 2.
10-15 dB
30 dB
40 dB
45-55 dB
65 dB
70 dB
90 dB
110 dB
120 dB
A quiet whisper 3 feet away
A house in the country
A house in the city
The noise level in an average restaurant
Ordinary conversation 3 feet away
City traffic
A jackhammer cutting up the street 10 feet away
A hammer striking a steel plate 2 feet away (very loud)
The threshold of physical pain from loudness
1. Loudness problems
Solve these loudness problems using the table of decibel scale. The first problem is done for you:
1.
2.
Questions about city sounds:
a. How many decibels does city traffic have? Answer: According to the table, city traffic has a decibel
reading of 70 dB.
b. How many decibels would a sound have if its loudness was twice that of city traffic? Answer: Since every
20 dB increase in decibels sounds about twice as loud, the sound relating to 90 dB (70 dB + 20 dB) would
sound twice as loud.
c. Which sound corresponds to the decibel level determined in 1(b)? Answer: A jackhammer 10 feet away
corresponds to 90 dB.
How many decibels would a sound have if its loudness was four times that of a 10 dB quiet whisper?
3.
How much louder would a 30 dB sound be than a 10 dB quiet whisper?
1
Skill Sheet 15.1 Decibel Scale Problems
4.
A house in the country has a loudness of 30 dB. How many times louder does a 90 dB jackhammer 10 feet
away sound?
5.
How many decibels louder is a house in the city than a house in the country?
6.
How much louder would a house in the city sound than a house in the country?
2. Amplitude problems
Solve the following amplitude problems using the table of decibel scale. Remember that the amplitude of the
sound wave increases by a factor of 10 for every 20 dB difference. The first problem is done for you.
1.
How much greater is a jackhammer 10 feet away’s amplitude than city traffic?
Answer: Since every 20 dB increase in decibels has 10 times greater amplitude, a jackhammer 10 feet away
(90 dB) has an amplitude 10 times greater than city traffic (70 dB).
2.
How much larger in amplitude would a 30 dB sound be than a 10 dB quiet whisper?
3.
How much greater is the amplitude of the sound waves of the jackhammer 10 feet away than the sound
waves of a house in the country?
4.
An ordinary conversation is 65 decibels. In a restaurant, people talk more quietly at around 45 decibels.
What is the difference in amplitude of sound waves produced during normal conversation and sound
produced when people are talking in a restaurant?
2
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Skill Sheet 16.1
Light Intensity Problems
This skill sheet will give you more practice in solving problems involving the intensity of light.
Light is a form of energy. The intensity of light is the amount of energy per second falling on a surface, using
units of watts per meter squared (W/m2). Most light sources distribute their light equally in all directions,
producing a spherical pattern.
Light intensity follows an inverse square law; the intensity diminishes as the square of the distance from the
source. The area of a sphere is 4πr2, where r is the radius or the distance from the light source. For a light source,
the intensity is the power per area. The light intensity equation is:
P
I = P
--- = ----------2
A
4πr
Remember that the power in this equation is the amount of light emitted by the light source. When you think of a
‘100 watt’ light bulb, the number of watts represents how much energy the light bulb uses, not how much light it
emits. Most of the energy in an incandescent light bulb is emitted as heat, not light. That 100-watt light bulb may
emit less than 1 watt of light energy with the rest being lost as heat.
1. Solving problems
Solve the following problems using the intensity equation. The first problem is done for you.
1.
For a light source of 60 watts, what is the intensity of light 1 meter away from the source?
P
60 W = 4.8 W/m 2
I = P
--- = ---------- = ----------------------2
2
A
4πr
4π ( 1 m )
2.
For a light source of 60 watts, what is the intensity of light 10 meters away from the source?
3.
For a light source of 60 watts, what is the intensity of light 20 meters away from the source?
4.
If the distance from a light source doubles, how does light intensity change?
1
Skill Sheet 16.1 Light Intensity Problems
5.
Answer the following problems for a distance of 4 meters from the different light sources.
a. What is the intensity of light 4 meters away from a 1-watt light source?
b. What is the intensity of light 4 meters away from a 10-watt light source?
c. What is the intensity of light 4 meters away from a 100-watt light source?
d. What is the intensity of light 4 meters away from a 1,000-watt light source?
6.
What is the relationship between the watts of a light source and light intensity?
7.
If you sit 1 meter away from a 10-watt source, is the light more or less intense than sitting 5 meters away
from a 50-watt source?
8.
If you sit 5 meters away from a 10-watt source, is the light more or less intense than sitting 5 meters away
from a 50-watt source?
9.
If an incandescent 100-watt light bulb has an efficiency of producing light from energy of 1%, what would
the intensity of light be 5 meters from the bulb?
2
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Skill Sheet 17.1A
The Law of Reflection
The law of reflection works perfectly with light and the smooth surface of a mirror. However, you can apply
this law to other situations. For example, how would the law of reflection help you win a game of pool or pass
a basketball to a friend on the court?
In this skill sheet you will review the law of reflection and perform practice problems that utilize this law. Use
a protractor to make your angles correct in your diagrams.
1. What is the law of reflection?
The law of reflection states that when an object hits a surface, its angle of incidence will equal the angel of
reflection. This is true when the object is light and the surface is a flat, smooth mirror. When the object and the
surface are larger and lack smooth surfaces, the angles of incidence and reflection are close but not always exact.
Nevertheless, this law is very helpful in performing activities like bouncing a ball to someone or in playing pool.
1.
When we talk about angles of incidence and reflection, we often talk about the normal. The normal to a
surface is an imaginary line that is perpendicular to the surface.
a. Draw a diagram that shows a surface, with a normal line, and a ray of light hitting the surface at an angle
of incidence of 60 degrees.
b. In the diagram above, label the angle of reflection. How many degrees is this angle of reflection?
2.
Light strikes a mirror’s surface at 30 degrees to the normal. What will the angle of reflection be?
3.
The angle made by the angle of incidence and angle of reflection for a ray of light hitting a mirror is
90 degrees. What are the measurements of each of these angles?
1
Skill Sheet 17.1A The Law of Reflection
4.
In a game of basketball, the ball is bounced (with no spin) toward a player at an angle of 40 degrees to the
normal. What will the angle of reflection be? Draw a diagram that shows this play. Label the angles of
incidence and reflection and the normal.
2. Playing pool
Use a protractor to figure out the angles of incidence and reflection for the following problems.
1.
Because a lot of her opponent’s balls are in the way for
a straight shot, Amy is planning to hit the cue ball off
the side of the pool table so that it will hit the 8-ball
into the corner pocket. In the diagram, show the angles
of incidence and reflection for the path of the cue ball.
How many degrees does each angle measure?
2.
You and a friend are playing pool. You are
playing solids and he is playing stripes. You
have one ball left before you can try for the
eight ball. Stripe balls are in the way. You
plan on hitting the cue ball behind one of the
stripe balls so that it will hit the solid ball
and force it to follow the pathway shown in
the diagram. Use your protractor to figure
out what angles of incidence and reflection
are needed at points A and B to get the solid
ball into the far side pocket.
2
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Skill Sheet 17.1B
Refraction
When light rays cross from one material into the other they bend. This bending of light rays is called
refraction. This phenomenon is very important and useful. All kinds of optics, from your glasses to your
camera lens to your binoculars use this principle. In this skill sheet we will understand this phenomenon and
learn how to calculate the actual amount of bending as light goes from one material into the other. All we need
to know is the properties of the materials and some simple geometry.
1. Introduction to refraction
The principle is illustrated on the diagram. In this case we have air and water as the
two materials where light travels. A light ray making an angle θ1 with the vertical hits
the water surface. This ray will enter the water at a different angle from vertical θ2.
Once we know the path that a light ray takes as it enters the water we also have the
solution of the problem where the light ray starts from the water and enters the air.
A fish in the water does not appear to be where it actually is. A light ray that leaves the
fish enters our eyes after it as been refracted as shown on the diagram. So, if you are a
hunter trying to spear this fish you better know about this phenomenon or the fish will
get away. Very early humans realized this phenomenon and adjusted their aim with great success.
Here are two questions to consider:
1.
2.
Why does the light ray bend as it crosses from one material into another?
How much does it bend?
The answer to the first question is related to the properties of the materials.
For the purpose of refraction, the property of the material is represented by a
number called the “index of refraction” which is represented by the symbol n.
We call ni the index of refraction of the material from which the ray is coming
(incident material) and nr the index of refraction of the material to which the
ray is entering (refractive material). Also the corresponding angles are called
θi and θr. The general rules of thumb are:
•
If ni is less than nr, then θr is less than θi.
•
If ni is greater than nr, then θr is greater than θi.
The answer to the second
question is provided by Snell’s
law which describes the relation
between ni, nr, θr, and θi.
In working with this formula, the
incident and refractive angles are
measured from the normal to the
surface. Also assume that the
surface between the two materials
is smooth.
1
Skill Sheet 17.1B Refraction
2. Example problems
Solve the following problems using Snell’s law. The first problem is done for you as an example.
1.
Air has an index of refraction equal to 1.0 and glass has an index of refraction equal to 1.5. Light travels
from air into glass with an angle of incidence θi = 25°. What is the refractive angle θr?
We can solve for θr from the relation: nisinθi = nrsinθr
nr
1.0
1.0
sinθ r = ---- sinθ i = ------- sin25° = ------- ( 0.423 ) = 0.282
ni
1.5
1.5
2.
and the angle θr is given by the inverse sine of 0.285: θr = sin–1(0.282) = 16.38° = 16°
Light travels from glass into air. The incident angle θi = 30°. What is the angle of refraction?
3.
Light travels from air into glass. The angle of refraction is θi = 30°. What is the angle of incidence?
4.
Light travels from air into diamond. The index of refraction for diamond is 2.4 and the incident angle
θi = 30°. What is the angle of refraction?
5.
If the incident and refraction angles are 30 degrees and 45 degrees, respectively, which material has the
larger index of refraction and what is the ratio of the refraction indices?
6.
Air has an index of refraction equal to 1.0 and glass has an index of refraction equal to 1.5. Light travels
from glass into air. Calculate the incident angle for which the refractive angle equals 90 degrees.
7.
When the angle of refraction becomes 90 degrees we have a very special and interesting situation. The
incident angle that corresponds to this case is called the critical angle. When the incident angle becomes
greater that the critical angle the incident light will now be reflected at the surface. This is called total
internal reflection. Calculate the critical angle (the angle of incidence) as we did on problem 6 for light
travelling from diamond to air.
8.
Calculate the critical angle of refraction for the water-air interface. The index of refraction for water is 1.33.
2
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Skill Sheet 17.2
Ray Diagrams
Here you will get practice in making ray diagrams. A ray diagram helps you determine where an image
produced by a lens will form and whether the image will be upside down or right side up. For each question
on this skill sheet, read the directions carefully and plot your ray diagram in the space provided.
1. Getting started
1.
Of the diagrams below, which one correctly illustrates how light rays come off an object? Explain your
answer.
2.
Of the diagrams below, which one correctly illustrates how a light ray enters and exits a piece of thick glass?
Explain your answer.
1
Skill Sheet 17.2 Ray Diagrams
3.
In your own words, explain what happens to light as it enters glass from the air. Why does this happen? Use
the terms refraction and index of refraction in your answer.
4.
Of the diagrams below, which one correctly illustrates how parallel light rays enter and exit a converging
lens? Explain your answer.
5.
Draw a diagram of a converging lens that has a focal point of 10 centimeters. In your diagram, show three
parallel lines entering the lens and exiting the lens. Show the light rays passing through the focal point of the
lens. Be detailed in your diagram and provide labels.
2
Skill Sheet 17.2 Ray Diagrams
2. How to draw ray diagrams
A ray diagram helps you see where the image produced by a lens appears. The components of the diagram
include the lens, the principal axis, the focal point, the object, and three lines drawn from the tip of the object and
through the lens. These light rays meet at a point and intersect on the other side of the lens. Where the light rays
meet indicates where the image of the object appears.
Example: A lens has a focal length of 2 centimeters. An object is placed 4 centimeters to the left of the lens.
Follow the steps to make a ray diagram using this information. Trace the rays and predict where the image will
form.
Steps:
•
Draw a lens and show the principal axis.
•
Draw a line that shows the plane of the lens.
•
Make a dot at the focal point of the lens on the right and left sides of the lens.
•
Place an arrow (pointing upward and perpendicular to the principle axis) at 4 centimeters on the left side of
the lens.
•
Line 1: Draw a line from the tip of the arrow that is parallel to the principal axis on the left, and that goes
through the focal point on the right of the lens.
•
Line 2: Draw a line from the tip of the arrow that goes through the center of the lens (where the plane and
the principal axis cross).
•
Line 3: Draw a line from the tip of the arrow that goes through the focal point on the left side of the lens,
through the lens, and parallel to the principal axis on the right side of the lens.
•
Lines 1, 2, and 3 converge on the right side of the lens where the tip of the image of the arrow appears.
•
The image is upside down compared with the object.
3
Skill Sheet 17.2 Ray Diagrams
3. Drawing ray diagrams
1.
A lens has a focal length of 4 centimeters. An object is placed 8 centimeters to the left of the lens. Trace the
rays and predict where the image will form. Is the image bigger, smaller, or inverted as compared with the
object?
2.
Challenge question: An arrow is placed at 3 centimeters to the left of a converging lens. The image appears
at 3 centimeters to the right of the lens. What is the focal length of this lens? (HINT: Place a dot to the right
of the lens where the image of the tip of the arrow will appear. You will only be able to draw lines 1 and 2.
Where does line 1 cross the principal axis if the image appears at 3 centimeters?)
3.
What happens when an object is placed at a distance from the lens that is less than the focal length? Use the
term virtual image in your answer.
4
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Skill Sheet 17.3
Thin Lens Formula
Here you will become familiar and practice with a mathematical formula called “the thin lens formula.” This
formula gives scientists a way to calculate the location and the size of an image that is produced by a lens.
1. What is the thin lens formula?
When you use the thin lens formula, you assume that the thickness of the lens is very small compared with the
distance between the lens and the object or the image. The formula applies both to convex or converging lenses
and concave or diverging lenses. Converging lenses are thicker in the center than in the edges. Diverging lenses
are thinner in the center than in the edges. The thin lens formula is:
Some important rules in using the thin lens formula:
•
Object distance, d0, is positive to the left of the lens and negative to the right of the lens.
•
Image distance, di, is negative to the left of the lens and positive to the right of the lens.
•
Positive d0, di indicates real object or image.
•
Negative d0, di indicates virtual object or image.
1
Skill Sheet 17.3 Thin Lens Formula
2. Examples in using the thin lens formula
If you are using a convex lens, what happens an image when
the object is very far from the lens? In other words, what
happens when d0 is a large number?
Graphically, we see that as d0 increases, the image becomes
smaller as it gets closer to the focal point. We can see this by
using the thin lens formula with a lens that has a focal length
equal to 5 centimeters.
If d0 =
then...
di =
8 cm,
1
---- = 1--- – 1--- = 0.2 – 0.125 = 0.075
di
5 8
1 - = 13.3 cm
-----------0.075
100 cm,
1- = 1--- – -------1 - = 0.2 – 0.01 = 0.19
--di
5 100
1 - = 5.26 cm
--------0.19
1,000 cm,
1
1 - = 0.2 – 0.001 = 0.199
---- = 1--- – ------------di
5 1,000
1 - = 5.03 cm
-----------0.199
If you are using a concave lens, you use a negative value for
the focal length, f. In this case, the resulting image distance is a
negative number indicating a virtual image. The calculations
are shown below:
If d0 =
then...
di =
8 cm,
1- = – 1--- – 1--- = – 0.2 – 0.125 = – 0.325
--di
5 8
1 - = – 4.71 cm
--------------– 0.325
100 cm,
1- = – 1--- – -------1 - = – 0.2 – 0.01 = – 0.21
--di
5 100
1 - = 4.76 cm
-----------– 0.21
1,000 cm
1- = – 1--- – ------------1 - = – 0.2 – 0.001 = – 0.201
--di
5 1,000
1 - = – 4.98 cm
--------------– 0.201
2
Skill Sheet 17.3 Thin Lens Formula
3. Problems
1.
Calculate the location of the image if the object is 20 centimeters in front of a convex (converging) lens with
a focal length of 5 centimeters.
2.
The image of an object as seen by a converging lens is located at 8 centimeters. The object is 24 centimeters
in front of the lens. What is the focal length?
3.
For a converging lens with a focal length of 10 centimeters, calculate the location of the object when the
image appears 20 centimeters to the right.
4.
An object is 10 centimeters in front of a convex (converging) lens with a focal length of 15 centimeters.
a. Calculate the location of the image.
b. What kind of image appears in this situation? Where does it appear in relation to the lens?
5.
Calculate the location of the image if the object is 20 centimeters in front of a concave (diverging) lens with
a focal length of 5 centimeters.
6.
An object is at a distance of 3 centimeters from a lens with a focal length of 1 centimeter. The lens creates an
image on the same side of the object. What kind of lens is this? What is the image location? Is the image real
or virtual?
3
Skill Sheet 17.3 Thin Lens Formula
4. Problems that involve the height of images and objects
Sample problem: The object is 10 centimeters in height and is located at a distance of 25 centimeters from the
lens. The focal length is 8 centimeters. Find the image location di and the height h of the image.
Solution:
Since we know the height of the object and the object distance d0, we can calculate the angle α.
– 1 10
–1
α = tan  ------ = tan ( 0.4 ) = 21.8°
25
The image distance di can be found from the thin lens formula, and it is 11.7 centimeters.
Now, since we know the image distance and the angle α, we can calculate the height.
h = tanαd i = tan21.8°11.17 cm = 4.7 cm
So we see that the image is actually smaller than the object.
1.
Prove that the di in the problem above is 11.7 centimeters by using the thin lens formula.
2.
A 14-centimeter tall object seen through a lens with a focal length of 10 centimeters has an image half its
size.
a. Calculate the location of the image and the object. (HINT: Use proportions.)
b. Using your answers for 2(a) and the height of the object, find the angle α.
4
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Skill Sheet 18.1
The Speed of Light
The speed of light is a fundamental quantity in physics. Light is an electromagnetic wave. Waves are
characterized by their frequency and wavelength. In this skill sheet you will work with the formula that relates
the speed of light to the frequency and wavelength of light.
1. Speed, frequency, and wavelength of light
The speed of light is related to frequency f and wavelength λ by the formula below. Frequency is given in hertz
(Hz), or cycles per second, and wavelength is given in meters. The speed of light in a vacuum is 3 × 108 m/sec.
c = fλ
The different colors of light that we see correspond to different frequencies. The frequency of red light is higher
than the frequency of blue light. Their speed is, however, the same and equal to c, the speed of light. Therefore,
the wavelength of red light is higher than the wavelength of blue light. When we know the frequency of light, the
wavelength is given by:
λ = c-f
When we know the wavelength of light, the frequency is given by:
f = --cλ
2. Problems
Answer the following problems and show your work.
1.
Yellow light has a longer wavelength than green light. Which color of light has the higher frequency?
2.
Green light has a lower frequency than blue light. Which color of light has a longer wavelength?
3.
Calculate the wavelength of violet light with a frequency of 750 × 1012 Hz.
4.
Calculate the frequency of yellow light with a wavelength of 580 × 10–9 m.
5.
Calculate the wavelength of red light with a frequency of 460 × 1012 Hz.
1
Skill Sheet 18.1 The Speed of Light
6.
Calculate the frequency of green light with a wavelength of 530 × 10–9 m.
7.
One light beam has wavelength, λ1, and frequency, f1. Another light beam has wavelength, λ2, and
frequency, f2. Write a proportion that shows how the ratio of the wavelengths of these two light beams is
related to the ratio of their frequencies.
8.
The waves used by a microwave oven to cook food have a frequency of 2.45 gigahertz (2.45 × 109 Hz).
Calculate the wavelength of this type of wave.
9.
A radio station has a frequency of 90.9 megahertz (9.09 × 107 Hz). What is the wavelength of the radio
waves the station emits from its radio tower?
10. An x-ray has a wavelength of 5 nanometers (5.0 × 10-9 m). What is the frequency of x-rays?
11. The ultraviolet rays that cause sunburn are called UV-B rays. They have a wavelength of approximately 300
nanometers (3.0 × 10-7 m). What is the frequency of a UV-B ray?
12. Infrared waves from the sun are what make our skin feel warm on a sunny day. If an infrared wave has a
frequency of 3.0 × 1012 Hz, what is its wavelength?
13. Electromagnetic waves with the highest amount of energy are called gamma rays. Gamma rays have
wavelengths of less than 10-trillionths of a meter (1.0 × 10-11 m).
a. Determine the frequency that corresponds with this wavelength.
b. Is this the minimum or maximum frequency of a gamma ray?
14. Use the information from this sheet to order the following types of waves from lowest to highest frequency:
visible light, gamma rays, x-rays, infrared waves, ultraviolet rays, microwaves, and radio waves.
15. Use the information from this sheet to order the following types of waves from shortest to longest
wavelength: visible light, gamma rays, x-rays, infrared waves, ultraviolet rays, microwaves, and radio
waves.
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Skill Sheet 18.3
Albert Einstein
L
Albert Einstein revolutionized the way we view the physical world on an atomic scale.
Albert Einstein was born in Ulm, Germany, in 1879. He was a quiet child who enjoyed
spending hours building houses of cards and playing the violin. One story he liked to tell
about his youth was of his first encounter with a magnetic compass: The needle seemed
to him to be guided northward by an invisible force. He was convinced there had to be
“something behind things, something deeply hidden.” The search for that “something”
occupied him until his death in 1955.
Einstein was not fond of school until he entered secondary school in Aarau,
Switzerland. There he found first-rate laboratory facilities and teachers who nurtured his
interest in science. Einstein went on to attend the Zurich Polytechnic Institute and
graduated in 1900 with a teaching degree. His first job was as a technician in the patent
office in Bern, Switzerland. Einstein enjoyed evaluating patent claims, but the best part of the position was the
stability it provided in his life. He spent many free evenings reading and thinking about current issues in
theoretical physics.
In 1905, Einstein published three papers which radically changed the way scientists understand the physical
world. While most work in theoretical physics is accomplished through an extended dialogue among scientists,
Einstein’s three papers were written in relative isolation. His first paper described light as discreet bundles of
radiation. His description formed the basis for much of quantum mechanics. The second paper proposed his
theory of special relativity. While Einstein was not the first scientist to generate all of the pieces of this theory, he
was the first to unify them. Einstein’s third paper showed that Brownian motion (the erratic motion of
microscopic particles in a fluid) provided physical evidence for the existence of atoms and molecules. Until
Einstein published his paper, scientists had only theoretical evidence of these tiny particles.
Einstein earned respect as one of Europe’s leading scientific thinkers as a result of these papers. In 1909 he
became a professor first in Zurich, then Prague, and eventually back again in Zurich. In 1914 he was appointed
director of the Kaiser Wilhelm Physical Institute and professor at the University of Berlin.
Einstein’s interests turned toward a theory of general relativity, which showed how inertia and gravity are
connected. His theory predicted that light from distant stars should be bent by the curvature of space near the sun.
During a solar eclipse in 1919, his prediction was proven correct. In 1921, he was awarded the Nobel Prize in
Physics for his work on the photoelectric effect.
During World War I, Einstein, a pacifist, refused to support Germany’s war aims. In 1933, he left Germany to
become a professor at Princeton University. In 1939, concerned about the rise of fascism, he decided force was
necessary to face this threat. He sent a letter to President Roosevelt that urged the United States to develop an
atomic bomb before Germany did. After the war, Einstein was a strong supporter of nuclear disarmament.
Einstein’s scientific interests in his later years focused on finding a unified field theory, which he hoped could
integrate all the known forces in nature into a single equation that would show they were all manifestations of a
single fundamental force. While he never managed to find what he was looking for, his work fascinates
theoretical physicists to this day.
Questions
1.
2.
Find out how scientists tested Einstein’s theory of general relativity in 1919. Write a paragraph to explain
their method.
Research Brownian motion and prepare a demonstration for your classmates.
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Skill Sheet 19.2
Using an Electric Meter
What do you measure in a circuit and how do you measure it? This skill sheet provides useful tips to help you
use an electric meter and understand electrical measurements.
1. The digital multimeter
Most people who work with electric circuits use a digital multimeter to measure electrical quantities. These
measurements help them analyze circuits. Most multimeters measure voltage, current, and resistance. A typical
multimeter is shown below:
1
Skill Sheet 19.2 Using an Electric Meter
2. Using the digital multimeter
This table summarizes how to use and interpret any digital meter in a battery circuit. Note: A component is any
part of a circuit, such as a battery, a bulb, or a wire.
Measuring voltage
Measuring current
Measuring resistance
Circuit is ON
Circuit is ON
Circuit is OFF
Turn meter dial to voltage,
Turn meter dial to current,
Turn meter dial to resistance,
labeled Ω
Connect leads to meter, following
meter instructions
Connect leads to meter, following
meter instructions
Connect leads to meter, following
meter instructions
Place leads at each end of
component (leads are ACROSS
the component)
Break circuit and place leads on
each side of the break (meter is IN
the circuit)
Place leads at each end of
component (leads are ACROSS
the component)
Measurement in VOLTS (V)
Measurement in AMPS (A)
Measurement in OHMS (Ω)
Battery measurement shows
relative energy provided
Measurement shows the value
of current at the point where
meter is placed
Measurement shows the resistance
of the component
Component measurement shows
relative energy used by
that component
Current is the flow of charge
through the wire
2
When the resistance is too high,
the display shows OL (overload)
or ∝ (infinity)
Skill Sheet 19.2 Using an Electric Meter
3. Meter practice
Build a series circuit with 2 batteries and 2 bulbs.
1.
Measure and record voltage across each battery:
2.
Measure and record voltage across each bulb:
3.
Measure and record voltage across both batteries:
4.
Draw a circuit diagram or sketch that shows all the posts in the circuit (posts are where wires and holders
connect).
5.
Break the circuit at one post. Measure current and record the value below. Repeat until you have measured
current at every post.
3
Skill Sheet 19.2 Using an Electric Meter
6.
Create a set of instructions on how to use the meter to do a task. Find someone unfamiliar with the meter.
See if he or she can follow your instructions.
7.
A fuse breaks a circuit when current is too high. A fuse must be replaced when it breaks a circuit. Explain
how measuring the resistance of a fuse can tell you if it is defective.
8.
You suspect that a wire is defective but cannot see a break in it. Explain how measuring the resistance of the
wire can tell you if it has a break.
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Skill Sheet 19.3A
Ohm’s Law
Building and testing series circuits has helped you understand the relationship between voltage, resistance,
and current. You know that if the voltage (energy) in a circuit increases, so does the current (flow of charges).
You also understand that if the resistance increases, the current flow decreases. The German physicist
Georg S. Ohm developed this mathematical relationship, which is present in most circuits. This relationship is
known as Ohm’s law:
Voltage Current = -----------------------Resistance
This skill sheet will provide you with an opportunity to test your knowledge of Ohm’s law.
1. Using Ohm’s law to understand circuits
As you use Ohm’s law, remember that the unit for current is amperes or amps. The unit for voltage is volts, and
the unit for resistance is ohms (symbolized Ω).
To work through this skill sheet, you will need the symbols used to depict circuits
in diagrams. The symbols that are most commonly used for circuit diagrams are
provided at right.
All of the circuits discussed in this skill sheet are series circuits. This means the
current has only one path through the circuit. Later, you will learn about another
kind of circuit in which the current has more than one possible path. This type of
circuit is called a parallel circuit.
Note: For convenience, the symbol for battery is used to represent one or more
batteries. The batteries you have used to build circuits are 1.5 -volt batteries.
Dividing the total voltage by 1.5 volts will tell you the number of batteries present
in the circuit.
For example, the total voltage in the second diagram on the right
is 6 volts. Divide 6 volts by 1.5 to find the number of batteries in
the circuit (6 ÷ 1.5 = 4). There are four batteries in the circuit.
2. Solving problems
In this section, you will find some problems based on diagrams and others without diagrams. In all cases, you
should show your work.
1.
If a toaster produces 12 ohms of resistance in a 120-volt circuit, what is the amount of current in the circuit?
1
Skill Sheet 19.3A Ohm’s Law
2.
You have a large flashlight that requires 4 D-cell batteries. If the current in the flashlight is 2 amps, what is
the resistance of the light bulb? (HINT: One D-cell battery has 1.5 volts.)
3.
What is the voltage of a circuit with 15 amps of current and a toaster with 8 ohms of resistance?
4.
Use the diagram below to answer the following problems:
a. What is the total voltage in each circuit?
b. If the bulbs are identical, which circuit has a greater current? Explain your answer.
c. How does the brightness of the bulb in circuit A compare to the brightness in circuit B? Explain your
answer.
d. How much current would be measured in each circuit if the light bulb had a resistance of 6 ohms?
e. How much current would be measured in each circuit if the light bulb had a resistance of 12 ohms?
f. Suppose a second bulb is added to each of the circuits in series. Explain what would happen to the
current in each circuit.
2
Skill Sheet 19.3A Ohm’s Law
5.
Use the diagram below to answer the following problems:
a. How much current would be measured in each circuit if each light bulb had a resistance of 6 ohms?
b. How much current would be measured in each circuit if each light bulb had a resistance of 12 ohms?
c. What happens to the brightness of each bulb as you add bulbs to a series circuit? (HINT: Compare these
diagrams to the ones in question 4 above.)
6.
What happens to the amount of current in a series circuit as the number of batteries increases?
7.
What happens to the amount of current in a series circuit as the number of bulbs increases?
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Skill Sheet 19.3B
Ben Franklin
Benjamin Franklin overcame his lack of formal education to become a prominent businessman, community
leader, inventor, scientist, and statesman. His study of “electric fire” changed our basic understanding of how
electricity works.
Benjamin Franklin was born in Boston in 1706. He had just one year of schooling but
was an avid reader and writer. At age 12, he was apprenticed to his brother, a printer.
The siblings did not always see eye to eye, and at 17, Ben ran away to Philadelphia.
In his new city, Franklin developed his own printing and publishing business. He
became a community leader, starting the first library, fire department, hospital, and fire
insurance company. He loved gadgets and invented some of his own: the Franklin
stove, the glass armonica (a musical instrument), bifocal eyeglasses, and swim fins.
In 1746, Franklin saw some demonstrations of “electric fire.” These were static
electricity tricks, meant for entertainment. He was determined to figure out how they
worked. Undeterred by his lack of science education, Franklin began experimenting.
He generated his own static electricity using a glass rod and silk cloth, and then recorded how the charge could
attract and repel lightweight objects. He read everything he could about this “electric fire” and became convinced
that a lightning bolt was a large-scale example of the same phenomenon.
In June 1752, Franklin and his 21-year old son conducted an experiment to test his theory. Although there is
some debate about the details, most historians agree that Franklin flew a kite on a stormy day to collect static
charges. Franklin explained that he and his son constructed a kite of silk cloth and two cedar strips. They attached
a metal wire to the top. Hemp string was used to fly the kite. A key was tied near the lower end. A silk ribbon was
affixed to the hemp, below the key.
It is probable that Franklin and his son were under some sort of shelter, to keep the silk ribbon dry. They flew the
kite, and once it was high in the sky they held onto it by the dry silk ribbon, not the wet hemp. Nothing happened
for a while. Then they noticed that the loose threads of the hemp suddenly stood straight up. The kite was
probably not struck directly by lightning, but instead collected charge from the clouds. Franklin touched his
knuckle to the key and received a static electric shock. He had proven that lightning was a discharge of static
electricity.
Through his experiments, Franklin determined that the “electric fire” was a single “fluid” rather than two
separate fluids, as European scientists had thought. He proposed that this “fluid” exists in two states, which he
called “positive” and “negative.” Franklin was the first to explain that if there is an excess buildup of charge on
one item, such as a glass rod, it must be exactly balanced by a lack of charge on another item, like the silk cloth.
Therefore, electric charge is conserved. He also explained that when there is a discharge of static electricity
between two items, the charges become balanced again. Many of Franklin’s electrical terms are still used today,
including battery, charge, discharge, electric shock, condenser, conductor, plus and minus, positive and negative.
Questions
1. Ben Franklin’s kite experiment was dangerous. Explain why.
2. Among Franklin’s many inventions is the lightning rod. Find out how the device works, and create a model or
diagram to show how it functions.
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Skill Sheet 20.1
Parallel and Series Circuits
There are two major types of electrical circuits: series and parallel. In a series circuit, current follows only one
path. In a parallel circuit, the current has two or more possible paths. In both types of circuits, the current
travels from the positive end of the battery toward the negative end. The amount of energy used by a circuit
(series or parallel) must equal the energy supplied by the battery. In this way, electrical circuits follow the law
of conservation of energy. Understanding these facts will help you solve problems that deal with series and
parallel circuits.
1. Solving series circuit problems
Now it is time for you to test your knowledge of series and parallel circuits by answering the questions below.
You will have to use Ohm’s law to solve many of the problems, so remember:
Voltage (volts) Current (amps) = ------------------------------------------Resistance (ohms)
Some questions ask you to calculate a voltage drop. We often say that each resistor creates a separate voltage
drop. As current flows along a series circuit, each resistor uses up some energy. As a result, the voltage gets lower
after each resistor. If you know the current in the circuit and the resistance of a particular resistor, you can
calculate the voltage drop using Ohm’s law.
Voltage drop (volts) = Current (amps) × Resistance of one resistor (ohms)
1.
Use the series circuit pictured at right to answer questions
a-e below.
a. What is the total voltage of the circuit?
b. What is the total resistance of the circuit?
c. What is the current flowing through the circuit?
d. What is the voltage drop across each light bulb? (Remember that voltage drop is calculated by
multiplying current in the circuit by the resistance of a particular resistor: V = IR.)
e. Draw the path of the current flow on the diagram.
1
Skill Sheet 20.1 Parallel and Series Circuits
2.
Use the series circuit pictured at right to answer questions
a-c below. Consider each resistor equal to all others.
a. What is the resistance of each resistor?
b. What is the voltage drop across each resistor?
c. On the diagram, show the amount of voltage in the circuit before and after each resistor.
3.
Use the series circuit pictured at right to answer questions a-d
below.
a. What is the resistance of the circuit?
b. What is the current flowing through the circuit?
c. What is the voltage drop across each resistor?
d. On the diagram, show the amount of voltage in the circuit before and after each resistor.
2. Solving parallel circuit problems
A parallel circuit has at least one point where the circuit divides, creating more than one path for current. Each
path is called a branch. The current through a branch is called branch current. Remember that if current flows
into a branch in a circuit, the same amount of current must flow out again. This rule is known as Kirchhoff’s
current law.
For example, suppose you have three light bulbs connected in parallel, and each has a current of 1 amp. The
battery must supply 3 amps since each bulb draws 1 amp. Before the first branch point, 3 amps are flowing. One
amp goes down the first branch to the first bulb, and 2 amps flow on to supply the next two bulbs.
2
Skill Sheet 20.1 Parallel and Series Circuits
1.
Use the parallel circuit pictured at right to answer
questions a-c below.
a. What is the total voltage for the circuit?
b. What is the current flow through each branch?
c. What is the voltage in each branch?
2.
Compare the circuits in Part 1, question 1 and Part 2, question 1. What is the current flow through each bulb
in the series circuit compared with the current flow through each bulb in the parallel circuit? Which bulbs
would be brighter? Explain your reasoning.
3.
Use the parallel circuit pictured at right to
answer questions a-d below.
a. What is the voltage through each branch?
b. What is the current flow through each
branch?
c. What is the power of each resistor? (Remember that power is current multiplied by voltage.)
d. What is the relationship between current and power?
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Skill Sheet 20.2
Network Circuits
Network circuits are combinations of series and parallel circuits. Therefore, solving problems that involve
networks circuits require using one or two formulas. In this skill sheet, you will practice solving these kinds of
problems.
1. The formulas for solving network circuit problems
When solving resistor network circuits we have to remember the following formulas:
1.
2.
Ohm’s law: V = IR
Combining resistors: Rtotal equals total resistance for the circuit:
a. In series: Rtotal = R1 + R2 + . . .
1 = ----1- + ----1- + . . .
b. In parallel: ----------R total
R1 R2
3.
4.
Kirchhoff’s voltage law: The sum of voltages around a loop is zero.
Kirchhoff’s current law: The sum of currents into a node equals the sum of currents out of the node.
Example problem:
For the following circuit, calculate the total resistance and the total current drawn from the 6-volt battery.
First notice that resistors R5 and R6 are in parallel and that they combine to give a resulting resistance of
1.5 ohms.
Next, notice that resistors R3 and R4 are in series resulting in a resistance of 2 ohms. The resulting 2 ohms
resistance is connected in parallel with another 2-ohm resistor. Thus the combined resistance of R3, R4, and R2 is
1 ohm. Now the circuit looks like:
We see now that the 1-ohm resistors are connected in series. Therefore, they represent a 2-ohm resistor connected
in parallel with the 1.5-ohm resistor.
The 2-ohm resistor in parallel with the 1.5 ohm gives a total resistance of 6⁄7 or 0.86 ohms.
The total current drawn from the battery can be now found by applying Ohm’s law:
volts = 7 A
I = V
--- = 6---------------R
6/7Ω
1
Skill Sheet 20.2 Network Circuits
2. Problems
1.
A 12-volt battery is connected to the resistor network shown on the schematic.
Calculate the current through each resistor in the network.
2.
Combine resistances and calculate the total resistance.
3.
Calculate the total current drawn from the 9-volt battery.
4.
For the circuit at right, trace the path of the current through the circuit and
then answer the following questions:
a. Does current flow through R5? Why or why not?
b. Given your answer to (a), what is the total resistance for the circuit?
c. Given your answer to (a), what is the total current leaving the battery?
d. What is the current through resistor R3?
e. What is the voltage across R5? Explain your answer.
2
Skill Sheet 20.2 Network Circuits
5.
Find the total current drawn from the 6-volt battery. What is the voltage
across resistor R4?
6.
What is the voltage across the 5-ohm resistor?
7.
Calculate the voltage across points A and B (VAB) when the resistor R4 is
10 Ω. HINT: Find the voltage at A (VA) and the voltage at B (VB) and subtract.
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Skill Sheet 20.3
Electrical Power
Which appliance in your kitchen uses the most power? The equation for electrical power is the tool you can
use to answer this question. This skill sheet will help you sharpen your skills at calculating electrical power
and analyzing the voltage, current, and power associated with electric circuits. You will also review the
relationship between electrical power and Ohm’s law. As you work through the problems, you will practice
calculating the power used by common appliances in your home.
1. What is electrical power?
Power is the rate at which work is performed. Power tells us how quickly work is being done or how much
energy is being used per unit time. When you work with machines, power is calculated by dividing work by the
amount of time it takes to perform the work. For electrical systems, the equation for power is:
power = voltage × current
P = V×I
where P = power, V = voltage and I = current. This equation allows us to assess the rate at which an appliance or
other device is using energy or performing work.
The graphic below shows you why power is called a rate. The unit for voltage is joules per coulomb. The unit for
current is coulombs per second. When you multiply voltage times current, the coulombs cancel so that the unit
for power is joules per second.
2. The unit for electrical power
One joule per second is equal to one watt of power. The watt (W) is a familiar unit of power to most people. You
can look at any appliance and see how many watts of power it uses. For example, the heating element on a
coffeemaker uses 1,050 watts of power. We can also say that the coffeemaker has a power rating of 1,050 watts.
One watt of power represents one joule of energy being used per second of time.
power = joules
-------------sec
joule
1 watt = 1---------------sec
The more watts of power a circuit or appliance has, the more energy it uses or the more work it can do per second
of time. A 1,500-watt microwave oven can perform the same amount of work as a 900-watt microwave;
however, the rate at which the 1,500W microwave performs the work is faster. Also, the amount of energy used
per second of time is greater (1,500 joules in one second compared with 900 joules in one second). As a result,
higher-wattage appliances are more expensive to operate.
1
Skill Sheet 20.3 Electrical Power
3. Practice problems
Complete the following problems. Be sure to show your work. The first problem has been done for you.
1.
A toaster oven has a power rating of 750 W. If the oven heats a piece of pizza for 360 seconds, how many
joules of energy have been used by the toaster oven?
of energypower = joules
------------------------------------time
of energy750 W = joules
------------------------------------360 sec
750 W × 360 sec = joules of energy = 270,000 joules
2.
You use your 60-watt DVD player to watch your favorite movie. If the player uses 324,000 joules of energy
playing the film, what is the running time of the movie?
3.
The current flowing through an electrical circuit is 9 amps. If the voltage in the same circuit is 120 volts,
what is the power of the circuit?
4.
A 7,200 -watt electric clothes dryer operates with a current of 30 amps. What is the voltage associated with
this circuit?
5.
A CD player uses 85 joules of energy per second. If the voltage in the CD player is 170 V, what amount of
current is required for the operation of player?
6.
A girl wants to build a radio that operates using 9-volt batteries. If the girl wishes the radio to function with
75 watts of power, with what amount of current will she have to design her circuits?
2
Skill Sheet 20.3 Electrical Power
7.
Your stereo has a power rating of 150 watts. Your friend buys a stereo with a power rating of 300 watts. If
you both play your stereos for one hour, who will spend more money to listen to their music? Explain your
answer.
8.
The voltage supplied to household circuits is generally 120 V. However, individual circuits (circuits
supplying the kitchen as opposed to the living room, for example) differ in the amount of current they carry.
What does that tell you about the amount of power operating in different household circuits? Why do you
think household circuits are designed in this way?
4. Comparing electrical power and mechanical power
Power is a term that is used when you talk about machines that use electricity, like blenders and refrigerators, and
mechanical machines, like pulleys. Look up the equation for mechanical power. Use this equation to answer the
following questions.
1.
Power can be calculated for electrical systems and mechanical systems. Write the equation for mechanical
power. What are the units for mechanical power?
2.
Compare the equations for electrical power and mechanical power. How are they alike, and how different?
3
Skill Sheet 20.3 Electrical Power
5. How do you calculate electrical power?
In everyday life we hear the word watt mentioned in reference to things like light bulbs and electric bills. The
watt is the unit that describes how much power is used when electricity flows. Therefore, the definition of power
is the “rate at which energy is flowing.” And since energy is measured in joules, power is measured in joules per
second. In fact, one joule per second is equal to one watt.
You may also have heard of the word kilowatt. A kilowatt is 1,000 watts or 1,000 joules of energy flowing in one
second. This term kilowatt is most often used with electrical use in houses and other large facilities. And on an
electric bill you may have noticed the term kilowatt-hour. A kilowatt-hour means that one kilowatt of power has
been used for one hour.
We can calculate the amount of electrical power used by an appliance or other electrical component by
multiplying the voltage by the current.
voltage × current = power, or P = VI
6. Solving problems
Solve the following problems using the power equation and Ohm’s law.
voltage (volts) current (amps) = ----------------------------------------resistance (ohms)
Remember, power is measured in watts.
1.
Your hair dryer has a power rating of 1,200 watts.
a. How many kilowatts is this?
b. If the hair dryer is used for 20 minutes per day, how many kilowatt-hours (kWh) per day is this?
(HINT: Convert 20 minutes to hours.)
c. Find the kilowatt-hours used by the hair dryer each month (assume 30 days/month).
d. If your town charges $0.15/kWh, what is the cost to use the hair dryer per month?
2.
Using the formula for power, calculate the amount of current through a 75-watt light bulb that is connected
to a 120-volt circuit in your home.
4
Skill Sheet 20.3 Electrical Power
3.
What is the power rating of a home appliance (in kilowatts) that uses 8 amps of current when plugged into a
120-volt outlet?
4.
The following questions refer to the diagram at right.
a. What is the total voltage for the circuit?
b. What is the total resistance for the circuit in ohms (Ω)?
c. What is the current that will flow through the circuit?
d. What is the power in watts for this circuit?
5.
A toaster is plugged into a 120-volt household circuit. It draws 5 amps of current.
a. What is the resistance of the toaster in ohms (Ω)?
b. What is the toaster’s power in watts? What is that power stated in kilowatts?
6.
A clothes dryer in a home has power of 4,500 watts and runs on a special 220-volt household circuit.
a. What is the current traveling through the dryer?
b. What is the clothes dryer’s resistance in ohms (Ω)?
5
Skill Sheet 20.3 Electrical Power
7.
A hair dryer is connected to a 120-volt household circuit. The current through the dryer is 10 amps.
a. What is the resistance of the hair dryer?
b. What is the hair dryer’s power rating in kilowatts?
c. If the dryer is used for 30 minutes per day, how many kilowatt-hours are used by the dryer each day?
d. How many kilowatt-hours are used per month? (Assume 1 month = 30 days.)
e. If the town charges 14 cents per kWh, what is the cost to run the hair dryer per month?
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Skill Sheet 21.2
Coulomb’s Law
In this skill sheet, you will work with Coulomb’s law. There are many similarities and some differences
between the equation of universal gravitation and the equation for Coulomb’s law. They are both inverse
square law relationships, and they both have similar arrangements of variables.
1. Introduction to Coulomb’s law
When two charges q1 and q2 are separated by a distance r, there exists a force between them that is given by:
where F equals the force in newtons and K is a constant equal to 9 × 109 N-m2/C2. The units of q1 and q2 are the
coulombs (C). Distance is given in meters. Here are a few important points about the relationships of the
variables in Coulomb’s law.
1.
2.
3.
4.
Force is inversely proportional to the square of the distance between the charges. Therefore, if the distance
increases by a factor of 2, the force decreases by a factor of 4.
Force is proportional to the strength of each charge.
When the two charges have the same sign (positive or negative), the force between them is repulsive because
like charges repel.
When the charges have opposite signs, the force between them is attractive because unlike charges attract.
2. Problems
1.
What happens to the force between two charges if the distance between them is tripled?
2.
What happens to the force between two charges if the distance between them is quadrupled?
3.
What happens to the force between two charges if the distance between them is cut in half?
4.
What happens to the force between two charges if the magnitude of one charge is doubled?
5.
What happens to the force between two charges is the magnitude of both charges is doubled?
1
Skill Sheet 21.2 Coulomb’s Law
6.
What happens to the force between two charges if the magnitude of both charges is doubled and the distance
between them is doubled?
7.
What happens to the force between two charges if the magnitude of both charges is doubled and the distance
between them is cut in half?
8.
Two particles, each with a charge of 1 C, are separated by a distance of 1 meter. What is the force between
the particles?
9.
Two equal charges separated by a distance of 0.5 meter experience a repulsive force of 1,000 newtons. What
is the strength in coulombs for each charge?
10. Two particles are each given a charge of 5 × 10-5 C. What is the force between the charged particles if the
distance between them is 2 meters? How does this force compare with the weight of a 5-kilogram cat?
11. The force between a pair of charges is 100 newtons. The distance between the charges is 1 centimeter. If one
of the charges is 2 × 10-10 C, what is the strength of the other charge?
12. The force between a pair of 0.001 C charges is 200 N. What is the distance between them?
13. The force between two charges is 1000 N. One has a charge of 2 × 10-5 C, and the other has a charge of
5 × 10-6 C. What is the distance between them?
14. The force between two charges is 2 newtons. The distance between the charges is 2 × 10-4 m. If one of the
charges is 3 × 10-6 C, what is the strength of the other charge?
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Skill Sheet 22.3
Magnetic Earth
An important and useful property of Earth is its magnetism. In this skill sheet, you will review concepts you
read about in 22.3 The Magnetic Field of Earth.
1. Earth is magnetic
The magnetic field of Earth is very weak compared with the strength of the field on the surface of the ceramic
magnets you probably have in your classroom. A small ceramic permanent magnet has a field of a few hundred
up to 1,000 gauss at its surface. The gauss is a unit used to measure the strength of a magnetic field. At Earth’s
surface, the magnetic field averages about 0.5 gauss. Of course, the field is much stronger nearer to the core of
the planet.
1.
What is the source of Earth’s magnetic field according to what you have read in section 22.3?
2.
Today, Earth’s magnetic field is losing approximately 7 percent of its strength every 100 years. If the
strength of Earth’s magnetic field at its surface is 0.5 gauss today, what will it be 100 years from now?
3.
If Earth’s magnetic field continues to lose strength, what might happen? Describe this event.
4.
The graphic at right illustrates one piece of evidence that proves the reversal of
Earth’s poles during the past millions of years. The ‘crust’ of Earth is a layer of
rock that covers Earth’s surface. There are two kinds of crust—continental and
oceanic. Oceanic crust is made continually (but slowly) as magma from Earth’s
interior erupts at the surface. Newly formed crust is near the site of eruption and
older crust is at a distance from the site. Based on what you know about
magnetism, why might oceanic crust rock be a record of the reversal of Earth’s
magnetic field? (HINT: What happens to materials when they are exposed to a
magnetic field?)
2. How does a compass work?
1.
The terms magnetic south pole and geographic north pole refer to locations on Earth. The magnetic south
pole is the point on the surface above Earth’s south pole is if you think of Earth as a giant bar magnet.
Geographic north is the point on the surface that we think of as ‘north.’ Explain these terms by answering the
following questions.
a. Are the locations of the magnetic south pole and the geographic north pole near Antarctica or the Arctic?
1
Skill Sheet 22.3 Magnetic Earth
b. How far is the magnetic south pole from the geographic north pole?
c. In your own words, define the difference between the magnetic south pole and the geographic north pole.
2.
A compass is a magnet and Earth is a magnet. How does the magnetism of a compass work with the
magnetism of Earth so that a compass is a useful tool for navigating?
3.
The directions—north, east, south, and west—are arranged on a compass so that they align with 360 degrees.
This means that zero degrees (0°) and 360° both represent north. For each of the following directions by
degrees, write down the direction in words. The first one is done for you.
a. 45° Answer: The direction is northeast.
b. 180°
c. 270°
d. 90°
e. 135°
f. 315°
3. Magnetic declination
Earth’s geographic north pole (true north) and magnetic south pole are located near each other, but they are not at
the same exact location. Because a compass needle is attracted to the magnetic south pole, it points slightly east
or west of true north. The angle between the direction a compass points and the direction of the geographic north
pole is called magnetic declination. Magnetic declination is measured in degrees and is indicated on
topographical maps.
1.
Let’s say you were hiking in the woods and relying on a compass to navigate. What would happen if you
didn’t correct your compass for magnetic declination?
2.
Are there places on Earth where magnetic declination equals 0°? If so, research on the Internet where on the
globe you would have to be to have 0° magnetic declination.
2
Name:
Skill Sheet 23.1
Magnetic Fields and Forces
The movement of charge in space creates some very interesting and useful phenomena. When current passes
through a wire, it creates a magnetic field around the wire. Magnetic field, like current, has a direction. This
skill sheet gives you practice calculating the strength or force of a magnetic field near a wire and at the center
of a coil of wire when current is being conducted.
1. Magnetic field near a wire
The direction of the magnetic field around a wire can be found by pointing the thumb of your right hand in the
direction of the current. The direction that your fingers curl is equivalent to the direction of the magnetic field.
The strength of the magnetic field decreases as the distance (R) from the wire increases and it is given by:
Magnetic field is a vector and, as such, has magnitude and direction. The unit of the magnetic field is the tesla
(T). Another common unit is the gauss. The relationship between the tesla and the gauss is:
1 tesla = 10,000 gauss . Earth’s magnetic field is about 0.5 gauss.
1
Skill Sheet 23.1 Magnetic Fields and Forces
2. Magnetic field at the center of a coil
When you take a wire and wrap it around a cylinder and you look at a cross-section of the wire, the total current
(ITotal) is equal to the current (I) in the wire multiplied by the number of turns (N).
I total = NI
The magnetic field at the center of the coil is given by:
When a current carrying wire is placed in a magnetic field, it experiences a
force. This magnetic force is perpendicular to both the direction of the magnetic
field and the direction of the current.
The direction of the force can be found by applying the following version of the
right-hand rule.
Point fingers in the direction of the current, wrap fingers in the direction of the
magnetic field, and the thumb points in the direction of the force.
The equations can be used to find the strength of a magnetic field around a wire
and through a loop. When I is given in amperes and R in meters, the magnetic
field is in tesla.
3. Problems
1.
A straight wire is carrying a current of 10 amperes. What is the magnitude of the magnetic field at a distance
of 0.5 meters from the wire? Express your answer in tesla and gauss.
2.
At what distance is the magnetic field due to the current in the wire of problem 1 equal to Earth’s magnetic
field?
2
Skill Sheet 23.1 Magnetic Fields and Forces
3.
On the diagram at right, draw an arrow to indicate the direction of the force
acting on the current-carrying wire.
The dot in the graphic represents the magnetic field vector ( B ) pointing
vertically out of the page. (HINT: Use the right-hand rule in Part 2.)
4.
Two parallel straight wires carry a current I in the same direction. The wires are separated by a distance r.
What is the direction of the force from each wire? Are the wires attracting or repelling each other?
5.
Referring to problem 4 above, if the distance between these wires doubles from R to 2R, by what factor has
the force of the magnetic field changed? Write the equation that helps you answer this question.
6.
A current of 10 amps flows in a coil made from 200 turns of thin wire. The diameter of the coil is
1.5 centimeters. Calculate the strength of the magnetic field (in tesla) at the center of the coil.
7.
The strength of the magnetic field at the center of a coil of thin wire is 1000 gauss. A current of 2 amps flows
in the coil and it has a diameter of 1 centimeter. How many turns of wire are in the coil?
3
Name:
Skill Sheet 23.3
Michael Faraday
Despite little formal schooling, Michael Faraday rose to become one of England’s top research scientists of
the nineteenth century. He is best known for his discovery of electromagnetic induction, which made possible
the large-scale production of electricity in power plants.
Michael Faraday was born in Surrey, England, in 1791, the son of a blacksmith. His
family soon moved to London, where Michael received a rudimentary education at a
local school. At age 14, he was apprenticed to a bookbinder. He enjoyed reading the
materials he was asked to bind, and found himself mesmerized by scientific papers that
outlined new discoveries. A wealthy client of the bookbinder noticed this voracious
young reader and gave him some tickets to hear Humphrey Davy, a prominent British
chemist, give a series of lectures to the public.
Faraday took detailed notes at each lecture. He bound his notes and sent them to Davy,
asking him for a job. In 1812, Davy hired him as a chemistry laboratory assistant at the
Royal Institution, London’s top scientific research facility.
Despite his lack of formal training in science or math, Faraday was an able assistant and soon began independent
research in his spare time. In the early 1820s, he discovered how to liquefy chlorine and became the first to
isolate benzene, an organic solvent with many commercial uses.
Faraday was also interested in electricity and magnetism. After reading about the work of Hans Christian
Oersted, the Danish physicist, chemist, and electromagnetist, Faraday repeated Oersted’s experiments and used
what he learned to build a machine that used an electromagnet to cause rotation—the first electric motor. Next,
he tried to do the opposite—use a moving magnet to cause an electric current. In 1831, he succeeded. His
discovery is called electromagnetic induction, and it is used by power plants to generate electricity even today.
Faraday first developed the concept of a field to describe magnetic and electric forces, and used iron filings to
demonstrate magnetic field lines. He also conducted important research in electrolysis and invented a voltmeter.
He was interested in finding a connection between magnetism and light. In 1845 he discovered that a strong
magnetic field could rotate the plane of polarized light. Today this is known as the Faraday effect.
Faraday was a teacher as well as a researcher. When he became director of the Royal Institution laboratory in
1825, he instituted a popular series of Friday Evening Discourses. Here paying guests (including Prince Albert,
Queen Victoria’s husband) were entertained with demonstrations of the latest discoveries in science. He also
instituted the Christmas Lectures for Children, which continue to this day.
Faraday continued his work at the Royal Institution until just a few years before his death in 1867. Two units of
measure have been named in his honor: the farad, a unit of capacitance, and the faraday, a unit of charge.
Questions
1.
2.
Name two ways that Michael Faraday’s work affects your own life in the twenty-first century.
Use iron filings and a magnet to demonstrate magnetic field lines, or prepare a simple demonstration of
electromagnetic induction for your classmates.
1
Name:
Skill Sheet 24.3
Binary Number Problems
This skill sheet will give you practice understanding and using the binary number system.
The digital number system uses Base 10 digits, “0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.” Each decimal number consists of
a series of digits in columns or places representing a power of 10. The column farthest to the right represents 100,
the next place to the left is 101, then 102. You recognize these power of 10 places as the “1’s” place, the “10’s”
place, and the “100’s” place.
The binary number system uses Base 2, or binary digits “1” and “0” to represent a number. Each binary digit is
called a bit (for binary digit) and each column or place in a binary number represents a power of 2. The column
farthest to the right represents 20, then 21, 22, and so on. A 4-bit binary number has the “1’s” place, the “2’s”
place, the “4’s” place, and the “8’s place”.
1. Decimal to binary problems
1.
Convert the decimal numbers 0-15 into 4-bit binary numbers in the table below. Four examples are done for
you. (HINT: It helps to use the power of 2 place holders for each of the four binary digits. This is why the
binary digits 8-4-2-1 are provided in the second row of the table.)
Number
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
8
0
0
Binary Digits
4
2
0
0
0
0
1
0
1
1
0
0
1
1
0
1
1
1
Skill Sheet 24.3 Binary Number Problems
2.
How many binary digits does it take to convert the decimal number 16 into a binary number?
3.
Convert the following decimal numbers into 8-bit binary numbers.
a. 99
b. 127
4.
What happens if you try to convert the decimal number 128 into a 6-bit binary number?
2. Binary to decimal problems
1.
Convert the following binary numbers into decimal numbers. The first one is done for you. (HINT: Use the
power of 2 binary digits as place holders to help you determine the digital number. In some cases you will
need eight binary digits or just two. However, eight binary digits are written down in each table.)
a. 11001100
Binary digits
Binary number
Addition
Digital number
128
1
64
32
16
8
4
2
1
0
0
1
1
0
128 + 64 + 0 + 0 + 8 + 4 + 0 + 0 = 204
204
1
0
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
b. 10101010
Binary digits
Binary number
Addition
Digital number
c. 1111 1111
Binary digits
Binary number
Addition
Digital number
2
Skill Sheet 24.3 Binary Number Problems
d. 001001
Binary digits
Binary number
Addition
Digital number
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
128
64
32
16
8
4
2
1
e. 110
Binary digits
Binary number
Addition
Digital number
f. 10
Binary digits
Binary number
Addition
Digital number
g. 11
Binary digits
Binary number
Addition
Digital number
h. 000010
Binary digits
Binary number
Addition
Digital number
3
Name:
Skill Sheet 25.2
Temperature Scales
Temperature, a measure of the average kinetic energy of the molecules of a substance, has an important role in
our daily lives. Whether we are cooking dinner, dressing for school, or suffering from a cold, we often wish to
know something about the temperature of our environment or some body of matter. To report values for
temperature, the Fahrenheit and Celsius scales are commonly used by scientists and society. In this exercise,
you will examine these scales and practice converting temperature values from one scale to the other.
1. Fahrenheit and Celsius
The Fahrenheit and Celsius temperature scales are the most commonly used scales for reporting temperature
values. Scientists use the Celsius scale almost exclusively, as do many countries of the world. Some countries,
such as the United States, still rely heavily on the Fahrenheit scale for reporting temperature information. When
the weatherman on television tells you the forecast for the week, the scale he is using for temperature is
Fahrenheit. Your oven is calibrated in Fahrenheit values, as is the thermometer your doctor uses to assess your
health. In the United States, we are comfortable with the Fahrenheit scale and design our appliances and tools
using this system of measurement.
However, if you were in Europe, you would see temperatures given in degrees Celsius. Scientists use the Celsius
scale for their experiments and report their results in degrees Celsius. Therefore, it may be necessary at some
point in time to convert information reported in degrees Celsius to degrees Fahrenheit or vice versa. To
accomplish this, we use conversion formulas.
2. Converting Fahrenheit values into Celsius
If you have been given temperature information in degrees Fahrenheit (°F) and need the values to be reported in
degrees Celsius (°C), you would use the following formula:
5
°C = --- ( °F – 32 )
9
Example:
What is the Celsius value for 65° Fahrenheit?
5
°C = --- ( 65°F – 32 )
9
°C = 18.3
1
Skill Sheet 25.2 Temperature Scales
3. Practice converting Fahrenheit to Celsius values
1.
The weatherman tells you that today will reach a high of 45°F. Your friend from Sweden asks what the
temperature will be in degrees Celsius. What value would you report to your friend?
2.
Your father orders a fancy oven from England. When it arrives, you notice that the temperature dial is
calibrated in degrees Celsius. You wish to bake a cake at 350°F. At what temperature will you have to set the
dial on this new oven?
3.
Your new German automobile's engine temperature gauge reads in Celsius, not Fahrenheit. You know that
the engine temperature should not rise above about 225°F. What is the corresponding Celsius temperature on
your new car's gauge?
4. Converting Celsius to Fahrenheit values
To convert Celsius temperature values to degrees Fahrenheit, you must again use a conversion formula:
9
°F =  --- × °C + 32
5
Example:
200°C is the same temperature as what value on the Fahrenheit scale?
9
°F = --- ( 200°C ) + 32
5
°F = 392
2
Skill Sheet 25.2 Temperature Scales
5. Practice converting Celsius to Fahrenheit values
1.
Your grandmother in Ireland sends you her favorite cookie recipe. Her instructions say to bake the cookies at
190.5°C. To what Fahrenheit temperature would you set the oven to bake the cookies?
2.
A scientist wishes to generate a chemical reaction in his laboratory. The temperature values in his laboratory
manual are given in degrees Celsius. However, his lab thermometers are calibrated in degrees Fahrenheit. If
he needs to heat his reactants to 232°C, what temperature will he need to monitor on his lab thermometers?
3.
You phone a friend who lives in Denmark and tell him that the temperature today only rose as high as 15°F.
He replies that you must have enjoyed the warm weather. Explain his answer using your knowledge of the
Fahrenheit and Celsius scales and conversion formulas.
6. Extension: the Kelvin temperature scale
For some scientific applications, a third temperature scale is used: the Kelvin scale. The Kelvin scale is calibrated
so that raising the temperature one degree Kelvin raises it by the same amount as one degree Celsius. The
difference between the scales is that 0°C is the freezing point of water, while 0 K is much, much colder. On the
Kelvin scale, 0 K (degree symbols are not used for Kelvin values) represents absolute zero. Absolute zero is the
temperature when the average kinetic energy of a perfect gas is zero—the molecules display no energy of
motion. Absolute zero is equal to -273°C, or -459°F. When scientists are conducting research, they often obtain
or report their temperature values in Celsius, and other scientists must convert these values into Kelvin for their
own use, or vice versa. To convert Celsius values to their Kelvin equivalents, you would use the formula:
K = °C + 273
Example:
Water boils at a temperature of 100°C. What would be the corresponding temperature for the Kelvin scale?
K = °C + 273
K = 100°C + 273
K = 373
3
Skill Sheet 25.2 Temperature Scales
To convert Kelvin values to Celsius, you would perform the opposite operation; subtract 273 from the Kelvin
value to find the Celsius equivalent.
Example:
A substance has a melting point of 625 K. At what Celsius temperature would this substance melt?
°C = K – 273
°C = 625 K – 273
°C = 352
Although we rarely need to convert between Kelvin and Fahrenheit, use the following formulas to do so:
5
K = --- ( °F + 460 )
9
9
°F =  --- × K – 460
5

Solving problems
1.
A gas has a boiling point of -175°C. At what Kelvin temperature would this gas boil?
2.
A chemist notices some silvery liquid on the floor in her lab. She wonders if someone accidentally broke a
mercury thermometer, but did not thoroughly clean up the mess. She decides to find out of the silver stuff is
really mercury. From her tests with the substance, she finds out that the melting point for the liquid is 275 K.
A reference book says that the melting point for mercury is -38.87°C. Is this substance mercury? Explain
your answer and show all relevant calculations.
3.
You are at a Science Camp in Florida. It’s August 1st. Today’s activity is an outdoor science quiz. The first
question on the quiz involves a thermometer that reports the current temperature as 90°. You need to state the
temperature scale in which this thermometer is calibrated: Kelvin, Fahrenheit, or Celsius. Which scale is
correct? Defend your answer with your knowledge of the temperature scales.
4
Name:
Skill Sheet 26.3
Heat Transfer
In this skill sheet, you will calculate the heat transfer in watts for conduction, convection, and radiation in
simple systems. You will make your calculations using the heat conduction equation, the convection equation,
and the Stefan-Boltzmann equation.
1. The heat conduction equation
Solve the following problems using the heat conduction equation. The first problem is done for you.
1.
A copper bar connects two beakers. The water in one beaker is 25°C. The water in the other beaker is 75°C.
The cross-sectional area of the 0.75-meter bar is 0.0004 m2. How many watts of heat are conducted through
the bar from the hot to the cold beaker? The thermal conductivity of copper is 401 W/m°C.
°
2
°
°
( 401 W/m C ) × ( 0.0004 m ) × ( 75 C – 25 C )- = 10.7 watts
PH = -------------------------------------------------------------------------------------------------------------0.75 m
2.
A copper bar that is 0.75 m long and has a cross section of 0.0004 m2 connects two beakers. If 20 watts of
heat are conducted through the bar, what is the temperature difference between the two beakers?
3.
A piece of aluminum and a piece of steel have the same length and the same cross-sectional area.
Aluminum’s thermal conductivity is 226 W/m°C and the thermal conductivity of steel is 43 W/m°C. The
piece of aluminum connects two regions with a temperature difference of 10°C. For the piece of steel to
conduct the same amount of heat as the aluminum, what would the temperature difference have to be
between two regions connected by the steel? (HINT: Plug the values into a heat equation for each material
and set the equations equal to each other.)
4.
The air temperature is 5°C and the temperature of coffee inside a styrofoam cup is 35°C. The thermal
conductivity of styrofoam is 0.025 W/m°C. Let’s say the length of the styrofoam is the thickness of the cup
or 0.003 meter (3 millimeters). The cross-sectional area that we will use is 0.0004 m2 (2 centimeters × 2
centimeters). How many watts of heat flow through the styrofoam coffee cup in this area?
1
Skill Sheet 26.3 Heat Transfer
5.
Write a short paragraph, describing an example (not mentioned in the text book) of how the heat conduction
equation is used in the real world. You may need to do some research in your library or on the Internet.
2. The convection equation
Solve the following problems using the convection equation. The first problem is done for you.
1.
A wind at 15°C blows on the surface of a window that is 21°C. The heat transfer coefficient is 70 W/m2-°C.
If the window is 0.5 meter2, what is the heat transfer in watts?
2°
2
°
°
P H = ( 70 W/m C ) × ( 0.5 m ) × ( 21 C – 15 C ) = 210 watts
2.
On a hot day, a window is 25°C and the air temperature outside 32°C. The heat transfer coefficient is 10
W/m2-°C. If the area of the window is 0.5 meter2, what is the heat transfer in watts?
3.
When comparing the free and forced convection of gases, oil, and water, water has the highest values. The
lowest values are for gases. The values for oil are closer to the values for gas than for water. Come up with a
hypothesis that explains why this may be so. (HINT: See Table 26.2 in chapter 26.)
4.
The range of heat transfer coefficients for water when convection is ‘free’ is 100 - 1,000 W/m2-°C. The
range when convection is forced is 100 - 20,000 W/m2-°C. Explain the large difference between these
values. What is the difference between free and forced convection?
5.
Write a short paragraph, describing an example (not mentioned in the text book) of how the convection
equation is used in the real world. You may need to do some research in your library or on the Internet.
2
Skill Sheet 26.3 Heat Transfer
3. The Stefan-Boltzmann equation
Solve the following problems using the Stefan-Boltzmann equation. The first problem is done for you.
1.
The power radiated by a filament is 0.3 watts. The diameter of the filament is 0.5 millimeters and it has a
length of 50 millimeters. If the surface area is 4 × 10-8 m2, what is the temperature?
P = 0.3 watts
–8
5.7 × 10 W–8 2
4
0.3 watts = ----------------------------×
(
4
×
10
m
)
×
(
Temperature
)
2
m K
4
14
1.3 × 10 = Temperature
3377 K = Temperature
2.
The surface area of a filament is 4 × 10-8 m2. What is the power in watts if the temperature is 1,727°C?
3.
The power radiated by a surface is 0.125 watts. The temperature of this surface is 1500°C. What is the area
of the surface?
4.
The area of a surface is one square meter and its temperature is 273 K. How much heat is this surface
radiating in units of watts of power? What is the temperature of this surface in degrees Celsius?
5.
Regarding radiation, is heat transfer at 0°C less than, greater than, or the same to heat transfer at absolute
zero? Explain your answer.
6.
Write a short paragraph, describing an example (not mentioned in the text book) of how the StefanBoltzmann conduction equation is used in the real world. You may need to do some research in your library
or on the Internet.
3
Name:
Skill Sheet 27.1
Stress and Strain
The stress in a material is the ratio of the force acting through the material divided by the cross-section area
through which the force is carried. The cross-section area is the area perpendicular to the direction of the
force. Dividing force by cross-section area (mostly) separates out the effects of size and shape from the
strength properties of the material itself. Stress (signified by the Greek letter sigma, σ) is force (F) divided by
cross-section area (A).
1. Introduction to stress
Stress (σ) is defined as force (F) divided by cross-section area (A).
Force is what we apply, while stress is what the material experiences.
For example, when a 10-kilogram ball hangs from a steel wire that has a cross-sectional area of 12 mm2, the
stress experienced by the wire is given below. One pascal (Pa) is equal to one newton of force per square meter of
area. A megapascal (MPa) is equal to one million pascals.
2
N4
10 kg ) ( 9.8 m/sec )- = -----------------------98 N - = 8.17 × 10 4 -----σ = (-----------------------------------------------= 8.17 × 10 Pa
2
12 mm
2
0.0012 m
2
m
The wire will break when the stress that it experiences exceeds the tensile strength of the material.
6 N
The tensile strength of steel is: 400 × 10 ------2m
Therefore, we can predict that a mass of 49,000 kilograms would break a steel wire with this cross-sectional area.
2
N6
49, 000 kg ) ( 9.8 m/sec -) = -----------------------4802 N - = 400 × 10 6 -----σ = (-----------------------------------------------------------= 400 × 10 Pa = 4.00 MPa
2
12 mm
2
0.0012 m
2
m
There are two ways to talk about the stress experienced by objects. First, stress can be tensile. For example, the
suspension cables in a bridge experience tensile, or pulling, stress. Second, stress can be compressive. For
example, the pillars supporting that bridge experience compressive stress.
1
Skill Sheet 27.1 Stress and Strain
2. Introduction to strain
When force is applied to a material or, equivalently, when a material is under stress, it deforms. When we pull a
rubber band it stretches. When we pull on a piece of metal wire, it also changes length, but not as noticeably as
does the rubber band. Also, when we compress a material, it deforms. The amount of deformation of a material
under stress is called strain. Strain (signified by the Greek letter epsilon, ε) is related to stress by the formula:
N
where E is called the modulus of elasticity and it is given in units of ------2- , or pascals (Pa).
m
Strain (ε) is a dimensionless quantity. It represents the amount of deformation of the material.
in length- = ∆l
ε = change
-----------------------------------------original length
l
As we know, materials expand and contract as temperature changes. This is called thermal expansion or
contraction, and it is also represented by the thermal strain (εθ), which is given by:
where α is the coefficient of thermal expansion and (T2– T1) represents the temperature change. Note that for this
1
equation, the units of α are given in -------- . If the value is negative, the material is contracting.
°C
These three formulas are very powerful and are used extensively in the design and evaluation of structures. We
will also use them in the examples to analyze and design some simple structural elements. The only information
that we need is properties of the materials used. These properties are represented by the parameters E, α, and the
tensile strength of the material.
2
Skill Sheet 27.1 Stress and Strain
3. Problems
1.
A cylindrical concrete pier with a radius of 0.25 meter is loaded with a force of 700 kilonewtons. Find the
stress (force per unit area) and the strain. Concrete has E = 17 × 109 N/m2. The formula for finding the area
2
of a circle is: πr
2.
Concrete is used in building many structures. Concrete has a coefficient of thermal expansion
α = 1 × 10-5/°C. Two pieces of concrete are used to build a bridge. The bridge has a total length of 50 meters,
and the two concrete slabs are anchored at the ends and meet in the middle of the bridge. The design is such
that the gap between the slabs closes in the summer when the temperature reaches 40°C. What is the
maximum gap in the winter when the temperature drops to -10°C? (HINT: Use the thermal strain formula,
and multiply your answer by the total length of the bridge to find the gap.)
3.
A weight of 1,500 kilograms is to be suspended by a wire. We have been asked to design the system using
steel wire. Steel has a tensile strength of 400 megapascals and a modulus of elasticity E = 200 × 109 Pa. We
will do the design in steps.
a. First, what is the minimum wire diameter that satisfies the tensile strength?
b. What is the minimum diameter if a safety factor of 10 is required? A safety factor of 10 means that you
design the wire to support 10 times the weight — so we need to design the wire to support 15,000 kg.
4.
If the weight is supported by 21.6-millimeter-diameter wire with a total length of 5 meters, what is the strain
on the wire? What is the change in length of the wire in meters?
3
Name:
Skill Sheet 27.2
Archimedes
Archimedes was a Greek mathematician who specialized in geometry. He figured out the value of pi and the
volume of a sphere, and has been called “the father of integral calculus.” During his lifetime, he was famous
for using compound pulleys and levers to invent war machines that successfully held off an attack on his city
for three years. Today he is best known for the Archimedean principle, which was the first explanation of how
buoyancy works.
Archimedes was born in Syracuse, on Sicily (then an independent Greek city-state), in
287 B.C. His letters suggest he studied in Alexandria, Egypt, as a young man.
Historians believe it was in Egypt that he invented a device for raising water known as
Archimedes’ screw. The device is still used in many parts of the world.
A famous Greek legend says that King Hieron II of Syracuse asked Archimedes to
figure out if his new crown was pure gold or if the craftsman had mixed some less
expensive silver into it. Archimedes had to determine the answer without destroying
the crown. He thought about it for days and then, as he lowered himself into a bath, the
answer struck him. The legend says Archimedes ran naked through the streets,
shouting “Eureka!”—meaning “I have found it.”
Archimedes realized that if he had equal masses of gold and silver, the denser gold would have a smaller volume.
Therefore, the gold would displace less water than the silver when submerged.
Archimedes found the mass of the crown, and then made a bar of pure gold with the same mass. He submerged
the gold bar and measured the volume of water it displaced. Next, he submerged the crown. He found that the
crown displaced more water than the gold bar had and, therefore, could not be pure gold. The gold had been
mixed with a less dense material. Archimedes had solved the mystery.
Archimedes wrote a treatise titled “On Floating Bodies,” further exploring density and buoyancy. He explained
that an object immersed in a fluid is buoyed upward by a force equal to the weight of the fluid displaced by the
object. Therefore, if an object weighs more than the fluid it displaces, it will sink. If it weighs less than the fluid
it displaces, it will float. This statement is known as the Archimedean principle. Although we commonly assume
the fluid is water, the statement holds true for any fluid, whether liquid or gas. A helium balloon floats because
the air it displaces weighs more than the balloon filled with lightweight gas.
Archimedes wrote several other treatises, including “On the Sphere and the Cylinder,” “On the Measurement of
the Circle,” “On Spirals,” and “The Sand Reckoner.’ In this last treatise, he devised a system of exponents that
allowed him to represent very large numbers on paper—large enough, he said, to count the grains of sand that
would be needed to fill the universe.
Archimedes was killed by a Roman soldier during an invasion of Syracuse in 212 B.C.
Questions
1. Draw a diagram showing how Archimedes solved the problem of determining the gold crown’s purity.
2. Research one of Archimedes’ inventions and create a poster that shows how the device worked.
1
Name:
Skill Sheet 27.3
Gas Laws
This skill sheet reviews the gas laws and includes practice problems that utilize these laws.
1. Boyle’s law: pressure and volume
The relationship between the volume of a gas and the pressure of a gas, at a constant temperature, is known as
Boyle’s law. The equation for Boyle’s law is:
Here’s how you solve a problem using this relationship:
A kit used to fix flat tires consists of an aerosol can containing compressed air and a patch to seal the hole in the
tire. Suppose 10.0 liters of air at atmospheric pressure (101.3 kilopascals, or kPa) is compressed into a 1.0-liter
aerosol can. What is the pressure of the compressed air in the can?
1. Identify what you know and what you are trying to find out from the information given.
P1 = 101.3 kPa
V1 = 10.0 L
P2 = unknown
V2 = 1.0 L
2. Rearrange the variables in the equation to solve for the unknown variable.
Divide each side by V2 to isolate P2 on one side of the equation. The final equation is:
P1 V1
P 2 = ----------V2
3. Plug in the values and solve the problem.
kPa × 10.0 L- = 1,013 kPa
P 2 = 101.3
--------------------------------------------1.0 L
The pressure inside the aerosol can is 1,013 kPa.
1
Skill Sheet 27.3 Gas Laws
2. Charles’ law: pressure and temperature
The French scientist Jacques Charles was a pioneer in hot-air ballooning. He investigated how changing the
temperature of a fixed amount of gas at constant pressure affected its volume. The Charles’ law equation is:
Charles’ law shows a direct relationship between the volume of a gas and the temperature of a gas when the
temperature is given in the Kelvin scale. Zero on the Kelvin scale is the coldest possible temperature, also known
as absolute zero. Absolute zero is equal to -273°C which is 273°C below the freezing point of water. Why do you
think this scale is used to solve these problems?
Converting from degrees Celsius to Kelvin is easy — you add 273 to the Celsius temperature. To convert from
Kelvins to degrees Celsius, you subtract 273 from the Kelvin temperature.
To solve problems with Charles’ law, you can follow the same problem-solving steps you learned for Boyle’s
law, except you use the equation for Charles’ law. You also need to convert degrees Celsius to Kelvin. To practice
both equations, do the problems below.
3. Practice problems with Bolyle’s and Charles’ laws
1.
A truck tire holds 25.0 liters of air at 25°C. If the temperature drops to 0°C, and the pressure remains
constant, what will be the new volume of the tire?
(HINT: Remember to convert degrees Celsius to Kelvins.)
2.
You pump 25.0 liters of air at atmospheric pressure (101.3 kPa) into a soccer ball that has a volume of
4.5 liters. What is the pressure inside the soccer ball if the temperature does not change?
3.
Hyperbaric oxygen chambers (HBO) are used to treat divers with decompression sickness. Research has
shown that HBO can also aid in the healing of broken bones and muscle injuries. As pressure increases
inside the HBO, more oxygen is forced into the bloodstream of the patient inside the chamber. To work
properly, the pressure inside the chamber should be three times greater than atmospheric pressure
(101.3 kPa). What volume of oxygen, held at atmospheric pressure, will need to be pumped into a 190-liter
HBO chamber to make the pressure inside three times greater than atmospheric pressure?
4.
A balloon holds 20.0 liters of helium at 10°C. If the temperature doubles, and the pressure does not change,
what will be the new volume of the balloon?
5.
A scuba tank holds 12.5 liters of oxygen at 101.3 kPa. If the oxygen pumped into the scuba tank was held at
a pressure of 202.6 kPa, what was the original volume of the gas?
2
Skill Sheet 27.3 Gas Laws
4. The ideal gas law
The ideal gas law combines the pressure, volume, and temperature relationships for a gas into one equation that
also includes the mass of the gas. The ideal gas law equation is:
In physics and engineering, mass (m) is used for the quantity of the gas in units of kilograms. The parameter R is
called the gas constant, and it has units of (J/kg-K). Table 27.7 on page 559 of the Student Text shows the value
for R for several common gases. Each different gas has its own value for R. The temperature, T, is given in
Kelvin. The pressure of the gas, P, is given in units of N/m2, also called pascals, and volume, V, is given in units
of m3.
To use the ideal gas law, pressure needs to be absolute pressure, not gauge pressure. Here is how to calculate
absolute pressure from gauge pressure:
absolute pressure = gauge pressure + atmospheric pressure (101,000 N/m2)
5. Practice problems using the ideal gas law
1.
A sample of helium gas occupies 5 liters at 22°C and a gauge pressure of 200,000 pascals. What is the mass
of the gas? (The gas constant for helium is 2.08 × 103 J/kg-K.)
2.
Helium gas of mass m occupies a volume of 2 liters at temperature of 30°C and atmospheric pressure. The
gas is then heated to 100°C and the pressure increases to 5 atmospheres. Find the mass of the gas and the
change in volume, if any.
3.
A gas of mass, m, occupies a volume, V1, at a pressure of 5 atmospheres and a temperature of 25°C. The gas
is heated to 200°C while the volume is increased by 20 percent. What is the final pressure? HINT: Solve the
problem using the formula:
P1
V
------ = -----2P2
V1
3
Name:
Skill Sheet 28.1A
The Structure of the Atom
You are probably familiar with the structure of an atom: Protons and neutrons are found in the nucleus, and
electromagnetic force loosely holds a number of electrons about the atom equal to the number of protons. The
number of protons in the nucleus is the atomic number and identifies the kind of atom, or element. When the
number of electrons and protons is equal, the atom is neutral. In this skill sheet, you will exercise your
understanding of the structure of an atom.
1. The anatomy of a nucleus
The graphic at right is an illustration of an atom of helium. This particular form
4
of helium is 2He , read “helium four.” We know it is a helium atom because it
has two protons; all helium atoms have two protons. The mass of this atom is
4 amu (atomic mass units). This number is equal to the number of protons and
neutrons in the nucleus. The number of electrons match the number of protons,
so the atom is neutral.
This information is easily determined for any atom if you know the following
rules.
•
The number of protons in a nucleus is the atomic number and identifies
an atom.
•
Protons and neutrons are much heavier than electrons and make up most of the atom’s mass.
•
Atoms with the same atomic number but different numbers of neutrons are called isotopes.
•
Unless otherwise stated, each atomic mass value shown on a periodic table of elements is based on the mass
number and abundance of naturally-occurring isotopes. For example, the atomic mass of carbon, 12.0111, is
based on the mass number and abundance of each of its isotopes (carbon-12, carbon-13, and carbon-14).
Because the mass number is close to the whole number “12,” we can assume that the most abundant isotope
of carbon is carbon-12 (whose mass number is 12).
•
Protons carry a charge of +1, neutrons carry no charge, and electrons carry a charge of -1.
•
The electromagnetic force loosely holds a number of electrons about the atom equal to the number of
protons. When the number of electrons and protons is equal, the atom is neutral.
2. Element identification
The following graphics represent the nuclei of atoms. Using the rules listed above and a periodic table of
elements, identify the element represented by each graphic.
1.
1
Skill Sheet 28.1A The Structure of the Atom
2.
3.
4.
5.
3. Estimating atomic mass
Estimate the atomic mass of each of the five atoms shown in Section 2. Each answer should be written to the
nearest whole number.
1.
2.
3.
4.
5.
6.
If you look at a periodic table, the atomic mass of hydrogen is 1.00794. Why is this number not rounded off
to 1?
2
Skill Sheet 28.1A The Structure of the Atom
4. Finding the number of protons and neutrons
The examples in this section are shown with their isotope numbers. This is the number of particles in the nucleus
of the atom shown. Find the number of protons and neutrons in each nucleus.
1.
2
1H
2.
45
21Sc
3.
27
13Al
4.
235
92U
5.
12
6C
5. Electrons and atomic structure
1.
Although electrons have mass, they are not considered in determining the mass number or atomic mass of an
atom. Why?
2.
A hydrogen atom has one proton, two neutrons, and no electrons.
a. Is this atom a stable isotope of hydrogen? Explain your answer.
b. Is this atom neutrally charged? Explain your answer.
3.
4.
How many electrons are in each of the following atoms? Assume that the atoms are neutrally charged.
2
a. 1H
b.
10
5B
c.
30
14Si
23
An atom of sodium ( 11Na ) has a positive charge of +1. Given this information, how many electrons does it
have? How many protons and neutrons does this atom have?
3
Name:
Skill Sheet 28.1B
The Periodic Table
Many science laboratories have a copy of the periodic table of the elements on display. This important chart
holds an amazing amount of information. In this skill sheet, you will use a periodic table to identify
information about specific elements, make calculations, and make predictions.
1. Periodic table primer
To work through this skill sheet, you will use the periodic table of the
elements. The periodic table shows five basic pieces of information.
Four are labeled on the graphic at right; the fifth piece of information is
the location of the element in the table itself. The location shows the
element group, chemical behavior, approximate atomic mass and size,
and other characteristic properties.
2. Atomic number
Write the name of the element that corresponds to each of the following atomic numbers.
1.
9
2.
18
3.
25
4.
15
5.
43
3. Symbol
For each of the following, write the element name that corresponds to the symbol. In addition, write the atomic
mass for each element.
1.
Fe
2.
Cs
3.
Si
4.
Na
5.
Bi
1
Skill Sheet 28.1B The Periodic Table
4. Number of protons and neutrons
Atomic mass is an average value based on the masses and abundance of the naturally occurring isotopes of each
element. For many of the elements, rounding the atomic mass to the nearest whole number gives us the number
of protons and neutrons in the nucleus of the most common isotope. Remember that electrons have so little mass
that we can ignore their contribution to the overall mass. Use this assumption and the atomic number to find the
number of protons and neutrons in the nucleus of each of these elements.
1.
Mg
2.
N
3.
K
5. Predicting radioactive decay products
In one form of radioactive decay, a nucleus loses a cluster of two protons and two neutrons, called an alpha
particle. The sum of protons and neutrons in the nucleus of each example below is shown with a superscript.
Subtract the alpha particle, and use the periodic table to determine the identity of the new element. Then compare
its new atomic mass with the value shown on the periodic table. If it doesn’t match, it’s probably also radioactive
and will decay further. If it does match, it’s probably stable and will endure for a long time. Here is an example:
235
92U
235
92U
– α particle →
231
90Th
– ( 2 protons + 2 neutrons ) →
231
90Th
First, subtract the 2 protons of the alpha particle (α) from the atomic number (92) to find the atomic number (90)
of the decay product. This is the atomic number of thorium. Next, subtract the whole alpha particle (2 protons +
2 neutrons = 4 particles) from the atomic mass of uranium to find the atomic mass (231) of the decay product.
The periodic table shows the atomic mass of thorium as 232 amu (atomic mass units). Therefore, we can assume
that thorium-231 is a radioactive isotope.
For each of the following radioactive isotopes, find the atomic number and atomic mass of the decay product
after an alpha particle has been emitted. Then state whether the decay product is stable or radioactive.
1.
211
84Po
–α→
2.
226
88Ra
–α→
3.
222
86Rn
–α→
2
Name:
Skill Sheet 28.1C
Lise Meitner
Lise Meitner identified and explained nuclear fission, proving it is possible to split an atom.
Lise Meitner was born into a liberal Jewish family in Vienna, Austria, in 1878. At age
13, she completed all of the schooling provided to girls in Vienna. Her father hired a
tutor to help her prepare for a university education, although women were not yet
allowed to attend.
The preparation was worthwhile. When the University of Vienna opened its doors to
women in 1901, Meitner was ready. She found a mentor there in physics professor
Ludwig Boltzmann, who encouraged her to pursue a doctoral degree. Meitner’s
nephew, Otto Frisch, wrote that “Boltzmann gave her the vision of physics as a battle
for ultimate truth, a vision she never lost.”
After earning her doctorate in 1906, Meitner went to Berlin. There she began a 30-year
collaboration with chemist Otto Hahn. Together, they studied radioactive substances. One of their first successes
was the development of a new technique for purifying radioactive material.
During World War I, Meitner volunteered as an x-ray nurse-technician with the Austrian army. She pioneered
cautious handling techniques for radioactive substances and, off duty, continued her work with Hahn.
In 1917, they discovered a new element, protactinium. Afterward, Meitner was appointed head of the physics
department at the Kaiser Wilhelm Institute for Chemistry in Berlin, where Hahn was head of the chemistry
department. The two continued their study of radioactivity, and Meitner became the first to explain how
conversion electrons were produced when gamma rays were used to remove orbital electrons.
When Enrico Fermi produced radioactive isotopes of uranium by neutron bombardment in 1934, he was puzzled
by the products. Meitner, Hahn, and another German scientist, Fritz Strassmann, began looking for answers.
Their research was interrupted when Nazi Germany annexed Austria in 1938. Meitner went into exile in Sweden.
She continued to correspond with her collaborators and suggested that they perform further tests on a product of
the uranium bombardment. When tests showed it was barium, the group was puzzled. Barium was so much
smaller than uranium. Hahn wrote to Meitner that uranium “can’t really break into barium … try to think of some
other possible explanation.” Meitner and Frisch worked on the problem and proved that splitting the uranium
atom was energetically possible. Using Neils Bohr’s model of the nucleus, they explained how the neutron
bombardment could cause the nucleus to elongate into a dumbbell shape. Occasionally, they explained, the
narrow center of the dumbbell could separate, leaving two nuclei. Meitner and Frisch called this process nuclear
fission.
Otto Hahn received the Nobel Prize in Chemistry in 1944 for the discovery of nuclear fission. Unfortunately,
Meitner’s vital role in the research was overlooked. However, in 1966, she was recognized along with Hahn and
Strassman when the three shared the Enrico Fermi Award, given by the president and the US Department of
Energy. Meitner died two years later, just days before her 90th birthday. In 1992, element 109 was named
meitnerium to honor her work.
Questions
1. Draw a diagram showing how fission occurs in a uranium nucleus.
2. Research and describe at least two ways nuclear fission was used in the twentieth century.
1
Skill Sheets
Name:
Skill Sheet 28.2
Niels Bohr
Danish physicist Niels Bohr first proposed the idea that electrons exist in specific orbits around the atom’s
nucleus. He showed that when an electron falls from a higher orbital to a lower one, it releases energy in the
form of visible light.
Niels Bohr was born in 1885 in Copenhagen, Denmark. His father was a physiology
professor at the University of Copenhagen. His parents often invited other professors
over for dinners and leisurely discussions. Niels and his sister and brother were
frequently invited to join this friendly exchange of ideas, and they learned to value the
search for deeper understanding of the world.
Bohr entered the University of Copenhagen in 1903 to study physics. Because the
university had no physics laboratory, Bohr conducted experiments in his father’s
physiology lab. He graduated with a doctorate in 1911.
In 1912, Bohr traveled to Manchester, England, to study under Ernest Rutherford.
Rutherford had recently published his new planetary model of the atom, which
explained that an atom contains a tiny dense core surrounded by orbiting electrons. Bohr began researching the
orbiting electrons, hoping to describe their behavior in greater detail.
Bohr studied the quantum ideas of Max Planck and Albert Einstein as he sought to describe the electron’s orbits.
In 1913 he published his results. He proposed that electrons traveled only in specific orbits. The orbits were like
rungs on a ladder—electrons could move up and down orbits, but did not exist in between the orbital paths. Bohr
explained that the outer orbits could hold more electrons than inner orbits, and that many chemical properties of
the atom were determined by the number of electrons in the outer orbit.
Bohr also described how atoms emit light. He explained that an electron needs to absorb energy to jump from an
inner orbit to an outer one. When the electron falls back to the inner orbit, it releases that energy in the form of
visible light.
In 1916, Bohr accepted a position as professor of physics at the University of Copenhagen, which created the
Institute of Theoretical Physics that Bohr directed for the rest of his life. In 1922, he was awarded the Nobel Prize
in Physics for his work in atomic structure and radiation.
In 1940, Germany occupied Denmark. Bohr’s mother came from a prominent Jewish family, and his own antiNazi sentiments made life difficult. In 1943, he escaped to Sweden in a fishing boat. The British rescued him and
brought him to England, where researchers were working on the atomic bomb. A few months later, the team went
to Los Alamos, New Mexico, to continue their work.
Although Bohr believed the creation of the atomic bomb was necessary to face the threat posed by the Nazi
regime, he was deeply concerned about its future implications. He actively promoted disarmament efforts
through the United Nations until his death in 1962.
Questions
1.
2.
Name two specific contributions Niels Bohr made to our understanding of atomic structure.
Create a drawing or three-dimensional model of Bohr’s model of the atom.
1
Name:
Skill Sheet 29.2
Dot Diagrams
You have learned that atoms are composed of protons, neutrons, and electrons. The electrons occupy energy
levels that surround the nucleus in the form of an “electron cloud”. The electrons that are involved in forming
chemical bonds are called valence electrons. Atoms can have up to eight valence electrons. These electrons
exist in the outermost region of the electron cloud often called the “valence shell”.
The most stable atoms have eight valence electrons. When an atom has eight valence electrons, it is said to
have a complete octet. Atoms will gain or lose electrons in order to complete their octet. In the process of
gaining or losing electrons, atoms will form chemical bonds with other atoms. The method we use to visually
represent an atom's valence state is called a dot diagram, and you will practice drawing these in the following
exercise.
1. What is a dot diagram?
Dot diagrams are composed of two parts—the chemical symbol for the element and dots surrounding
the chemical symbol. Each dot represents one valence electron. If an element, such as oxygen, has six
valence electrons, then six dots will surround the chemical symbol as shown to the right.
Boron has three valence electrons, so three dots surround the chemical symbol for boron as shown to the
right.
There can be up to eight dots around a symbol, depending on the number of valence electrons the atom
has. The first four dots are single, and then as more dots are added, they fill in as pairs.
2. Drawing dot diagrams
Using a periodic table, complete the following chart. With this information, draw a dot diagram for each element
in the chart. Remember, only the valence electrons are represented in the diagram, not the total number of
electrons.
Element
Chemical
symbol
Potassium
K
Nitrogen
N
Carbon
C
Beryllium
Be
Neon
Ne
Sulfur
S
Total number of
electrons
1
Number of valence
electrons
Dot diagram
Skill Sheet 29.2 Dot Diagrams
3. Using dot diagrams to represent chemical reactivity
Once you have a dot diagram for an element, you can predict how an atom will achieve a full valence shell. For
instance, it is easy to see that chlorine has one empty space in its valence shell. It is likely that chlorine will try to
gain one electron to fill this empty space rather than lose the remaining seven. However, potassium has a single
dot or electron in its dot diagram. This diagram shows how much easier it is to lose this lone electron than to find
seven to fill the seven empty spaces. When the potassium loses its electron, it becomes positively charged. When
chlorine gains the electron, it becomes negatively charged. Opposite charges attract, and this attraction draws the
atoms together to form what is termed an ionic bond, a bond between two charged atoms or ions.
Because chlorine needs one electron, and potassium needs to lose one electron, these two elements can achieve a
complete set of eight valence electrons by forming a chemical bond. We can use dot diagrams to represent the
chemical bond between chlorine and potassium as shown above.
For magnesium and chlorine, however, the situation is a bit different. By examining the electron or Lewis dot
diagrams for these atoms, we see why magnesium requires two atoms of chlorine to produce the compound,
magnesium chloride, when these two elements chemically combine.
Magnesium can easily donate one of its valence electrons to the chlorine to fill chlorine’s valence shell, but this
still leaves magnesium unstable; it still has one lone electron in its valence shell. However, if it donates that
electron to another chlorine atom, the second chlorine atom has a full shell, and now so does the magnesium.
The chemical formula for potassium chloride is KCl. This means that one unit of the compound is made of one
potassium atom and one chlorine atom.
The formula for magnesium chloride is MgCl2. This means that a one unit of the compound is made of one
magnesium atom and two chlorine atoms.
Now try using dot diagrams to predict chemical formulas. Fill in the table below:
Elements
Dot diagram
for each element
Dot diagram
for compound formed
Na and F
Br and Br
Mg and O
2
Chemical
formula
Name:
Skill Sheet 29.3
Chemical Equations
Chemical symbols provide us with a shorthand method of writing the name of an element. Chemical formulas
do the same for compounds. But what about chemical reactions? To write out, in words, the process of a
chemical change would be long and tedious. Is there a shorthand method of writing a chemical reaction so
that the all the information is presented correctly and is understood by all scientists? Yes! This is the function
of chemical equations. You will practice writing and balancing chemical equations in this skill sheet.
1. What are chemical equations?
Chemical equations show what is happening in a chemical reaction. They provide you with the identities of the
reactants (substances entering the reaction) and the products (substances formed by the reaction). They also tell
you how much of each substance is involved in the reaction. Chemical equations use symbols for elements and
formulas for compounds. The reactants are written to the left of the arrow. Products go on the right side of the
arrow.
H2 + O2 → H2 O
The arrow should be read as “yields” or “produces.” This equation, therefore, says that hydrogen gas (H2) plus
oxygen gas (O2) yields or produces the compound water (H2O).
2. Practice writing chemical equations
Write chemical equations for the following reactions:
Reactants
Products
Chemical Equation
Hydrochloric acid
HCl
and
Sodium hydroxide
NaOH
Water
H2O
and
Sodium chloride
NaCl
Calcium carbonate
CaCO3
and
Potassium iodide
KI
Potassium carbonate
K2CO3
and
Calcium iodide
CaI2
Aluminum fluoride
AlF3
and
Magnesium nitrate
Mg(NO3)2
Aluminum nitrate
Al(NO3)3
and
Magnesium fluoride
MgF2
1
Skill Sheet 29.3 Chemical Equations
3. Conservation of atoms
Take another look at the chemical equation for making water:
2H 2 + O 2 → 2H 2 O
Did you notice that something has been added?
The large number in front of H2 tells how many atoms of H2 are required for the reaction to proceed. The large
number in front of H20 tells how many molecules are formed by the reaction. These numbers are called
coefficients. Using coefficients, we can balance chemical equations so that the equation demonstrates
conservation of atoms. The law of conservation of atoms says that no atoms are lost or gained in a chemical
reaction. The same types and numbers of atoms must be found in the reactants and the products of a chemical
reaction.
Coefficients are placed before the chemical symbol for single elements and before the chemical formula of
compounds to show how many atoms or molecules of each substance are participating in the chemical reaction.
When counting atoms to balance an equation, remember that the coefficient applies to all atoms within the
chemical formula for a compound. For example, 5CH4 means that 5 atoms of carbon and 20 atoms of hydrogen
are contributed to the chemical reaction by the compound methane.
4. Balancing chemical equations
To write a chemical equation correctly, first write the equation using the correct chemical symbols or formulas
for the reactants and products.
If a reaction is to occur between sodium chloride and iodine to form sodium iodide and chlorine gas, we would
write:
NaCl + I 2 → NaI + Cl 2
•
•
Remember that certain elements, when found alone, are diatomic; iodine and chlorine are two examples.
Next, count the number of atoms of each element present on the reactant and product side of the chemical
equation:
Reactant Side of Equation
Element
Product Side of Equation
1
Na
1
1
Cl
2
2
I
1
For the chemical equation to be balanced, the numbers of atoms of each element must be the same on either
side of the reaction. This is clearly not the case with the equation above. We need coefficients to balance the
equation.
2
Skill Sheet 29.3 Chemical Equations
•
First, choose one element to balance. Let’s start by balancing chlorine. Since there are two atoms of chlorine
on the product side and only one on the reactant side, we need to place a “2” in front of the substance
containing the chlorine, the NaCl.
2NaCl + I 2 → NaI + Cl 2
•
This now gives us two atoms of chlorine on both the reactant and product sides of the equation. However, it
also give us two atoms of sodium on the reactant side! This is fine—often balancing one element will
temporarily unbalance another. By the end of the process, however, all elements will be balanced.
We now have the choice of balancing either the iodine or the sodium. Let's balance the iodine. (It doesn’t
matter which element we choose.)
There are two atoms of iodine on the reactant side of the equation and only one on the product side. Placing
a coefficient of “2” in front of the substance containing iodine on the product side:
2NaCl + I 2 → 2NaI + Cl 2
There are now two atoms of iodine on either side of the equation, and at the same time we balanced the
number of sodium atoms!
In this chemical reaction, two molecules of sodium chloride react with one diatomic molecule of iodine to
produce two molecules of sodium iodide and one diatomic molecule of chlorine. Our equation is balanced!
5. Polyatomic ions and balancing equations
Some compounds contain multiple numbers of polyatomic ions. For example, Mg(OH)2 is formed when two
complete hydroxide (-OH) groups attach to the element magnesium. Therefore, in a chemical equation:
3Mg ( OH ) 2
represents 3 atoms of magnesium, 6 atoms of oxygen, and 6 atoms of hydrogen. Each hydroxide group
contributes 2 oxygen atoms and 2 hydrogen atoms to the compound. There are 3 molecules of the compound in
the reaction, so there are 6 total atoms of oxygen and hydrogen in the reaction. Be aware of this as you balance
equations.
If your equation contains a polyatomic ion with a subscript, like SO4, you really have to watch your math! For
example, if you have 3 molecules of aluminum sulfate, 3Al2(SO4)3, you actually have nine atoms of sulfur and
thirty-six atoms of oxygen! Can you figure out why?
6. Practice balancing equations
Balance the equations on the next page using the appropriate coefficients. Remember that balancing one element
may temporarily unbalance another. You will have to correct the imbalance in the final equation. Check your
work by counting the total number of atoms of each element—the numbers should be equal on the reactant and
product sides of the equation.
1.
Al + O2 → Al2O3
3
Skill Sheet 29.3 Chemical Equations
2.
CO + H2 → H2O + CH4
3.
HgO → Hg + O2
4.
CaCO3 → CaO + CO2
5.
C + Fe2O3 → Fe + CO2
6.
N2 + H2 → NH3
7.
K + H2O → KOH + H2
8.
P + O2 → P2O5
9.
Ba(OH)2 + H2SO4 → H2O + BaSO4
10. CaF2 + H2SO4 → CaSO4 + HF
11. KClO3 → KClO4 + KCl
4
Name:
Skill Sheet 30.1A
Radioactivity
In this skill sheet we are reviewing the concepts of radioactivity, radiation, and half-life. This will give you
practice in solving problems that include these terms.
1. Introduction to radiation
There are three main types of radiation that involve the decay of the nucleus of an atom:
•
alpha radiation (α):
4
release of a 2He nucleus
(two protons and two
neutrons).
•
beta radiation (β): release
of an electron.
•
gamma radiation (γ):
release of an
electromagnetic wave.
Radiation can also be released when two nuclei react (collide). In this case, radiation is one of the products of the
reaction. Also, in many cases, the collision creates an unstable nucleus — think of this type of nucleus as having
extra energy it must get rid of — which after a very short time releases energy in the form of radiation.
Conservation of energy and the use of Einstein’s formula (E = mc2) allow us to determine the amount of energy
released during a nuclear reaction or during a radioactive decay. The graphic below illustrates the decay of
carbon-14 to nitrogen-14.
Another example:
2
3
Consider the reaction of deuterium ( 1H , a proton and a neutron) and tritium ( 1H , a proton and two neutrons):
2
1H
3
4
+ 1H → 2He + n + energy
The energy released in this reaction is 2.8 × 10-12 joules. This energy is carried by the products of the reaction,
4
( 2He and n). If we were to weigh the reactants and the products of this reaction, the difference in their mass
would be given according to Einstein’s formula by:
– 12
E- = -------------------------------------------2.8 × 10 joules = 3.1 × 10 –29 kg
+ m n ) = ---He
2
8 2 2
2
2
c
( 3 × 10 ) m /sec
∆m = ( m 2 + m 3 ) – ( m 4
1H
1H
1
Skill Sheet 30.1A Radioactivity
2. What is half-life?
The time it takes for half of the atoms in a sample to decay is called the half-life.
Example:
Four kilograms of a certain substance undergo radioactive decay. Let’s calculate the amount of substance left
over after 1, 2, and 3 half-lives.
•
After one half-life, the substance will be reduced by half, to 2 kilograms.
•
After two half-lives, the substance will be reduced by another half, to 1 kilogram.
•
After three half-lives, the substance will be reduced by another half, to 0.5 kilogram.
So, if we start with a sample of mass m that decays, after a few half-lives, the mass of the sample will be:
Number of half-lives
Mass left
1
1---m =
1
2
1--m
2
2
1---m =
2
2
1--m
4
3
1---m =
3
2
1--m
8
4
11
---m = ------ m
4
16
2
3. Problems
1.
The decay series for uranium-238 and plutonium-240 are listed below. Above each arrow, write “a” for alpha
decay or “b” for beta decay to indicate which type of decay took place at each step.
→
234 Th
90
→
234 Pa
91
→
234 U
92
→
230 Th
90
→
226 Ra
88
→
222 Rn
86
→
218 Po
84
→
214 Pb
82
→
214 Bi
83
→
214 Po
84
→
210 Pb
82
→
210 Bi
83
→
210 Po
84
→
206
Pb
82
240Am
95
→
236 Np
93
→
232 Pa
91
→
232 U
92
→
→
216 At
85
→
a. 238 U
92
Pu →
b. 240
94
2.
228 Bi
90
→
224 Ra
88
→
224 Ac
89
→
220 Fr
87
212 Bi
83
→
212 Po
84
→
208 Pb
82
→
208 Bi
83
A large bomb releases about 5 × 1012 joules of energy. How much mass is converted into energy during such
an explosion?
2
Skill Sheet 30.1A Radioactivity
18
3.
Fluorine-18 ( 9F ) has a half-life of 110 seconds. This material is used extensively in medicine. If the
18
hospital laboratory starts the day (9 a.m.) with 10 grams of 9F , how many grams would be left after one
hour? Do you see a problem with this? How does the lab manage inventory?
4.
The isotope
5.
An isotope decreased to one-fourth its original amount in 18 months. What is the half-life of this radioactive
isotope?
14
6C
has a half-life of 5,730 years. What is the fraction of
14
6C
in a sample after 30,000 years?
4. The intensity of radiation
This diagram illustrates a formula that is used to calculate the intensity of radiation from a radioactive source.
Radiation “radiates” from a source into a spherical area. Therefore, you can calculate intensity using the area of a
2
sphere ( 4πr ). Use the formula and the diagram to help you answer the questions below.
1.
A radiation source with a power of 1,000 watts is located at a point in space. What is the intensity of
radiation at a distance of 10 meters from the source?
2.
The fusion reaction 1H + 1H → 2He + n + energy releases 2.8 × 10-12 joules of energy. How many such
reactions must occur every second in order to light a 100-watt light bulb? Note that one watt equals one joule
per second.
2
3
4
3
Name:
Skill Sheet 30.1B
Marie and Pierre Curie
Marie and Pierre Curie’s pioneering study of radioactivity had a profound impact on the development of
twentieth-century science. Madame Curie’s bold assertion that uranium rays seemed to be an intrinsic part of
uranium atoms encouraged physicists to explore the possibility that atoms might have an internal structure.
Out of this idea, the field of nuclear physics was born. Together the Curies discovered two radioactive
elements, polonium and radium. Through Pierre Curie’s study of how living tissue responds to radiation, a
new era in cancer treatment was born.
Marya Sklodowska was born in 1867, in Russian-occupied Warsaw, Poland. Marya
loved school, especially math and science courses, but higher education opportunities
for women in Poland were limited. At age 17, she and her older sister enrolled in an
illegal, underground “floating university” in Warsaw.
After completing her studies, she worked for three years as a governess. Her employer
allowed her to teach reading classes to the children of peasant workers at his beetsugar factory, although this was forbidden under Russian rule. At the same time, she
took chemistry lessons from the factory’s chemist, studied independently, and took
math lessons from her father by mail.
By fall 1891, she had saved enough money to enroll in a master’s degree program at the University of Paris (also
called the Sorbonne). She earned two degrees, in physics and mathematics. A Polish friend introduced Marie, as
she was called in French, to Pierre Curie, the laboratory chief at the Sorbonne’s Physics and Industrial Chemistry
Schools. Pierre Curie’s early research centered on properties of crystals. He and his brother Jacques discovered
the piezoelectric effect, which describes how a crystal will oscillate when electric current is applied. The
oscillation of crystals is now used to precisely control timing in computers, watches, and many other devices.
Pierre Curie and Marie Sklodowska found that despite their different nationalities, they had the same passion for
scientific research and a desire to use their discoveries to promote humanitarian causes. Their friendship
deepened and they married in 1895.
Pierre continued his pioneering research in crystal structures while Marie decided to
pursue a physics doctorate. As her research topic, she chose uranium rays, recently
discovered by French physicist Henri Becquerel. After reading Becquerel’s report that
uranium compounds emitted some sort of ray that fogged photographic plates, Marie
decided to research the effect that these rays had on the air’s ability to conduct
electricity. To measure this effect, she adapted a device that Pierre and Jacques Curie
had invented 15 years earlier.
Madame Curie confirmed that the electrical effects of uranium rays were similar to the
photographic effects that Becquerel reported—both were present whether the uranium
was solid or powdered, pure or in compound, wet or dry, or exposed to heat or light.
She concluded that the emission of rays by uranium was not the product of a chemical reaction, but could be
something built into the very structure of the uranium atoms.
Madame Curie’s idea was revolutionary because atoms were still believed to be tiny, featureless particles. She
decided to test every known element to see if others would also improve the air’s ability to conduct electricity.
She found that the element thorium had this property.
1
Skill Sheet 30.1B Marie and Pierre Curie
Pierre Curie put aside his research on crystals and joined Marie after she found that two different uranium ores
(raw materials gathered from uranium mines) caused the air to conduct electricity much better than even pure
uranium or thorium. They wondered if an as-yet undiscovered element might be mixed into each ore. They
worked to separate the chemicals in the ores, and found two substances that were responsible for the increased
conductivity. They called these elements polonium, in honor of Poland, Marie’s native country, and radium, from
the Greek word for ray.
While Madame Curie searched for ways to extract these pure elements from the ores, Pierre turned his attention
to the properties of the rays themselves. He tested the radiation on his own skin and found that it damaged living
tissue. As he published his findings, a whole new field of medicine developed, using targeted rays to destroy
cancerous tumors and cure skin diseases. Unfortunately, both Curies became ill from overexposure to radiation.
In June 1903, Madame Curie defended her thesis and became the first woman in France to receive a doctorate. In
December of that year, the Curies and Becquerel shared the Nobel Prize in Physics, the Curies for their work in
the spontaneous radiation Becquerel had discovered (and which Madame Curie called “radioactivity.”)
Tragedy struck in April 1906, when Pierre was killed by a horse-drawn wagon in Paris. A month later, the
Sorbonne asked Madame Curie to take over her husband’s position there. She agreed, in hopes of creating a stateof-the art research center in her husband’s memory. The mother of two young daughters, she threw herself into
the busy academic schedule of teaching and conducting research (she was the first woman to lecture, the first to
be named professor, and the first to head a laboratory at the Sorbonne), but managed to find time to work on
raising money for the new center. The Radium Institute of the University of Paris opened in 1914 and Madame
Curie was named director of its Curie Laboratory.
In 1911, Madame Curie received a second Nobel Prize, this time in chemistry, for her work in finding new
elements and determining the atomic weight of radium.
When World War I broke out in 1914, she turned her attention to the use of radiation to alleviate the suffering of
wounded soldiers. Assisted by her teenage daughter, Irene, she created a fleet of 20 mobile x-ray units to help
medics quickly treat injuries in the field. Next, she set up nearly 200 x-ray labs in hospitals and trained 150
women to operate the equipment.
After the war, Madame Curie went back to direct the Radium Institute, which grew to two centers, one devoted to
research and the other to treatment of cancer. She died of radiation-induced leukemia in July 1934. The next year,
Irene Joliot-Curie and her husband, Frederic Joliot-Curie, were awarded the Nobel Prize in Chemistry for their
discovery of artificial radiation.
Questions
1. What fundamental change in our understanding of the atom was brought about by the work of Marie Curie?
2. Describe how Marie and Pierre Curie discovered two new elements.
3. Name at least three new fields of science that stem from the work of Marie and/or Pierre Curie.
2
Skill Sheets
Name:
Skill Sheet 30.3
Ernest Rutherford
Ernest Rutherford initiated a radical new view of the atom. He explained the mysterious phenomenon of
radiation as the spontaneous disintegration of atoms. He was the first to describe the atom’s internal structure
and performed the first successful nuclear reaction.
Ernest Rutherford was born in rural New Zealand in 1871. His father was a Scottish
immigrant, his mother from England. Both valued education and instilled a strong
work ethic in their 12 children. Ernest enjoyed the family farm, but was encouraged by
his parents and teachers to pursue a university scholarship. He had to take the
qualifying exam twice, but in 1890 was admitted on scholarship to the University of
New Zealand.
After earning three degrees in his home country, Rutherford traveled to Cambridge,
England, to pursue graduate research under the guidance of the man who first
discovered the electron, J.J. Thomson. Through his research with Thomson,
Rutherford became interested in studying radioactivity. In 1898 he described two
different kinds of particles emitted from radioactive atoms, calling them alpha and beta particles. He also coined
the term half-life to describe the amount of time taken for the radioactivity to decrease to half its original level.
Rutherford accepted a professorship at McGill University in Montreal, Canada, in 1898. It was there that he
proved that atoms of a radioactive element could spontaneously decay into another element by expelling a piece
of the atom. This was surprising to the scientific community—the idea that atoms could change into other atoms
had been scorned as alchemy. Rutherford received the 1908 Nobel Prize in Chemistry for “his investigations into
the disintegration of the elements and the chemistry of radioactive substances.” Rutherford considered himself a
physicist and joked that, “of all the transformations I have seen in my lifetime, the fastest was my own
transformation from physicist to chemist.”
Rutherford had returned to Manchester, England, in 1907. There, he and two students bombarded gold foil with
alpha particles. Most of the particles passed through the foil, but a few bounced back. They reasoned these
particles must have hit denser areas of foil. Rutherford hypothesized that the atom must mostly empty space,
through which the alpha particles passed, with a tiny dense core he called the nucleus, which some of the
particles hit. From this experiment he developed a new “planetary model” of the atom. The inside of the atom,
Rutherford suggested, contained electrons orbiting a small nucleus the way the planets of our solar system orbit
the sun.
In 1917, Rutherford made another discovery. He bombarded nitrogen gas with alpha particles, and found that
occasionally an oxygen atom was produced. He concluded that the alpha particles must have knocked a
positively charged particle (which he named the proton) from the nucleus. Rutherford called this “playing with
marbles” but word quickly spread that he had become the first to split an atom. Rutherford returned to Cambridge
in 1919 to head the laboratory where he began his research in radioactivity. He remained there until his death in
1937.
Questions
1.
2.
Name three important contributions Rutherford made to our understanding of the atom.
Compare and contrast Rutherford’s “planetary model” of the atom with our current understanding of an
atom’s internal structure.
1
Name:
Skill Builder
Internet Research Skills
The Internet is a valuable tool for finding answers to questions about the world. This skill sheet will introduce
you to some basic techniques for finding the information you need quickly. It also provides some questions
you should ask as you evaluate whether or not the source of the information is reliable.
1. What is a search engine?
A search engine is like an online index to information on the World Wide Web. There are many different search
engines to choose from. Some search engines index sites geared toward children. Others focus on educational
sites, science-oriented sites, or government sites. Search engines differ in how often they are updated, how many
documents they contain in their index, and how they search for information. Your teacher may suggest several
search engines for you to try.
2. Defining your topic
Search engines ask you to type a word or phrase into a box known as a field. Knowing how search engines work
can help you pinpoint the information you need. However, if your phrase is too vague, you may end up with a lot
of unhelpful information.
How could you find out who was the first woman aboard a space shuttle flight?
First, put key phrases in quotation marks. You want to know about the “first woman” on a “space shuttle.”
Quotation marks tell the engine to search for those words together.
Second, if you only want websites that contain both phrases, use a + sign between them. Typing “first woman”
+ “space shuttle” into a search engine will limit your search to websites that contain both phrases.
If you want to broaden your search, use the word or between two terms. For example, if you type “first female”
or “first woman” + “space shuttle” the search engine will list any website that contains either of the first two
phrases, as long as it also contains the phrase “space shuttle.”
You can narrow a search by using the word not. For example, if you wanted to know about marine mammals
other than whales, you could type “marine mammals” not “whales” into the field. Please note that some search
engines use the minus sign (-) rather than the word not.
1.
If you wanted to find out about science museums in your state that are not in your own city or town (or in the
nearest city or town), what would you type into the search engine?
2.
If you wanted to find out which dog breeds are not expensive and are easy to care for, what would you type
into the search engine?
3.
How could you research alternatives to producing electricity through the combustion of coal or natural gas?
1
Skill Builder Internet Research Skills
3. Evaluating information found on the Internet
The quality of information found on the Internet varies widely. This section will give you some things to think
about as you decide which sources to use in your research.
1.
2.
3.
4.
Authority: How well does the author know the subject matter? If you search for “Newton’s laws” on the
Internet, you may find a science report written by a fifth-grade student, and a study guide written by a
college professor. Which website is the most authoritative source?
Museums, national libraries, government sites, and big, well-known encyclopedia sources are good places to
look for authoritative information.
Bias: Think about the author’s purpose. Is it to inform or to persuade? Is it to get you to buy something?
Comparing several authoritative sources will help you get a more complete understanding of your subject.
Target audience: For whom was this website written? Avoid using sites designed for students well below
your grade level. You need to have an understanding of your subject matter at or above your own grade
level. Even authoritative sites for younger students (children’s encyclopedias, for example) may leave out
details and simplify concepts in ways that would leave gaps in your understanding of your subject.
Is the site up-to-date, clear, and easy to use? Try to find out when the website was created, and when it
was last updated. If the site contains links to other sites, but those links don’t work, you may have found a
site that is infrequently or no longer maintained. It may not contain the most current information about your
subject. Is the site cluttered with distracting advertisements? You may wish to look elsewhere for the
information you need.
4. Putting ideas into practice
1.
What is your favorite sport or activity? Search on the web for information about it. List two sites that are
authoritative and two sites that are not authoritative. Explain the reasons for your judgments of the websites.
Finally, write down the best site for finding information about your favorite sport or activity.
2.
Search the Internet for information about a physical science topic of your choice (e.g., “simple machines,”
“Newton’s laws,” “Galileo”). Find one source that you would consider NOT authoritative. Write the key
words you used in your search, the web address of the source, and a sentence explaining why this source is
not authoritative.
2
Skill Builder Internet Research Skills
3.
Find a different source that is authoritative, but intended for a much younger audience. Write the web
address and a sentence describing who you think the intended audience is.
4.
Find three sources on the Internet that you would consider good choices for your research here. Write a twoto three-sentence description of each site. Describe the author, the intended audience, the purpose of the site,
and any special features not found in other websites.
Extra space for notes:
3
Name:
Skill Builder
Lab Report Format
Use this skill sheet to help you write a lab report. You can think of a lab report as a document that tells a story
about an experiment you performed. The story itself is the experiment. There is always a beginning to the
story (the Introduction) and an ending (the Conclusions). The details you provide about this “story” help
others learn from what you did. As a way to share scientific knowledge, lab reports are great contributions to
our progress in understanding of how the world works.
1. What is a lab report?
A lab report is an explanation of your findings from an investigation. If your lab report is written clearly, anyone
who reads it will be able to understand what you were trying to learn. They will see why and how you performed
your experiment.
When you tell a reader how to repeat your experiment, you are giving another person the tools to evaluate your
work. It they can get data that is similar to yours and come to similar conclusions, then you have support for your
ideas.
Because Isaac Newton’s findings from his experiments were repeatable, what started out as ideas are now
scientific laws. For example, Newton’s thoughts about a falling apple led to the law of universal gravitation. This
law is an equation that explains why the moon and each of us experience gravitational attraction to the Earth.
Wow!
2. What are the parts of a lab report?
On a cover sheet or at the top of your report include your name, the title of the lab, the date of completion, and
your lab partners’ names.
These are the main parts of a lab report along with a description of each part.
•
Research question: What are you trying to find out through this experiment?
•
Introduction: This paragraph describes the topic you are studying and how it relates to your experiment.
State your hypothesis at the end of the introduction.
•
Procedure: This paragraph is a description of the experiment you performed to test your hypothesis. You
may wish to include a sketch of the apparatus you used. Be sure to name the experimental variable and list
the variables that you controlled in the experiment.
•
Results: In this paragraph, you describe your data. Often you will include a graph. Write a short description
of the data, but do not draw any conclusions in this paragraph.
•
Conclusions: Your conclusions about your experiment are described in this part of the lab report. The
conclusions paragraph describes what happened in your experiment, and whether or not your hypothesis was
correct.
1
Skill BuilderLab Report Format
3. A sample lab report
Use this sample lab report as a guide for writing your own lab reports. Keep in mind that you are telling a story
about something you did. Write clearly so that others can easily understand what you accomplished.
2
Name:
Skill Builder
Graphing Skills
This skill sheet will help you master graphing skills. There are three main kinds of graphs: line graphs, bar
graphs, and pie graphs. On this skill sheet, we will focus on line graphs. Line graphs represent data as a series
of points. A line is drawn through the points to show the pattern made by the points.
1. The parts of a graph
A graph is a picture of information. The labels on a graph provide important information. The diagram below
shows the labels you should put on a graph and how these labels should look.
When you are plotting your data on a graph, be sure to use all of the available space. Avoid plotting all your data
in one corner of the graph.
1
Skill BuilderGraphing Skills
2. Data sets
Tables 1 and 2 below each contain a data set. A data set is organized into pairs of values. For every value in the
“x” column, there is a value in the “y” column. Each pair of values can be represented by writing (x, y).
A pair of values (x,y) represents a certain location or point on a graph. The x and y values are the coordinates of
the point. A graph of a data set represents a “picture” of the points.
Fill in the third column of each data set by writing the pairs of x and y values. The first row in each data table has
been done for you.
Table 1: A car wash is being held to raise money for a school trip. The data set shows the relationship between
the amount of money in the cash box and the number of hours spent washing cars. Why is $20 in the cash box at
the beginning of the car wash?
Table 1: Money in cash box vs. number of hours washing cars
Number of hours washing cars
Amount of money in box
Coordinates
(hours)
($)
(x,y)
x
y
0
20
1
35
2
50
3
65
4
80
5
95
(0,20)
Table 2: If you could measure how much water gets splashed out of a swimming pool as the number of people in
the pool increases, the data might look like the following:
Table 2: Number of people in a swimming pool vs. amount of water splashed
out of the pool
Number of people
Amount of water
Coordinates
in a swimming pool
splashed out of the pool
(x,y)
x
(gallons)
y
0
0
1
1
2
4
3
9
4
16
5
25
6
36
2
(0,0)
Skill BuilderGraphing Skills
3. Making graphs from data sets
Now make a graph of each data set. Use the data in Table 1 to make graph 1, and the data in Table 2 to make
graph 2. The labels for graph 1 have been added for you. You will need to add labels, including the title of the
graph, to graph 2.
Steps for making a graph:
•
Label the graph with the titles from the headers of the data tables. The y-value header comes first in the title.
•
Label the horizontal axis (the x-axis) with the label from the header for the “x” column. Label the vertical
axis (the y-axis) with the label from the header for the “y” column.
•
Decide what value each box on each axis will represent. Base your decision on the range of numbers for the
x-values and the y-values. When the data is plotted on graph 1 and graph 2, the data points should spread out
over the entire graph rather than cluster in a corner of the graph. Example: The range of numbers for the xvalues for graph 1 is 0 to 5. There are 12 boxes on the x-axis of graph 1. To spread out the points along this
axis, each box represents 0.5 or every two boxes represent 1.
•
Plot the coordinates (the pairs of x- and y-values). Each coordinate represents a point.
•
Draw a line through the points to represent the pattern (or trend) that you see. Do not connect the points
“dot-to-dot.” Look at these examples:
3
Skill BuilderGraphing Skills
4. Interpreting graphs
1.
Write a short description of each graph. Describe the shape of each graph as if you were explaining them to
someone who had not seen them. How are graph 1 and graph 2 alike? How are they different?
2.
Write one conclusion you based on the data in graph 1. For example, what can you conclude about the
relationship between the number of hours spent washing cars and the amount of money in the cash box?
3.
Write one conclusion based on the data in graph 2. For example, what do you conclude about the
relationship between the number of people in a swimming pool and the amount of water splashed out of the
pool?
5. Challenge questions
1.
If you were looking at a graph with the title “Distance walked vs. time,” would the graph look more like
graph 1 or graph 2?
2.
If you were looking at a graph with the title “Number of bacteria in a Petri dish vs. time,” would the graph
look more like graph 1 or graph 2? To answer this question, think about how populations of organisms
increase in number.
4
Name:
Skill Builder
Safety Skills
Science equipment and supplies are fun to use. However, this equipment and carrying out procedures in an
Investigation always require safety. Here you will learn how to be safe in a science lab. When you use safe
practices, you create an environment in which everyone can be safe.
1. Safety icons
Throughout the CPO Science Investigations, safety icons and words and phrases like “caution” and “safety tip”
are used to highlight important safety information. Read the description of each icon carefully and look out for
them when reading the Student Text and the Investigations.
i
Use extreme caution: Follow all instructions carefully to avoid injury to yourself or others.
n
Electrical hazard: Follow all instructions carefully while using electrical components to avoid injury to
yourself or others.
s
Wear safety goggles: Requires you to protect your eyes from injury.
e
Wear a lab apron: Requires you to protect your clothing and skin.
t
Wear gloves: Requires you to protect your hands from injury due to heat or chemicals.
d
Cleanup: Includes cleaning and putting away reusable equipment and supplies, and disposing of
leftover materials.
2. Safety guidelines
Read these safety guidelines before each Investigation.
1.
Always prepare for each Investigation.
a. Read the Investigation sheets carefully.
b. Take special note of safety instructions.
2. i Pay close attention to your teacher’s instructions before, during, and after the Investigation. Take notes to
help you remember what your teacher has said. Special safety instructions from your teacher will include
information about:
a. Working with hot items or solutions.
b. Working with electrical components.
c. How to use your equipment and supplies.
d. How to dispose of chemicals and trash.
3. est If the Investigation requires protective devices or clothing (a lab apron, goggles, gloves), gather
these at the beginning of the Investigation.
4. Emphasize teamwork. Help each other. Watch out for one another’s safety.
5. d Always clean up after an Investigation. Your teacher will give you special instructions for disposing of
your materials.
1
Skill BuilderSafety Skills
3. Special instructions about safety topics
1.
2.
3.
4.
5.
t Thermal hazards.
a. Always carry or hold hot items with a hot pad. Never use your bare hands.
b. Move carefully when you are near hot items or solutions. Sudden movements could cause you to burn
yourself by touching or spilling something hot.
c. Inform others if they are near hot items or liquids.
i Electrical hazards.
a. Always keep electrical components away from water.
b. Avoid creating a short circuit with electrical components. Short circuits could cause the components to
heat up or spark.
c. Remove metal accessories (watches and jewelry) when working with electrical components.
Chemical hazards.
a. st Use goggles and gloves when working with chemicals whether you are inside the lab or outside
doing a field experiment.
b. When making an acidic or basic solution, be sure to pour the acid or base into water so that it is safely
diluted. NEVER pour water into an acid or base.
d Disposal of materials and supplies.
a. Generally, liquid household chemicals can be poured into a sink. Completely wash the chemical down the
drain with plenty of water.
b. Generally, solid household chemicals can be placed in a trash can.
c. Any liquids or solids that should not be poured down the sink or placed in the trash have special disposal
instructions. Follow your teacher’s instructions. Special disposal instructions can be found on the
Materials Safety Data Sheet for a chemical.
d. If an item breaks, do not use your bare hands to pick up the pieces. Use a dustpan and a brush to clean up.
“Sharps” trash (trash that has pieces of glass) should be well labeled. The best way to throw away broken
glass is to seal it in a labeled cardboard box.
What to do in case of danger or an emergency:
a. If you are concerned about your safety or the safety of others, talk to your teacher immediately. Talk to
your teacher immediately when:
* You smell chemical or gas fumes. This might indicate a chemical or gas leak.
* You smell something burning.
* You injure yourself or see someone else who is injured.
* You are having trouble using your equipment.
* You do not understand the instructions for the Investigation.
b. Listen carefully to your teacher’s instructions.
c. Carefully follow your teacher’s instructions.
2
Skill BuilderSafety Skills
4. Safety quiz
1.
Draw a diagram of your science lab in the space below. Include in your diagram the following items. Include
notes that explain how to use these important safety items.
•
Exit/entrance ways
•
Eye wash and shower
•
•
Fire extinguisher(s)
•
•
•
Fire blanket
•
Location of eye goggles and lab
aprons
First aid kit
Location of special safety
instructions
Sink
•
Trash cans
2.
How many fire extinguishers are in your science lab?
3.
List the steps that your teacher and your class would take to safely exit the science lab and the building in
case of a fire or other emergency.
4.
Before beginning certain Investigations, why should you first put on protective goggles and clothing?
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Skill BuilderSafety Skills
5.
Why is teamwork important when you are working in a science lab?
6.
Why should you always clean up after every Investigation?
7.
For one Investigation, you need to combine 40 milliliters of a 1 percent acid solution and 160 milliliters of
water to make 200 milliliters of a 0.2 percent acid solution. Which of the following procedures do you
follow and why?
a. You pour 160 mL of water into a beaker. Then you carefully add 40 mL of the 1% acid solution to the
water in the beaker.
b. You pour 40 mL of the 1% acid solution into a beaker. Then you carefully add 160 mL of water to the
acid in the beaker.
8.
List at least three things you should you do if you sense danger or see an emergency in your classroom or
lab?
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Skill BuilderSafety Skills
9.
Five lab situations are described below. What would you do in each situation? Explain how you would be
careful and safe in each situation.
a. You accidentally knock over a beaker and it breaks on the floor.
b. You accidentally spill a large amount of water on the floor.
c. You suddenly you begin to smell a “chemical” odor that gives you a headache.
d. You hear the fire alarm while you are working in the lab. You are wearing your goggles and lab apron.
e. While your lab partner has her lab goggles off, she gets some liquid from the experiment in her eye.
f. A fire starts in the lab.
Safety in the science lab is the responsibility of everyone! Help create a safe environment in your science
lab by following the safety guidelines from your teacher as well as the guidelines discussed in this
document.
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Skill BuilderSafety Skills
5. Safety contract
Keep this contract in your notebook at all times.
By signing it, you agree to follow all the steps necessary to be safe in your science class and lab.
I, ____________________, (Your name)
;Have read the safety guidelines on this Safety Skill Sheet.
;Understand the safety information presented.
;Will ask questions when I do not understand safety instructions.
;Pledge to follow all of the safety guidelines that are presented on the Safety Skill Sheet at all times.
;Pledge to follow all of the safety guidelines that are presented on Investigation sheets for every
Investigation.
;Will always follow the safety instructions that my teacher provides.
Additionally, I pledge to be careful about my own safety and to help others be safe. I understand that I am
responsible for helping to create a safe environment in the classroom and lab.
Signed and dated,
______________________________
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Name:
Skill Builder
Significant Digits
Francisco is training for a 10-kilometer run. Each morning, he runs a loop around his neighborhood. To find
out exactly how far he’s running, he asks his older sister to drive the loop in her car. Using the car’s trip
odometer, they find that the route is 7.2 miles long.
To find the distance in kilometers, Francisco looks in the reference
section of his science text and finds that 1.000 mile = 1.609
kilometers. He multiplies 7.2 miles by 1.609 km/mile. The answer,
according to his calculator, is 11.5848 kilometers.
Francisco wonders what all those numbers after the decimal point
really mean. Can a car odometer measure distances as small as
0.0008 kilometer? That’s less than one meter!
This skill sheet will help you answer Francisco’s question. It will
also help you figure out which digits in your own calculations are
significant.
1. What are significant digits?
Significant digits are the meaningful digits in a number. Scientists have agreed upon a number of rules to
determine which numbers in a measurement are significant. The rules are:
1.
2.
3.
4.
5.
6.
Non-zero digits in a measurement are always significant. This means that the distance measured by the
car odometer, 7.2 miles, has two significant digits.
Zeros between two significant digits in a measurement are significant. This means that the measurement
of kilometers per mile, 1.609 kilometers, has four significant digits.
All final zeros to the right of a decimal point in a measurement are significant. This means that the
measurement 1.000 miles has four significant digits.
If there is no decimal point, final zeros in a measurement are NOT significant. This means that the
number 10 in the phrase “10-kilometer run” has one significant digit.
In a measurement, zeros that exist only to put the decimal point in the right place are NOT significant.
This means that the number 0.0008 in the phrase “0.0008 kilometer” has one significant digit.
A number that is found by counting rather than measuring is said to have an infinite number of
significant digits. For example, the race officials count 386 runners at the starting line. The number 386, in
this case, has an infinite number of significant digits.
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Skill BuilderSignificant Digits
2. Practice finding the number of significant digits
Table 1: Number of significant digits
Value
How many significant digits does this value have?
36.33 minutes
100 miles
120.2 milliliters
0.0074 kilometers
0.010 kilograms
42 students
3. Using significant digits when doing calculations
Taking measurements and recording data are often a part of science classes. When you use the data in
calculations, keep in mind that the answer cannot be more precise than the least precise measurement.
For example, say you are using a ruler to measure the length of each side of a
rectangle.The ruler is marked in tenths of a centimeter. This means that you can
estimate the distance between two 0.1-centimeter marks and make
measurements that are to two places after the decimal. For example, you
measure the two short sides of the triangle and find that they each have a length
of 12.25 centimeters. The long sides each have a length of 20.75 centimeters.
The distance around the rectangle is 12.25 centimeters + 20.75 centimeters +
12.25 centimeters + 20.75 centimeters, or 66.00 centimeters. The two zeros to the right of the decimal point show
that you measured with a precision of 0.01 centimeter.
The area of the rectangle is found by multiplying the length of the short side by the length of the long side.
12.25 cm × 20.75 cm = 254.1875 cm
2
The answer you get from your calculator has seven significant digits. This incorrectly implies that your ruler can
measure to ten-thousandths of a centimeter. Your ruler cannot measure distances that small!
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Skill BuilderSignificant Digits
Follow these steps for determining the right answer for your calculation:
•
When multiplying or dividing measurements, find the measurement in the calculation with the least number
of significant digits. After doing your calculation, round the answer to that number of significant digits.
In the example, each measurement has 4 significant digits. Therefore, the answer should be rounded to four
significant digits. The answer should be reported as 254.2 centimeters2.
•
When you are finding the average of several measurements, remember that numbers found by counting have
an infinite number of significant digits.
For example, a student measures the distance between two magnets when their attractive force is first felt.
He repeats the experiment three times. His results are: 23.25 centimeters, 23.30 centimeters, 23.20
centimeters.
To find the average distance, he adds the three measurements and divides the sum by three. “Three” is the
number of times the experiment is repeated.
23.25
cm + 23.30 cm + 23.20 cm = 23.25 cm
------------------------------------------------------------------------------3
In this equation, the number 3 is found by counting the number of times the experiment is repeated, not by
measuring something. Therefore, it is said to have an infinite number of significant digits. That is why the
answer has four significant digits, not just one.
4. Practice doing calculations with significant digits
Have you ever participated in a road race? The following problems are all related to a road race event. Can you
come up with some other problems that you might have to solve if you were running in or volunteering for a road
race?
1.
The banner over the finish line of a race is 4.00 meters long and 0.75 meters high. What is the area of the
banner?
2.
Heidi stops at three water stations during the race. She drinks 0.25 liters of water at the first stop, 0.3 liters at
the second stop, and 0.37 liters at the third stop. How much water does she consume throughout the race?
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Skill BuilderSignificant Digits
3.
The race has been held annually for ten years. The high temperatures for the race dates (in degrees Celsius)
are listed in the table below. What is the average temperature for the day of the race based on the
temperatures for the past 10 years? Write your answer in the bottom row in the table.
Table 2: Annual race-day temperatures
Year of race
Race-day temperature (°C)
1
27.2
2
18.3
3
28.9
4
22.2
5
20.6
6
25.5
7
21.1
8
23.9
9
26.7
10
27.8
Average temperature
4.
Challenge! Ji-Sun has participated in the race for the past four years. His times, reported in
minutes:seconds, were 40:30, 43:40, 39:06, and 38:52. What is his average time to complete the race? Give
your answer in minutes:seconds.
5.
Come up with one more problem that uses information that is related to a road race. Write your problem in
the space below and come up with the answer. Be sure to write your answer with the correct number of
significant digits.
4
Name:
Skill Builder
Student Evaluation Sheet
Evaluate your experience and progress during the Investigation using this sheet.
1. Investigation information
1.
What were the number and title of this Investigation?
2.
What was the key question?
2. How did it go?
1.
Did you follow the safety guidelines that were given by your teacher and listed on the Investigation sheet?
2.
Did you follow the Investigations procedures?
3.
Did you care for the materials in the Investigation? Did anything break? Explain.
4.
Did you clean up your work area after the Investigation was completed?
5.
Did you demonstrate your skills in making accurate measurements, data collection, and graphing? Explain
and give examples.
6.
Did your group work well together? Give examples.
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Skill BuilderStudent Evaluation Sheet
3. What did you learn?
1.
In the space below, write a short paragraph that answers the key question of the Investigation. You may also
use diagrams to help you answer the question.
2.
Did you ask any questions during the Investigation? What did you ask? Was your question answered?
3.
List three things you learned from performing this Investigation.
4. Your comments
2