Download Physics 11 Reading Booklet

Physics11ReadingBooklet
In Order complete the Physics 11 Substantive Assignment, you must read and complete
the self-marking exercises in this booklet.
1.
2.
3.
4.
5.
Read all the information provided.
Complete the Practice and Self Check assignments.
Mark and correct your Practice questions.
As an online student it is an important skill to learn to self-assess throughout your work.
As an online student, you should be able to find answers outside the provided course
materials
6. Complete the Substantive Assignment – you must show all your work.
7. When you have completed all your work, scan it and attach it to your Online
Registration.
8. You may use a calculator for any part of the assignment.
Students should be able to demonstrate the following Prescribed Learning Outcomes following the
completion of this First Assignment:







Correctly determine the number of significant digits in a number
Perform calculations using the rules for significant digits
Combine significant digits and scientific notation in calculations
Solve any equation for a given variable
Use the sides and angles of a right triangle to determine the unknown sides and angles.
Apply mathematical models to solve a variety of problems
Use appropriate units and metric prefixes.
Physics 11 Reading Booklet
1.1 Significant Figures
Not all numbers are created equal. In fact even numbers which represent the same value
may not be equal to one another, based on our level of accuracy in measuring them. As
scientists, and science students, it is important to be able to express to those reading our
work the level of accuracy portrayed in our measurement. Significant digits are the rules
followed to show this accuracy.
There are two kinds of numbers in the world
•
exact:
o
o
•
Example: There are exactly 12 eggs in a dozen.
Example: Most people have exactly 10 fingers and 10 toes.
inexact numbers:
o Example: Any measurement. If I quickly measure the width of a piece of
notebook paper, I might get 220 mm (2 significant figures). If I am more
precise, I might get 216 mm (3 significant figures). An even more precise
measurement would be 215.6 mm (4 significant figures).
o Measurements of any physical information are only as good as the
instrument used to measure the information and the level of sensitivity to
those measurements. The inaccuracy stems from the precision of the
measuring device and any error introduced by the person measuring.
Because a measurement contains some degree of inaccuracy, the numbers
of digits that are valid for the measurement are also limited. So when
using any instrument for measurement we are able to determine some
place values easily. But there is one place value we must estimate. This is
the last place value we have any confidence in, and even that is a best
guess.
Precision versus Accuracy
Accuracy refers to how closely a measured value agrees with the correct value.
Precision refers to how closely individual measurements agree with each other.
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Physics 11 Reading Booklet
accurate (the
average is
accurate)
not precise
precise
not accurate
accurate
and
precise
In any measurement, the number of significant figures is critical. The number of
significant figures is the number of digits believed to be correct by the person doing the
measuring. It includes one estimated digit.
How can we determine how many significant numbers a measurement has?
The following rules apply to determining the number of significant figures in a measured
quantity:
1. All nonzero digits are significant.
•
•
457 cm (three significant figures)
0.25 g (two significant figures).
An example can be seen measuring the fish below using the meter stick. We can
see that it is bigger than the 40 cm mark, counting shows that it is just a bit longer
than 46 cm. From the angle we are viewing it is very difficult to tell but it looks to
be fractionally larger than 46 cm, so we are able to estimate a single decimal place
beyond the ones to 46.1 cm. As the 4 and 6 are known, and the 1 is an estimate.
We say there are 3 significant digits in this measurement. Any digit that is NOT
ZERO is significant.
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Physics 11 Reading Booklet
2. Zeros between nonzero digits (Captive Zeros) are significant.
•
•
1005 kg (four significant figures)
1.03 cm (three significant figures).
In the number 10.02 the zeros are captured between the 1 and the 2. Since 1 and
2 are significant any digit between them is significant.
3. Zeros to the left of the first nonzero digits in a number are not significant;
they merely indicate the position of the decimal point.
.
•
•
0.02 g (one significant figure)
0.0026 cm (two significant figures).
Red blood cells are about 0.000 006 meters in diameter. All the zeros in the front
of the 6 represent decimal places and are not significant.
4. When a number ends in zeros that are to the right of the decimal point, they
are significant.
•
•
0.0200 g (three significant figures)
3.0 cm (two significant figures).
Occasionally a scientist will express a number like 10.00 this is very different
from the number 10. When a scientist takes the effort to purposely record place
values with zeros after a decimal the scientist is implying that those values were
measured. Otherwise the number 10 would have been recorded instead. Zeros at
the end of a number, after a decimal, are significant.
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Physics 11 Reading Booklet
5. When a number ends in zeros that are to the left of a decimal point, the zeros
are not necessarily significant.
•
130 cm (two or three significant figures)
•
10,300 g (three, four, or five significant figures)
The population of the earth is 6 446 000 000 people according to the CIA fact
book dated 20 April, 2006. How is this number arrived at, and how accurate is it?
Did the CIA go out and count everyone? There is some ambiguity in this number.
It seems reasonable to assume there to be greater than 6 billion but less than 7
billion people so the first 6 is significant. If the CIA collected its data carefully
from governments, and simply added these together we might assume that the 4,
4, and 6 to also be relatively accurate, but how many people will be living in India
and China who do not fill out their countries' census? Can we really predict the
world population to the nearest million? Perhaps 6 400 000 000 is a more accurate
estimate. But are all those zeros significant?
We are unsure and so as scientists we have to assume they ARE NOT! That means
the only digits we have confidence in is the 6 and the 4 is an estimate. Any zero
digit at the end of a number greater than one is assumed to not be significant.
So how do scientists record really large numbers like the population of earth, and
really small ones like the diameter of a blood cell without including many
insignificant zeros? Use of standard exponential notation avoids the potential
ambiguity of whether the zeros at the end of a number are significant. For
example, a mass of 10,300 g can be written in exponential notation showing three,
four, or five significant figures:
1.03 x 104 g
1.030 x 104 g
1.0300 x 104 g
(three significant figures)
(four significant figures)
(five significant figures)
In these numbers all the zeros to the right of the decimal point are significant
(rules 2 and 4).
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Physics 11 Reading Booklet
Practice and Self-Check A
State the number of significant digits in each measurement:
a) 2804 m
b) 0.0045 m
c) 0.002 089
d) 3.06 x 105 m
e) 1.0003
f) 6.504 x 10
g) 24400
Ans) a) 4 b) 2 c) 4 d) 3 e)5 f)4 g)3
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Physics 11 Reading Booklet
Rule for Multiplication and Division
In multiplication and division of measurements, the result must be reported as having the
same number of significant figures as the least precise measurement.
If you multiply 21.3 cm by 9.80 cm, the answer is 209, not 208.74.
Since the less precise measurement,
21.3 cm, has only three significant figures, the product has three significant digits.
Example 1
A rectangle is measured to be 3.22 cm by 2.1 cm. What is the area of the rectangle shown
in the appropriate significant figures? Multiply 3.22 cm by 2.1 cm.
Solution: 3.22 cm × 2.1 cm= 6.762 cm 2
The less precise factor, 2.1 cm, contains two significant digits.
Therefore the product has only two significant digits. The answer
is then best stated as 6.8 cm 2 .
Example 2
A cyclist rides a distance of 36.5 m over a time of 3.414 s. How fast did he ride?
Divide 36.5 m by 3.414 s.
Solution:
The less precise factor, 36.5 m, contains three significant digits.
Therefore the division has only three significant digits. The answer
is then best stated as 10.7 m/s.
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Physics 11 Reading Booklet
Rule for Addition and Subtraction
In addition and subtraction, the result of your calculation can never be more precise than
the least precise measurement. The result of the calculation must be reported to the same
number of decimal places as that of the term with the least number of decimal places.
Example 1
Add 24.686 m + 2.343 m + 3.21 m
Solution:
24.686 m
2.343 m
3.21 m
30.239 m
=30.24 m
Note that 3.21 m is the least precise measurement. Therefore round off the
result to the nearest hundredth on one meter and the answer is best stated as 30.24 m.
You will follow the same rule for subtraction.
Rounding Rules
1. Look at the leftmost digit to be removed.
• If less than 5: truncate (preceding number is unchanged)
• If 5 or greater: increase by 1(round up)
2. When a calculation involves two or more steps, retain at least two additional digits
beyond the correct number of significant figures for intermediate calculations. Round
off to the correct number of significant figures only for the answer you want to report
(which is usually the final answer). This helps to avoid accumulated round-off error.
Significant Figures for Combined Math Operations
Click on the link below to watch the YouTube video:
https://www.youtube.com/watch?v=L829hEsgNxM#t=130
Check out other videos as well on how to do these operations.
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Physics 11 Reading Booklet
Practice and Self-Check C
Answer the following questions, and mark your practice work before completing the First
Assignment .
1. Express the following measurements in scientific notation.
a) 156.90
a)
b) 12 000
b)
c) 0.0345
c)
d) 0.008 90
d)
2. Give the number of significant digits in the following measurements.
a) 2.09910 m
a)
b) 5600 km
b)
c) 0.006 70 kg
c)
d) 809 g
d)
3. Solve the following problems and give the answer in the correct number of significant
digits.
a) 2.674 m ÷ 2.0 m =
a)
b) 5.25 L x 1.3 L =
b)
c) 9.0 cm + 7.67 cm + 5.44 cm =
c)
d) 10.07 g – 3.1 g =
d)
Answers:
Self Check A (a) 4 (b) 2 (c) 4 (d) 3
Self Check B(a) 0.00572 kg = 5.72 x 10-3 kg (b) 520 000 000 000 km = 5.2 x 1011 km (c) 300 000 000 m/s = 3.0 x 108 m/s (d)
0.00000000000000000016 C = 1.6x10-19C (e) 118.70004 g = 1.1870004 x 102 g
Self Check C1a)1.569 x 102 b)1.2 x 104 c)3.45 x 10-2 d) 8.9 x 10-3 2a)6 b)2 c)3 d)3 3a)1.3 b)6.8 L2 c)22.1 cm d)7.0
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Physics 11 Reading Booklet
1.2 Scientific Notation
Scientific Notation is a method of writing numbers which are very large, and numbers
which are very small. In physics it is common to have to work with masses of planets and
stars (very large numbers), charges on electrons and protons (very small numbers).
Scientific notation is a method of expressing those numbers in a short way rather than
having to write them out in full. For example, one ampere of electric current is the same as
6 240 000 000 000 000 000 electrons passing a point in a wire in one second.
This same number can be written, in scientific notation, as 6.24 x 1018 electrons/second.
This means 624 followed by 16 zeros.
Any number can be expressed in scientific notation. Here are some examples:
0.10 = 1.0 x 10-1
1.0 = 1.0 x 100
10.0 = 1.00 x 101
100.0 = 1.000 x 102
1 000.0 = 1.0000 x 103
Power-of-Ten Notation
0.000001 = 10-6
0.00001 = 10-5
0.0001 = 10-4
0.001 = 10-3
0.01 = 10-2
0.1 = 10-1
1 = 10
10 = 101
100 = 102
1 000 = 103
10 000 = 104
100 000 = 105
Practice and Self-Check B
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Physics 11 Reading Booklet
Write the following measurements in scientific notation.
(a) 0.00572 kg
(b) 520 000 000 000 km
(c) 300 000 000 m/s
(d) 0.000 000 000 000 000 000 16 C
(e) 118.70004 g
Ans) a) 5.72* 10 b)5.2*10
c) 3*10 d ) 1.6 * 10
e) 1.1870004 * 10 11
Physics 11 Reading Booklet
1.3 Equation Solving
By completion of this lesson students should be able to: Solve any equation for a given
variable
How do we find values for measured physical values?
Many of the physical phenomenon in the world we measure through experiment. Others
we can calculate with the use of equations which describe the relationship between
variables. In order to do this we often have to solve equations for particular variables and
that is the focus in this lesson. To solve equations we isolate the variable for which we
are looking.
This is done by first moving any term that does not have the variable in it to the other
side of the equation by opposite operations (addition or subtraction for other terms).
Example 1 Solve for “c” given this equation
We move the “d” to the other side of the equation and then divide by “x”
Therefore
Example 2 Solve for “k” given this equation
We multiply both sides by
Therefore k =

and then divide by
by both sides
∗
View this you-tube link for more guidance
http://www.youtube.com/watch?v=38Eeqbh3Cwc
Or search “solving equations Physics”
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Physics 11 Reading Booklet
Sometimes the equation is more complex and involves more intricate mathematical
functions; examples of this would include squares, square roots, sines, cosines and other
functions. In this case the opposite operations required to find a variable are also more
complex. Two examples follow.
Example 1 of complex functions.
1
Example 2 of complex functions.
The previous examples were somewhat artificial in that only one variable was found in
each equation. In most physics equations many variables are present and solving is done
without simplification after each step. In this case the terms and variables remain but are
written on the side of the equation away from the variable you are solving for. Examples
ensue.
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Physics 11 Reading Booklet
Practice and Self-Check D
Given the following equations solve the required variable
solve for
a) Given
b Given
c
solve for
Ans) a)
Given
b)
∗
solve for t
c)
√2
Often a list of values can be found in reading any problem in physics and these values
may be substituted into an equation at any time allowing the student to simplify. It is a
valuable skill to be able to manipulate equations as many physics problems deal with
general cases rather than those with specific values.
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Physics 11 Reading Booklet
1.4 Trigonometry
By completion of this lesson students should be able to: Use the sides of a right triangle
to determine angles


Use the angles in a right triangle to calculate sides
Solve word problems using trigonometry
What method is used to determine sides in triangles?
Due to the nature of having to determine directions of many variables in physics it is
important to be able to find the missing sides and angles in right triangles. Trigonometry
is a method of using ratios to determine these unknowns. You will make use of the SIN,
COS and TAN functions on your calculator in this section.
To begin a right triangle is one with one 90 angle in it. The side across from this angle is
known as the hypotenuse (h). Another angle in the triangle is reference angle and is often
given the variable (theta). The side opposite the reference angle is known as the
OPPOSITE (o) and the side touching the reference angle the ADJACENT (a).
h
o
reference angle theta Ɵ
a
The diagram above shows the right angle, hypotenuse (h), opposite (o) and adjacent
(a).
Trig Ratios: Sine (sin), cosine (cos) and tangent (tan) are ratios which compare the
various sides of right triangles. The ratios are as follows:
Sin
Cos
Or the acronym that will help you recall is SOHCAHTOA where O, A, H refer to
opposite, adjacent, hypotenuse.
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Physics 11 Reading Booklet
A ratio is a general comparison of any two sides to get an angle an example can be
seen just below. Sin, Cos and Tan are simply mathematical functions like those
seen in the previous lesson. They can be moved from one side of an equation to
another by performing an inverse sine (or cosine or tangent) as needed.
In any right triangle missing angles may be found using the trigonometric functions when
2 sides are known. If asked to find another angle the fastest way is to use the total angles
in triangles. Simply subtract the right angle (90 ) and the one you have found from 180
to get the third angle.
Trigonometry can also be used to find the length of unknown sides based on a known
angle in a right triangle. Here we use the same trig functions as above but solve for any
lengths. Two examples follow:
Example 1 of using angles to find sides.
Given
h
20
a
If the reference angle is 30
We can easily calculate h by stating the ratio
Sin 30 =
Therefore hyp =
= 40 = h
We can calculate a by stating the ratio
Tan 30 =
Therefore a =
=
= 34.6
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Physics 11 Reading Booklet
Example 2 of using sides to find angles.
B
5
3
4
A
Determine Angle A we can use any of the information we have
We can use “3” as the opposite side and the hypotenuse “5”
Sin A =
Therefore angle A is
= 37 (using sine inverse on your calculator)
And B can be easily found by subtracting 180 with 90 and 37 equaling 53
Trigonometry has many useful applications for determining distances if an angle is
known or angles if a distance is known.
The key to any trigonometry problem is being able to visualize the problem and draw it
out. All physics students should get into the habit of diagramming every problem.
Click on this YouTube link for video lesson on How to Solve Right Angle.
https://www.youtube.com/watch?t=294&v=p6hcLw4lzTQ
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Physics 11 Reading Booklet
Practice and Self-Check E
1. Solve for side “x”
2. Solve for side x
3. Solve for angle x
Ans) 1)7 2) 5.6 3) 460
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Physics 11 Reading Booklet
Solving Trigonometry Word Problems
Trigonometry is often expressed as an image representing the angles, circles and other trigonometric
concepts involved. It is therefore best to translate your word problem into a picture that represents the
word problem. Use the words to visualize and draw the image involved to represent the problem.
Angle of Elevation:
Angle of Depression:
What do we do with angle of depression that is outside of our triangle?
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Physics 11 Reading Booklet
Example 1:
Two players stand on a basketball court. The angle of elevation from the foot of each player to the 10”
high basket are 400 and 500.
How far apart are the players?
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Physics 11 Reading Booklet
Example 2:
Two towers face each other separated by a distance d = 15 m. As seen from the top of the first tower, the
angle of depression of the second tower's base is 60o and that of the top is 30o. What is the height (in
meters) of the second tower?
Example 3:
Two men on opposite sides of a TV tower of height 30 m notice the angle of elevation of the top of this
tower to be 45o and 60o respectively. Find the distance (in meters) between the two men.
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Example 4
Physics 11 Reading Booklet
Elise is standing on top of a 50 foot building and sees her friend, Molly. If Molly is 30 feet away from
the base of the building, what is the angle of depression from Elise to Molly? Elise’s eye height is 4.5
feet.
Solutions
Click on the YouTube link to watch the video solution
https://www.youtube.com/watch?v=CyKRsz_RYVI
Click on this YouTube link for a video lessons on Solving Trigonometry Word Problems.
https://www.youtube.com/watch?t=182&v=-QOEcnuGQwo
You should be able to do your own research for any of the topic in this Booklet.
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Physics 11 Reading Booklet
Practice and self-check
1. Harry is flying a lite. The kite string makes an angle 430 with the ground. If Harry is standing 100,
from a point on the ground directly below the kite, find the length of the string kite.
2. A ladder leans against a building. The top of the ladder reaches a point on the building which is 18 m
from the ground. The foot of the ladder is 7 m from the building. Find the angle the ladder makes
with the ground.
3. Two men on the same side of a tall building notice the angle of elevation to the top of the building to
be 300 and 600 respectively. If the height of the building is known to be h =60 m, find the distance (in
meters) between the two men.
Answers:
1. 137 m
2. 690
3. 69 m
Now that you have completed this Reading Booklet, go and complete the Physics 11 Substantive
Assignment
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