Download Measurement Systems Physical sciences like physics, chemistry

Name _________________________
Date: ________
Measurement Systems
Physical sciences like physics, chemistry, and astronomy are highly quantitative.
The laws and theories that make up the foundation of these sciences are the result of
experimentation and verification over many years, decades, or even centuries. The
experiments done by scientists in these fields rely on measurements of a wide variety of
properties that are found in nature. It is then the job of the scientist to determine which
properties are connected and which ones aren’t. In addition to making measurements, it is
equally important to understand the ways in which our measurements are inaccurate.
Professional scientists learn many sophisticated methods for analyzing the validity of
their experimental results, many of which are far beyond the scope of this course.
However, we will learn basic error analysis with the hopes that it will give you a sense of
the accuracy of scientific measurements.
Measurements Systems
Since the late 19th century measurements of the properties of nature has been
standardized under the definitions provided within the International System of
Measurement (referred to as the ‘SI System’ due to it being established as the System
International d’Unites in France). The SI System identifies seven (7) basic properties of
nature and has established a base unit of measurement for each (Table 2.1).
Basic Physical Property
Distance
Mass
Time
Temperature
Electric Current
Amount of a Substance
Intensity of Light
Base Unit
Symbol of Base Unit
Meter
Kilogram
Second
Kelvin
Ampere
Mole
Candela
m
kg
s
K
A
mol
cd
Table 2.1: The seven basic physical properties of nature and their designated base units.
Distance, mass, time, and temperature are the basic properties that we will be
concerned with in this class, though you should be familiar with the other properties and
their units as well.
1. What base unit from Table 2.1 would you use to express the following quantities?
How far is it to the Moon?
_______________
How long does it take Venus to orbit the Sun?
_______________
How many atoms are in a rock sample?
_______________
How heavy is the Pluto Express satellite?
_______________
How tall is the Martian volcano Olympus Mons?
_______________
How bright does the star Sirius appear?
_______________
What is the life expectancy of the Voyager spacecrafts?
_______________
Dimensional Analysis
In nature there are countless physical properties beyond the base unit that require
measurement. These additional properties are a combination of the elementary properties
found in Table 2.1 and will therefore have units that are a combination of the base units.
The way these new units are determined is through a process called dimensional analysis.
By knowing how the elementary properties combine through some definition or law of
nature, we can substitute the base units to determine the appropriate unit. Below are a
couple of examples of how dimensional analysis is used to determine these new units.
Speed
The speed of an object is defined as the amount of distance covered within a
measured amount of time:
speed =
distance
time
If we wanted to know the numerical value of an object’s speed we would simply
substitute the known quantities for the distance and time (the speed of a car that goes 150
! = 50). To determine the unit of measurement for speed
miles in 3 hours; speed = 150/3
we would instead substitute the base units of the properties used in the definition for
speed:
base unit for speed =
!
base unit for distance
base unit for time
base unit for speed =
meters
= m s ("meters - per - second")
second
Acceleration
! object’s acceleration of a object is defined as its speed changes over a known
An
period of time:
acceleration =
speed
time
An example of acceleration would be a car slowing down to a stop for a red light. To
determine the unit for acceleration we substitute the base unit for speed (m/s) and the
base unit for time (s) to !
get:
base unit for acceleration =
base unit for speed
base unit for time
" m%
$ '
m 1
! for speed = # s & =
base unit
(
= m 2 ("meters - per - square - second")
s
s
s
s
Force
!
An object is accelerated whenever a force is applied to it. The amount of force felt by
an object is defined through Newton’s 2nd Law of Motion as being equal to the mass of
an object multiplied by its acceleration (F = ma). Using dimensional analysis we find
that the base unit for force is:
base unit for force = (base unit of mass) " (base unit of accelration)
!
kg # m
base unit for force = (kg) " (m 2 ) =
s
s2
The base unit for force is cumbersome to write and to read (“kilogram-meter-persquare-second). It is common to rename such units for the sake of simplicity. The base
! has been renamed the Newton (N) which is defined in the following way:
unit for force
$ kg # m '
1 Newton (N) " 1 &
)
% s2 (
!
2. Determine the base unit of the following physical properties by employing
dimensional analysis. Show your work. Once you have determined the base
units, research the units to determine if any have more commonly accepted
names. There are a couple of items you should make not of when performing
these exercises:
(a) The physical property of velocity has the same units of speed (m/s)
(b) Numerical factors, such as ½ and π, are dimensionless and therefore
do not effect the units of a physical property
Area of a square = length " width
!
!
Volume of a sphere =
Density =
4
" (radius)3
3
mass
Volume
Base unit for Area ≡ _______
Base unit for Volume ≡ _______
Base unit for Density ≡ _______
!
Momentum = mass " velocity
Base unit for Momentum ≡ _______
!
Torque = Force " length
!
Kinetic Energy =
!
Luminosity =
!
1
(mass) " (velocity)2
2
Energy
time
Base unit for Torque ≡ _______
Base unit for Kinetic Energy ≡ _______
Base unit for Luminosity ≡ _______
Beyond the Base Unit
There are many instances when the base unit is insufficient when describing a
physical quantity. People don’t typically refer to being alive for 662,691,456 seconds as
a birthday milestone, but that is what happens when someone turns 21 years old. In other
words, it is quite impractical to express one’s age in terms of the base unit of time
(secodns). A much larger unit, the year, being roughly 31,500,000 seconds long, is more
practical when expressing ones age. As such, each of the base units of the SI system can
be built up to describe large physical properties or cut down to describe those that are
small.
The simplest way to create new units of measurement beyond a base unit is to rely on
building up or breaking down the base unit by factors of ten. Doing so allows us to
visualize scales more easily because we are not required to memorize random conversion
factors that are found within the Imperial System of measurements used in daily
experience (e.g. 12 inches = 1 foot; 5,280 feet = 1 mile; 16 oz. = 1 pound). Making the
process easier is that each factor of ten has a unique prefix that can be added to any base
unit in question. Table 2.2 lists the prefixes and the corresponding powers of ten that are
used in the SI system.
Power of Ten
Prefix
Abbreviation
1024
1021
1018
1015
1012
109
106
103
102
101
100
10-1
10-2
10-3
10-6
10-9
10-12
10-15
10-18
10-21
10-24
YotaZettaExaPetaTeraGigaMegaKiloHectoDeca-DeciCentiMilliMicroNanoPicoFemtoAttoZeptoYocto-
Y
Z
E
P
T
G
M
k
h
da
(Base Unit)
d
c
m
µ*
n
p
f
a
z
y
* The symbol for micro- is the Greek letter, “mu,” because “m” is already used for milli-.
Table 2.2: SI prefixes and their powers of ten.
The prefixes listed in Table 2.2 are added to the base unit when referring to a specific
factor of the base unit. One thousand (103) meters is referred to as one kilometer. Onehundredth (10-2) of a gram is called a centigram. Similarly, to obtain the correct
abbreviation for any of these units, the prefix symbol is added before the base unit
abbreviation. One kilometer is written as 1 km while one centigram is written as 1 cg.
When presented with measurements written out completely it is useful to write the
numbers in scientific notation in order to determine the correct prefix. For example:
2,500 meters = 2.5 " 10 3 meters = 2.5 kilometers = 2.5 km
0.0041 seconds = 4.1 " 10 -3 seconds = 4.1 milliseconds = 4.1 ms
In!practice measurements are often not written in proper scientific notation where the
decimal point would be moved so that there is a single, non-zero digit to the left. Rather,
the!decimal point is moved so that that mantissa will be followed by a factor of ten whose
power is a multiple of three (e.g. 103, 106, 10-9, etc.). To give an example, the number
56,300 grams is typically written as follows:
56,300 grams = 56.3 " 10 3 grams = 56.3 kilograms = 56.3 kg
3. Fill in the blank spaces of Table 2.3 below by expressing each measurement in the
! desired form.
Column 1: measurements written in terms of the base unit
Column 2: measurements from Column 1 written in terms more
appropriate units by making use of prefixes
Column 3: measurements from Column 2 written with the use of the
appropriate symbol.
Base Unit Measurement
E.g. – 0.002 grams
1 Billionth of a meter
7.52 x 10-15 meters
Measurement with Prefix
Measurement with Symbol
2.0 milligrams
2.0 mg
84.4 kiloseconds
1.21 GW
0.0058 Amperes
6.5 Megayears
16,000,000,000,000 Bytes
2.15 ZN
Table 2.3
Sometimes when we calculate a physical quantity we end up with a value that would
be clearer if we were to express it in more convenient units. There are several ways to go
about converting measurements within the metric system, but the method described
below will always work when carried out properly:
Known Measurement
Desired Measurement's Power of 10
=
Desired Measurement
Known Measurement's Power of 10
Let’s consider a case where the mass of an object was calculated to be 140,000
centigrams (cg). If we wanted to express that measurement in terms of kilograms, we
! perform the conversion method from above once we define each part of the
could
proportion:
Known measurement = 140,000 cg
Desired measurement = x kg
Desired measurement’s (kg) power of ten = 103
Known measurement’s (cg) power of ten = 10-2
We begin the conversion by substituting our values into the proper places in the
proportion:
140,000 cg
10 3
=
x kg
10 -2
Next we cross-multiply:
! 10 3 " x kg = 140,000 cg " (10 -2 )
The we divide to get x by itself:
!
(140,000 cg " 10 -2 )
x kg =
10 3
Calculating for x we end with the result:
!
x = 1.4 kg
4. Perform the following conversions using the method outlined above. Be sure to
! the steps in your calculation.
include a page showing
Convert 4 m into cm.
4 m = __________ cm
Convert 4 m into mm.
4 m = __________ mm
Convert 185 mg into cg.
185 m = __________ cm
Convert 3 x 103 ns into µs.
3 x 103 ns = __________ µs
Convert 0.5 MW into PW.
0.5 MW = __________ PW
Convert 620 dA into daA.
620 dA = __________ daA
Convert 1 Em into fm.
1 Em = __________ fm
Standard Astronomical Measurements
From the extraordinarily far distances to galaxies to the enormous masses of stars
astronomers contend with excessively large numbers on a regular basis. In an effort to
make these outrageous numbers more comprehensible astronomers have developed a
system of measurements that is exclusive to the scales involved in their field. Listed
below are some examples of standard astronomical measurements that are used
frequently.
Astronomical Unit (AU) – the average distance between the Earth and the Sun. It has a
measured value of:
1 AU = 1.50 x 1011 m = 1.50 x 10 8 km
The Astronomical Unit is typically used to describe the distances between objects
within our solar system, the distance an exoplanet orbits its parent star, and the distances
! make up a binary star system.
between stars that
Parsec (pc) – the distance a star would need to be in order for it to have a parallactic shift
of one arc-second. The name is derived from combination of the parallax-second. The
parsec has been determined to have a distance of:
1 pc = 3.09 x 1016 m = 3.09 x 1013 km
The parsec is the unit from which all astronomical distance equations have been
derived and is used frequently when reporting distances to objects beyond our solar
system as well!as the sizes of individual galaxies.
Light-Year (LY) – defined as the distance light travels in one year’s time. It is a
calculated distance that is found by multiplying the measured speed of light (3.00 x 108
m/s) by the number of seconds in one year (3.15 x 107 s). Thus the light-year is defined
numerically as:
1 LY = 9.46 x 1015 m = 9.46 x 1012 km
1 pc = 3.26 LY
! can be used to describe the same distance/size measurements as the
The light-year
parsec. Because the light-year is based on the speed of light, it provides an added
component of understanding.! If a star were exactly one light-year away from Earth, then
it would take the light emitted exactly one year to make its trip to Earth. Similarly if a
different star were 8 LY from Earth then the light from that star would take exactly 8
years to reach Earth. Knowing the number of light-years an object is away from Earth
provides a direct measurement of how “old” the light from that object is.
Solar Mass (MSun) – defined as the amount of mass contained within our Sun. It has a
measured value of:
1 MSun = 1.99 x 10 30 kg
The solar mass is one of several solar units that are used throughout astronomy. Other
solar units include the solar luminosity (LSun), the solar radius (RSun), and the solar
! solar units are useful when discussing the properties of stars
lifespan (τ Sun). In general,
because they provide a direct comparison to the star we know most about.
Converting from standard metric measurements to standard astronomical
measurements (or vice versa) requires setting up proportions. The two key things to
remember to properly setting up the proportions are:
(1) Make sure you have the proper conversion factor.
(2) Make sure the units are the same on each side of the equal sign.
As an example, let’s determine the distance from the Sun to Jupiter (7.8 x 108 km) in
Astronomical Units (AU).
First, we write out the problem algebraically:
7.80 x 10 8 km = x AU
Next we set up the proportion using the conversion factor, 1 AU = 1.50 x 108 km, making
sure to keep the units the same on each side of the equal sign:
!
7.80 x 10 8 km
x AU
=
8
1.50 x 10 km
1 AU
Solving for x yields the result:
!
x = 5.20 AU
5. Use any resource other than another student’s lab (internet, textbook, or other) to
! listed in the table below. Pay close attention to the units
determine the distances
that are asked for in each example. Once you are finished, convert each
measurement into the units of kilometers.
Distances
Measurements Found
Converted Measurements
Distance from NYC to LA
m
Diameter of the Earth
m
Distance from Earth to Moon
m
Distance from Sun to Venus
AU
Distance from Sun to Saturn
AU
Distance from Sun to
Proxima Centaruri
Diameter of the Milky Way
Galaxy
Distance to the Andromeda
Galaxy (M31)
Distance to the farthest
quasar
Estimated size of the
observable universe
pc
pc
LY
LY
LY
Ratios and Scale
Comparing measurements is an essential form of analysis in the sciences. A simple
glance will tell whether one measurement is larger or smaller than another, but scientists
are more interested in a more accurate analysis. To them it’s more important to know
how much one quantity is larger than another. The most effective way quantities are
compared is through the use of ratios.
Any measurement with which we are familiar can use that as a basis for comparison.
Living on in central Suffolk County on Long Island, we are most likely familiar with the
fact that a car trip to New York City takes roughly one hour. We could use this
information to estimate how long a trip to Boston would take by using ratios.
A search on MapQuest reveals that the distance form Selden to New York City is 58
miles and the distance from Selden to Boston is 252 miles. Comparing the distances
using a ratio would yield the following:
Distance to Boston
252
=
= 4.34
Distance to NYC
58
This comparison says “the distance from Selden to Boston is 4.34 times the distance
from Selden to NYC.” From there we can use the knowledge that if it’s an hour long trip
! a trip to Boston would be a little over 4.3 hours (4 hrs 18 min).
to NYC by car then
Ratios are the means for creating model scales so that accurate representations of the
physical world can be created to aid in visualization. The most common form of model
scale is the map scale. When reading a map, a scale is typically provided in the maps’
legend that lets the reader know what a unit distance on the map equates to in reality. So
a map scale of 1 cm = 25 km means, “for every centimeter’s distance that is measured on
the map, the real distance is 25 kilometers.” If the map scale is not provided then it can
be determined so long we know one real distance.
As an example, let’s assume we are presented with the map below that shows New
York City, Nassau County, and central Suffolk County and we want to determine the map
scale. Earlier we saw that the distance from Selden (indicated by the star) to NYC is 92.8
kilometers (58 miles).
Using a ruler we find a map distance of 11.6 cm directly between Selden and NYC.
The 11.6 cm distance on the map represents 92.8 km in true distance. Using the ratio of
the two measurements we can find the map scale.
True Distance
92.8 km
x km
=
=
Map Distance
11.6 cm
1 cm
The map scale is therefore, 1 cm = 8 km. This “map conversion factor” can be used to
compute the true distance between any two points on the map.
!
Example:
What is the true distance between Selden and Port Jefferson?
True Distance to Port Jefferson
z km
8 km
=
=
Map Distance to Port Jefferson
1.4 cm
1 cm
After cross-multiplying and dividing we find:
!
z = 1.4 " 8 = 11.2 km
6. Find the true distances between the locations that are given in the table below.
Point A
!
Point B
Selden
Patchogue
Huntington
West Gilgo Beach
Rocky Point
Wading River
Manorville
Queens
Map Distance
True Distance