Unit 1 – Physical Quantities and Measurements

Beta Keramati

Introduction to Physics

Last Update: 08/18/2021

Physical Quantities and Units

We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other measurements. For example, we define distance and time by specifying methods for measuring them, whereas we define average speed by stating that it is calculated as distance traveled divided by time of travel.

Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length of a race, which is a physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners). Without standardized units, it would be extremely difficult for scientists to express and compare measured values in a meaningful way.

There are two major systems of units used in the world: SI units (also known as the metric system) and English units (also known as the customary or imperial system). English units were historically used in nations once ruled by the British Empire and are still widely used in the United States. Virtually every other country in the world now uses SI units as the standard; the metric system is also the standard system agreed upon by scientists and mathematicians. The acronym “SI” is derived from the French Système International.

SI Units: Fundamental and Derived Units

The table below gives the fundamental SI units that are used throughout this textbook.

Fundamental SI Units
Length	Mass	Time	Electric Current
meter (m)	kilogram (kg)	second (s)	ampere (A)

In this textbook, the fundamental physical quantities are taken to be the length, mass, time, and electric current. All other physical quantities, such as force and electric charge, can be expressed as algebraic combinations of length, mass, time, and current (for example, speed is length divided by time); these units are called derived units.

Metric Prefixes

SI units are part of the metric system. The metric system is convenient for scientific and engineering calculations because the units are categorized by factors of 10. The table below gives metric prefixes and symbols used to denote various factors of 10.

Metric systems have the advantage that conversions of units involve only powers of 10. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. In nonmetric systems, such as the system of U.S. customary units, the relationships are not as simple—there are 12 inches in a foot, 5280 feet in a mile, and so on. Another advantage of the metric system is that the same unit can be used over extremely large ranges of values simply by using an appropriate metric prefix. For example, distances in meters are suitable in construction, while distances in kilometers are appropriate for air travel, and the tiny measure of nanometers are convenient in optical design. With the metric system, there is no need to invent new units for particular applications.

The term order of magnitude refers to the scale of a value expressed in the metric system. Each power of 10 in the metric system represents a different order of magnitude. For example, 10¹, 10², 10³, and so forth are all different orders of magnitude. All quantities that can be expressed as a product of a specific power of 10 are said to be of the same order of magnitude. For example, the number 800 can be written as 8×10², and the number 450 can be written as 4.5×10²Thus, the numbers 800 and 450 are of the same order of magnitude: 10² Order of magnitude can be thought of as a ballpark estimate for the scale of a value. The diameter of an atom is on the order of 10^-9m while the diameter of the Sun is on the order of 10⁹m.

Metric Prefixes for Powers of 10 and their Symbols
Prefix	Symbol	Value¹	Example (some are approximate)
exa	E	10¹⁸	exameter	Em	10¹⁸ m	the distance light travels in a century
peta	P	10¹⁵	petasecond	Ps	10¹⁵ s	30 million years
tera	T	10¹²	terawatt	TW	10¹² w	powerful laser output
giga	G	10⁹	gigahertz	GHz	10⁹ Hz	a microwave frequency
mega	M	10⁶	megacurie	MCi	10⁶ Ci	high radioactivity
kilo	k	10³	kilometer	km	10³ m	about 6/10 mile
hecto	h	10²	hectoliter	hL	10² L	26 gallons
deka	da	10¹	dekagram	dag	10¹ g	teaspoon of butter
—	—	10⁰(=1)
deci	d	10^-1	deciliter	dL	10^-1 L	less than half a soda
centi	c	10^-2	centimeter	cm	10^-2 m	fingertip thickness
milli	m	10^-3	millimeter	mm	10^-3 m	flea at its shoulders
micro	µ	10^-6	micrometer	µm	10^-6 m	detail in microscope
nano	n	10^-9	nanogram	ng	10^-9 g	small speck of dust
pico	p	10^-12	picofarad	pF	10^-12 F	small capacitor in radio
femto	f	10^-15	femtometer	fm	10^-15 m	size of a proton
atto	a	10^-18	attosecond	as	10^-18 s	time light crosses an atom

Unit Conversion and Dimensional Analysis

Let us consider a simple example of how to convert units. Let us say that we want to convert 80 meters (m) to kilometers (km).

The first thing to do is to list the units that you have and the units that you want to convert to. In this case, we have units in meters and we want to convert them to kilometers.

Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how many of one unit is equal to another unit. For example, there are 12 inches in 1 foot, 100 centimeters in 1 meter, 60 seconds in 1 minute, and so on. In this case, we know that there are 1,000 meters in 1 kilometer.

Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor so that the units cancel out, as shown:

$80m \times \frac{1 km}{1000m} = 0.080km$

Note that the unwanted m unit cancels, leaving only the desired km unit. You can use this method to convert between any type of unit.

Example1.1 – Unit Conversion – A Short Drive Home

Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km/h) and (b) in meters per second (m/s). (Note: Average speed is the distance traveled divided by time of travel.)

Strategy

First, we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct conversion factor and multiplying by it. The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its place.

Solution for (a)

(1) Calculate average speed. Average speed is the distance traveled divided by time of travel. In equation form,

$average\ speed = \frac{distance}{time}$

(2) Substitute the given values for distance and time.

$average\ speed = \frac{10.0 km}{20.0 min} = 0.500 \frac{km}{min}$

(3) Convert km/min to km/h: multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is 60 min/hour. Thus,

$average\ speed =(0.500 \frac{km}{min})(\frac{60 min}{1h})=30.0 \frac{km}{h}$

Discussion for (a)

To check your answer, consider the following:

(1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows:

$(\frac{km}{min})(\frac{1 hr}{60min}) = \frac{1}{60} \frac{km.hr}{min^2}$

which are obviously not the desired units of km/h.

(2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have indeed obtained these units.

(3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km/hr does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined to be 60 minutes, so the precision of the conversion factor is perfect.

(4) Next, check whether the answer is reasonable. Let us consider some information from the problem—if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable.

Solution for (b)

There are several ways to convert the average speed into meters per second.

(1) Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed—one to convert hours to seconds, and another to convert kilometers to meters.

(2) Multiplying by these yields

$average\ speed = (30.0 \frac{km}{h})(\frac{1h}{3600s})(\frac{1000m}{1 km})$

$average\ speed = 8.33\frac{m}{s}$

Discussion for (b)

If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m/s.

Example 1.2 – Dosage Calculation

An MD orders 300 mg of Ibuprofen to be taken by a 6 kg infant every 4 hours. The label
shows 75 – 150 mg/kg per day max. Is the physician’s order within normal range?

First let’s calculate the max dosage for this infant:

$150 mg/kg/day = (150 \frac{mg}{kg.day})(6kg)=900(\frac{mg}{day})$

Now let’s calculate the ordered dose.

$300 mg/(4 hrs) = (300 \frac{mg}{4 hrs})(\frac{24 hrs}{1 day})=1800 \frac{mg}{day}$

The dosage is not within the normal range.

Accuracy, Precision, and Uncertainty of a Measurement

Science is based on observation and experiment—that is, on measurements. Accuracy is how close a measurement is to the correct value for that measurement. For example, let us say that you are measuring the length of standard computer paper. The packaging in which you purchased the paper states that it is 11.0 inches long. You measure the length of the paper three times and obtain the following measurements: 11.1 in., 11.2 in., and 10.9 in. These measurements are quite accurate because they are very close to the correct value of 11.0 inches. In contrast, if you had obtained a measurement of 12 inches, your measurement would not be very accurate.

The precision of a measurement refers to how close the agreement is between repeated measurements (which are repeated under the same conditions). Consider the example of the paper measurements. The precision of the measurements refers to the spread of the measured values. One way to analyze the precision of the measurements would be to determine the range, or difference, between the lowest and the highest measured values. In that case, the lowest value was 10.9 in. and the highest value was 11.2 in. Thus, the measured values deviated from each other by at most 0.3 in. These measurements were relatively precise because they did not vary too much in value. However, if the measured values had been 10.9, 11.1, and 11.9, then the measurements would not be very precise because there would be significant variation from one measurement to another.

The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let us consider an example of a GPS that is attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bulls-eye target, and think of each GPS attempt to locate the restaurant as a black dot. In figure 1.1, you can see that the GPS measurements are spread out far apart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target. This indicates a low precision, but high accuracy. However, in figure 1.2, the GPS measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high precision, low accuracy measuring system.

Figure 1.1

Low Precision-High Accuracy		High Precision – Low Accuracy
Figure 1.1		Figure 1.2 (credit: Dark Evil)

The degree of precision of a measurement is related to the uncertainty in the measurements. Uncertainty is a quantitative measure of the spread of your measurements. If your measurements are not very precise, then the uncertainty of your values will be very high. In more general terms, uncertainty can be thought of as a disclaimer for your measured values. For example, if someone asked you to provide the mileage on your car, you might say that it is 45,000 miles, plus or minus 500 miles. The plus or minus amount is the uncertainty in your value. That is, you are indicating that the actual mileage of your car might be as low as 44,500 miles or as high as 45,500 miles, or anywhere in between. All measurements contain some amount of uncertainty. In our example of measuring the length of the paper, we might say that the length of the paper is 11.0 in., plus or minus 0.2 in. The uncertainty in a measurement, A, is often denoted as ∂A (“delta A), so the measurement result would be recorded as A±∂A. In our paper example, the length of the paper could be expressed as11.0in±0.2in.

The factors contributing to uncertainty in a measurement include:

Limitations of the measuring device,
The skill of the person making the measurement,
Irregularities in the object being measured,
Any other factors that affect the outcome (highly dependent on the situation).

In our example, such factors contributing to the uncertainty could be the following: the smallest division on the ruler is 0.1 in., the person using the ruler has bad eyesight, or one side of the paper is slightly longer than the other. At any rate, the uncertainty in measurement must be based on careful consideration of all the factors that might contribute and their possible effects.

Percent Uncertainty

One method of expressing uncertainty is as a percent of the measured value. If a measurement A is expressed with uncertainty, ∂A, the percent uncertainty (%unc) is defined to be

$\% unc = \frac{\partial A}{A} \times 100 \%$

Example 1.3 – Calculating Percent Uncertainty: A Bag of Apples

A grocery store sells 5lb bags of apples. You purchase four bags over the course of a month and weigh the apples each time. You obtain the following measurements:

Week 1 weight: 4.8 lb
Week 2 weight: 5.3 lb
Week 3 weight: 4.9 lb
Week 4 weight: 5.4 lb

You determine that the weight of the 5-lb bag has an uncertainty of ±0.4lb. What is the percent uncertainty of the bag’s weight?

Strategy

First, observe that the expected value of the bag’s weight, A, is 5 lb. The uncertainty in this value ∂A is 0.4 lb. We can use the following equation to determine the percent uncertainty of the weight:

$\% unc =\frac{\partial A}{A}\times 100$

Solution

Plug the known values into the equation:

$\% unc =\frac{0.4lb}{5lb}\times 100\% =8\%$

Discussion

We can conclude that the weight of the apple bag is 5lb±8%. Consider how this percent uncertainty would change if the bag of apples were half as heavy, but the uncertainty in the weight remained the same. Hint for future calculations: when calculating percent uncertainty, always remember that you must multiply the fraction by 100%. If you do not do this, you will have a decimal quantity, not a percent value.