Introduction
Chapter Review

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1-1 - 1-2  Physics and the Laws of Nature & Units of Length, Mass, and Time

The study of physics deals with the fundamental laws of nature and many of their applications. These laws govern the behavior of all physical phenomena. We describe the behavior of physical systems using various quantities that we create for this purpose; however, there are three quantities — length, mass, and time — that we take as fundamental quantities and we use these three to create other quantities.

We define a system of units for these quantities so that we can specify how much length, mass, or time we have. The system of units used in this book is the SI, which stands for Système International. In this system the unit of length is the meter (m), the unit of mass is the kilogram (kg), and the unit of time is the second (s). This system of units is still sometimes referred to by its former name, the mks system.

SI units are based on the metric system. An important aspect of this system is its hierarchy of prefixes used for quantities of different magnitudes. Certain of these prefixes are used very frequently in physics, so you should become very familiar with them. Some of the more common ones are listed here:

PowerPrefixSymbol
10-15 femto f
10-12 pico p
10-9 nano n
10-6 micro µ
10-3 milli m
10-2 centi c
103 kilo k
106 mega M

Exercise 1-1 Metric Prefixes Write the following quantities using a convenient metric prefix. (a) 0.00025 m, (b) 25,000 m, (c) 250 m, (d) 250,000,000 m, (e) .0000025 m

Solution (a) 0.25 mm (b) 25 km (c) 0.25 km (d) 250 Mm (e) 2.5 µm


Practice Quiz

1.
Which of the following quantities is not one of the fundamental quantities?
(a) length
(b) speed
(c) time
(d) mass

sorry, try again

your answer: speed

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sorry, try again

1-3 Dimensional Analysis

In physics we derive the physical quantities of interest from the set of fundamental quantities of length, mass, and time. The dimension of a quantity tells us what type of quantity it is. When indicating the dimension of a quantity only, we use capital letters enclosed in brackets. Thus, the dimension of length is represented by [L], mass by [M], and time [T].

We use many equations in physics, and these these equations must be dimensionally consistent. It is extremely useful to perform a dimensional analysis on any equation about which you are unsure. If the equation is not dimensionally consistent, it cannot be a correct equation. The rules are simple:

  • Two quantities can be added or subtracted only if they are of the same dimension.

  • Two quantities can be equal only if they are of the same dimension.

Notice that only the dimension needs to be the same, not the units. It is perfectly valid to write 12 inches = 1 foot because both of them are lengths, [L] = [L], even though their units are different. However, it is not valid to write x inches = t seconds because they have different dimensions, [L] [T].


Example 1-2 Checking the Dimensions Given that the quantities x (m), v (m/s), a (m/s2), and t (s) are measured in the units shown in parentheses, perform a dimensional analysis on the following equations.

(a) x = t (b) x = 2vt (c) v = at + t/x (d) x = vt + 3at2

Picture the Problem There is no picture.

Strategy Write each equation in terms of its dimensions and check if the equation obeys the preceding rules.

Solution

Part (a)
1. Write the equation with dimensions only: Since these dimensions are not the same, the equation is not valid.

Part (b)

[L] = [T]
2.

Write out the dimensions of this equation:
The right-hand-side dimension is equal to the dimension on the left, so the equation is dimensionally correct.

Part (c)

3. Write out the dimension of this equation:
The first and second terms on the right-hand-side do not have equal dimension and cannot be added. This is not a valid equation.

Part (d)

4. Write out the dimension of this equation: Here, both terms on the right have the same dimension, which is also equal to the dimension on the left. This equation is dimensionally correct.

Insight  Notice that in the dimensional analysis purely numerical factors are ignored because they are dimensionless. Since there are dimensionless quantities, dimensional consistency does not guarantee that the equation is physically correct, but it makes for a quick and easy first check.


Practice Quiz

2.
Which of the following expressions is dimensionally correct?
(a) [L] = [M] x [T]
(b) [T] = [L]/[T]
(c)
(d)

3.
If speed v has units of m/s, distance d has units of m, and time t has units of s, which of the following expressions is dimensionally correct?
(a) v = t/d
(b) t = vd
(c) d = v/t
(d) t = d/v

sorry, try again

sorry, try again

your answer:

sorry, try again

sorry, try again

sorry, try again

sorry, try again

your answer: t = d/v



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