Chapter 1: Introduction
Chapter Review


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Chapter Review




1-1 - 1-2 Basics Physical Quantities and Standard Prefixes

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.

Along with the quantities must come a way to qualify them, that is, a way to specify how much length, mass, or time you have. This quantification is accomplished with a system of units. An officially adopted system of units starts out by specifying the units for the three fundamental quantities. 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). The system of units based on these choices is called SI units which stands for System International. This system of units is still sometimes referred to by its old name, the mks system.

As you probably noticed, SI units is 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 below:

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

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


Practice Quiz

 
Which of the following quantities is not one of the fundamental quantities?
length
speed
time
mass

sorry, try again

your answer: speed

sorry, try again

sorry, try again

1-3 Dimensional Analysis

As stated above, in physics we derive the physical quantities of interest from the set of fundamental quantities of length, mass, and time. When a quantity is broken down to its makeup in terms of the fundamental quantities we call this breakdown its dimension. The dimension, therefore, represents the fundamental type of the quantity. When indicating the dimension of a quantity only, we use capital letters and enclose them in brackets. Thus, the dimension of length is represented by [L], mass by [M], and time [T].

We use a lot of equations in physics; 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 dimensional consistency is not there, it cannot be a correct equation. The rules are simple:

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 for this problem.

Strategy Write each quantity in terms of its dimension and check if the equation obeys the above rules.

Solution
(a) Write the equation with dimensions only: Since these dimensions are not the same the equation is not valid. [L] = [T]
(b) Write out the dimensions of this equation:
Once broken down, the right-hand-side is of equal dimension to the left so the equation is dimensionally correct.
(c) 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.
(d) Write out the dimension of this equation: Here, both terms on the right have the same dimension which also equals that of the left-hand-side. 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

 
Which of the following expressions is dimensionally correct?
[L] = [M] x [T]
[T] = [L]/[T]



 
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?
v = t/d
t = vd
d = v/t
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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