TEST BANK FOR Serway & Jewett’s Physics for Scientists and Engineers 9th Ed By Raymond A. Serway

1. Physics and Measurement 1
2. Motion in One Dimension 14
3. Vectors 40
4. Motion in Two Dimensions 59
5. The Laws of Motion 89
6. Circular Motion and Other Applications
of Newton’s Laws 120
7. Energy of a System 143
8. Conservation of Energy 167
9. 195
10. Rotation of a Rigid Object About a Fixed Axis 225
11. Angular Momentum 259
12. Static Equilibrium and Elasticity 282
13. Universal Gravitation 303
14. Fluid Mechanics 326
15. Oscillations and Mechanical Waves 348
16. Wave Motion 377
17. Sound Waves 395
18. Superposition and Standing Waves 416
Linear Momentum and Collisions
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iv Table of Contents
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19. Temperature 441
20. The First Law of Thermodynamics 462
21. The Kinetic Theory of Gases 486
22. Heat Engines, Entropy, and the
Second Law of Thermodynamics 508
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1
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1
Physics and Measurement
EQUATIONS AND CONCEPTS
The density of any substance is
defined as the ratio of mass to volume.
Density is an example of a derived
quantity.
ρ ≡ m
V
(1.1)
The SI units of density are kg/m3.
SUGGESTIONS, SKILLS, AND STRATEGIES
A general strategy for problem solving will be described in Chapter 2.
Appendix B of your textbook includes a review of mathematical techniques
including:
• Scientific notation: using powers of ten to express large and small
values.
• Basic algebraic operations: factoring, handling fractions, and solving
quadratic equations.
• Fundamentals of plane and solid geometry: graphing functions,
calculating areas and volumes, and recognizing equations and graphs
of standard figures (e.g., straight line, circle, ellipse, parabola, and
hyperbola).
• Basic trigonometry: definition and properties of functions (e.g., sine,
cosine, and tangent), the Pythagorean Theorem, and basic
trigonometry identities.
REVIEW CHECKLIST
You should be able to:
• Describe the standards which define the SI units for the fundamental
quantities length (meter, m), mass (kilogram, kg), and time (second, s).
Identify and properly use prefixes and mathematical notations such as
the following: ∝ (is proportional to), < (is less than), ≈ (is
approximately equal to), Δ (change in value), etc. (Section 1.1)
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2 Physics and Measurement
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• Convert units from one measurement system to another (or convert
units within a system). Perform a dimensional analysis of an equation
containing physical quantities whose individual units are known.
(Sections 1.3 and 1.4)
• Carry out order-of-magnitude calculations or estimates. (Section 1.5)
• Express calculated values with the correct number of significant figures.
(Section 1.6)
ANSWERS TO SELECTED OBJECTIVE QUESTIONS
3. Answer each question yes or no. Must two quantities have the same
dimensions (a) if you are adding them? (b) If you are multiplying them? (c) If
you are subtracting them? (d) If you are dividing them? (e) If you are equating
them?
Answer. (a) Yes. Three apples plus two jokes has no definable answer. (b) No.
One acre times one foot is one acre-foot, a quantity of floodwater. (c) Yes. Three
dollars minus six seconds has no definable answer. (d) No. The gauge of a rich
sausage can be 12 kg divided by 4 m, giving 3 kg/m. (e) Yes, as in the examples
given for parts (b) and (d). Thus we have (a) yes, (b) no, (c) yes, (d) no, and (e)
yes.
☐ ☐ ☐ ☐
ANSWERS TO SELECTED CONCEPTUAL QUESTIONS
3. Suppose the three fundamental standards of the metric system were length,
density, and time rather than length, mass, and time. The standard of density in
this system is to be defined as that of water. What considerations about water
would you need to address to make sure the standard of density is as accurate as
possible?
Answer. There are the environmental details related to the water: a standard
temperature would have to be defined, as well as a standard pressure. Another
consideration is the quality of the water, in terms of defining an upper limit of
impurities. A difficulty with this scheme is that density cannot be measured
directly with a single measurement, as can length, mass, and time. As a
combination of two measurements (mass and volume, which itself involves three
measurements!), a density value has higher uncertainty than a single
measurement.
☐ ☐ ☐ ☐
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Chapter 1 3
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SOLUTIONS TO SELECTED END-OF-CHAPTER PROBLEMS
9. Which of the following equations are dimensionally correct?
(a) vf = vi + ax (b) y = (2 m)cos(kx), where k = 2 m–1
Solution
Conceptualize: It is good to check an unfamiliar equation for dimensional
correctness to see whether it can possibly be true.
Categorize: We evaluate the dimensions as a combination of length, time, and
mass for each term in each equation.
Analyze:
(a) Write out dimensions for each quantity in the equation vf = vi + ax.
The variables vf and vi are expressed in units of m/s, so [vf] = [vi] = LT –1
The variable a is expressed in units of m/s2: [a] = LT –2
The variable x is expressed in meters. Therefore [ax] = L2 T –2
Consider the right-hand member (RHM) of equation (a): [RHM]=LT –1+L2 T –2
Quantities to be added must have the same dimensions. Therefore, equation
(a) is not dimensionally correct. 
(b) Write out dimensions for each quantity in the equation y = (2 m) cos (kx)
For y, [y] = L
for 2 m, [2 m] = L
and for (kx),
[kx] = 2 m–1 ( )x ⎡⎣
⎤⎦
= L–1L
Therefore we can think of the quantity kx as an angle in radians, and we can
take its cosine. The cosine itself will be a pure number with no dimensions.
For the left-hand member (LHM) and the right-hand member (RHM) of the
equation we have
[LHM] = [y] = L [RHM] = [2m][cos (kx)] = L
These are the same, so equation (b) is dimensionally correct. 
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4 Physics and Measurement
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Finalize: We will meet an expression like y = (2 m)cos(kx), where k = 2 m–1, as
the wave function of a wave.
15. A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From
these data, calculate the density of lead in SI units (kilograms per cubic meter).
Solution
Conceptualize: From Table 14.1, the density of lead is 1.13 × 104 kg/m3, so we
should expect our calculated value to be close to this value. The density of water
is 1.00 × 10 3 kg/m3, so we see that lead is about 11 times denser than water,
which agrees with our experience that lead sinks.
Categorize: Density is defined as ρ = m/V. We must convert to SI units in the
calculation.
Analyze:
ρ =
23.94 g
2.10 cm3
⎛⎝ ⎜
⎞⎠ ⎟
1 kg
1 000 g
⎛⎝ ⎜
⎞⎠ ⎟
100 cm
1 m ( )3
=
23.94 g
2.10 cm3
⎛⎝ ⎜
⎞⎠ ⎟
1 kg
1 000 g
⎛⎝ ⎜
⎞⎠ ⎟
1 000 000 cm3
1 m3 ( ) = 1.14 × 104 kg/m3

Finalize: Observe how we set up the unit conversion fractions to divide out the
units of grams and cubic centimeters, and to make the answer come out in
kilograms per cubic meter. At one step in the calculation, we note that one
million cubic centimeters make one cubic meter. Our result is indeed close to the
expected value. Since the last reported significant digit is not certain, the
difference from the tabulated values is possibly due to measurement
uncertainty and does not indicate a discrepancy.
17. A rectangular building lot has a width of 75.0 ft and a length of 125 ft.
Determine the area of this lot in square meters.
Solution
Conceptualize: We must calculate the area and convert units. Since a meter is
about 3 feet, we should expect the area to be about A ≈ (25 m)(40 m) = 1 000 m2.
Categorize: We will use the geometrical fact that for a rectangle Area = Length
× Width; and the conversion 1 m = 3.281 ft.
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Chapter 1 5
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Analyze: A =  × w = (75.0 ft)
1 m
3.281 ft
⎛
⎝ ⎜
⎞
⎠ ⎟
(125 ft)
1 m
3.281 ft
⎛
⎝ ⎜
⎞
⎠ ⎟
= 871 m2 
Finalize: Our calculated result agrees reasonably well with our initial estimate
and has the proper units of m2. Unit conversion is a common technique that is
applied to many problems. Note that one square meter is more than ten square
feet.
25. One cubic meter (1.00 m3) of aluminum has a mass of 2.70 × 103 kg, and the
same volume of iron has a mass of 7.86 × 10 3 kg. Find the radius of a solid
aluminum sphere that will balance a solid iron sphere of radius 2.00 cm on
an equal-arm balance.
Solution
Conceptualize: The aluminum sphere must be larger in volume to compensate
for its lower density. Its density is roughly one-third as large, so we might guess
that the radius is three times larger, or 6 cm.
Categorize: We require equal masses: mA1 = mFe or ρA1VA1 = ρFeVVe
Analyze: We use the volume of a sphere. By substitution,
ρA1
4
3
π rA1
3 ⎛
⎝ ⎜
⎞
⎠ ⎟
= ρFe
4
3
π (2.00 cm)3 ⎛
⎝ ⎜
⎞
⎠ ⎟
Now solving for the unknown,
rA1
3 =
ρFe
ρA1
⎛
⎝ ⎜
⎞
⎠ ⎟
(2.00 cm)3 = 7.86 × 103 kg/m3
2.70 × 103 kg/m3
⎛
⎝ ⎜⎞
⎠ ⎟
(2.00 cm)3 = 23.3 cm3
Taking the cube root, rA1= 2.86 cm. 
Finalize: The aluminum sphere is 43% larger than the iron one in radius,
diameter, and circumference. Volume is proportional to the cube of the linear
dimension, so this excess in linear size gives it the (1.43)(1.43)(1.43) = 2.92 times
larger volume it needs for equal mass.
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6 Physics and Measurement
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27. One gallon of paint (volume = 3.78 × 10–3 m3) covers an area of 25.0 m2. What
is the thickness of the fresh paint on the wall?
Solution
Conceptualize: We assume the paint keeps the same volume in the can and on
the wall.
Categorize: We model the film on the wall as a rectangular solid, with its
volume given by its “footprint” area, which is the area of the wall, multiplied by
its thickness t perpendicular to this area and assumed to be uniform.
Analyze:
V = At gives t = V
A = 3.78 × 10–3 m3
25.0 m2 = 1.51 × 10–4 m 
Finalize: The thickness of 1.5 tenths of a millimeter is comparable to the
thickness of a sheet of paper, so this answer is reasonable. The film is many
molecules thick.
29. (a) At the time of this book’s printing, the U.S. national debt is about $16
trillion. If payments were made at the rate of $1 000 per second, how many years
would it take to pay off the debt, assuming no interest were charged? (b) A dollar
bill is about 15.5 cm long. How many dollar bills attached end to end would it
take to reach the Moon? The front endpapers give the Earth-Moon distance. Note:
Before doing these calculations, try to guess at the answers. You may be very
surprised.
Solution
(a) Conceptualize: $16 trillion is certainly a large amount of money, so even at
a rate of $1 000/second, we might guess that it will take a lifetime (~100
years) to pay off the debt.
Categorize: The time interval required to repay the debt will be calculated
by dividing the total debt by the rate at which it is repaid.
Analyze: T = $16 trillion
$1000 / s
= $16 × 1012
($1000 / s)(3.156 × 107 s/yr) = 507 yr 
Finalize: Our guess was a bit low. $16 trillion really is a lot of money!
(b) Conceptualize: We might guess that 16 trillion bills would reach from the
Earth to the Moon, and perhaps back again, since our first estimate was low.
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Chapter