Solution Manual for Physics (5th Edition) by James S. Walker

All 32 Chapters
Covered




SOLUTIONS

, Chapter 1: Introduction to
Physics

Answers to Even-Numbered
Conceptual Questions

2. The quantity T + d does not make sense physically, because it adds together variables that
have different physical dimensions. The quantity d/T does make sense, however; it could
represent the distance d traveled by an object in the time T.
4. The frequency is a scalar quantity. It has a numerical value, but no associated direction.
7 17 8 9
6. (a) 10 s; (b) 10,000 s; (c) 1 s; (d) 10 s; (e) 10 s to 10 s.



Solutions to Problems and Conceptual Exercises
1. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
1 gigadollars
Solution: (a) Convert $152, 000, 9

the units: 000  110 dollars 0.152
1 teradollars 4
110 dollars 1.5210

12
(b) Convert the units $152, 000,
again: 000 
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.

2. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the given number by conversion 6 factors to obtain the desired units.
1.010 m
85 m   8.5105
Solution: (a) Convert m 6
the units: 1.010 m 1000 mm
85  m   
m 1m 0.085
(b) Convert the units
again:
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.

3. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
9
Gm 110 m
Solution: Convert 0.3  8
 310
the units: s Gm
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.




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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Chapter 1: Introduction to Physics James S. Walker, Physics, 5 Edition


4. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the given number by conversion factors to obtain
12
the desired units.
9
teracalculation 110 calculations 110 s
Solution: Convert 136.  
the units: 8 s teracalculation ns
5
 136,800 calculations/ns 1.36810
calculations/ns
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.

5. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
Solution: 1. (a) Substitute 1 2
x
dimensions for the variables: 2 at

1 [L] 2
[L] 2 [T]2 [T] [L]The equation is dimensionally consistent.

2. (b) Substitute v
t 
dimensions x
for the variables: TLT  1
L T  Not dimensionally consistent
2x
t
a
3. (c) Substitute
dimensions for the
variables:

 T    L    T 2  T  Dimensionally consistent
LT2
Insight: The number 2 does not contribute any dimensions to the problem.

6. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
x 1
Solution: 1. (a) Substitute   L  T
dimensions for the variables: v  L  T  1 T Y
a LT2 1   T   T  1
2. (b) Substitute dimensions for the   
variables: v LT 1 T T N

3. (c) Substitute dimensions for the 2x   L 
1
 T2  T
variables: a LT2 1 T 2 Y

L 
2
v
2
LT L T
2 2 2

4. (d) Substitute dimensions for the  LT2  LT2  L N
variables: a L
Insight: When squaring the velocity you must remember to square the dimensions of both
the numerator (meters) and the denominator (seconds).




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Chapter 1: Introduction to Physics James S. Walker, Physics, 5 Edition


7. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
L
Solution: 1. (a) Substitute
dimensions for the variables: vt    T Y  L
  T L   2
1 2
at 1  T L
2. (b) Substitute dimensions for the
2 Y
variables:
T 
2 2

 L 
2at 2   T   
L
3. (c) Substitute dimensions for the T 2
 T N
variables: 2 
v
LT2 L2 T2 L
2



LT2  LT2 
4. (d) Substitute dimensions for the
variables:  L Y
a
Insight: When squaring the velocity you must remember to square the dimensions of both
 L 
the numerator (meters) and the denominator (seconds).

8. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
L 2
1 2
at 1  T  L
Solution: 1. (a) Substitute 2 2 N
T2 
dimensions for the variables:
L
at   T  
L
2. (b) Substitute dimensions for the Y
variables:
T2  T

3. (c) Substitute dimensions for the 2x  2  L T
variables: a LT2 N

 L   L
4. (d) Substitute dimensions for the 2  L  L2 
2a x
variables:  T2 Y
T T
2


Insight: When taking the square root of dimensions you need not worry about the positive
and negative roots; only the positive root is physical.

9. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
2 p
v2 2a x
Solution: Substitute dimensions for
the variables:  L  p
 L   L
  
T2 
 T   p 1
2
L L therefore
p
Insight: The number 2 does not contribute any dimensions to
the problem.




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Chapter 1: Introduction to Physics James S. Walker, Physics, 5 Edition


10. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
p
a 2xt
Solution: Substitute [L]
dimensions for the
variables: p
[T] [L][T]
2


[T] [T] therefore p
2 p



Insight: The number 2 does not contribute any dimensions to the problem.

11. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
Solution: Substitute dimensions p 2
for the t h
g
variables:
1  T
T[ [ 1 2
L
2

L]
p  L   T 
L]
p



T L 1 2 T therefore p  1
p 

2

Insight: We conclude the h belongs inside the square root, and the time to fall from 2h g .
rest a distance h is t 


12. Picture the Problem: This is a dimensional analysis question.
Strategy: Rearrange the expression to solve for the force F, and then substitute the
appropriate dimensions for the corresponding variables.
Solution: Substitute dimensions for F ma 
the variables, using [M] to represent
the dimension of mass: [M]
[L]
2
Insight: This unit, kg· m/s , will later be given the name “Newton” and abbreviated as N.

13. Picture the Problem: This is a dimensional analysis question.
Strategy: Rearrange the expression to solve for the force constant k, and then substitute the
appropriate dimensions for the corresponding variables.
m 2 2 4m
square both T 4 or k 
Solution: 1. Solve T m
2 sides:
2
for k:
2
k k T
2. Substitute the dimensions, [M]
using [M] to represent the k 
[T]2
dimension of mass:
2
Insight: This unit will later be renamed “Newton/meter.” The 4 does not contribute any
dimensions.

14. Picture the Problem: This is a significant figures question.
Strategy: Follow the given rules regarding the calculation and display of significant figures.
Solution: Round to the 8
2.997910 m/s 
rd
3 digit: 3.0010
8


Insight: It is important not to round numbers off too early when solving a problem because
excessive rounding can cause your answer to significantly differ from the true answer,
especially when two large values are subtracted to find a small difference between them.


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Chapter 1: Introduction to Physics James S. Walker, Physics, 5 Edition


15. Picture the Problem: The parking lot is a rectangle.
124.3 m
Strategy: The perimeter of the parking lot is the sum of
the lengths of its four sides. Apply the rule for addition
of numbers: the number of decimal places after addition 41.06 m 41.06 m
equals the smallest number of decimal places in any of
the individual terms. 124.3 m


Solution: 1. Add the numbers: 124.3 + 41.06 + 124.3 + 41.06 m = 330.72 m
2. Round to the smallest number of decimal
330.72 m
places in any of the individual terms:  330.7
Insight: Even if you changed the problem to 2124.3 m   2 41.06 m , you’d still have
to report 330.7 m as the answer; the 2 is considered an exact number so it’s the “124.3 m”
value that limits the number of significant digits.

16. Picture the Problem: The weights of the fish are added.
Strategy: Apply the rule for addition of numbers, which states that the number of decimal
places after addition equals the smallest number of decimal places in any of the individual
terms.
Solution: 1. Add the numbers: 2.77 + 14.3 + 13.43 lb = 30.50 lb
2. Round to the smallest number of decimal
30.50 lb
places in any of the individual terms:  30.5
Insight: The 14.3-lb rock cod is the limiting figure in this case; it is only measured to within an
accuracy of 0.1 lb.

17. Picture the Problem: This is a significant figures question.
Strategy: Follow the given rules regarding the calculation and display of significant figures.
Solution: 1. (a) The leading zeros are not significant: 0.0000 3 0 3 has
2. (b) The middle zeros are significant: 5
6.2 0 1×10 has 3 significant
4 significant
Insight: Zeros are the hardest part of determining significant figures. Scientific notation can
remove the ambiguity of whether a zero is significant because any zero written to the right of
the decimal point is significant.

18. Picture the Problem: This is a significant figures question.
Strategy: Apply the rule for multiplication of numbers, which states that the number of
significant figures after multiplication equals the number of significant figures in the least
accurately known quantity.
2
A r 11.37 m 406.13536
2 2
m 
406.1
Solution: 1. (a) Calculate the
area and round to four
2
significant figures: A r 
2
 6.8 m 145.2672443
2
m2 
1.510
2. (b) Calculate the area and
round to two significant
figures:
Insight: The number is considered exact so it will never limit the number of significant digits
you report in an answer. If we present the answer to part (b) as 150 m the number of
significant figures is ambiguous, so we present the result in scientific notation to clarify that
there are only two significant figures.




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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Chapter 1: Introduction to Physics James S. Walker, Physics, 5 Edition


19. Picture the Problem: This is a significant figures question.
Strategy: Follow the given rules regarding the calculation and display of significant figures.
rd
Solution: (a) Round to the 3 digit: 3.14159265358979 
th
(b) Round to the 5 digit: 3.14159265358979  3.
th
(c) Round to the 7 digit: 3.14159265358979 3.14

Insight: It is important not to round numbers off too early when solving a problem because
3.1415
excessive rounding can cause your answer to significantly differ from the true answer.

20. Picture the Problem: This problem is about the conversion of units.
Strategy: Convert each speed to m/s units to compare their magnitudes.
Solution: 1. (a) The speed is already va 0.25 m/s
in m/s units:
 km   1000 m    1 h 
2. (b) Convert the speed to vb   0.21 m/s
m/s units: 0. 3600 s
75
 h    1 km   
 ft   1m 
3. (c) Convert the speed to vc m/s s 3.7
  3.281 ft 
m/s units: 12

 cm   1m 
4. (d) Convert the speed to vd   0.16
m/s s   100 cm 
m/s units: 16


5. Rank the four speeds: vd vb va
Insight: To one significant digit the speeds in (b) and (d) are identical (0.2 m/s), but it is
ambiguous how to round the
0.25 m/s of (a) to one significant digit (either 0.2 or 0.3 m/s). Notice that it is impossible to
compare these speeds without converting to the same unit of measure.

21. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion
17.7 factors
 to change the units.
Solution:
in   1 ft 1. Find the length in feet:  2.5 cubit  3.68 ft
   
1 cubit
17.7 12in
 
2. Find the
in    1 ft width and height in feet: 1.5 cubit 2.21 ft
   
1 cubit  12 in
V LWH 3.68 ft2.21
3.
Find the volume in 18
ft2.21 ft
cubic feet:
Insight: Conversion factors are conceptually equal to one, even though numerically they
often equal something other than one. They are often helpful in displaying a number in a
convenient, useful, or easy-to-comprehend fashion.

22. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate
  conversion factors to change the units.
6 mi  1.609 km 
Solution: 109 km/h 
to km/h: Convert mi/h 8 
   1 mi 1.110
2
 h  
Insight: The given 68 mi/h has only two significant figures, thus the answer is limited to two
significant figures. If we present the answer as 110 km/h the zero is ambiguous, thus we use
scientific notation to remove the ambiguity.



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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Chapter 1: Introduction to Physics James S. Walker, Physics, 5 Edition


23. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution:
 Convert feet to kilometers: 3212 1 1.609 km 
ft   mi  
5280 ft 1 mi 0.9788
  
Insight: Conversion factors are conceptually equal to one, even though numerically they
often equal something other than one. They are often helpful in displaying a number in a
convenient, useful, or easy-to-comprehend fashion.

24. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.

Solution: 1 msg   3600 s   24 h   7 d msg
to weeks: Convert seconds       
67,  200  d wk wk 4 msg
7 10
 9 s h    wk
Insight: In this problem there is only one significant figure associated with the phrase, “every 9
seconds.”

25. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
  1 m 
Solution: Convert feet to meters: 108 ft  
 
3.281 ft 32.9

Insight: Conversion factors are conceptually equal to one, even though numerically they
often equal something other than one. They are often helpful in displaying a number in a
convenient, useful, or easy-to-comprehend fashion.

26. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution:
 0.20 Convert
g  carats to pounds: 530.2  2.21 lb 

ct  1
    kg  
ct 1000 g kg 0.23
   
Insight: Conversion factors are conceptually equal to one, even though numerically they
often equal something other than one. They are often helpful in displaying a number in a
convenient, useful, or easy-to-comprehend fashion.

27. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
2
Solution: Convert m/s to feet per second  m   3.28 ft 
per second:
 98.12     ft
 s  1m 322
Insight: Conversion factors are conceptually equal to one, even though s 2 numerically they
often equal something other than one. They are often helpful in displaying a number in a
convenient, useful, or easy-to-comprehend fashion.


28. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: 1. (a) The speed 55 km/h because 1 mi/h = 1.609 km/h.
must be greater
2. (b) Convert the miles to 
kilometers: mi   1.609 km 
55 
 km
  mi 88
 h   h
Insight: Conversion factors are conceptually equal to one, even though numerically they
often are equal to something other than one. They often help to display a number in a
convenient, useful, or easy-to-comprehend fashion.


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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Chapter 1: Introduction to Physics James S. Walker, Physics, 5 Edition


29. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution:
per second:1. (a) Convert to feet   m   3.28 ft 
 23    ft
 s  1 m  75
s
2.
per(b) Convert to miles
hour:  m 1  3600 s 
23 
     mi   mi
s 1609 m 1 hr 51
    h
Insight: Mantis shrimp have been known to shatter the glass walls of the aquarium in which
they are kept.

30. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units. In
this problem, one
“jiffy” corresponds to the time in seconds that it takes light to travelone centimeter.
s
Solution: 1. (a): Determine the magnitude of a jiffy:1 1m
  311
.3357 jiffy
s 10 
1
 8  




2.9979 10 m 100 cm
 
1 jiffy  cm cm
2. (b) Convert
 60 sminutes
 to jiffys: 1 1 jiffy3.3357
 1011

minute
  
11 1.7987 10
12
1 min 3.3357 10 s
 
Insight: A jiffy is 33.357 billionths of a second. In other terms 1 jiffy = 33.357 picosecond (ps).

31. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
 28.3 L   1 mutchkin 

Solution:
Convert 1. (a) 
1
3
ft

cubic feet to mutchkins:
ft3


0.42 L

67


 0.28  mutchkin 

2. (b) Convert noggins to gallons: 1 noggin   1 gal 

0.42 
   L  
noggin mutchkin 3.785 L 0.031
   
Insight: To convert noggins to gallons, multiply the number of noggins by 0.031 gal/noggin.
Conversely, there are 1 noggin/0.031 gal = 32 noggins/gallon. That means a noggin is about
half a cup. A mutchkin is about 1.8 cups.

32. Picture the Problem: A cubic meter of oil is spread out into a slick that is one molecule thick.
Strategy: The volume of the slick equals its area times its thickness. Use this fact to find the
area. 3
V 1.0 m 1 m 
A   6 

Solution: Calculate the 2.010
6
h 0.50  m 110 m
area for the known
volume and thickness:
Insight: Two million square meters is about 772 square miles!

33. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
 m   3.28 ft 
Solution: Convert meters 9.81 
to feet:
 2     32.2
s 1m
  

Copyright © 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–8

, th
Chapter 1: Introduction to Physics James S. Walker, Physics, 5 Edition
Insight: Conversion factors are conceptually equal to one, even though numerically they
often are equal to something other than one. They often help to display a number in a
convenient, useful, or easy-to-comprehend fashion.




Copyright © 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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