College Physics – Instructor’s Solutions Manual,

,© Copyright 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
Chapter 1: Introduction to Physics
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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, 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. (a) 107 s; (b) 10,000 s; (c) 1 s; (d) 1017 s; (e) 108 s to 109 s.



Solutions to Problems and Conceptual Exercises
1. Picture the Problem: This is simply a units conversion problem.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
1 gigadollars
Solution: (a) Convert the units: $114,000,000  = 0.114 gigadollars
1109 dollars
1 teradollars
(b) Convert the units again: $114, 000,000  = 1.14 10−4 teradollars
11012 dollars
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.

2. Picture the Problem: This is simply a units conversion problem.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
1.0 10−6 m
Solution: (a) Convert the units: 70 m  = 7.0 10−5 m
m
1.0 10−6 m 1 km
(b) Convert the units again: 70 m   = 7.0 10−8 km
m 1000 m
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.

3. Picture the Problem: This is simply a units conversion problem.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
Gm 1109 m
Solution: Convert the units: 0.3  = 3108 m/s
s Gm
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.

4. Picture the Problem: This is simply a units conversion problem.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
teracalculation 11012 calculations 110−6 s
Solution: Convert the units: 70.72  
s teracalculation s
= 7.072 107 calculations/s

Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.




© Copyright 2010 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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, Chapter 1: Introduction to Physics James S. Walker, Physics, 4th Edition


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 dimensions x = vt
forthe variables: m
 s ( )
m= s = m  The equation is dimensionally consistent.
 
2. (b) Substitute dimensions x = 21 at 2
for the variables: m
  2
m= 1
2 2 
( s ) = m  dimensionally consistent
s 
3. (c) Substitute dimensions 2x m
t=  s= = s2 = s  dimensionally consistent
for the variables: a ms 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.
m
 s ( )
Solution: 1. (a) Substitute dimensions vt = s = m Yes
for the variables:  
m 2
2. (b) Substitute dimensions for the variables:
1
at2 = 1
2 2 
( s ) = m Yes
 
2
s
m m
3. (c) Substitute dimensions for the variables: 2at = 2 (s ) =
 s2  No
s

= ( m s) = m
2
v2
4. (d) Substitute dimensions for the variables: Yes
a m s2
Insight: When squaring the velocity you must remember to square the dimensions of both the numerator (meters) and
the denominator (seconds).



7. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.
m 2
Solution: 1. (a) Substitute dimensions 1
at2 = 1
2 2 
( s ) = m No
for the variables: 2
 s 
m m
2. (b) Substitute dimensions for the variables: at = (s) =
 2 Yes
s  s
2x 2m
3. (c) Substitute dimensions for the variables: = = s No
a m s2
m m
4. (d) Substitute dimensions for the variables: 2ax = 2  2  = Yes
s  s
Insight: When taking the square root of dimensions you need not worry about the positive and negative roots; only the
positive root is physical.


Copyright © 2010 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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