College Physics – Instructor’s Solutions Manual, 10th Edition

SOLUTION MANUAL
All Chapters Included

, SOLUTIONS MANUAL COLLEGE PHYSICS 10TH Ed

1
Introduction

ANSWERS TO WARM-UP EXERCISES

1. (a) The number given, 568 017, has six significant figures, which we will retain in converting the number to scientific nota-
5
tion. Moving the decimal five spaces to the left gives us the answer, 5.680 17 × 10 .

(b) The number given, 0.000 309, has three significant figures, which we will retain in converting the number to scientific
–4
notation. Moving the decimal four spaces to the right gives us the answer, 3.09 × 10 .

2. We first collect terms, then simplify:
2 2
[M ][L]2 . [T ] [T ] = [M ][L] [T ] = [M ][L]
[T ]3 [L] [T ]3[L] [T ]

As we will see in Chapter 6, these are the units for momentum.

3. Examining the expression shows that the units of meters and seconds squared (s2) appear in both the numerator and the de-
nominator, and therefore cancel out. We combine the numbers and units separately, squaring the last term before doing so:
2
 m•  1.00 km •  60.0 s •
7.00 2 
 s   1.00 103 m   1.00 min 
 1.00
= (7.00)  •  3600•m • km•  s 2 •
2
3 
1.00 10   1.00  s 2  m min
   
km
= 25.2
min2

4. The required conversion can be carried out in one step:

 •
h = (2.00 m ) 1.00 cubitus =
  4.49 cubiti
 0.445 m 

5 The area of the house in square feet (1 420 ft2) contains 3 significant figures. Our answer will therefore also contain three
2
significant figures. Also note that the conversion from feet to meters is squared to account for the ft units in which the area
is originally given.

 1.00 m • 2
(
A = 1 420 ft2 )  3.281 ft  = 131.909 m2 = 132 m
2




6. Using a calculator to multiply the length by the width gives a raw answer of 6 783 m2. This answer must be rounded to con-
tain the same number of significant figures as the least accurate factor in the product. The least accurate factor is the length,
which contains 2 significant figures,
3
since the trailing zero is not significant (see Section 1.6). The correct answer for the
area of the airstrip is 6.80 × 10 m2.

7. Adding the three numbers with a calculator gives 21.4 + 15 + 17.17 + 4.003 = 57.573. However, this answer must be
rounded to contain the same number of significant figures as the least accurate number in the sum, which is 15, with two
significant figures. The correct answer is therefore 58.

8. The given Cartesian coordinates are x = –5.00 and y = 12.00. The least accurate of these coordinates contains 3 significant
figures, so we will express our answer in three significant figures. The specified point, (–5.00, 12.00), is in the second quad-
rant since x < 0 and y > 0. To find the polar coordinates (r,  ) of this point, we use


1

,r = x2 + y2 = (5.00)2 + (12.00)2 = 13.0




2

, and
 y•  12.00 •
 = tan−1 = tan
−1
= –67.3
   
x –5.00

Since the point is in the second quadrant, we add 180° to this angle to obtain  = −67.3 + 180 = 113. The polar coordi-
nates of the point are therefore (13.0, 113°).

9. Refer to ANS. FIG 9. The height of the tree is described by the tangent of the 26° angle, or

h
tan 26 =
45 m

from which we obtain

h = ( 45 m) tan 26 = 22 m
ANS. FIG 9

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS




3