Fundamentals of Physics Extended 10th Edition – Complete Solutions Manual by Halliday, Resnick & Walker (All Chapters Included)

SOLUTION MANUAL
All Chapters Included

,Chapter 1

1. THINK In this problem we’re given the radius of Earth, and asked to compute its
circumference, surface area and volume.

EXPRESS Assuming Earth to be a sphere of radius

RE 6.37 106 m103 km m6.37 103 km,

the corresponding circumference, surface area and volume are:
4
C 2R , A 4R2 , V  R3 .
E E E
3
The geometric formulas are given in Appendix E.

ANALYZE (a) Using the formulas given above, we find the circumference to be

C 2RE 2(6.37 103 km) 4.00104 km.

(b) Similarly, the surface area of Earth is

A 4RE2 46.37 103 km 5.10 108 km2 ,
2


(c) and its volume is
4
V  R3 4 6.37 103 km 1.08 1012 km3.
3

E 
3 3
2 3
LEARN From the formulas given, we see that C RE , A RE , and V RE . The ratios
of volume to surface area, and surface area to circumference are V / A RE / 3 and
A / C 2RE .

2. The conversion factors are: 1 gry 1/10 line, 1 line 1/12 inch and 1 point = 1/72
inch. The factors imply that

1 gry = (1/10)(1/12)(72 points) = 0.60 point.

Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry 2 = 0.18 point 2 .

3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside
front cover of the textbook (see also Table 1–2).



1

,
, 2 CHAPTER 1


(a) Since 1 km = 1 103 m and 1 m = 1 106 m,

1km 103 m 103 m106 m m109 m.

The given measurement is 1.0 km (two significant figures), which implies our result
should be written as 1.0 109 m.

(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 102 m,

1cm = 102 m = 102 m 106 m m104 m.

We conclude that the fraction of one centimeter equal to 1.0 m is 1.0 104.

(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,

1.0 yd = 0.91m106 m m9.1 105  m.

4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we
obtain
 6 picas 
0.80 cm = 0.80 cm  1 inch   1.9 picas.
2.54 cm 1 inch
  
(b) With 12 points = 1 pica, we have
1 inch
0.80 cm = 0.80 cm 6 picas 12
  points  23 points.
2.54 cm 1 inch 1 pica
   

5. THINK This problem deals with conversion of furlongs to rods and chains, all of
which are units for distance.

EXPRESS Given that 1 furlong  201.168 m, 1 rod 5.0292 m and 1 chain 20.117 m,
the relevant conversion factors are
1 rod
1.0 furlong 201.168 m (201.168 m) 40 rods,
5.0292 m
and
1 chain
1.0 furlong 201.168 m (201.168 m) 10 chains .
20.117 m
Note the cancellation of m (meters), the unwanted unit.

ANALYZE Using the above conversion factors, we find

40
(a) ) the distance d in rods to be d 4.0 furlongs 4.0 furlongs 160 rods,
rods
1 furlong