Fundamentals of Physics Extended – Chapters 1–10 | 10th Edition | Halliday, Resnick & Walker | Complete Solution Manual

INSTRUCTOR
SOLUTIONS
MANUAL

,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:
C  2R , A  4R2 , 4 3
V R .
E E
3 E
The geometric formulas are given in Appendix E. algebra. Mathematics is crucial in many industries, including engineering, finance, data
science, and technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their problem-solving abilities



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R2E  4 6.37  103 km  5.10  108 km2 ,
2


(c) and its volume is
4 3 4 6.37  103 km3  1.08  1012 km3.
V R
3 E  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 gry2 = 0.18 point2 .

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

algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and technology. The primary goal of mathematics exams is to test students' understanding of
mathematical concepts, their problem-solving abilities

(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2m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

0.80 cm = 0.80 cm 1 inch  6 picas 
 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 1chain  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  160 rods,
(a) ) the distance d in rods to be d  4.0 furlongs 4.0 furlongs
rods
1 furlong