Complete Solutions Manual for Principles of Physics, Extended, International Adaptation, 12th Edition by David Halliday, Robert Resnick, Jearl Walker ; ISBN13: 9781119820628. (Full Chapters included Chapter 1 to 44)....1. Measurement.
2. Motion Along a Straight Line.
3. Vectors.
4. Motion in Two...
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Principles of Physics, 12e, International Adaptation Solutions Manual
Solutions Manual
for
Principles of Physics
Twelfth Edition
By
David Halliday, Robert Resnick, Jearl Walker
CHAPTER 1
Complete Chapter Solutions Manual
are included (Ch 1 to 44)
** Immediate Download
** Swift Response
** All Chapters included
,Principles of Physics, 12e, International Adaptation Solutions Manual
1. From Fig. 1.1, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is
equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z.
(a) In units of W, we have
258 W
75.0 S = ( 75.0 S) = 91.27 W
212 S
(b) In units of Z, we have
156 Z
75.0 S = ( 75.0 S) = 65.0 Z
180 S
2. 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 3
C = 2 RE , A = 4 RE2 , V= RE .
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
( )
2
A = 4 RE2 = 4 6.37 103 km = 5.10 108 km2 ,
, Principles of Physics, 12e, International Adaptation Solutions Manual
LEARN From the formulas given, we see that C RE , A RE2 , and V RE3 . The ratios of
volume to surface area, and surface area to circumference are V / A = RE / 3 and
A / C = 2RE .
3. The metric prefixes (micro (), pico, nano, …) are given in Table 1.1.2.
100 y 365 day 24 h 60 min
(
(a) 1 century = 10−6 century ) = 52.6 min.
1 century 1 y 1 day 1 h
(b) The percent difference is therefore
52.6 min − 50 min
= 4.9%.
52.6 min
4. 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 rods
(a) the distance d in rods to be d = 3.0 furlongs = ( 3.0 furlongs ) = 120 rods,
1 furlong
10 chains
(b) and in chains to be d = 3.0 furlongs = ( 3.0 furlongs ) = 30 chains.
1 furlong
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