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Solutions for Principles of Physics, Extended, International Adaptation, 12th Edition Halliday (All Chapters included)

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  • Physics - General Relativity
  • Institution
  • Physics - General Relativity

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 ,

(c) and its volume is

4 3 4
( )
3
V= RE = 6.37  103 km = 1.08  1012 km3 .
3 3

, 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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