Test Bank for Fundamentals of Physics Extended 10th Edition – Complete Solution Manual | Verified Answers | All Chapters Included | Updated for 2026

1




Soluton_ManualFundamentals_of_Physics_Extended_10th_E
dition
aW




1. THINK aWIn aWthis aWproblem aWwe’re aWgiven aWthe aWradius aWof aWEarth, aWand aWasked aWto
aWcompute aWits aWcircumference, aWsurface aWarea aWand aWvolume.




EXPRESS aWAssuming aWEarth aWto aWbe aWa aWsphere aWof aWradius



R a W =EaW(6.37 aWaW106 aWm)(10−3 aWkm m)aW= aW6.37 aWaW103 a W km,



the aWcorresponding aWcircumference, aWsurface aWarea aWand aWvolume aWare:
4 a W
C aW= aW2aWR A aW= aW4aWR2aW, V a W = aW
W
a
a W , R3 aW.
E E
3 E
The aWgeometric aWformulas aWare aWgiven aWin aWAppendix aWE.




ANALYZE (a) Using the formulas given above, we find the
aW aW aW aW aW aW aW aW aW




circumference to be
aW aW aW




C aW= aW2aWRE =aW2aW(6.37 aW aW103 a W km) aW= aW4.00104 a W km.


(b) Similarly, aWthe aWsurface aWarea aWof aWEarth aWis

A aW= aW4 E = aW4 aW(6.37 aWaW103 a W km)
2 aW
= aW5.10 aWaW108 a W km2aW,
2
aWR
(c) and aWits aWvolume
aWis

,2


a 4 a W
(6.37 aW103 aWkm)
3 aW

V a W = aW
W
R3 aW = aW1.08 aWaW1012 a W km3.
a 4
= aW E
a W W


3 3


LEARN aWFrom aWthe aWformulas aWgiven, aWwe aWsee aWthat a W C , a W A R2 aW, aWand a W V R3 aW. aWThe
aWratios aWof aWvolume RE
E E

to aWsurface aWarea, aWand aWsurface aWarea aWto aWcircumference aWare a W V aW/ aWAaW= aWRE a W /aW3 a W and a W AaW/aWC
aW= aW2RE a W .




2. The aWconversion aWfactors aWare: aW1 aWgry aW=aW1/10 aW line aW, aW1 aWline aW=aW1/12 aW inch aW and aW1
aW point aW= aW1/72 aWinch. aWThe aWfactors aWimply aWthat



1 aWgry aW= aW(1/10)(1/12)(72 aWpoints) aW= aW0.60 aWpoint.



Thus, aW1 aWgry2 aW= aW(0.60 aWpoint)2 aW= aW0.36 aWpoint2, aWwhich aWmeans aWthat a W 0.50 a W gry2aW= a W 0.18
2
a W point aW.




3. The aWmetric aWprefixes aW(micro, aWpico, aWnano, aW…) aWare aWgiven aWfor aWready aWreference aWon aWthe
aWinside aWfront aWcover aWof aWthe aWtextbook aW(see aWalso aWTable aW1–2).




(a) Since aW1 aWkm aW= aW1 aW aW103 aWm aWand aW1 aWm aW= aW1 aW aW106 aWm,



1km aW= aW103 aWm aW= aW(103 aWm m)aW= m.
m)(106
9
aW10
aW




The aWgiven aWmeasurement aWis aW1.0 aWkm aW(two aWsignificant aWfigures), aWwhich aWimplies aWour aWresult
aWshould aWbe aWwritten aWas aW1.0 aW aW10 aWm.
9




(b) We aWcalculate aWthe aWnumber aWof aWmicrons aWin aW1 aWcentimeter. aWSince aW1 aWcm aW= aW10−2 aWm,


1cm aW= aW10−2 aWm aW= aW(10−2m)(106 aWaWm m) m.
aW= aW10
4

, 3



We aWconclude aWthat aWthe aWfraction aWof aWone aWcentimeter aWequal aWto aW1.0 aWm aWis aW1.0 aW aW10−4.

, 4




(c) Since aW1 aWyd aW= aW(3 aWft)(0.3048 aWm/ft) aW= aW0.9144 aWm,


1.0aWyd aW= aW(0.91m)(106 aW aWm m)aW= aW9.1aWaW105 a W m.

4. (a) aWUsing aWthe aWconversion aWfactors aW1 aWinch aW= aW2.54 aWcm aWexactly aWand aW6 aWpicas aW= aW1 aWinch,
a W we aWobtain
 a W 1aWinch a W  aW6 a W picas aW a W
0.80 aWcm aW= a W (0.80 aWaW cm) aW  a W   aW1.9 a W picas.
2.54 aW cmaW 1 aWinch a W
  
(b) aWWith aW12 aWpoints aW= aW1 aWpica, aWwe aWhave

 a W 1aWinch a W  aW6 a W picas aW aW12 a W points aW a W
0.80 aWcm aW= a W (0.80 aW cm) aW  a W 23
aW points.
 aW  a W  a W a W 
2.54 aW cm aW 1 aWinch a W 1 aWpica
   


5. THINK aWThis aWproblem aWdeals aWwith aWconversion aWof aWfurlongs aWto aWrods aWand aWchains, aWall
aWof aWwhich aWare a W units aWfor aWdistance.




EXPRESS a W Given a W that a W 1 = a W 201.168 aWm, aW1aWrod aW= and a W 1aWchain aW= aW20.117 aW maW,
aWfurlong aWrelevant aWconversion aW5.0292 aW m a W the


aWfactors aWare


1 aWrod
1.0 aW furlong a W =aW201.168 aWm aW= =aW40 a W rods,
aW(201.168 aWmaW)
5.0292 m

and
1 =10 aW chains aW.
1.0 aW furlong a W =aW201.168 aWm aW= chainm
aW

aW(201.168 aWmaW)

20.117

Note aWthe aWcancellation aWof aWm aW(meters), aWthe aWunwanted aWunit.



ANALYZE aWUsing aWthe aWabove aWconversion aWfactors, aWwe aWfind



(a) the aWdistance aWd aWin aWrods aWto aWbe a W d a W = a W 4.0 aW furlongs a W =(4.0 = aW160 a W rods,
40 aWrods
aWfurlongs) a W

1 aWfurlong
10 aW chains a W
(b) and aWin aWchains aWto d a W = a W 4.0 a W furlongs a W =(4.0 aWfurlongs) = a W 40 aW chains.
aWbe
1 aWfurlong