Test bank for fundamentals of physics 9th edition by resnick walker and halliday instructor solution manual_compressed

TEST,BANK,For,Fundamentals,of,Physics,10th,
Edition,By,Resnick,,Walker,and,Halliday,Chapt
ers,1,-,44

,Chapter,1

1. Various,geometric,formulas,are,given,in,Appendix,E.

(a) Expressing,the,radius,of,the,Earth,as

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

its,circumference,is, s,,2,R,,2,(6.37,,103, km),,4.00,104, km.

(b) The,surface,area,of,Earth,is,A,,4,R2 ,4,  6.37,,103,,km  , 5.10,,108 km2.
2




R ,,,,  6.37,,103,,km  ,1.08,,1012,,km 3 .
4,, 3 4, 3,,,
(c),The,volume,of,Earth,is,V, ,
3 3

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).

(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

,
, 2 CHAPTER,1



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


0.80,cm,=, 0.80,cm, , 1,inch ,6,
, picas,,12,
,, points,,

2.54,cm, 1,inch, 1,pica  , 23, points.
   


5. Given,that,1,furlong , 201.168,m,,,1,rod,,5.0292,m and,1,chain,,20.117,m,,,we,find
the,relevant,conversion,factors,to,be
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., Using, the, given, conversion,
factors,,we,find

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

(b) and,that,distance,in,chains,to,be

10,chains,
d, ,4.0,furlongs,4.0,furlongs ,40,chains.
1,furlong

6. We,make,use,of,Table,1-6.

(a) We,look,at,the,first,(“cahiz”),column:,1,fanega,is,equivalent,to,what,amount,of,cahiz?,
We,note,from,the,already,completed,part,of,the,table,that,1,cahiz,equals,a,dozen,fanega.,
1
Thus,,1,fanega,=, 12, cahiz,,or,8.33,,102,cahiz., Similarly,,“1,cahiz,=,48,cuartilla”,(in,the
already,completed,part),implies,that,1,cuartilla,=,1,cahiz,,or,2.08,,
48
102, cahiz.,Continuin
g, in, this, way,, the, remaining, entries, in, the, first, column, are, 6.94, , 103,and
3.47103,.