Test Bank For Fundamentals of Physics 7th Edition by Brownstein Halliday, Walker, REsnick, Christma ISBN NO-10,0471470627 ISBN NO-13,978-0471470625 Complete Guide A++.pdf

Editedv by v Foxitv Reader
Copyright(C)vby vFoxitvSoftwarevCompany,2005-
2007v Forv Evaluationv Only.


Chapter v 1: MEASUREMENT

1. The vSIvstandard vofvtime vis vbased von:
A. the vdailyvrotationvofvthe vearth
B. the vfrequencyvofvlightvemitted vbyvKr86
C. the vyearlyvrevolutionvofvthe vearthvaboutvthe vsun
D. avprecisionvpendulum vclock
E. none vofvthese
vAns: v E



2. Avnanosecond vis:
A. 109vs
B. 10−9vs
C. 10−10 v s
D. 10−10 v s
E. 10−12
Ans:v B

3. The vSIvstandard vofvlengthvis vbased von:
A. the vdistance vfrom vthe vnorthvpole vtovthe vequatorvalongvavmeridianvpassing vthroughvParis
B. wavelength vofvlightvemitted vbyvHg198
C. wavelength vofvlightvemitted vbyvKr86
D. avprecisionvmetervstickvinvParis
E. the vspeedvofvlight
vAns: v E



4. Inv1866,vthe vU.vS.vCongressvdefinedvthevU.vS.vyardvas vexactlyv3600/3937 v internationalvmeter.v
This vwas vdone vprimarilyvbecause:
A. lengthvcanvbe vmeasured vmore vaccuratelyvinvmeters vthanvinvyards
B. the vmetervis vmore vstable vthanvthe vyard
C. this vdefinitionvrelates vthe vcommonvU.vS.vlengthvunits vtovavmore vwidelyvused vsystem
D. there vare vmore vwavelengths vinvavyardvthanvinva vmeter
E. the vmembersvofvthis vCongressvwerevexceptionallyvintelligentv
Ans:v C

5. Whichvofvthe vfollowing vis vclosestvtovavyard vinvlength?
A. 0.01vm
B. 0.1m v
C. 1vm
D. 100vm
E. 1000vm
Ans:v C




Chapterv1: MEASUREMENT 1

, 6. There vis vnovSIvbase vunitvforvarea vbecause:
A. anvarea vhas vnovthickness; vhence vnovphysicalvstandard vcanvbe vbuilt
B. we vlive vinvavthree v(notvavtwo)vdimensionalvworld
C. itvis vimpossible vtovexpress vsquare vfeetvinvterms vofvmeters
D. area vcanvbe vexpressed vinvterms vofvsquare vmeters
E. area vis vnotvanvimportantvphysicalvquantityv
Ans:v D

7. The vSIvbase vunitvforvmass vis:
A. gram
B. pound
C. kilogram
D. ounce
E. kilopound
vAns: v C



8. Av gram v is:
A. 10−6v kg
B. 10−3v kg
C. 1v kg
D. 103v kg
E. 106v kg
Ans:v B

9. Whichvofvthe vfollowing vweighs vaboutva vpound?
A. 0.05v kg
B. 0.5vkg
C. 5v kg
D. 50vkg
E. 500vkg
Ans:v D

10. (5.0v ×v 104)v ×v (3.0v ×v 106)v =v
A. 1.5v×v 109
B. 1.5v×v1010
C. 1.5v×v1011
D. 1.5v×v1012
E. 1.5v×v1013
Ans:v C

11. (5.0v ×v 104)v ×v (3.0v ×v 10−6)v =v
A. 1.5v×v10−3
B. 1.5v×v10−1
C. 1.5v×v101
D. 1.5v×v103
E. 1.5v×v105
Ans:v B

2 Chapterv1: MEASUREMENT

,12. 5.0v×v 105v +v3.0v×v 106v =
A. 8.0v×v 105
B. 8.0v×v 106
C. 5.3v×v105
D. 3.5v×v105
E. 3.5v×v106
Ans:v E

13. (7.0v ×v 106)/(2.0v ×v 10−6)v =v
A. 3.5v×v10 −12
B. 3.5v×v10− 6
C. 3.5
D. 3.5v×v106
E. 3.5v×v1012
Ans:v E

14. The vnumbervofvsignificantvfigures vinv0.00150vis:
A.v v 2
B.v v 3
C.v v 4
D.v v 5
E.v v 6
Ans:v B

15. The vnumbervofvsignificantvfigures vinv15.0vis:
A.v v 1
B.v v 2
C.v v 3
D.v v 4
E.v v 5
Ans:v C

16. 3.2v ×v 2.7 v =v
A. 9
B. 8
C. 8.6
D. 8.64
E. 8.640
Ans:v C




Chapterv1: MEASUREMENT 3

, 17. 1.513v+v27.3v=v
A. 29
B. 28.8
C. 28.9
D. 28.81
E. 28.813
( )Ans:v B
18. 1vmivis vequivalentvtov1609 vm vsov55vmphvis:
A. 15v m/s
B. 25vm/s
C. 66 vm/s
D. 88vm/s
E. 1500vm/s
vAns: v B




19. Avsphere vwithvavradius vofv1.7 vcm vhas va vvolume vof:
A. 2.1v×v10 −5 v m 3
B. 9.1v×v10 − 4 v m 3
C. 3.6 v×v10− 3 vm 3
D. 0.11vm 3
E. 21v m 3
Ans:v A

20. Avsphere vwithvavradius vofv1.7 vcm vhas va vsurface varea vof:
A. 2.1v×v10 −5 v m 2
B. 9.1v×v10 − 4 v m 2
C. 3.6 v×v10− 3 vm 2
D. 0.11vm 2
E. 36 vm 2
Ans:v C

21. Avrightvcircularvcylindervwithvavradius vofv2.3vcm vand vavheightvofv1.4 vm vhas vavvolume vof:
A. 0.20vm 3
B. 0.14 vm 3
C. 9.3v×v10− 3 vm 3
D. 2.3v×v10 − 3 vm 3
E. 7.4 v×v10−4 v m 3
Ans:v D

22. Avrightvcircularvcylindervwithvavradius vofv2.3vcmvandva vheightvofv1.4vcmvhasvavtotalvsurfacevarea v
of:
A. 1.7 v×v 10 −3 v m 2
B. 3.2v×v10 − 3 vm 2
C. 2.0v×v 10−3v m 3
D. 5.3v×v10 − 3v m 2
E. 7.4 v×v10−3 v m 2
Ans:v D

4 Chapterv1: MEASUREMENT