PHY 126 Final Exam Problem 1 (6+6+6+6+6 points)

F

F

image of arrow by mirror

final image 2

20.0 cm 85.0 cm20.0 cm

(a) (6 points)

lens. theofright theto

cm, )0.86(0.43'cm )0.64(0.32

1

'

1

cm (250) 0.125

11

'

11 from

obtained is ' distance image final The lens. theofobject thebecomesit Then lens. the

ofleft the tocm (250) 125 85.0(170) 40.0(80.0) formed ismirror by the image The

points) (6 A(B) (b)

ssfss

s

(3 points for the mirror image, 3 points for the final image)

40.0 cm 40.0 cm 170.0 cmAB

cm/s. 0)0.200(0.45

is image final theof speed theTherefore sec. 2.00in cm 42.6(85.1) tocm (86.0)

43.0 from cm 0)0.400(0.90 of distance a moves image final The cm. 2.00(4.00)by

move tosec 2.00 it takes cm/s, (2.00) 1.00 of speed aat moving ismirror theSince

lens. theofright the tocm, )1.85(6.42'

cm )0.64(0.32

1

'

1

cm )258(129

11

'

11 from obtained is ' distance

image final The lens. theofobject thebecomesit Then lens. theofleft theto

cm 129.0(258) 85.0(170) 44.0(88.0) formed ismirror by the image The

s

sfsss

(6 points)

(6 points)

(d)

(e)

(c) Real (3 point) and inverted (3 point).

Problem 2 (15+15 points)

(a)Since the density of concrete is larger than that of water, the object is completely submerged in the water. Therefore the apparent mass of the object is: (5 points).0.1)( kgVm waterconcretec

As the net torque around the pivot is zero:).2/(30sin LMgLmLT c

.29)/8.9)(0.10.12()2(2/

)2/(NkgNkgkggMm

L

gLMmT c

c

(b) Since the density of ice is smaller than that of water, the object floats. Therefore there is no weight contribution from the ice object.

As the net torque around the pivot is zero:

).2/(30sin LMgLT

.8.9)/8.9)(0.1(2/

)2/(NkgNkgMg

L

gLMT

(5 points)

(5 points)

(5 points)

(5 points)

(5 points)

.39)/8.9)(0.20.12()2(2/

)2/(NkgNkgkggMm

L

gLMmT c

c

A:

B:

A:

.19)/8.9)(0.2(2/

)2/(NkgNkgMg

L

gLMT B:

Problem 3 (3+3+3+3+4+4+10 points)

(a) ab: For isothermal process T=0 and U=0. W=nRTCln(Vb/Va)=-nRTCln(r) and Q=W=-nRTCln(r). (1 point for each quantity)

(b) bc: For isochoric process, V=0 and W=0; Q=U=nCV(TH-TC). (1 point for each)

(d) da: For isochoric process, V=0 and W=0; Q=U=nCV(TC-TH). (1 point for each)

(e) The net work for one cycle is Wnet=nR(TH-TC)ln(r). (4 points)

(f) The net heat added for one cycle Qadded is the same as Qcd=nRTHln(r). (4 points).

(e) The efficiency is Wnet/Qadded=1-(TC/TH) (5 points). This is the same as the efficiency of a Carnot cycle engine operating between the two temperatures (5 points).

(c) cd: For isothermal process T=0 and U=0. W=nRTHln(Va/Vb)=nRTHln(r) and Q=W=nRTHln(r). (1 point for each)