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Winter 2017
Section A/B/C
PHYSICS 115 FINAL EXAM
PRACTICE - SOLUTION
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1. [6 pts] A research submarine has a 20-cm-diameter window 8.0 cm thick. The manufacturer says the
window can withstand net forces up to 1.0 x 106 N. What is the submarine’s maximum safe depth in
seawater? The pressure inside the submarine is maintained at 1.0 atm. (𝜌#$%&%'$( = 1030kg/m2 )
A) 1.3 km
B) 1.7 km
C) 2.3 N
D) 2.8 km
E) 3.2 km
𝐹5$' = 𝑃5$' 𝐴 = 𝑃85#89$ − 𝑃;<'#89$ 𝐴 = 𝑃%'= − 𝑃%'= + 𝜌𝑔𝑑 𝐴
𝐹5$' = 𝜌𝑔𝑑𝐴
𝑑=
𝐹5$'
4𝐹5$'
=
= 3150m ≈ 3.2km
𝜌𝑔𝐴 𝜌𝑔𝜋𝑑 C
2. [6 pts] Two cylindrical containers, 1 and 2, contain different unknown
liquids. The heights of the liquids in containers 1 and 2 are equal.
The masses of the liquids in containers 1 and 2 are also equal. (Note
that the diameter of container 2 is greater than that of container 1.) Is
the pressure at point A greater than, less than, or equal to that at
point B?
A
B
1
2
A) Greater than
B) Less than
C) Equal to
D) Not enough information to answer.
The masses of the liquids are the same but the volume of container 2 is larger, so the density of the
liquid in container 2 must be smaller than the density of the liquid in container 1. Points A and B are
at the same depth and the surfaces of the two liquids are at atmospheric pressure. The increase in
pressure as depth increases is proportional to the density of the liquid, so the pressure at A must be
larger than the pressure at B since liquid 1 is more dense.
3. [6 pts] Block A of mass m and volume V floats in a
fluid of density 𝜌 as shown. Half of block A is
submerged. Block B of mass 2m and 3V is placed
in a container with the same fluid density. Which
of the blocks, 1-5 best represents the location of
block B in its container? 𝐹H = 𝑊
A) 1
B) 2
C) 3
D) 4
E) 5
ObjectA
𝐹H = 𝑊
𝜌J 𝑔𝑉#<L = 𝜌M 𝑔𝑉M
𝑉M
𝑉#<L =
2
𝑉M
𝜌J 𝑔 = 𝜌M 𝑔𝑉M
2
𝜌J = 2𝜌M
Physics 115 – Winter 2017
ObjectB
𝐹H = 𝑊
𝜌J 𝑔𝑉#<L = 𝜌H 𝑔𝑉H
𝜌J = 2𝜌M
2𝜌M 𝑔𝑉#<L
𝑉#<L =
C
𝜌H = 𝜌M
2
2
= 𝜌M 𝑔𝑉H
3
1
𝑉
3 H
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4. [7 pts] The 3.0-cm diameter water line shown at right splits into
two 1.0-cm diameter pipes. All pipes are circular and at the
same elevation. At point A, the water speed is 2.0 m/s and the
gauge pressure is 50 kPa. What is the gauge pressure at point B?
A) 12 kPa
B) 22 kPa
C) 29 kPa
D) 38 kPa
E) 42 kPa
𝐴M 𝑣M = 2 𝐴H 𝑣H
𝑣H =
𝐴M 𝑣M
2𝐴H
1
1
𝑃M + 𝜌𝑣MC = 𝑃H + 𝜌𝑣HC
2
2
1
1 𝐴MC 𝑣MC
𝑃M + 𝜌𝑣MC = 𝑃H + 𝜌
2
2 4𝐴CH
1
1 𝐴MC 𝑣MC
1 C
𝐴MC
𝑃H = 𝑃M + 𝜌𝑣MC − 𝜌
=
𝑃
+
𝜌𝑣
1
−
= 11,500Pa ≈ 12kPa
M
2
2 4𝐴CH
2 M
4𝐴CH
5. [6 pts] The viscosity of blood at normal hematocrit is about 4.75 mPa-s, compared to the viscosity of
water, which is 1 mPa-s. In an adult human the coronary arteries have a diameter of about 4 mm.
Suppose you are able to produce a flow rate Q of water in a tube of a given diameter. Which of the
following could you do to produce that same flow rate if you were to use blood instead?
A) Increase the diameter of the tube to 19 mm but keep the pressure difference the same.
B) Increase the diameter of the tube to 5.9 mm but keep the pressure difference the same.
C) Decrease the pressure difference by a factor of 4.75 but keep the diameter the same.
D) Increase the pressure difference by a factor of 22.5 but keep the diameter the same.
E) I don’t need to change anything; the flow rates will be the same.
DpA2
Volume flow rate 𝑄 = Q =
, so if hblood = 4.75hwater then DpA2
= 4.75 DpA2
S
blood
water
8phl
to keep the same flow rate. Since A2 µ d 4 , to increase A2 by a factor of 4.75 we must increase the
ST
(
)
(
diameter of the tube by a factor of (4.75)1/4 = 1.48, to 1.48 x 4 mm = 5.9 mm.
Physics 115 – Winter 2017
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)
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6. [6 pts] At 18.75°C a brass sleeve has a diameter of 2.21988 cm and a steel shaft has a diameter of
2.22258 cm. To what temperature must the sleeve be heated in order for it to slip over the shaft?
𝛼L(%## = 19×10XY K X[ , 𝛼#'$$\ = 12×10XY K X[
A) 64°C
B) 83°C
C) 140°C
D) 101°C
E) 120°C
∆𝑙 = 𝛼𝑙_ 𝑇J − 𝑇8
𝑇J =
∆𝑙
+ 𝑇8 = 83℃
𝛼𝑙_
7. [7 pts] A laboratory technician drops an 85.0 g solid sample of unknown material at a temperature of
100.0°C into a well-insulated copper can. The copper can has a mass of 0.150 kg and contains
0.200 kg of water, and both the can and the water are initially at 19.0°C. The final temperature of the
system is measured to be 26.1°C. Compute the specific heat capacity of the sample.
(ccopper = 390 J/kg-K, cwater = 4190 J/kg-K)
A) 2.51×103 J/kg-K
B) 1.01×103 J/kg-K
C) 0.912×103 J/kg-K
D) 1.62×103 J/kg-K
E) 2.11×103 J/kg-K
Qnet = 0
Qsample + Qwater + Qcopper = 0
ms cs ΔTs + mw cw ΔTw + mc cc ΔTc = 0
cs =
−mw cw ΔTw − mc cc ΔTc −(0.2kg)(4190J/kg-K)(7.1K) − (0.15kg)(390J/kg-K)(7.1K)
=
ms ΔTs
(0.085kg)(−73.9K)
cs = 1.01×103 J/kg-K
8. [6 pts] A window has dimensions of 1.40 m x 2.50 m and is made of glass with a thickness d. On a
winter day, the outside temperature is -20.0°C, while the inside temperature is a comfortable 19.56°C.
The rate of heat loss by conduction is known to be 2.13 x 104 W. Determine the thickness of the
window. (kglass = 0.8 W/m-K)
A) 3.49 mm
𝑄 = 𝑘𝐴
B) 4.31 mm
C) 5.20 mm
D) 6.71 mm
E) 7.48 mm
∆𝑇
∆𝑙
∆𝑙 = 𝑘𝐴
∆𝑇
= 5.20mm
𝑄
Physics 115 – Winter 2017
Practice Final Exam
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The pV diagram at right illustrates two different processes in
which an ideal gas changes from state 1 to state 2.
9. [6 pts] Determine the ratio of the work done on the gas in
process A (1-A-2) to the work done on the gas in
process B (1-B-2)? Explain.
A) 6
B) 4
C) 3
D) 2
E) 1
The work done on the gas is equal to the negative of the area under the P-V curve.
Process A – area = (3P)(3V-V) = 6PV
Process B – area = (P)(3V-V) = 2PV
𝑊M −6𝑃𝑉
=
=3
𝑊H −2𝑃𝑉
10. [7 pts] Suppose that the energy of the gas in state 1 is U. In terms of U, how much heat is added in
process B (1-B-2)?
A) 6U
B) 7.5U
C) 9.3U
D) 12U
E) 14U
2
The first law of thermodynamics states: 𝑊 + 𝑄 = 𝛥𝑈 = 𝑁𝑘H 𝛥𝑇
C
Each point on a P-V diagram corresponds to a particular temperature since T = PV/nR, and in these
processes n is a constant. In process B, the gas goes from a temperature of PV/nR to 9PV/nR. Thus
the absolute temperature increases by a factor of 9. The thermal energy is proportional to the change
in temperature. Since the temperature increases by a factor of 9, the final internal energy must be 9
times larger than the initial thermal energy (9U). The change in thermal energy is therefore 8U.
The initial thermal energy is equal to 3/2NkBTi, which is equivalent to 3/2PV and U as stated in the
question. The work in process B is -2PV from question 9 above. The work is thus equal in magnitude
to -1.33U. The heat transferred is equal to DU – W and equal to 8U –(-1.33U) = 9.3U.
Physics 115 – Winter 2017
Practice Final Exam
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11. [6 pts] An ideal engine does 10 J of work and exhausts 15 J of waste heat. If the cold reservoir
temperature is 20°C, what is the temperature in degrees of the hot reservoir?
A) 100°C
B) 180°C
C) 220°C
D) 350°C
E) 490°C
𝑄h = 𝑊 + 𝑄i = 25J
𝑄h 𝑇h
=
𝑄i 𝑇i
𝑇h = 𝑇i
𝑄h
25J
= 293𝐾
= 488K = 215℃
𝑄i
15J
12. [6 pts] How far above a proton would an electron have to be held to exert an electrostatic force that
just equals the weight of the proton? (mp = 1.67 x 10-27 kg, me = 9.11 x 10-31 kg)
A) 14 mm
B) 21 mm
C) 58 mm
D) 0.12 m
E) 8.4 m
FCoulomb =
Kqe q p
r
2
= mp g ® r =
Kqe q p
mp g
= [(9 x 109 N-m2/C2)(1.6 x 10-19 C)(1.6 x 10-19 C)/(1.67 x 10-27 kg)(9.8 m/s2)]1/2 = 0.12 m
13. [6 pts] In case C, two positive point charges +2Q are each a
distance s away from a third positive point charge +q. In case D,
four positive point charges +Q are each a distance s away from a
fifth positive point charge +q. (The angle shown is the same in
both cases.) Is the net electric force on the +q charge in case C
greater than, less than, or equal to net electric force on the +q
charge in case D?
A) Greater than
B) Less than
C) Equal to
D) Not enough information given to answer.
The net force by the upper and lower +Q charges in case D will be half as large as the net force
applied by the two charges in case C. However, the two inner charges in case D will exert a net force
on the +q charge that is greater than half of net force in case C. Thus the combined force from all
the charges in case D is greater than that in case C.
Physics 115 – Winter 2017
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14. [6 pts] Consider a system of three point charges on the x-axis. Charge 1 is at x = 0, charge 2 and
charge 3 are located to the right of charge 1 at x = r, and at x = 2r, respectively. In addition, the
charges have the following values q1 = -q, q2 = q3 = q. The electric field is zero at some point along
the x-axis between x = r and x = 2r. Is the point of zero field:
A) At x = 1.5r
B) To the left of x = 1.5 r
C) To the right of x = 1.5 r.
D) Not possible to tell.
At x = 1.5r the contributions to the electric field from charge 2 (to the right) and 3 (to the left) are
equal and opposite and the contribution from charge 1 points to the left. Therefore this is not a valid
answer.
Moving to the left of x = 1.5r decreases the contribution from charge 3 and increases the contribution
from charges 1 and 2. Since E µ 1/ r 2 , the contribution from charge 2 (which is to the right)
changes more rapidly than the contributions from the other two charges, and the net change is to
make the magnitude of the left-pointing field at x = 1.5r get smaller as you move to the left.
To the right of 1.5r, the electric field from charge 3 is greater in magnitude than that from charge 2.
The electric field from charge 1 also points to the left. To the right of 1.5r, the net field cannot be
zero.
15. [6 pts] The charge q1 in the figure at right is positively charged. When
a small test charge, +q0, travels from point A to point B, it is known
that the net work done by the electric field on the charge is negative.
Determine the sign of q2 and its relative magnitude in comparison to
q 1.
A) q2 is negative, 𝑞C < 𝑞[
B) q2 is negative, 𝑞C > 𝑞[
C) q2 is negative, 𝑞C = 𝑞[
D) q2 is positive, 𝑞C < 𝑞[
E) q2 is positive, 𝑞C > 𝑞[
The work done on a charge is given as:
𝑊$\$i = -𝛥𝑉𝑞
Since the charge moving through the potential is positive, and the work done is negative, the change
in potential must be positive. For the potential at point B to be more positive than at point A, q2 must
be positive and have a larger magnitude than q1.
Physics 115 – Winter 2017
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16. [6 pts] A point charge, +Q, is fixed in place as shown at right. Points
a, c, and e are equidistant from the +Q charge, as are points b, d and f.
shown in the diagram at right. A small negative test charge, -q0,
travels along three different paths, 1 (a to b), 2 (d to a) and 3 (c to e to
f). Rank the paths according to the work done by the electric field on
the negative test charge. Rank positive work values as larger than
negative work values.
A) 3 > 2 > 1
B) 3 = 2 = 1
C) 2 > 1 = 3
D) 1 = 2 = 3
E) 3 > 1 = 2
In path 1, the force on the charge is to the left and the displacement is to the right. The work done by
the field is negative. In path 2, a component of the force is in the same direction as the displacement
of the charge. The work done by the field is positive. In path 3, no work is done when the charge
moves from point c to point e, since these two points are at the same potential. In moving from point e
to point f, the force opposes the displacement and the work is negative. Since the force and
displacement have the same magnitude as that along path 1, the works for path 1 and 3 are both
negative and have the same magnitude.
17. [7 pts] A parallel-plate capacitor is constructed with plates of a particular shape. The plates are
separated by 0.25 mm, and the space between the plates is filled with a dielectric with dielectric
constant of value 4.6. When the charge on the capacitor is 1.2 µC the potential difference between
the plates is 750 V. Which of the following choices best describes the shape of the plates.
A) Square plates with length 0.056 m.
B) Rectangular plates with length 0.056 m and width 0.112 m.
C) Circular plates with radius 0.056 m.
D) None of the above.
𝐶=
𝜅𝜀_ 𝐴
𝑑
𝐴=
𝐶𝑑
𝑄𝑑
=
= 9.83×10X2 mC
𝜅𝜀_ 𝑉𝜅𝜀_
Physics 115 – Winter 2017
Practice Final Exam
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18. [6 pts] Consider the two cases at right. Is the
magnitude of the flux through the Gaussian surface in
case A greater than, less than, or equal to that through
the surface in case B?
A) Greater than
B) Less than
C) Equal to
D) Not enough information given to answer.
There is no change enclosed by the surface in case A, so the magnitude of the flux is zero. The
magnitude of the flux in case B is non-zero since the surface encloses a +4Q charge.
19. [7 pts] Consider the circuit at right, where R1 = R, R2 = ½ R, and R3 = 2R. The
battery has a voltage V. What is the voltage across the resistor, R1?
A. 5V/7
B. 3V/5
D. 8V/11
E. 11V/13
C. 5V/6
R1
R2
R3
The resistance of the parallel combination is
-1
-1
æ 1 1 ö
2R
æ2 1 ö
R23 = ç
The voltage across R1 andt he parallel combination are
+ ÷ =ç +
÷ =
R
R
R
2
R
5
è
ø
3 ø
è 2
V
R
R
5
proportional to their resistances: 1 = 1 =
= . V1 + V23 = V, so V1 = 5V/7
V23 R23 2 R
2
5
and V23 = 2V/7.
20. [6 pts] Consider the circuit shown at right. The current flowing
through the battery is 0.92 A. Determine the potential difference
across the 12-Ω resistor.
A. 15 V
B. 4.9 V
C. 3.2 V
D. 4.3 V
E. 3.7 V
The potential drop across the 11 Ω resistor = (0.92 A)(11 Ω) = 10.1 V
Thus the potential drop VBA = 15 V – 10.1 V = 4.9 V. The 6.2 Ω and 12 Ω resistors are in series, so
V6.2 + V12 = 4.9 V. The potential drop across each resistor is proportional to its resistance:
V6.2 6.2
=
= 0.52 . Thus V12 = 4.9 - V6.2 = 4.9 - 0.52V12 ®
V12 12
V12 = 3.2 V
Physics 115 – Winter 2017
Practice Final Exam
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21. [7 pts] A flash unit in a camera consists of an RC circuit. The capacitor has a capacitance of
1500 µF. What resistance is needed in this RC circuit if the flash is to charge to 90% of its full charge
in 21 s?
A. 4900 Ω
B. 5400 Ω
C. 5800 Ω
D. 6100 Ω
E. 6700 Ω
As the flash charges up the voltage across the RC combination changes with time according to
V = V0 (1 - e-t / RC ) . We want V/V0 = 0.9 when t = 21 s.
1-
æ V ö
V
t
= e-t / RC ® ln ç1 - ÷ = ®R=V0
RC
è V0 ø
t
æ V ö
C ln ç1 - ÷
è V0 ø
= - (21 s)/(1500 x 10-6 F)ln(0.1)
= 6100 Ω
22. [6 pts] The time constant for the RC circuit at right is 3.5 ms. Calculate
the current in the circuit 7.0 ms after the switch is closed. Assume that
the capacitor is uncharged initially and that the emf of the battery is
9.0 V.
A) 11 mA
I = I 0e −t / RC =
B) 9.2 mA
C) 8.4 mA
145 Ω
9.0 V
+
D) 6.8 mA
E) 5.6 mA
I 0 −t / RC
e
R
τ = RC
9.0 V −0.007s/0.0035s
I=
e
145Ω
I = 8.4 mA
23. [6 pts] In a mass spectrometer, charged particles pass through a velocity selector with electric and
magnetic fields at right angles to each other, as shown. If the electric field has a magnitude of
450 N/C and the magnetic field has a magnitude of 0.18 T, what speed must the particles have to pass
through the selector undeflected?
A. 1500 m/s
𝑣=
B. 1800 m/s
C. 2000 m/s
D. 2200 m/s
E. 2500 m/s
𝐸 450N/C
=
= 2500m/s
𝐵
0.18T
Physics 115 – Winter 2017
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24. [6 pts] An solenoid with 500 turns has a diameter of 10 cm, is 40 cm long, and carries a current of
30 A. What is the magnitude of the magnetic field at the center of this solenoid?
A) 7.5 mT
B) 19 mT
C) 47 mT
D) 190 mT
B=
µ0 NI
L
= (1.26 x 10 -6 T-m/A)(500)(30 A)/(0.4 m) = 4.7 x 10-2 T
25. [7 pts] Two long straight wires carry current in the positive xdirection. At an instant in time, three charges, q1, q2, and q3 are
moving in the positive x-direction at the points shown. Rank the
magnitude of the net magnetic force exerted on each charge due
to the current-carrying wires.
A) 𝐹[ > 𝐹C > 𝐹2
B) 𝐹C > 𝐹[ > 𝐹2
C) 𝐹C > 𝐹[ = 𝐹2
D) 𝐹[ > 𝐹2 > 𝐹C
E)
𝐹C = 𝐹[ > 𝐹2
𝐵∝
𝐼
𝑟
The fields from both wires point in the same direction between the wires (right hand rule). The field
will be strongest at this location. The field at q1 will be greater than that at the location of q3 as the
field is proportional to the current through the wire. The force exerted on the charges is proportional
to the field strength at their location and all charges have the same speed.
Physics 115 – Winter 2017
Practice Final Exam
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26. [7 pts] Consider the set-up at right that was presented in the
Magnetic Interactions tutorial. The wire is observed to move out of
the page when the wire carries a current. Which of the following
choices is consistent with this observation?
Score: _______
X
A) Pole X is a north pole and current travels right to left through the
horizontal section of the wire.
B) Pole X is a north pole and current travels left to right through the
horizontal section of the wire.
C) Pole X is a south pole and current travels left to right through the
horizontal section of the wire.
D) Descriptions A and C are both correct.
E) Descriptions B and C are both correct.
In choices A and C, the magnetic force on the horizontal segment is into the page.
27. [6 pts] A conducting rectangular loop of wire is near a long straight
wire. The long straight wire carries a constant current I in the direction
shown in the figure at right. If the loop moves with a speed v in the
direction shown, what is the direction of the current in the loop?
A) clockwise
v
I
B) counterclockwise
C) There is no current.
D) There is not enough information to tell.
At the loop the magnetic field points into the page, and its magnitude decreases with increasing
distance from the wire. Thus the flux through the loop decreases, and a current is induced to keep the
flux constant by producing a field pointing into the page. That requires a clockwise current.
Physics 115 – Winter 2017
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28. [6 pts] An emf is induced in a conducting loop of wire 1.22 m long as its
shape is changed from square to circular. The area enclosed by the loop is in
the plane of the page, and the loop has a resistance of 0.62 Ω. Find the
magnitude of the induced current and its direction, if the change in shape
occurs in 4.25 s and the local 0.125-T magnetic field is perpendicular to the
plane of the loop and points out of the page.
A) 1.2 mA, counterclockwise
B) 1.2 mA, clockwise
C) 2.4 mA, counterclockwise
D) 2.4 mA, clockwise
E) 3.1 mA, clockwise
Csquare = 4d = 1.22 m
Since the area
increases, the flux
increases. The
induced magnetic
field opposes this
change, and thus
points into the page.
This is consistent
with a clockwise
current in the loop.
1.22m
= 0.305 m
4
Asquare = d 2 = (0.305 m)2 = 0.093 m 2
d=
Ccircle = 2π r = 1.22 m
1.22m
= 0.194 m
2π
Acircle = π r 2 = π (0.194 m)2 = 0.118 m 2
" 0.118 m 2 − 0.093 m 2 %
dΦ
dA
−4
EMF =
=B
= (0.125T ) $
' = 7.4 ×10 V
dt
dt
4.25s
#
&
r=
I=
EMF 7.4 ×10−4 V
=
= 1.2 mA
R
0.62Ω
vo
29. [6 pts] The observer on the right observes a counter-clockwise current
in loop 2. Which of the following statements is consistent with this
observation?
X
Eye
A) The pole labeled with an X is the north pole of the magnet.
B) The pole labeled with an X is the south pole of the magnet.
C) The pole labeled X could be a north or south pole; it depends on the magnet’s velocity.
D) There is not enough information to answer.
The current in the loop is counter-clockwise from the perspective of the observer, which means the
induced magnetic field points to the right. The flux through the loop is increasing, since the magnet
is moving toward the loop. The induced magnetic field opposes the change in the flux, therefore the
magnetic field of the magnet must point to the left. This means that the pole label with an X is a south
pole.
Physics 115 – Winter 2017
Practice Final Exam
Name ____________________________________
Last
First
Student ID: _______________
Score: _______
30. [6 pts] A conducting stationary loop (copper loop) is placed in a uniform
external magnetic field as shown. A clockwise current is induced in the
loop. While the current is flowing through the loop, how is the external
magnetic field changing?
A) The magnetic field is increasing.
B) The magnetic field is decreasing.
C) The magnetic field remains constant.
D) Not enough information to tell.
A clockwise current around the loop produces a magnetic field that points into the page at the center
of the loop. Lets call the flux positive if the field lines point through the loop into the page. This would
make the external flux and the induced flux positive. The induced flux always opposes the change in
external flux. Since the induced flux is positive, the change in external flux must be negative. This
means that the field must be decreasing in strength, such that the change in external flux is negative.
31. The figure at right shows a zero-resistance rod sliding to the
right on two zero-resistance rails separated by the distance
L = 0.500 m. The rails are connected by a 10.0 ohm resistor,
and the entire system is in a uniform magnetic field with a
magnitude of 0.750 T. Find the speed at which the bar must be
moved to produce a current of 0.175 A in the resistor?
A) 4.67 m/s
B) 3.75 m/s
C) 3.13 m/s
D) 2.29 m/s
E) 1.58 m/s
ℇ = 𝐵𝑣𝑙
𝑣=
ℇ
𝐼𝑅
0.175A (10Ω)
=
=
= 4.67m/s
𝐵𝑙 𝐵𝑙 (0.750T)(0.5m)
Physics 115 – Winter 2017
Practice Final Exam
Name ____________________________________
Last
First
Student ID: _______________
Score: _______
32. [6 pts] An ac generator with a maximum voltage of 12.0 V and a
frequency of 60.0 Hz is connected to a resistor with a resistance
R =150 Ohm. Determine the average power dissipated in the
resistor.
A) 0.057 W
B) 0.48 W
C) 1.1 W
D) 4.3 W
E) 8.5 W
𝑃%•‚
𝑉=%ƒ
2
=
=
𝑅
𝑅
C
𝑉(=#
Physics 115 – Winter 2017
C
12.0V
2
=
150Ω
C
= 0.48W
Practice Final Exam