# Answer: D For dipoles, the force decays faster than 1/R2. Answer

```STUDENT ID: ______________________________NAME: __________________________________
1010, Fall 2006, Exam 3: YELLOW
Write the color of your exam paper in the large blank area of your scantron sheet. DO NOT WRITE ON
THE MARGIN OF THE SCANTRON SHEET.
Be sure to turn your exam in to the proper pile and to return both the answer sheet and the exam.
This test is closed book, but you may use a single 3x5 card with your own notes written on it. You may
also use a calculator.
The exam is worth 40 pts total. All multiple choice questions are worth the same number of points, but
not all of them are of the same difficulty. If you have difficulty answering a question, move on to the
next one and come back to the difficult ones later. The last (long hand) question is worth 10 points.
Make sure you leave enough time for it.
Answer the last (long hand) question on the exam sheet, not the scantron sheet.
Conversions you may or may not need: 1 pound = 4.45 N, 1 slug = 14.594 kg, 1 mph = 0.447 m/s.
Speed of sound = 330 m/s. Coulomb’s constant, k = 9 x 109 N m2/C2. The elementary charge is
e = 1.60 x 10-19 C. The surface of a sphere of radius R is given by As = 4πR2. The surface of a circle of
radius R is given by Ac = πR2. The Stefan-Boltzman constant is σ = 5.67 x 10-8 J K-4 m-2 s-1
Wires and batteries are taken to be perfect conductors unless otherwise stated in the problem.
Multiple Choice and True/False Questions: 30 pts total
1. Two bar magnets, each 4 cm long, feel a force of magnitude 5 N when they are 1 cm apart. When
the distance between them is increased to 1 m, the magnitude of the force between them will be:
a. greater than 5&times;104 N
b. 5&times;104 N
c. 5&times;10-4 N
d. less than 5&times;10-4 N
For dipoles, the force decays faster than 1/R2.
2. What is the wavelength in air of a sound wave of frequency 220 Hz?
a. 1.5 m
b. 1.0 m
c. 0.5 m
d. 0.1 m
V=λ/T = λf so λ = v/f = (330 m/s)/(220Hz) = 1.5m.
STUDENT ID: ______________________________NAME: __________________________________
3. A battery converts
a. Potential energy to heat
b. Electrostatic potential energy to heat
c. Chemical energy to electrostatic potential energy
d. Electrostatic potential energy to kinetic energy.
4. Burning converts
a. Kinetic energy to gravitational potential energy
b. Chemical energy to gravitational potential energy
c. Kinetic energy to thermal energy
d. Chemical energy to thermal energy
5. Two steel balls each have 1,000 extra electrons on them. The balls are 0.5 m apart. The force
between them is
a. 9.2&times;10-22 N and attractive
b. 9.2&times;10-22 N and repulsive
c. 9.2&times;10-19 N and attractive
d. 9.2&times;10-19 N and repulsive
F = k (q1*q2)/R2 = 9*109 Nm2/C2 (1000*1.6*10-19 C)2 / 0.52 = 9.2*10-22 N
The force is repulsive, because both balls are negatively charged.
6. Which of the following statements is true for an object that is a perfect conductor?
a. The current through the object is limited to 1 A.
b. The voltage across the object is limited to 1V.
c. The current through the object is zero.
d. The voltage across the object is zero.
e. The current can go through the object in only one direction.
STUDENT ID: ______________________________NAME: __________________________________
7. The north pole for the coil at right is
a. up
b. left
c. down
d. right
Right hand rule.
8. In a transformer, 5,000 volts is found on the output, while 100 volts is on the input. The input has 20
coil loops. How many loops on the output?
a. 10
b. 100
c. 1,000
d. 5,000
V1/n1 = V2/n2 so 100/20 = 5000/ n2 =&gt; n2 = 1000.
9. When a single electron passes through a 1V battery, it gains energy equal to
a. 1 joule
b. 1 watt
c. 1.6&times;10-19 joule
d. 6.2&times;1018 joule
E = Vq = Ve = 1 V * 1.6*10-19 C = 1.6*10-19 J.
10. A camper trailer runs off of 12V. All of the electrical appliances in the camper together add up to
200 W. What current must the wires in the camper trailer be capable of carrying?
a. 0.06 A
b. 0.6 A
c. 1.7 A
d. 17 A
P = IV =&gt; I = P/V = 200W / 12V = 16.67A.
STUDENT ID: ______________________________NAME: __________________________________
In the circuit on the right, the voltage across the prongs of the plug is 120V and all light bulbs
have resistance 10 Ohms.
11. The voltage between B and C is:
a. 0 V
b. 40 V
c. 80 V
d. 120 V
All resistors are the same and the current through all of
them is the same (resistors are in series) so the voltage across each one is the
same. The voltage across all three is 120V, so the voltage across each one is
120/3 V = 40V
12. The current at point D is:
a. 36. A
b. 12. A
c. 8.0 A
d. 4.0 A
The voltage across each resistor is 40V, and each resistor is 10 Ohms, so the
current is I = V/R = 40 V / 10 Ohm = 4A.
13. The current at point B is:
a. 36. A
b. 12. A
c. 8.0 A
d. 4.0 A
The current is the same at all points of the circuit (resistors are in series).
14. The total power consumed by the circuit is:
a. 160 W
b. 480 W
c. 1440 W
d. 1200 W
STUDENT ID: ______________________________NAME: __________________________________
P = IV = 4A * 120V = 480W.
15. The black body equilibrium temperature of a planet is given by T=(P/σ4πR2)1/4. In this formula, P is
a. The total power radiated by the sun
b. The total solar power that falls on the planet
c. The total power radiated by the planet
16. In the formula in the previous question, R is the radius of
a. The sun
b. The planet
c. The planet’s orbit around the sun
In the circuit on the right, the voltage across the prongs of the plug is 120V and all light bulbs
have resistance 10 Ohms.
17. The voltage between F and I
is:
a. 0 V
b. 40 V
c. 80 V
d. 120 V
Resistors are in parallel.
18. The current between the plug
and point E is:
a. 36. A
b. 12. A
c. 8.0 A
d. 4.0 A
The current across each resistor is given by I = V/R = 120V / 10 Ohm = 12A.
All three currents will add up between E and the plug, so the total current is
36A.
STUDENT ID: ______________________________NAME: __________________________________
19. The current at point G is:
a. 36. A
b. 12. A
c. 8.0 A
d. 4.0 A