Download PHYSICS 214 TEST 1 12 February 2008

PHYSICS 214
TEST 2
Tues. 31 March 2009
Fill in on the OPSCAN sheet:
1) Name
2) Student Identification number
3) Exam number = 02 [there is no section number]
4) YOU MUST SIGN THE OPSCAN SHEET
This test is worth 100 points – it consists of 16 multiple choice problems, each of equal value, 6.25
points.
You may keep your copy of the exam questions, with your answers.
INSTRUCTIONS: For each problem choose the one answer that is correct or most nearly correct.
Make a small mark, for your eyes only, near the letter of your choice. Be sure to transfer all your
answers to your OPSCAN sheet. Then, until you go to hand in the OPSCAN sheet, turn it over and
leave it face down to prevent others from seeing your answers.
Only answers on the OPSCAN sheet will be graded – it will not be returned (but you can examine it in
Room 144 at a later date, with help from the secretaries), so be sure you have recorded your answer on
the test paper that you will keep.
A key of the correct answers will be displayed on the course web page, and you can find your score for
this exam on CHIP.
This is a closed book exam, but a crib sheet is provided and you can use your own crib sheet. You may
also use a calculator.
Any form of cheating will result in severe penalties, which will include a score of zero for this exam and
may result in a grade of F for the course and referral to the Dean of Students.
All cell phones, blackberries, pagers, etc. must be turned off and put out of sight during the exam.
The only items you are allowed are the exam, the OPSCAN sheet, your crib sheet, a pencil, and a
calculator.
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FORMULAE AND CONSTANTS
Conversion Factors
1 inch = 2.54 cm
1 ft = 30.5 cm
1 m = 3.281 ft
1 km = 0.621 miles
1 mile = 5280 ft
1 nautical mile = 1.1508 miles
1 kg = 2.205 lbs (where g = 9.8 m/s2)
Equations
s = d/t speed
s = d/Δt “
v = d/Δt (vector velocity)
a = Δv/Δt (vector acceleration)
v = v0 + at
d = v0t + ½at2
v2 = v02 + 2ad
d = ½(v + v0)t
F = ma (vector) “N2“
F12 = - F21
“N3“
W = mg
P = 1015
T = 1012
G = 109
M = 106
k = 103
m = 10-3
μ = 10-6
n = 10-9
p = 10-12
f = 10-15
Pressure = Force/Area
Peta
Tera
Giga
Mega
kilo
milli
micro
nano
pico
femto
[N/m2 = Pa = Pascal]
1 Atmosphere = 101.3 kPa = 760mmHg = 33 ft H2O
P = ρgd where d is depth below surface (or height of
a liquid column
Buoyant force = weight of displaced liquid
Circumference = 2R
2
g = 9.8 m/s
G = 6.67  10-11N·m2/kg2
 = 3.14159
 = θ/t = θ/t
e = 1.9x10-19 C
the size of the charge on an electron or a proton
-31
me = 9.1x10 kg
the mass of the electron
mp = 1836 me
the mass of the proton
cH2O= 1 Cal/gm
heat capacity of water
Lf = 80 Cal/gm
latent heat of fusion of water
Lv = 540 Cal/gm
latent heat of vaporization of water
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DEFINITIONS
a = v2/R
F = mv2/R
F = GM1M2/R2
F = kq1q2/R2
W = Fd
P = W/t = Fv
KE = ½ mv2
PE = mgh
PE = ½ kx2
P = mv
Ft = Pf – Pi = p
Fexternal = 0, p = 0
W = qΔV
P = I ΔV
E = ΔV/d
centripetal acceleration
centripetal force
gravitational force G = 6.61x10-11Nm2/kg2
electrostatic (Coulomb) force k = 9x109 Nm2/C2
work
power
Kinetic energy
potential energy
potential energy
momentum
impulse
momentum conservation
work or energy, electrical
power, electrical
for uniform electric field, where d is distance along
the voltage gradient
ε = (Thot – Tcold)/Thot Carnot efficiency of an ideal heat engine
PV = NkT
Ideal gas law where P,V,T are pressure, volume,
and Kelvin temperature and k is Boltzmann’s constant
2
I = mR
Moment of rotational inertia of a point mass a
distance R from the axis of rotation
Torque = r x F
Torque is product of Force x “Lever arm” distance
to the axis of rotation (the component of distance at right angle to F)
Torque = I α
α
angular acceleration, in radians/s2
KE = 0.5 I ω2
Kinetic energy of rotation of a solid body
2
I = MR
for hoop
2
I = 0.5 MR
for disk
2
I = 0.4 MR
for sphere
2
I = (1/12)ML
for a rod of length L
KEPLER’s Laws
Elliptical orbit sweeps out equal areas in equal time intervals
T2 = r3
for all satellites of a common attracting body, where T is the
period and r is the average radius of the orbit
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1. A billiard ball (1) moving at 5 m/s hits a second billiard ball (2) moving at -5 m/s (meaning in the
exact opposite direction) head-on. The balls have identical masses and collide elastically. After the
collision, which statement is true:
A. (1) recoils backward, at -5 m/s and (2) moves forward at 5 m/s
B. (1) recoils backward, at -5 m/s and (2) remains at rest
C. (1) is at rest; (2) moves backward, at -5 m/s
D. (1) is at rest; (2) moves forward at 5 m/s
E. Overall, the final kinetic energy is half the initial kinetic energy.
2. A 50 kg mass going 100 m/s collides INELASTICALLY with a stationary 30 kg mass (they stick
together.) After the collision, what is the speed [in m/s] of the two masses?
A. 43.3
B. 62.5
C. 50
D. 100
E. 30
3. How much Kinetic Energy [in J] is lost during the collision of problem 2. ? [Be careful to
subtract any residual KE that may be present after the collision.]
A. 5.15 x 104
B. 1.56 x 104
C. 2.50 x 105
D. 6 x 105
E. 9.38 x 104
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4. A vinyl LP record is spinning at 33.3 rpm (rev/minute). What is its angular velocity [in
radians/s] ?
A 1.7
B. 3.5
C. 10.2
D 22.8
E. 77.3
5. The same record is now spun up to an angular velocity of (328.3 rad/s). What is its Kinetic
Energy, if its mass is 90 grams and its radius is 15 cm? [You will need to use the formula for the
moment of inertia of a disc, and be careful to convert all units to MKS (SI).]
A. 18.9 J
B. 2.4 kJ
C. 4.4 MJ
D. 7.3 GJ
E. 7.3 MJ
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6. You dive 49.5 feet deep in a fresh-water lake. What is the extra pressure on your body, over and
above the pressure of the atmosphere?
A. 1 Atmosphere
B. 1.5 Atm.
C. 2 Atm.
D. 2.5 Atm.
E. 5 Atm.
7. An oil tanker has a mass of 750,000 kg. 60% of the volume of the tanker is ABOVE the water
level. What is the buoyant force of the water on the tanker?
A. 750 kN
B. 7.35 MN C. 0
D. 980,000 N
E. 2.94 MN
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8. A hemisphere of radius 0.5 m is held against a smooth surface by a perfect vacuum inside it,
surrounded by normal atmosphere. What force, in Newtons, is needed to pull the hemisphere free?
[1 Atm. = 101.3 kPa]
A. 1013 kN
B. 6.26 MN
C. 4.88 GN
D. 7.96 MN
E.
3.14 kN
9. You have a fever of 106 degrees Farenheit. What is this temperature in Celsius degrees?
A.
37
B. 221
C. 303
D. 41
E. 28
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10. A mass of water at 80 oC is added to twice as much mass of a substance with three
times the specific heat capacity of water, which is at 20 oC. If the combined system is
perfectly insulted from the surroundings, what will be the final equilibrium temperature?
A. 50.0 oC
B. 28.6 oC
C. 34.1 oC
D. 67.9 oC
E. 75.5 oC
11. What is the theoretical maximum efficiency (the Carnot efficiency) of a solar thermal
engine that operates with hot gas at 800 K and releases heat to the environment at
350 K?
A. 100%
B. 33.3%
C. 56.3%
D. 70%
E. 43.8%
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12. What is the electrical force between an electron and a proton separated by 2x10-13m?
A. 0.0058 N
B. 3.2 x 10-5 N
C. 0.00043 N
D. 6.56 x 10-38 N
E. 4.6 x 10-6 N
13. How far apart are two wide parallel metal plates if the magnitude of the electric field
between the two plates is 15 kV/m when there is a potential difference of 25 volts
between them?
A. 1.67 mm
B. 15 mm
C. 25 cm
D. 40 m
E. 2.88 μm
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14. The plates are now disconnected from the battery, and then pulled apart to a
separation five times greater than in problem 13. Since the field is caused by the
charges on the plates, which do not change, now what is the potential difference
between the plates?
A. 100 V
B. 125 V
C. 250 V
D. 330 V
E. 2500 V
15. Two electric charges, +4mC and -1mC are located on the x-axis at x = 0 and x = 1
m, respectively. Where is the Electric field equal to zero? To help you solve it,
draw a picture and point your vectors carefully. Hint: The location cannot be
between the two charges since the electric fields from the two charges must
cancel. Use the inverse square law to compensate for the difference in
magnitudes of the two charges.
A. x = 0
B. x = 1 m
C. x= 2 m
D. x= 4m
E. someplace NOT on the x-axis
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16.
A helium balloon with volume 5000 m3 is launched, and floats up to a height in the
atmosphere where the pressure is 25 kPa. What is the approximate new volume
of the balloon, if the Temperature is kept constant?
A. 10,000 m3
B. 35,000 m3
C. 7,500 m3
D. 25,000 m3
E. 20,000 m3
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