Q3ExRev

Transcription

Q3ExRev
Physics (H)
3rd Quarter Exam Review
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Quarter Exam Review
1. An ideal gas confined in a box initially has pressure p. If the absolute temperature
of the gas is doubled and the volume of the box is quadrupled, the pressure is
(A) p/8
(B) p/4
(C) p/2
(D) p
(E) 2p
C) p/2
2. A mass m of helium gas is in a container of constant volume V. It is initially at
pressure p and absolute (Kelvin) temperature T. Additional helium is added,
bringing the total mass of helium gas to 3m. After this addition, the
temperature is found to be 2T. What is the gas pressure?
(A) 2/3 p
E) 6P
(B) 3/2 p
(C) 2 p
(D) 3 p
(E) 6P
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3. A monatomic ideal gas at pressure P = 105 Pa is in a container
of volume V = 12 m3 while at temperature T = 50ºC. How many
molecules of gas are in the container?
(A) 447 (B) 2888 (C) 6.02 × 1023 (D) 2.83 × 1026 (E) 1.74 × 1027
D) 2.83 x 10 26
4. A steel wire, 150m in length at 10 oC, has a coefficient of
linear expansion of 11 x 10 –6 /C o. Give its change in length as the
temperature increases to 45 oC.
a. 0.65 cm
b. 1.8 cm
c. 5.8 cm
d. 12 cm
C) 5.8 cm
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5. One mole of an ideal gas has a temperature of 100°C. If this gas fills the 10.0m3
volume of a closed container, what is the pressure of the gas?
(A) 0.821 Pa
(B) 3.06 Pa
(C) 83.1 Pa
(D) 310 Pa
(E) 1.84 × 1024 Pa
(D) 310 Pa
6. What is the Kelvin scale temperature that is equivalent to
10 degrees on the Fahrenheit scale?
(A) 283 K (B) 273 K (C) 261 K (D) 10 K (E) 10 K
(C) 261 K
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7. An inventor develops a stationary cycling device by which an individual,
while pedaling, can convert all of the energy expended into heat for warming water.
How much mechanical energy is required to increase the temperature of 300. g of water
from 20 C to 95 C? c = 4186 J/kg-C
a. 94000 J
b. 22000 J
c. 5400 J
d. 14 J
a. 94000 J
8. A 2.00 kg copper rod is 50.00 cm long at 23 C. If 40000 J are transferred to the rod
by heat, what is its changed in length? c = 387 J/kg-C,  = 17 x 10 –6/ C
a. 0.022 cm
b. 0.044 cm
c. 0.059 cm d. more information is needed
b. 0.044 cm
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9. Iced tea is made by adding ice to 1.8 kg of hot tea at 80 C.
How many kg of ice at 0 C are required to bring the mixture to 10 C?
Lf = 3.33 x 10 5 J/kg-C, c = 4186 J/kg- C
a. 1.8 kg
b. 1.6 kg
c. 1.4 kg
d. 1.2 kg
c. 1.4 kg
10. How much energy must be removed from 100 g of oxygen at 22 C to liquefy it at
–183 C? c = 912.5 J/kg-C, Lv = 2.13 x 10 5J/kg
a. 57348J
b. 40007 J
c. 21903 J
d. 9042 J
b. 40007 J
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11. The period of a spring-mass system undergoing simple harmonic motion
is T. If the amplitude of the spring-mass system’s motion is doubled, the period will be:
(A) ¼ T
(B) ½ T
(C) T
(D) 2T
(E) 4T
(C) T
12. A particle oscillates up and down in simple harmonic motion.
Its height y as a function of time t is shown in the diagram. At what time t
does the particle achieve its maximum positive acceleration?
(A) 1 s
(B) 2 s
(C) 3 s
(D) 4 s
E) None of the above
(A) 1 s
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13. Two objects of equal mass hang from independent springs of unequal spring
constant and oscillate up and down. The spring of greater spring constant must have the
(A) smaller amplitude of oscillation
(C) shorter period of oscillation
(E) lower frequency of oscillation
(B) larger amplitude of oscillation
(D) longer period of oscillation
(A) smaller amplitude of oscillation
14. String L and string H have the same tension and length. String L has mass m and
string H has mass 4m. If the speed of the waves in string L is v, the speed of the waves
in string H is
A) v/2
B) v
C) 1.4 v
D) 2v
E) 4v
A) v/2
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15. Assume that waves are propagating in a uniform medium. If the frequency of the
wave source doubles then
A) The speed of the waves doubles B) the wavelength do the waves doubles
C) the speed of the waves halves D) the wavelength of the waves halves
E) none of the above
D) the wavelength of the waves halves
. 16. What is the electric force of two electrons 10 –10 m apart?
a. 2.3 x 10 –8 N
b. 4.6 x 10 –8 N
c. 9.2 x 10 –8 N d. 2.3 x 10 –9 N e. 4.6 x 10 –10 N
a. 2.3 x 10 –8 N
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17. The electric field halfway between two positive point charges of 1.7 x 10 –5 C is:
a. 34 N/C
b. 17 N/C
c. 0 N/C
d. –17 N/C
e. –34 N/C
c. 0 N/C
18. What is the electric field halfway between two collinear negative charges
separated by 6 cm. One charge is – 2.1 C, whereas the other is – 1.3 C.
a. 2.6 x 10 5 N/C
b. 6.3 x 10 5 N/C
c. 3.2 x 10 6 N/C
d. 8.0 x 10 6 N/C
e. 1.1 x 10 7 N/C
d. 8.0 x 10 6 N/C
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19. A positive charge of 10–6 coulomb is placed on an insulated solid conducting sphere.
Which of the following is true?
(A) The charge resides uniformly throughout the sphere.
(B) The electric field inside the sphere is constant in magnitude, but not zero.
(C) The electric field in the region surrounding the sphere increases with increasing
distance from the sphere.
(D) An insulated metal object acquires a net positive charge when brought near to,
but not in contact with, the sphere.
(E) When a second conducting sphere is connected by a conducting wire to the first
sphere, charge is transferred until the electric potentials of the two spheres are equal.
19. Charges flow when there is a difference in potential. Analyzing
the other choices: A is wrong because the charge resides on the
surface. For B, E = 0 in a charged conducting sphere. E = kQ/r2
eliminates choice C. And for D, charge separation will occur, but
the object will not acquire any charge.
E
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20. The following configurations of electric charges are located at the vertices
of an equilateral triangle. Point P is equidistant from the charges.
In which configuration is the electric field at P equal to zero?
(A)A
(B) B
(C) C
(D) D
(E) E
E points away from + charges and toward – charges. Use symmetry.
A
3rd Quarter Exam Review
Free Response
1.Two point charges, Q1 and Q2, are located a distance 0.20 meter apart,
as shown above. Charge Q1 = +8.0C. The net electric field is zero at point P, located
0.40 meter from Q1 and 0.20 meter from Q2.
a. Determine the magnitude and sign of charge Q2.
b. Determine the magnitude and direction of the net force on charge Q1
c. Determine the coordinate of the point R on the x–axis between the two charges at which the
electric field is zero.
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a. E = kQ/r2 and since the field is zero E1 + E2 = 0 giving k(Q1/r12 + Q2/r22) = 0
This gives the magnitude of Q2 = Q1(r22/r12) = 2C and since the fields must point
in opposite directions from each charge at point P, Q2 must be negative.
b. F = kQ1Q2/r2 = 3.6 N to the right (they attract)
c. between the charges we have a distance from Q1 of x and from Q2 of (0.2 m – x)
E = kQ1/x + kQ2/(0.2 m – x) = 0, solving for x gives x = 0.16 m
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2. A charge Q1 = –1.6 x 10–6 coulomb is fixed on the x–axis at +4.0 meters,
and a charge Q2 = + 9 x 10–6 coulomb is fixed on the y–axis at +3.0 meters, as
shown on the diagram above.
a. Calculate the magnitude of the electric field E1 at the origin O due to charge Q1
b. Calculate the magnitude of the electric field E2 at the origin O due to charge Q2.
a.
E = kq/r2 = 900 N/C
b.
E = kq/r2 = 9000 N/C
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3. An ideal spring of unstretched length 0.20 m is placed horizontally on a
frictionless table as shown above. One end of the spring is fixed and the other
end is attached to a block of mass M = 8.0 kg. The 8.0 kg block is also
attached to a mass less string that passes over a small frictionless pulley. A block
of mass m = 4.0 kg hangs from the other end of the string. When this springand-blocks system is in equilibrium, the length of the spring is 0.25 m and the 4.0
kg block is 0.70 m above the floor.
(a) Calculate the tension in the
string.
(b) Calculate the force constant of
the spring.
The string is now cut at point P.
(c) Calculate the time taken by the
4.0 kg block to hit the floor.
(d) Calculate the frequency of
oscillation of the 8.0 kg block.
(e) Calculate the maximum speed
attained by the 8.0 kg block
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Quarter Exam Review
(a) Calculate the tension in the string.
(b) Calculate the force constant of the spring.
The string is now cut at point P.
(c) Calculate the time taken by the 4.0 kg block to hit the floor.
(d) Calculate the frequency of oscillation of the 8.0 kg block.
(e) Calculate the maximum speed attained by the 8.0 kg block
a) Simply isolating the 4 kg mass at rest. Fnet = 0 Ft – mg = 0
Ft = 39 N
b) Tension in the string is uniform throughout, now looking at the 8 kg mass,
Fsp = Ft = k∆x
39 = k (0.05)
k = 780 N/m
T  2
m
k
 2
8
780
 0.63s
d) 4 kg mass is in free fall. D = vit + ½ g t2
– 0.7 = 0 + ½ (– 9.8)t2
t = 0.38 sec
e) First find the period. then the frequency is given by
f = 1/T = 1.6 Hz
The 8 kg block will be pulled towards the wall and will reach a maximum speed when it
passes the relaxed length of the spring. At this point all of the initial stored potential energy is
converted to kinetic energy Usp = K ½ k ∆x2 = ½ mv2
½ (780) (0.05)2 = ½ (8) v2
v = 0.49 m/s