Q. An object is thrown horizontally from the top of a tower 80 m high. How long will it take to hit the ground?
A.
4 s
B.
5 s
C.
3 s
D.
2 s
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Solution
Using the equation h = (1/2)gt², where h = 80 m and g = 9.8 m/s², we have 80 = (1/2)*9.8*t². Solving gives t² = 16.33, so t ≈ 4 s.
Correct Answer: B — 5 s
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Q. An object is thrown upwards with a speed of 20 m/s. How long will it take to reach the maximum height? (Assume g = 10 m/s²)
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Time to reach maximum height = initial velocity / g = 20 m/s / 10 m/s² = 2 s.
Correct Answer: B — 2 s
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Q. An object is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
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Solution
Maximum height (H) = (u²)/(2g) = (20²)/(2*9.8) ≈ 20.4 m.
Correct Answer: C — 30 m
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Q. An object is thrown vertically upward with a speed of 30 m/s. How high will it rise before coming to a momentary stop?
A.
45 m
B.
60 m
C.
90 m
D.
135 m
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Solution
Using the formula: h = (v² - u²) / (2g), where v = 0 m/s (at the highest point), u = 30 m/s, g = 9.8 m/s². h = (0 - 30²) / (2 * -9.8) = 45.92 m, approximately 45 m.
Correct Answer: B — 60 m
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Q. An object is thrown vertically upward with a speed of 30 m/s. How high will it rise before coming to rest?
A.
45 m
B.
90 m
C.
135 m
D.
180 m
Show solution
Solution
Using the formula: h = (v² - u²) / (2g), where v = 0 m/s, u = 30 m/s, g = 9.8 m/s². h = (0 - 30²) / (2 * -9.8) = 45.92 m, approximately 45 m.
Correct Answer: B — 90 m
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Q. An object is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.
11.25 m
B.
22.5 m
C.
15 m
D.
20 m
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Solution
Maximum Height (h) = v^2 / (2g) = (15^2) / (2 * 10) = 11.25 m
Correct Answer: B — 22.5 m
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Q. An object is thrown vertically upwards with a speed of 20 m/s. How long will it take to reach the maximum height?
A.
1 s
B.
2 s
C.
3 s
D.
4 s
Show solution
Solution
Time to reach maximum height (t) = u/g = 20/9.8 = 2.04 s.
Correct Answer: B — 2 s
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Q. An object is thrown vertically upwards with a speed of 30 m/s. How high will it rise before coming to rest?
A.
45 m
B.
60 m
C.
90 m
D.
120 m
Show solution
Solution
Using the formula: h = (v² - u²) / (2g), where v = 0 m/s, u = 30 m/s, g = 9.8 m/s². h = (0 - 30²) / (2 * -9.8) = 45.92 m, approximately 45 m.
Correct Answer: B — 60 m
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Q. An object is thrown vertically upwards with a speed of 30 m/s. How high will it rise before coming to rest momentarily?
A.
45 m
B.
60 m
C.
90 m
D.
135 m
Show solution
Solution
Using the formula: h = (v² - u²) / (2g), where v = 0 m/s (at the highest point), u = 30 m/s, g = 9.8 m/s². h = (0 - 30²) / (2 * -9.8) = 45.92 m, approximately 45 m.
Correct Answer: B — 60 m
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Q. An object moves in a circular path with a radius of 10 m at a speed of 5 m/s. What is the period of the motion?
A.
2π s
B.
4π s
C.
10 s
D.
20 s
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Solution
Period (T) = (2πr)/v = (2π * 10)/5 = 4π s.
Correct Answer: A — 2π s
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Q. An object moves in a straight line with a velocity of 10 m/s. If it accelerates at 2 m/s² for 5 seconds, what will be its final velocity?
A.
20 m/s
B.
30 m/s
C.
10 m/s
D.
15 m/s
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Solution
Final velocity (v) = u + at = 10 + (2 * 5) = 20 m/s.
Correct Answer: B — 30 m/s
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Q. An object moves in a straight line with an acceleration of 2 m/s². If its initial velocity is 5 m/s, what will be its velocity after 3 seconds?
A.
5 m/s
B.
8 m/s
C.
11 m/s
D.
14 m/s
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Solution
Final velocity (v) = u + at = 5 + (2 * 3) = 11 m/s.
Correct Answer: C — 11 m/s
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Q. An object moves in a straight line with an acceleration of 2 m/s². If its initial velocity is 3 m/s, what will be its velocity after 5 seconds?
A.
8 m/s
B.
10 m/s
C.
13 m/s
D.
15 m/s
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Solution
Final velocity (v) = u + at = 3 + (2 * 5) = 13 m/s.
Correct Answer: C — 13 m/s
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Q. An object moves in a straight line with an acceleration of 2 m/s². If its initial velocity is 10 m/s, what will be its velocity after 5 seconds?
A.
10 m/s
B.
20 m/s
C.
30 m/s
D.
40 m/s
Show solution
Solution
Final velocity (v) = u + at = 10 + 2*5 = 20 m/s.
Correct Answer: C — 30 m/s
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Q. An object moves in a straight line with an initial velocity of 5 m/s and accelerates at 2 m/s². What will be its velocity after 4 seconds?
A.
5 m/s
B.
13 m/s
C.
15 m/s
D.
20 m/s
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Solution
Using the formula: final velocity = initial velocity + acceleration * time = 5 + 2 * 4 = 13 m/s.
Correct Answer: B — 13 m/s
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Q. An object moves with a constant speed of 15 m/s for 10 seconds. What distance does it cover?
A.
100 m
B.
150 m
C.
200 m
D.
250 m
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Solution
Distance = speed * time = 15 m/s * 10 s = 150 m.
Correct Answer: B — 150 m
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Q. An object moves with a uniform speed of 15 m/s. How far will it travel in 20 seconds?
A.
150 m
B.
300 m
C.
450 m
D.
600 m
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Solution
Distance = speed * time = 15 m/s * 20 s = 300 m.
Correct Answer: B — 300 m
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Q. At absolute zero, the kinetic energy of gas molecules is:
A.
Maximum
B.
Zero
C.
Minimum
D.
Undefined
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Solution
At absolute zero, the kinetic energy of gas molecules is zero, as they are at their lowest energy state.
Correct Answer: B — Zero
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Q. At absolute zero, what is the expected volume of an ideal gas?
A.
Zero
B.
Infinite
C.
Constant
D.
Undefined
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Solution
At absolute zero, the volume of an ideal gas is expected to be zero according to Charles's Law.
Correct Answer: A — Zero
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Q. At absolute zero, what is the theoretical volume of an ideal gas?
A.
Zero
B.
Infinite
C.
Constant
D.
Undefined
Show solution
Solution
At absolute zero, the volume of an ideal gas is theoretically zero according to Charles's Law.
Correct Answer: A — Zero
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Q. At constant pressure, if the temperature of a gas is increased, what happens to its volume?
A.
It decreases
B.
It increases
C.
It remains constant
D.
It becomes zero
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Solution
According to Charles's Law, at constant pressure, the volume of a gas increases with an increase in temperature.
Correct Answer: B — It increases
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Q. At constant pressure, what is the relationship between volume and temperature of a gas?
A.
Directly proportional
B.
Inversely proportional
C.
Independent
D.
Exponential
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Solution
According to Charles's Law, volume and temperature are directly proportional at constant pressure.
Correct Answer: A — Directly proportional
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Q. At what distance from a mass M does the gravitational potential become zero?
A.
At infinity
B.
At the surface of the mass
C.
At 1 meter
D.
At 2 meters
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Solution
The gravitational potential becomes zero at infinity, as V approaches zero as r approaches infinity.
Correct Answer: A — At infinity
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Q. At what point between two equal masses does the gravitational force become zero?
A.
At the midpoint
B.
At a point closer to one mass
C.
At a point closer to the other mass
D.
It never becomes zero
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Solution
The gravitational forces exerted by both masses cancel each other out at the midpoint.
Correct Answer: A — At the midpoint
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Q. At what point in a gravitational field is the gravitational potential constant?
A.
At the center of the mass
B.
At infinity
C.
Along an equipotential surface
D.
At the surface of the mass
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Solution
The gravitational potential is constant along an equipotential surface.
Correct Answer: C — Along an equipotential surface
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Q. At what point in space is the gravitational potential energy of an object zero?
A.
At the center of the Earth
B.
At infinity
C.
At the surface of the Earth
D.
At the Moon
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Solution
The gravitational potential energy is considered zero at infinity.
Correct Answer: B — At infinity
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Q. At what point in space is the gravitational potential zero?
A.
At the center of the Earth.
B.
At infinity.
C.
At the surface of the Earth.
D.
At the center of the Moon.
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Solution
The gravitational potential is considered zero at infinity.
Correct Answer: B — At infinity.
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Q. At what temperature (in Kelvin) does the volume of a gas become zero according to Charles's Law?
A.
0 K
B.
273 K
C.
100 K
D.
32 K
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Solution
According to Charles's Law, the volume of a gas approaches zero at absolute zero, which is 0 K.
Correct Answer: A — 0 K
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Q. At what temperature does the volume of a gas theoretically become zero?
A.
0°C
B.
0 K
C.
273 K
D.
100 K
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Solution
According to Charles's Law, the volume of a gas approaches zero at absolute zero, which is 0 K.
Correct Answer: B — 0 K
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Q. At what temperature will the RMS speed of a gas be 1000 m/s if its molar mass is 0.044 kg/mol? (R = 8.314 J/(mol K))
A.
500 K
B.
600 K
C.
700 K
D.
800 K
Show solution
Solution
Using v_rms = sqrt(3RT/M), we solve for T: T = (v_rms^2 * M) / (3R) = (1000^2 * 0.044) / (3 * 8.314) = 700 K.
Correct Answer: C — 700 K
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