Q. What is the gravitational potential energy of a 2 kg mass at a height of 10 m above the ground? (g = 9.8 m/s^2)
A.
196 J
B.
98 J
C.
20 J
D.
0 J
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Solution
Gravitational potential energy (U) is given by U = mgh = 2 kg * 9.8 m/s^2 * 10 m = 196 J.
Correct Answer: A — 196 J
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Q. What is the gravitational potential energy of a 2 kg mass at a height of 10 m above the ground? (g = 9.8 m/s²)
A.
196 J
B.
98 J
C.
20 J
D.
0 J
Show solution
Solution
Gravitational potential energy (U) is calculated as U = mgh = 2 kg * 9.8 m/s² * 10 m = 196 J.
Correct Answer: A — 196 J
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Q. What is the gravitational potential energy of a 2 kg mass at a height of 10 m?
A.
20 J
B.
15 J
C.
10 J
D.
5 J
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Solution
Gravitational potential energy (U) = mgh = 2 kg * 9.8 m/s^2 * 10 m = 196 J.
Correct Answer: A — 20 J
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Q. What is the gravitational potential energy of a 2 kg mass at a height of 10 m? (g = 9.8 m/s²)
A.
196 J
B.
98 J
C.
20 J
D.
0 J
Show solution
Solution
U = mgh = 2 * 9.8 * 10 = 196 J.
Correct Answer: A — 196 J
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Q. What is the gravitational potential energy of a 2 kg mass at a height of 5 m?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Using U = mgh, U = 2 kg * 9.8 m/s^2 * 5 m = 98 J.
Correct Answer: B — 20 J
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Q. What is the gravitational potential energy of a 2 kg mass at a height of 5 m? (g = 9.8 m/s²)
A.
19.6 J
B.
39.2 J
C.
49 J
D.
9.8 J
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Solution
Potential Energy PE = m * g * h = 2 kg * 9.8 m/s² * 5 m = 98 J
Correct Answer: B — 39.2 J
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Q. What is the gravitational potential energy of a 5 kg mass at a height of 10 m above the ground?
A.
490 J
B.
500 J
C.
510 J
D.
450 J
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Solution
Gravitational potential energy U = mgh = 5 * 9.8 * 10 = 490 J.
Correct Answer: B — 500 J
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Q. What is the gravitational potential energy of a mass 'm' at height 'h' above the Earth's surface?
A.
mgh
B.
mg/h
C.
gh/m
D.
mgh^2
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Solution
The gravitational potential energy U at height h is given by U = mgh.
Correct Answer: A — mgh
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Q. What is the gravitational potential energy of a mass m at a height h above the Earth's surface?
A.
U = mgh
B.
U = mg/h
C.
U = mgh²
D.
U = gh/m
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Solution
The gravitational potential energy is given by the formula U = mgh, where g is the acceleration due to gravity.
Correct Answer: A — U = mgh
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Q. What is the gravitational potential energy of a mass of 10 kg at a height of 5 m above the ground? (2000)
A.
500 J
B.
1000 J
C.
1500 J
D.
2000 J
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Solution
Gravitational potential energy U = mgh = 10 kg * 9.8 m/s² * 5 m = 490 J, approximately 500 J.
Correct Answer: B — 1000 J
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Q. What is the gravitational potential energy of a mass of 10 kg at a height of 5 m in a gravitational field of strength 9.8 N/kg?
A.
490 J
B.
50 J
C.
98 J
D.
0 J
Show solution
Solution
Gravitational potential energy U = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer: A — 490 J
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Q. What is the gravitational potential energy of a mass of 10 kg at a height of 5 m above the ground? (g = 9.8 m/s²)
A.
490 J
B.
98 J
C.
588 J
D.
0 J
Show solution
Solution
U = mgh = 10 * 9.8 * 5 = 490 J
Correct Answer: A — 490 J
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Q. What is the gravitational potential energy of a mass of 2 kg at a height of 10 m in a gravitational field of 9.8 m/s²?
A.
19.6 J
B.
39.2 J
C.
78.4 J
D.
98 J
Show solution
Solution
Gravitational potential energy (U) = mgh = 2 kg * 9.8 m/s² * 10 m = 196 J.
Correct Answer: B — 39.2 J
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Q. What is the gravitational potential energy of a mass of 2 kg at a height of 10 m? (g = 9.8 m/s²)
A.
19.6 J
B.
39.2 J
C.
78.4 J
D.
98 J
Show solution
Solution
PE = mgh = 2 * 9.8 * 10 = 196 J
Correct Answer: B — 39.2 J
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Q. What is the gravitational potential energy of a mass of 2 kg at a height of 10 m above the ground? (g = 9.8 m/s^2)
A.
196 J
B.
98 J
C.
20 J
D.
10 J
Show solution
Solution
Gravitational potential energy (U) = mgh = 2 kg * 9.8 m/s^2 * 10 m = 196 J.
Correct Answer: A — 196 J
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Q. What is the gravitational potential energy of a satellite of mass m at a height h above the Earth's surface?
A.
-GMm/R
B.
-GMm/(R+h)
C.
-GMm/(R-h)
D.
-GMm/(R^2 + h^2)
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Solution
The gravitational potential energy of a satellite at height h is given by U = -GMm/(R+h), where R is the radius of the Earth.
Correct Answer: B — -GMm/(R+h)
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Q. What is the gravitational potential energy of an object of mass m at a height h above the Earth's surface?
A.
mgh
B.
gh/m
C.
mg/h
D.
mgh^2
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Solution
The gravitational potential energy is given by the formula U = mgh.
Correct Answer: A — mgh
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Q. What is the gravitational potential energy of two masses m1 and m2 separated by a distance r?
A.
-G * (m1 * m2) / r
B.
G * (m1 * m2) / r
C.
G * (m1 + m2) / r
D.
-G * (m1 + m2) / r
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Solution
The gravitational potential energy is given by U = -G * (m1 * m2) / r.
Correct Answer: A — -G * (m1 * m2) / r
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Q. What is the gravitational potential energy U of two masses m1 and m2 separated by a distance r?
A.
U = -G * (m1 * m2) / r
B.
U = G * (m1 * m2) / r
C.
U = (m1 + m2) * r
D.
U = G * (m1 - m2) / r
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Solution
The gravitational potential energy is given by U = -G * (m1 * m2) / r.
Correct Answer: A — U = -G * (m1 * m2) / r
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Q. What is the half-life of a radioactive isotope?
A.
Time taken for half of the sample to decay
B.
Time taken for the entire sample to decay
C.
Time taken for the sample to double
D.
None of the above
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Solution
The half-life is defined as the time taken for half of the radioactive sample to decay.
Correct Answer: A — Time taken for half of the sample to decay
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Q. What is the half-life of a radioactive substance if it takes 10 years for half of the substance to decay?
A.
5 years
B.
10 years
C.
20 years
D.
30 years
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Solution
The half-life is defined as the time required for half of the radioactive substance to decay, which is given as 10 years.
Correct Answer: B — 10 years
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Q. What is the half-life of a radioactive substance?
A.
The time taken for half of the substance to decay
B.
The time taken for the entire substance to decay
C.
The time taken for the substance to double
D.
The time taken for the substance to reach equilibrium
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Solution
The half-life of a radioactive substance is defined as the time taken for half of the substance to decay.
Correct Answer: A — The time taken for half of the substance to decay
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Q. What is the heat required to raise the temperature of 250 g of aluminum from 25°C to 75°C? (Specific heat of aluminum = 0.9 J/g°C)
A.
4500 J
B.
5000 J
C.
6000 J
D.
7000 J
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Solution
Q = m*c*ΔT = 250 g * 0.9 J/g°C * (75°C - 25°C) = 4500 J.
Correct Answer: A — 4500 J
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Q. What is the ideal gas equation?
A.
PV = nRT
B.
PV = nRT^2
C.
PV = nR/T
D.
PV = nRT^3
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Solution
The ideal gas equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Correct Answer: A — PV = nRT
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Q. What is the image distance for a concave lens with a focal length of -12 cm when the object is placed at 24 cm?
A.
-8 cm
B.
8 cm
C.
-12 cm
D.
12 cm
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Solution
Using the lens formula, 1/f = 1/v - 1/u, we find v = -8 cm.
Correct Answer: A — -8 cm
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Q. What is the image distance for a convex lens if the object is placed at 10 cm and the focal length is 5 cm?
A.
-15 cm
B.
15 cm
C.
5 cm
D.
10 cm
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Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 15 cm.
Correct Answer: A — -15 cm
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Q. What is the image distance for a convex lens if the object is placed at 30 cm and the focal length is 10 cm?
A.
15 cm
B.
20 cm
C.
25 cm
D.
30 cm
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Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 15 cm.
Correct Answer: B — 20 cm
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Q. What is the image distance for a convex lens with a focal length of 10 cm when the object is placed 30 cm from the lens?
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
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Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 20 cm.
Correct Answer: D — 20 cm
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Q. What is the impedance of a series circuit containing a resistor (R) and an inductor (L)?
A.
R
B.
√(R^2 + (ωL)^2)
C.
R + ωL
D.
R - ωL
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Solution
The impedance Z = √(R^2 + (ωL)^2) in a series R-L circuit.
Correct Answer: B — √(R^2 + (ωL)^2)
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Q. What is the impedance of a series circuit containing a resistor and an inductor?
A.
R
B.
√(R^2 + (ωL)^2)
C.
R + jωL
D.
R - jωL
Show solution
Solution
The impedance Z in a series R-L circuit is given by Z = √(R^2 + (ωL)^2).
Correct Answer: B — √(R^2 + (ωL)^2)
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