Q. If a 12 kg object is moving with a constant velocity, what can be said about the net force acting on it?
A.
It is zero
B.
It is equal to its weight
C.
It is equal to the applied force
D.
It is increasing
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Solution
An object moving with constant velocity has a net force of zero according to Newton's first law.
Correct Answer: A — It is zero
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Q. If a 12 kg object is pushed with a force of 48 N, what is the acceleration of the object?
A.
2 m/s²
B.
4 m/s²
C.
6 m/s²
D.
8 m/s²
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Solution
Using F = ma, acceleration a = F/m = 48 N / 12 kg = 4 m/s².
Correct Answer: B — 4 m/s²
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Q. If a 12V battery is connected across a 4 ohm resistor, what is the current flowing through the resistor?
A.
2 A
B.
3 A
C.
4 A
D.
6 A
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Solution
Using Ohm's law, I = V/R = 12V / 4Ω = 3 A.
Correct Answer: A — 2 A
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Q. If a 12V battery is connected across a 4 ohm resistor, what is the power dissipated in the resistor?
A.
12 W
B.
24 W
C.
36 W
D.
48 W
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Solution
Power (P) can be calculated using P = V^2 / R = 12V^2 / 4Ω = 144 / 4 = 36 W.
Correct Answer: B — 24 W
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Q. If a 12V battery is connected across a 4Ω resistor, what is the current flowing through the resistor?
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Solution
Using Ohm's law, I = V/R = 12V / 4Ω = 3A.
Correct Answer: B — 2A
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Q. If a 12V battery is connected to a 4 ohm resistor, what is the power dissipated by the resistor?
A.
12 W
B.
24 W
C.
36 W
D.
48 W
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Solution
Power (P) can be calculated using P = V^2 / R = 12V^2 / 4Ω = 36 W.
Correct Answer: B — 24 W
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Q. If a 1500 kg car is moving at a speed of 30 m/s, what is its kinetic energy?
A.
675,000 J
B.
450,000 J
C.
225,000 J
D.
900,000 J
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Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1500 kg × (30 m/s)² = 675,000 J
Correct Answer: B — 450,000 J
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Q. If a 15Ω resistor is connected in parallel with a 5Ω resistor, what is the equivalent resistance? (2023)
A.
3.75Ω
B.
4Ω
C.
5Ω
D.
10Ω
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Solution
1/R_eq = 1/15 + 1/5 = 1/15 + 3/15 = 4/15; R_eq = 15/4 = 3.75Ω.
Correct Answer: A — 3.75Ω
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Q. If a 2 kg object is acted upon by a net force of 6 N, what is the object's acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, we find a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. If a 2 kg object is dropped from a height of 5 m, what is its speed just before it hits the ground? (g = 10 m/s²)
A.
10 m/s
B.
5 m/s
C.
15 m/s
D.
20 m/s
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Solution
Using energy conservation: PE_initial = KE_final; m * g * h = 1/2 * m * v^2; v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s
Correct Answer: A — 10 m/s
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Q. If a 2 kg object is dropped from a height of 5 m, what is its velocity just before it hits the ground? (g = 10 m/s²)
A.
10 m/s
B.
5 m/s
C.
15 m/s
D.
20 m/s
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Solution
Using energy conservation, v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s
Correct Answer: A — 10 m/s
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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground? (g = 9.8 m/s², h = 20 m) (2023)
A.
14 m/s
B.
19.6 m/s
C.
20 m/s
D.
28 m/s
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Solution
Using v² = u² + 2gh, where u = 0, we find v = √(2 * 9.8 * 20) = 19.6 m/s.
Correct Answer: B — 19.6 m/s
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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground (g = 10 m/s²)? (2023)
A.
10 m/s
B.
20 m/s
C.
30 m/s
D.
40 m/s
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Solution
Using v = √(2gh), for h = 10 m, v = √(2 * 10 m/s² * 10 m) = √200 = 20 m/s.
Correct Answer: B — 20 m/s
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Q. If a 2 kg object is lifted to a height of 10 m, what is the potential energy gained by the object? (g = 9.8 m/s²)
A.
19.6 J
B.
98 J
C.
39.2 J
D.
78.4 J
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Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 10 m = 196 J.
Correct Answer: B — 98 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (g = 9.8 m/s²)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
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Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer: A — 58.8 J
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Q. If a 2 kg object is moving with a velocity of 15 m/s, what is its kinetic energy?
A.
225 J
B.
150 J
C.
300 J
D.
450 J
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Solution
Kinetic Energy = 1/2 * m * v^2 = 1/2 * 2 kg * (15 m/s)^2 = 225 J.
Correct Answer: A — 225 J
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Q. If a 2 kg object is moving with a velocity of 3 m/s and a force of 6 N is applied in the opposite direction, what will be its final velocity after 2 seconds?
A.
0 m/s
B.
1 m/s
C.
2 m/s
D.
3 m/s
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Solution
Net force = -6 N, acceleration = F/m = -6 N / 2 kg = -3 m/s². Final velocity = initial velocity + at = 3 m/s + (-3 m/s² * 2 s) = 0 m/s.
Correct Answer: B — 1 m/s
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Q. If a 2 kg object is moving with a velocity of 3 m/s, what is its kinetic energy?
A.
9 J
B.
6 J
C.
3 J
D.
12 J
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Solution
Kinetic Energy = 0.5 * m * v² = 0.5 * 2 kg * (3 m/s)² = 9 J.
Correct Answer: A — 9 J
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Q. If a 2 kg object is pulled with a force of 10 N and experiences a frictional force of 4 N, what is its acceleration?
A.
3 m/s²
B.
5 m/s²
C.
2 m/s²
D.
1 m/s²
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Solution
Net force = applied force - friction = 10 N - 4 N = 6 N. Acceleration a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer: A — 3 m/s²
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Q. If a 2 kg object is subjected to a net force of 10 N, what will be its acceleration? (2022)
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
20 m/s²
Show solution
Solution
Using F = ma, a = F/m = 10 N / 2 kg = 5 m/s².
Correct Answer: B — 5 m/s²
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Q. If a 2 kg object is subjected to a net force of 6 N, what is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we have a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. If a 3 kg object is in free fall, what is the force acting on it due to gravity?
A.
3 N
B.
9 N
C.
30 N
D.
None of the above
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Solution
Force due to gravity F = mg = 3 kg * 9.8 m/s² = 29.4 N, approximately 9 N for simplification.
Correct Answer: B — 9 N
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by friction?
A.
-24 J
B.
-48 J
C.
0 J
D.
24 J
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Solution
Initial kinetic energy = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer: B — -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by the friction?
A.
-24 J
B.
-48 J
C.
-12 J
D.
-36 J
Show solution
Solution
Initial kinetic energy = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer: B — -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s and comes to a stop, what is the work done by the friction force?
A.
-24 J
B.
-48 J
C.
-72 J
D.
-96 J
Show solution
Solution
Initial kinetic energy = 0.5 × 3 kg × (4 m/s)² = 24 J. Work done by friction = -24 J.
Correct Answer: B — -48 J
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Q. If a 3 kg object is moving with a speed of 4 m/s, what is its kinetic energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
KE = 0.5 × m × v² = 0.5 × 3 kg × (4 m/s)² = 0.5 × 3 × 16 = 24 J.
Correct Answer: B — 24 J
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Q. If a 3 kg object is moving with a speed of 4 m/s, what is the total mechanical energy if it is at a height of 2 m?
A.
30 J
B.
40 J
C.
50 J
D.
60 J
Show solution
Solution
Total mechanical energy = K.E + P.E = 0.5 × 3 kg × (4 m/s)² + 3 kg × 9.8 m/s² × 2 m = 24 J + 58.8 J = 82.8 J.
Correct Answer: C — 50 J
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Q. If a 3 kg object is moving with a velocity of 4 m/s, what is its momentum?
A.
12 kg m/s
B.
6 kg m/s
C.
8 kg m/s
D.
10 kg m/s
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Solution
Momentum (p) = m * v = 3 kg * 4 m/s = 12 kg m/s
Correct Answer: A — 12 kg m/s
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Q. If a 4 kg object is acted upon by a net force of 16 N, what is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we have a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer: C — 4 m/s²
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Q. If a 4 kg object is at rest and a net force of 8 N is applied, what will be its velocity after 2 seconds?
A.
4 m/s
B.
2 m/s
C.
1 m/s
D.
0 m/s
Show solution
Solution
Using F = ma, a = F/m = 8 N / 4 kg = 2 m/s². Velocity after 2 seconds = a * t = 2 m/s² * 2 s = 4 m/s.
Correct Answer: B — 2 m/s
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