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Q. A force of 10 N is applied at a distance of 0.5 m from the pivot point. What is the torque about the pivot?
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A.
2.0 Nm
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B.
5.0 Nm
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C.
10.0 Nm
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D.
20.0 Nm
Solution
Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.
Correct Answer: A — 2.0 Nm
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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot point?
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A.
5 Nm
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B.
10 Nm
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C.
20 Nm
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D.
15 Nm
Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot?
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A.
5 Nm
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B.
10 Nm
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C.
20 Nm
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D.
15 Nm
Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 10 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
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A.
10 Nm
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B.
17.32 Nm
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C.
20 Nm
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D.
5 Nm
Solution
Torque = Force × Distance × sin(60°) = 10 N × 2 m × (√3/2) = 17.32 Nm.
Correct Answer: B — 17.32 Nm
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Q. A force of 10 N is applied to a 2 kg object at rest. What is the final velocity of the object after 5 seconds?
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A.
2 m/s
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B.
5 m/s
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C.
10 m/s
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D.
15 m/s
Solution
Using F = ma, a = F/m = 10 N / 2 kg = 5 m/s². Final velocity v = u + at = 0 + 5 * 5 = 25 m/s.
Correct Answer: B — 5 m/s
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Q. A force of 10 N is applied to a 2 kg object. What is the net force acting on the object if it is also experiencing a frictional force of 4 N?
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A.
6 N
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B.
10 N
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C.
14 N
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D.
4 N
Solution
Net force = applied force - frictional force = 10 N - 4 N = 6 N.
Correct Answer: A — 6 N
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Q. A force of 10 N is applied to a 2 kg object. What is the object's acceleration?
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A.
5 m/s²
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B.
10 m/s²
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C.
2 m/s²
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D.
20 m/s²
Solution
Using F = ma, acceleration a = F/m = 10 N / 2 kg = 5 m/s².
Correct Answer: A — 5 m/s²
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Q. A force of 10 N is applied to a 2 kg object. What is the resulting acceleration?
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A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
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D.
5 m/s²
Solution
Using F = ma, we have a = F/m = 10 N / 2 kg = 5 m/s².
Correct Answer: A — 2 m/s²
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Q. A force of 15 N is applied at an angle of 30° to the horizontal while moving an object 4 m. What is the work done by the force?
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A.
30 J
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B.
60 J
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C.
90 J
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D.
120 J
Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(30°) = 15 N × 4 m × (√3/2) = 60 J.
Correct Answer: C — 90 J
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Q. A force of 15 N is applied at an angle of 60 degrees to a lever arm of 1 m. What is the torque?
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A.
7.5 Nm
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B.
12.5 Nm
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C.
15 Nm
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D.
25 Nm
Solution
Torque = Force × Distance × sin(θ) = 15 N × 1 m × sin(60°) = 15 N × 1 m × (√3/2) = 12.5 Nm.
Correct Answer: B — 12.5 Nm
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Q. A force of 15 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m. What is the work done?
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A.
30 J
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B.
60 J
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C.
45 J
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D.
75 J
Solution
Work Done = F * d * cos(θ) = 15 N * 4 m * cos(60°) = 30 J
Correct Answer: C — 45 J
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Q. A force of 15 N is applied to a 3 kg object. What is the net force acting on the object if it is also experiencing a frictional force of 5 N?
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A.
10 N
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B.
15 N
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C.
20 N
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D.
5 N
Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer: A — 10 N
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Q. A force of 15 N is applied to a 3 kg object. What is the object's acceleration?
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A.
3 m/s²
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B.
5 m/s²
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C.
7 m/s²
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D.
10 m/s²
Solution
Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: B — 5 m/s²
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Q. A force of 15 N is applied to a 3 kg object. What is the resulting acceleration?
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A.
3 m/s²
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B.
5 m/s²
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C.
7 m/s²
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D.
10 m/s²
Solution
Using F = ma, we find a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: B — 5 m/s²
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Q. A force of 15 N is applied to a 5 kg object. What is the object's acceleration?
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A.
3 m/s²
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B.
2 m/s²
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C.
1 m/s²
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D.
4 m/s²
Solution
Using F = ma, acceleration a = F/m = 15 N / 5 kg = 3 m/s².
Correct Answer: A — 3 m/s²
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Q. A force of 15 N is applied to move an object 3 m at an angle of 60 degrees to the horizontal. What is the work done by the force?
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A.
15 J
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B.
20 J
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C.
30 J
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D.
45 J
Solution
Work done = F × d × cos(θ) = 15 N × 3 m × cos(60°) = 15 N × 3 m × 0.5 = 22.5 J.
Correct Answer: C — 30 J
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Q. A force of 15 N is applied to move an object 4 m in the direction of the force. What is the work done?
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A.
30 J
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B.
60 J
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C.
75 J
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D.
90 J
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer: D — 90 J
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Q. A force of 20 N is applied at an angle of 30 degrees to the lever arm of 1 m. What is the torque about the pivot?
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A.
10 Nm
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B.
17.32 Nm
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C.
20 Nm
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D.
5 Nm
Solution
Torque = Force × Distance × sin(30°) = 20 N × 1 m × 0.5 = 10 Nm.
Correct Answer: B — 17.32 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
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A.
10 Nm
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B.
20 Nm
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C.
30 Nm
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D.
40 Nm
Solution
Torque = Force × Distance × sin(θ) = 20 N × 2 m × sin(60°) = 20 N × 2 m × (√3/2) = 20√3 Nm.
Correct Answer: B — 20 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of length 0.5 m. What is the torque about the pivot?
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A.
5 Nm
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B.
10 Nm
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C.
8.66 Nm
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D.
17.32 Nm
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 0.5 m × sin(60°) = 20 N × 0.5 m × (√3/2) = 8.66 Nm.
Correct Answer: C — 8.66 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
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A.
10 Nm
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B.
20 Nm
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C.
17.32 Nm
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D.
34.64 Nm
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 17.32 Nm
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Q. A force of 25 N is applied at an angle of 30 degrees to the horizontal while moving an object 10 m. What is the work done?
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A.
100 J
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B.
125 J
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C.
150 J
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D.
175 J
Solution
Work done = Force × Distance × cos(θ) = 25 N × 10 m × cos(30°) = 216.5 J.
Correct Answer: B — 125 J
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Q. A force of 25 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m. What is the work done?
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A.
37.5 J
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B.
50 J
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C.
75 J
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D.
100 J
Solution
Work done = Force × Distance × cos(θ) = 25 N × 3 m × cos(60°) = 37.5 J.
Correct Answer: A — 37.5 J
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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?
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A.
15 Nm
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B.
30 Nm
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C.
60 Nm
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D.
52 Nm
Solution
Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.
Correct Answer: D — 52 Nm
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Q. A force of 30 N is applied to a 5 kg object. What is the object's acceleration?
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A.
3 m/s²
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B.
6 m/s²
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C.
9 m/s²
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D.
12 m/s²
Solution
Using F = ma, a = F/m = 30 N / 5 kg = 6 m/s².
Correct Answer: B — 6 m/s²
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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?
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A.
10 Nm
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B.
15 Nm
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C.
20 Nm
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D.
25 Nm
Solution
Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
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A.
20 Nm
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B.
40 Nm
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C.
34.64 Nm
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D.
69.28 Nm
Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 34.64 Nm
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Q. A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?
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A.
25 Nm
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B.
43.3 Nm
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C.
50 Nm
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D.
0 Nm
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 m × (√3/2) = 43.3 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?
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A.
25 Nm
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B.
43.3 Nm
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C.
50 Nm
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D.
86.6 Nm
Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?
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A.
25 N·m
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B.
50 N·m
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C.
86.6 N·m
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D.
100 N·m
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.
Correct Answer: C — 86.6 N·m
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