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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?

A. 2.0 Nm
B. 5.0 Nm
C. 10.0 Nm
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?

A. 5 Nm
B. 10 Nm
C. 20 Nm
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?

A. 5 Nm
B. 10 Nm
C. 20 Nm
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?

A. 10 Nm
B. 17.32 Nm
C. 20 Nm
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?

A. 2 m/s
B. 5 m/s
C. 10 m/s
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?

A. 6 N
B. 10 N
C. 14 N
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?

A. 5 m/s²
B. 10 m/s²
C. 2 m/s²
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?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
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?

A. 30 J
B. 60 J
C. 90 J
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?

A. 7.5 Nm
B. 12.5 Nm
C. 15 Nm
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?

A. 30 J
B. 60 J
C. 45 J
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?

A. 10 N
B. 15 N
C. 20 N
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?

A. 3 m/s²
B. 5 m/s²
C. 7 m/s²
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?

A. 3 m/s²
B. 5 m/s²
C. 7 m/s²
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?

A. 3 m/s²
B. 2 m/s²
C. 1 m/s²
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?

A. 15 J
B. 20 J
C. 30 J
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?

A. 30 J
B. 60 J
C. 75 J
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?

A. 10 Nm
B. 17.32 Nm
C. 20 Nm
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?

A. 10 Nm
B. 20 Nm
C. 30 Nm
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?

A. 5 Nm
B. 10 Nm
C. 8.66 Nm
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?

A. 10 Nm
B. 20 Nm
C. 17.32 Nm
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?

A. 100 J
B. 125 J
C. 150 J
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?

A. 37.5 J
B. 50 J
C. 75 J
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?

A. 15 Nm
B. 30 Nm
C. 60 Nm
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?

A. 3 m/s²
B. 6 m/s²
C. 9 m/s²
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?

A. 10 Nm
B. 15 Nm
C. 20 Nm
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?

A. 20 Nm
B. 40 Nm
C. 34.64 Nm
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?

A. 25 Nm
B. 43.3 Nm
C. 50 Nm
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?

A. 25 Nm
B. 43.3 Nm
C. 50 Nm
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?

A. 25 N·m
B. 50 N·m
C. 86.6 N·m
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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