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Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be:
Options
(a) Mv newton
(b) 2 Mv newton
(c) Mv/2 newton
(d) zero
Correct Answer:
Mv newton
Explanation:
F = d(Mv) / dt = M . dv / dt + v . dM / dt
v is constant,
F = vdM / dt But dM / dt = Mkg / s
F = vM newton.
Related Questions: - At the first minimum adjacent to the central maximum of a single-slit diffraction
- Acceleration of a particle, executing SHM, at its mean position is
- The distance travelled by a particle starting from rest and moving with an acceleration
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- At the first minimum adjacent to the central maximum of a single-slit diffraction
- Acceleration of a particle, executing SHM, at its mean position is
- The distance travelled by a particle starting from rest and moving with an acceleration
- A body undergoes no change in volume. Poisson’s ratio is
- If F⃗ is the force acting on a particle having position vector r⃗ and ? ⃗ be the torque
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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