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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: - A black body at a temperature of 227⁰C radiates heat at a rate of 20 cal m⁻²s⁻¹.
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A black body at a temperature of 227⁰C radiates heat at a rate of 20 cal m⁻²s⁻¹.
- The flux linked with a circuit is given by φ=t³+3t-7. The graph between time
- A block of 2 kg is kept on the floor. The coefficient of static friction is 0.4
- A metal rod of length l cuts across a uniform magnetic field B with a velocity v.
- A L-C-R circuit with L=1.00 mH, C=10μF and R=50Ω, is driven with 5 V AC voltage.
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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