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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: - An astronomical telescope arranged for normal adjustment has a magnification of 6.
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An astronomical telescope arranged for normal adjustment has a magnification of 6.
- Solar radiation is
- The speed of light in media M₁ and M₂ are 1.5×10⁸ ms⁻¹ and 2×10⁸ ms⁻¹ respectively.
- The value of coefficient of volume expansion of glycerin is 5 x 10⁻⁴ k⁻¹.
- On adjusting the P-N junction diode in forward bias,
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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