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If a force on rocket having exhaust velocity of 300m/s is 210 N, then rate of combustion of fuel isA) 0.7 Kg/sB) 1.4 Kg/sC) 0.07 Kg/sD) 10.7 Kg/s |
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Answer» A) 0.7 kg/s Explanations: F = d/dt (mv) = v dm/dt ⇒ 210 = 300 * dm/dt ⇒ dm/dt = 0.7 Kg/s |
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