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InterviewSolution
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If the force on a rocket, moving with a velocity of300 m s' is 210 N, then the rate of combustion ofthe fuel is(a) Ow7kgst (b) 1.4 kgs(c) 0.7 kg s-(d) 10.7 kg s- |
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Answer» (C) is the correct answer rate of combustion = f/v = 210/300 = 0.7kg/s (c) is correct answer. and it comes from Newton second law f=mdv/dt Force on rocket = udm/dt∴ rate ofd combustion of fuel [dm/dt] = F/u = 210/300 = 0.7 kg/s F = vdm/dt dm/dt = F/v= 210 / 300= 0.7 kg/s Rate of consumption of fuel is 0.7 kg/s as from Newton second law of motion dp/dt=f and dm /dt = f/v so dm /dt = 210/300 = 0.7kg/s. so option c is right |
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