A) 0.7 kg/s
B) 1.4 kg/s
C) 0.07 kg/s
D) 10.7 kg/s
Correct Answer: A
Solution :
Thrust force on the rocket \[{{F}_{t}}={{v}_{r}}\,\left( -\frac{dm}{dt} \right)\] (upwards) Rate of combustion of fuel \[-\frac{dm}{dt}=\frac{{{F}_{t}}}{{{v}_{r}}}\] Given, \[{{F}_{t}}=210\,N,\,\,{{v}_{r}}=300\,m/s\] \[\therefore -\frac{dm}{dt}=\frac{210}{300}=0.7\,kg/s\]You need to login to perform this action.
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