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If the force on a rocket ejecting gases with a relative velocity of $300\,\text{m/s}$ is $210\,\text{N}$, then the rate of combustion of the fuel is:

1. 10.7 kg/sec
2. 0.07 kg/sec
3. 1.4 kg/sec
4. 0.7 kg/sec

Correct answer: 0.7 kg/sec

Solution

The thrust on a rocket is given by $F=\dot{m}v_{rel}$, where $\dot{m}$ is the rate of fuel combustion. Substituting $F=210\,\text{N}$ and $v_{rel}=300\,\text{m/s}$ gives $\dot{m}=210/300=0.7\,\text{kg/s}$.

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