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A machine gun fires 300 bullets per minute each with a velocity of $500 \mathrm{~ms}^{-1}$. If the mass of each bullet is $4 \mathrm{~g}$, the power of the machine gun is
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The correct answer is:
$2.5 \mathrm{~kW}$
Power of the machine gun,
$$
\begin{aligned}
& \mathrm{P}=\frac{\text { total K.E. of bullet fired }}{\text { time }} \\
& =\frac{\mathrm{n} \times \frac{1}{2} \mathrm{mv}^2}{\mathrm{t}}=\frac{\mathrm{n}}{\mathrm{t}} \times \frac{1}{2} \mathrm{mv}^2 \\
& =\frac{300}{60} \times \frac{1}{2} \times \frac{4}{1000} \times(500)^2 \\
& =\frac{300}{60} \times \frac{2}{1000} \times 500 \times 500 \\
& \therefore P=\frac{300}{60} \times 500=2.5 \mathrm{KW} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{P}=\frac{\text { total K.E. of bullet fired }}{\text { time }} \\
& =\frac{\mathrm{n} \times \frac{1}{2} \mathrm{mv}^2}{\mathrm{t}}=\frac{\mathrm{n}}{\mathrm{t}} \times \frac{1}{2} \mathrm{mv}^2 \\
& =\frac{300}{60} \times \frac{1}{2} \times \frac{4}{1000} \times(500)^2 \\
& =\frac{300}{60} \times \frac{2}{1000} \times 500 \times 500 \\
& \therefore P=\frac{300}{60} \times 500=2.5 \mathrm{KW} \\
&
\end{aligned}
$$
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