A machine gun fires 300 bullets per minute, e | vedclass

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(D) The power $P$ of the machine gun is defined as the total kinetic energy delivered per unit time.$P = \frac{	ext{Total Kinetic Energy}}{	ext{Time

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$2.5$

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(D) The power $P$ of the machine gun is defined as the total kinetic energy delivered per unit time.$P = \frac{	ext{Total Kinetic Energy}}{	ext{Time}}$$P = \frac{n 	imes (\frac{1}{2}mv^2)}{t} = \frac{n}{t} 	imes \frac{1}{2}mv^2$Given:Number of bullets $n = 300$Time $t = 60 \,s$Mass of each bullet $m = 4 \,g = 4 	imes 10^{-3} \,kg$Velocity $v = 500 \,ms^{-1}$Substituting the values:$P = \frac{300}{60} 	imes \frac{1}{2} 	imes (4 	imes 10^{-3}) 	imes (500)^2$$P = 5 	imes \frac{1}{2} 	imes 0.004 	imes 250000$$P = 5 	imes 0.002 	imes 250000$$P = 5 	imes 500 = 2500 \,W$$P = 2.5 \,kW$

$A$ machine gun fires $300$ bullets per minute, each with a velocity of $500 \,ms^{-1}$. If the mass of each bullet is $4 \,g$, the power of the machine gun is: (in $\,kW$)

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