A machine gun fires a bullet of mass 40,g wit | vedclass

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(D) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets fired.$F = \frac{dp}{dt} = v \cdot \frac{dm}{dt}$He

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$3$

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(D) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets fired.$F = \frac{dp}{dt} = v \cdot \frac{dm}{dt}$Here,$v = 1200\,m/s$ is the velocity of the bullet.Let $n$ be the number of bullets fired per second.The mass of one bullet is $m = 40\,g = 40 	imes 10^{-3}\,kg = 0.04\,kg$.The total mass fired per second is $\frac{dm}{dt} = n 	imes m$.Given the maximum force $F = 144\,N$,we have:$144 = 1200 	imes (n 	imes 0.04)$$144 = 1200 	imes 0.04 	imes n$$144 = 48 	imes n$$n = \frac{144}{48} = 3$.Therefore,the man can fire at most $3$ bullets per second.

$A$ machine gun fires a bullet of mass $40\,g$ with a velocity $1200\,m/s$. The man holding it can exert a maximum force of $144\,N$ on the gun. How many bullets can he fire per second at the most?

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