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

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Question

(B) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets.$F = n \cdot \frac{\Delta p}{\Delta t} = n \cdot m

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets.$F = n \cdot \frac{\Delta p}{\Delta t} = n \cdot m \cdot v$Where $n$ is the number of bullets fired per second, $m$ is the mass of one bullet, and $v$ is the velocity of the bullet.Given: $m = 40 	ext{ g} = 0.04 	ext{ kg}$, $v = 600 	ext{ m/s}$, and $F = 168 	ext{ N}$.Substituting the values:$168 = n \cdot (0.04 	ext{ kg}) \cdot (600 	ext{ m/s})$$168 = n \cdot 24$$n = \frac{168}{24} = 7$Therefore, the number of bullets that can be fired per second is $7$.

$A$ machine gun fires a bullet of mass $40$ g with a speed of $600$ m/s. The person holding the gun can bear a maximum force of $168$ $N$ on it. The number of bullets that can be fired from the gun per second is

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