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Problem 71 Easy Difficulty

ssm A 15 -g bullet is fired from a rifle. It takes $2.50 \times 10^{-3}$ s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 $\mathrm{m} / \mathrm{s}$ . Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

Answer

4290 $\mathrm{N}$

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Video Transcript

at the beginning, the bullet was at rest. Therefore, its initial velocity is it close to zero meters per second. Then it reaches this velocity in nine interval of time of 2.5 times 10 to the minus three seconds. So we can use the following equation to calculate its acceleration. Its velocity is equal streets initial velocity, plus its acceleration times, the interval of time they're afford its final velocity is equal streets initial velocity. Plus it's acceleration times the interval off time. Then we can solve this equation forward acceleration to that. That acceleration is equal street 715 divided by troop on five times Stan to the minor street, the Hendrick and complex than that force that acts on the bullet By using Newton's second law, which says that the net force is equal to the mass times declaration afford and that force his equals truth 15 grams or 15 times Stan to the minor street kilograms times 715 divided by 3.5 times. Stand to them on the street. There is a simplification that can be done here, and these results in the net force off 4290 Newtons

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