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Alik [6]
3 years ago
12

A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a sp

eed of 345 m/s. The bullet embeds itself into the ground a distance of 8.9 cm before coming to a stop. What force does the ground exert on the bullet? Express your answer in newtons.
Physics
1 answer:
Y_Kistochka [10]3 years ago
4 0

Answer:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

      v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

     0² = 345²+2 x a x 0.089

       a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

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b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

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Rewriting the equation with the given values:

5400000 J=(12000 kg)(9.8 m/s^{2})h

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