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3 years ago

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The bullet starts at rest in the gun. An 8.6 g bullet leaves the muzzle of a rifle with a speed of 430.1 m/s. What constant forc

e is exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle?

1 answer:

xenn [34]3 years ago

3 0

Answer:

The constant force exerted on the bullet is 1590.87 N.

Explanation:

It is given that,

Mass of the bullet, m = 8.6 g

Initial speed of the bullet, u = 0

Final speed of the bullet, v = 430.1 m/s

We need to find the force exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle. Let a is the acceleration of the bullet. So,

$v^2-u^2=2ad$

$v^2=2ad$

$a=\dfrac{v^2}{2d}$

$a=\dfrac{(430.1)^2}{2\times 0.5}$

$a=184986.01\ m/s^2$

Let F is the force exerted. It is given by :

$F=ma$

$F=8.6\times 10^{-3}\times 184986.01$

F = 1590.87 N

So, the constant force exerted on the bullet is 1590.87 N. Hence, this is the required solution.

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