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2 years ago

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Certain rifles can fire a bullet with a speed of 950 m/s just as it leaves the muzzle (this speed is called the muzzle velocity)

. The muzzle is 75.0 cm long and the bullet is accelerated uniformly from rest within it. What is the acceleration (in {g}'s) of the bullet in the muzzle? If, when this rifle is fired vertically, the bullet reaches a maximum height {H}, what would be the maximum height (in terms of H) for a new rifle that produced half the muzzle velocity of this one?

1 answer:

TiliK225 [7]2 years ago

4 0

Answer:

a) By $v^2 = u^2 + 2as$ => a= 70291.70.

(b)By $v = u + at$ => t= 1.58 ms.

(c)By $v^2 = u^2 - 2gh$ => H = 46045.92 m.

Explanation:

a) By $v^2 = u^2 + 2as$

$(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70$

(b)By $v = u + at$

$950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms$

(c)By $v^2 = u^2 - 2gh$

$0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m$

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