If the pressure force of the powder gases in the bore is 4 N, then at what speed will a bullet weighing 7.9 g fly out of a 45 cm long barrel?
Data: F (force of pressure of powder gases in the bore) = 4 N; m (bullet weight) = 7.9 g (7.9 * 10 ^ -3 kg); L (barrel length) = S (distance traveled by the bullet) = 45 cm (0.45 m); V0 (muzzle velocity) = 0 m / s.
1) The acceleration with which the bullet moved in the bore: a = F / m = 4 / (7.9 * 10 ^ -3) = 506.3 m / s2.
2) The speed with which the bullet flew: S = (V ^ 2 – V0 ^ 2) / 2a = V ^ 2 / 2a, whence V = √ (S * 2a) = √ (0.45 * 2 * 506.3 ) = 21.35 m / s.
Answer: The bullet will fly out of the barrel at a speed of 21.35 m / s.
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