The bullet at the barrel of a Kalashnikov assault rifle moves with an acceleration of 616 km / s². What is the bullet velocity if the barrel length is 41.5 cm?
To calculate the bullet exit velocity, apply the formula: S = V ^ 2 / 2a, whence V = √ (2a * S).
Values of variables: a – bullet acceleration (a = 616 km / s2; in SI a = 616 * 10 ^ 3 m / s); S – barrel length, bullet path (S = 41.5 cm; in SI S = 0.415 m).
Let’s calculate: V = √ (2a * S) = √ (2 * 616 * 10 ^ 3 * 0.415) = 715 m / s.
Answer: The bullet departure speed, according to the calculation, is 715 m / s.
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