Asked by haziq
A 4 g bullet leaves the muzzle of a riffle with a speed of 350 m s-1. What force is exerted on the bullet when it is travelling down the 0.85 m long barrel of the rifle?
Answers
Answered by
Henry
F = m*a
V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d
a = ((350)^2-0)/1.7 = 72,059 m/s^2
F = 0.004 * 72059 = 288 N.
V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d
a = ((350)^2-0)/1.7 = 72,059 m/s^2
F = 0.004 * 72059 = 288 N.
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