posted by annie .
A 6.5 g bullet leaves the muzzle of a rifle with
a speed of 347 m/s.
What total constant force is exerted on the
bullet while it is traveling down the 0.83 m
long barrel of the rifle?
a = (V^2-Vo^2) / 2d.
a = ((347)^2-0) / 1.66. = 72536 m/s^2.
F = ma = 0.0065kg * 72,536 = 472 N.