Asked by POPPY
a bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20 m/s. If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the
1) magnitude of the
2) Distance moved by the bullet in the wood.?
1) magnitude of the
2) Distance moved by the bullet in the wood.?
Answers
Answered by
oobleck
a = ∆v/∆t = -20/0.1 = -200 m/s^2
s = vt + 1/2 at^2 = 20*0.1 - 100*0.1^2 = ?
s = vt + 1/2 at^2 = 20*0.1 - 100*0.1^2 = ?
Answered by
henry2,
1. V = Vo + a*t = 0.
20 + a*0.1 = 0,
a = -200 m/s^2.
F = M*a = 0.120 * (-200) = -24 N. = opposing resistance.
20 + a*0.1 = 0,
a = -200 m/s^2.
F = M*a = 0.120 * (-200) = -24 N. = opposing resistance.
Answered by
Anonymous
Thanks but can u break it down for me
Answered by
Anonymous
Thanks but can you break it for me please
Answered by
Anonymous
Tnx to but I need it in details please
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