Asked by Anonymous
A .22 rifle bullet, traveling at 373m/s, strikes a block of soft wood, which it penetrates to a depth of 0.125m. The block of wood is clamped in place and doesn't move. The mass of the bullet is 1.90. Assume a constant retarding force.How much time is required for the bullet to stop? and What is the magnitude of the force exerted on the bullet by the wood while the bullet is being stopped?
Answers
Answered by
Mak
Magnitude of the force (F):bullet mass: 1.9g
1/2 x mv^2 = F x d
1/2 x 1.9 g x 373^2 = F x 0.125
F = 1057.4 N
Time (t):
1.9g x 373 =1057.4 x t
t = 6.7 x 10^-4 s
1/2 x mv^2 = F x d
1/2 x 1.9 g x 373^2 = F x 0.125
F = 1057.4 N
Time (t):
1.9g x 373 =1057.4 x t
t = 6.7 x 10^-4 s
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