A 10g bullet moving at 200m/s stops after penetrating 5cm of wooden plank. The average force exerted on the bullet is.?
Answers
Answered by
3
by putting the equation of motion v2-u2=2as u will get 4×10^5m/s2 acceleration and then put the formula F=ma nd get this answer4×10^6N
Answered by
1
hey,
your answer is below...
Acceleration = (v^2/2d)
= (200^2/0.1m)
= -400,000m/sec.
F = (ma)
= 0.01kg*400,000
= 4,000N.
I hope this information is useful for you.
your answer is below...
Acceleration = (v^2/2d)
= (200^2/0.1m)
= -400,000m/sec.
F = (ma)
= 0.01kg*400,000
= 4,000N.
I hope this information is useful for you.
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