A bullet of mass 20 g moving with a velocity of 120 m/s hits a thick muddy wall and penetrates it. It takes 0.05s to stop in the wall. Find (a) Acceleration of the bullet in the wall. (b) Force exerted by the wall on the bullet.
Class 9th please.
Answers
Answered by
3
Explanation:
Mass of the bullet =20g or 0.02kg
initial velocity of the bullet(u)=120m/s
final velocity(v)=0m/s (because the bullet eventually stops)
a) Acceleration = (v-u)÷t
= (0-120)÷0.03
= -120÷0.03
= -4000m/s²→Acceleration
Or it can be said that the bullet is retarding at 4000m/s²
b) Force=mass × acceleration
Force exerted by wall on bullet= 0.02kg×4000m/s²
=80 Newton
c)Answer will be the same...80 Newton. Every action has an equal and opposite reaction. Just the direction will be opposite.
d) Distance covered = speed×time taken
=120m/s×0.03 seconds
=3.6 m
Hence, the dist. covered by the bullet =3.6m
Hope it helps :)
Answered by
0
Answer:
d=speed*time
120 *0.03=3.6
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