A different bullet has a mass of 0.09 kg. Starting from rest, after its gun's trigger is pulled, a constant force acts on the bullet for the next 0.025 seconds until the bullet leaves the barrel of the gun with a speed of 1,346 m/s. What force acts on this bullet?

The force acts on this bullet : 4.8456 N

Given

m=0.09 kg

Δt=0.025 s

vo=0(from rest)

vt=1.346 m/s

Required

Force

Solution

Impulse is a change in momentum

I=ΔP

F.Δt=m(vt-vo)

Input the value

F x 0.025 = 0.09(1.346-0)

F=4.8456 N

0.863333

86%

hope this

a and b sometimes have a positive value