Answer:
289.05N
Step-by-step explanation:
The mass of the bullet is 4.2-g bullet, if we convert it to "kg" we have
4.2-g bullet/1000
= 0.0042kg
The speed of the bullet leaving the missle is 338 m/s.
The Lenght of the barrel= 0.83-m-long
From kinematics, v²= u² +2al
V= final speed when it leaves the barell=
338 m/s
U= intial speed=0
L= length of the barrel travelled by the bullet
V²= 2al( initial speed=0)
If we make "a" subject of the formula we have
a=v²/2l
a= (338)²/(2×0.83)
=68821.687 m/s²
We can now calculate the force acting on the body using the Expression below
F=ma
Where F= force acting on the body
m= mass
a= Acceleration
F= 0.0042kg×68821.687
= 289.05N
Therefore, the force exerted on the bullet while it is traveling down the 0.83-m-long barrel of the rifle is 289.05N