As a small bullet leaves a gun, a net force is applied to shoot it through a barrel. if the bullet has a mass of 34.67 grams and leaves the barrel at a speed of 249.7 m/s. if the barrel of the gun is only 13 cm, what force is applied to the bullet as it leaves the gun?

F = 333.1 N

Explanation:

Force: This is defined as the product of mass of a body and its acceleration. The S.I unit of force is Newton (N).

From Newton's equation of motion.

v² = u² + 2as Equation 1

Where v = final velocity of the bullet, u = initial velocity of the bullet, s = length of the gun, a = acceleration of the bullet

Making a the subject of formula in the equation above

a = (v² - u²)/2s Equation 2

Given: v = 249.7 m/s, u = 0 m/s ( The bullet was at rest before the gun was shot), s = 13 cm = (13/100) m = 0.13 m.

Substituting these values into equation 2

a = (2497.7-0)/(2×0.13)

a = 2497.7/0.26

a = 9606.54 m/s²

Force = Mass × acceleration

F = M a Equation 3

Where M = mass of the bullet, F = force applied to the bullet

Given: m = 34.67 g = (34.67/1000) = 0.03467 kg, a = 9606.54 m/s²

Substituting these values into Equation 3

F = 0.03467 × 9606.54

F = 333.1 N

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force