Question

A 0.025-kilogram bullet is fired from a rifle by an unbalanced force of 200. newtons. if the force acts on the bullet for 0.1 second, what is the maximum speed attained by the bullet?

Answer 1

Answer: 800m/s

Explanation: First, calculate the acceleration of the bullet by making use of the second law of motion by Newton which can be translated in equation as,

F = m x a

Solving for a,

a = F/m

Substituting the known values,

a = (200 N)/(0.025 kg) = 8000 m/s²

The maximum velocity of the bullet is the product of the acceleration and the given time.

v = (8000 m/s²)(0.1 s) = 800 m/s

Answer 2

First, calculate the acceleration of the bullet by making use of the second law of motion by Newton which can be translated in equation as,

F = m x a

Solving for a,
a = F/m

Substituting the known values,
a = (200 N)/(0.025 kg) = 8000 m/s²

The maximum velocity of the bullet is the product of the acceleration and the given time.

v = (8000 m/s²)(0.1 s) = 800 m/s

Answer: 800 m/s