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?

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