Acannonball with a mass of 1.0 kilogram is fired horizontally from a 500.- kilogram cannon, initially at rest, on a horizontal, frictionless surface. the cannonball is acted on by an average force of 8.0x10^3 newtons for 1.0x10^-1 second. (a) what is the magnitude of the average net force acting on the cannon? (b) determine the acceleration of the cannon ball.

(a) According to Newton’s third law, for every action there is an equal yet opposite reaction. In this case, the cannonball is affected by 8.0x10^3 newtons in one direction. Now the cannon must also be affected by the same amount of net force on the opposite direction.

- 8.0x10^3 Newtons (the negative symbol only shows that it is opposite direction)

(b) We solve this using the formula:

F = m a

8.0x10^3 N = (1.0 kg) a

a = 8,000 m/s^2

In this exercise we will use the knowledge of Newton's laws to be able to answer about the force, in this way:

A) opposite direction to the movement.

B) a = 8,000 m/s^2

So, recalling Newton's concept of force, we have:

One newton corresponds to the force exerted on a body of mass equal to 1 kg that induces an acceleration of 1 m/s² in the same direction as the force.

A)The magnitude of a force is linked to its direction, so we will have that force is in the opposite direction to the movement.

B) We solve this using the formula:

See more about force at link

neither accurate nor precise.