Question

The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in

Answer 1

Since g is constant,  the force the escaping gas exerts on the rocket will be 10.4 N

What is Escape Velocity ?

This is the minimum velocity required for an object to just escape the gravitational influence of an astronomical body.

Given that the velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in 0.60s. The gravitational field intensity is 9.8N/kg.

To calculate the force the escaping gas exerts of the rocket, let first highlight all the given parameters

• Mass (m) of the rocket 0.25 Kg

• Initial velocity u = 15 m/s

• Final Velocity v = 40 m/s

• Time t = 0.6s

• Gravitational field intensity g = 9.8N/kg

The force the gas exerts of the rocket = The force on the rocket

The rate change in momentum of the rocket = force applied

F = ma

F = m(v - u)/t

F = 0.25 x (40 - 15)/0.6

F = 0.25 x 41.667

F = 10.42 N

Since g is constant,  the force the escaping gas exerts on the rocket is therefore 10.4 N approximately.

Learn more about Escape Velocity here: brainly.com/question/13726115

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