Since g is constant, the force the escaping gas exerts on the rocket will be 10.4 N
<h3>
What is Escape Velocity ?</h3>
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
- 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
#SPJ1
You would want to increase the temperature.
Answer:
5 .07 s .
Explanation:
The child will move on a circle of radius r
r = 1.5 m
Let the velocity of rotation = v
radial acceleration = v² / r
v² / r = 2.3
v² = 2.3 r = 2.3 x 1.5
= 3.45
v = 1.857 m /s
Time of revolution = 2π r / v
= 2 x 3.14 x 1.5 / 1.857
= 5 .07 s .
The rock is kicked horizontally off the cliff at 20 m/s.
There's no air resistance, and gravity has no effect on horizontal motion. There's no horizontal force acting on the rock to make it move horizontally slower or faster than 20 m/s, so it keeps moving horizontally at 20 m/s.
It's in the air for 7 seconds before it hits the ground. Moving horizontally at 20 m/s for 7 seconds, it sails (20 x 7) = 140 meters horizontally away from the cliff.
