Answer:
muscular dystrophy and myasthenia gravis
Answer:

Explanation:
Given that:
- mass of meteoroid,

- radial distance from the center of the planet,

- mass of the planet,

<u>For gravitational potential energy we have:</u>

substituting the respective values:


Answer:
Potential energy
Explanation:
Before release, the catapult has potential energy stored in a tension of torsion device in it. Normally a flexible bow like object that could be made of wood or of metal.
Answer:
A body becomes weightless in a zero-gravity scenario and when a force is applied to a body that is equal and opposite to the force of gravity.
- Coefficient of static friction = 0.5
- Coefficient of Kinetic friction = 0.3
- Angular velocity = 500 RPMs
<h3>The Radius of the System</h3>
Let R be the radius of cylinder

The angular velocity is 500 RPMs

The normal force

Since the radius is very little for two block to execute circular motion so system will slide down.
Learn more on coefficient of static friction here;
brainly.com/question/11841776
brainly.com/question/25772665
brainly.com/question/26400616