Answer:
both
Explanation:
and I'm serious he is literally both
Answer: D. Energy released from atoms.
So, the work was done by that hot air-balloon is <u>30,000 J or 30 kJ</u>.
<h3>Introduction</h3>
Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

With the following condition :
- W = work (J)
- F = force (N)
- s = shift or displacement (m)
Now, the s (displacement) can be written as ∆h (altitude change) because the object move to vertical line. The formula can also be changed to:

With the following condition :
- W = work (J)
- F = force (N)
= change of altitude (m)
If an object has mass, then the object will also be affected by gravity. Always remember that F = m × g. So that :


With the following condition :
- W = work (J)
- m = mass of the object (kg)
- g = acceleration of the gravity (m/s²)
= change of altitude (m)
<h3>Problem Solving</h3>
We know that :
- F = force = 100 N
= change of altitude 300 m
What was asked :
Step by step :



<h3>Conclusion</h3>
So, the work was done by that hot air-balloon is 30,000 J or 30 kJ.
<h3>See More :</h3>
In empty space probably means, there is no force on the ball.
(This assumption is not quite correct since there is still the force of gravity between the ball and the astronaut, but this force is very very small and can be neglected.)
Assuming there is no force on the ball, Newtown's 1st law says: When viewed in an internal frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.
This means:
If there is no force on the ball, there will be no acceleration on the ball either.
If the acceleration is zero, the velocity of the ball never changes.
Friction , as the angle gets huger and higher , this is less and less normal force into the inclined plane .