The answer is C. Potential energy is when an object is at rest.
Kinetic is when an object is in motion.
Answer : Cardio training
Explanation : Cardio exercise helps improve lung function , strengthens your heart and improve your endurance and increases lung capacity. The peoples do cardio is burn calories and lose the weight.
Cardio helps decrease your heart rate and blood pressure and improve your breathing.
A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.
<h3>Why does the first ball comes to rest after collision ?</h3>
Let m be the mass of the two identical balls.
u1 = velocity before the collision of ball 1
u2 = 0 = velocity of second ball that is at rest
v1 and v2 are the velocities of the balls after the collision.
From the conservation of momentum,
∴ mu1 + mu2 = mv1 + mv2
∴ mu1 = mv1 + mv2
∴ u1 = v1 + v2
In an elastic collision, the kinetic energy of the system before and after collision remains same.

∴ 
∴ 
∴
₁
₂ = 0
- It is impossible for the mass to be zero.
- Because the second ball moves, velocity v2 cannot be zero.
- As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>
An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.
Can learn more about elastic collision from brainly.com/question/12644900
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Answer:0.000002
Explanation: I Looked It Up lol
Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''


Vy in this problem will follow this equation =

where g is the gravity acceleration

This is equation (1)
For Y(t) :

We suppose yi = 0

This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s

The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m