<span>Variations in Earth-Sun orbital relationships.</span>
Answer: velocity = -0.65 speed =0.65
Explanation:
Velocity =speed+direction speed =distance/time
Answer:
Acceleration, 
Explanation:
Given that,
Height from a ball falls the ground, h = 17.3 m
It is in contact with the ground for 24.0 ms before stopping.
We need to find the average acceleration the ball during the time it is in contact with the ground.
Firstly, find the velocity when it reached the ground. So,

u = initial velocity=0 m/s
a = acceleration=g

It is in negative direction, u = -18.41 m/s
Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

So, the average acceleration of the ball during the time it is in contact is
.
I = MR^2
The Attempt at a Solution:::
I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2
I total = 3ML^2/2
It says the answer is 3ML^2/4 though.
⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔
mark it as brainliest.... ✌✌✌