Answer:
9:00 AM
Explanation:
I took the test and that was the answer
Answer:
The average acceleration of the bearings is 
Explanation:
Given that,
Height = 1.94 m
Bounced height = 1.48 m
Time interval 
Velocity of the ball bearing just before hitting the steel plate
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



Negative as it is directed downwards
After bounce back,
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



We need to calculate the average acceleration of the bearings while they are in contact with the plate
Using formula of acceleration

Put the value into the formula



Hence,The average acceleration of the bearings is 
Answer:
The power will be "3.92×10⁹ Watts". A further explanation is given below.
Explanation:
The given values as per the question,
Rate,
= 8 million kg
Distance,
= 50 m
Gravity,
= 9.8 m/s²
As we know,
The power will be:
⇒ 
On putting the values, we get
⇒ 
⇒ 
Answer:
The potential energy increases and the kinetic energy decreases
Answer:
The answer is 4 pounds
Explanation:
The explanation is that 1 kilogram is equal to 2 pounds so multiply the kilogram with the 1 pound