If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
Answer:
Δe=0.578 kJ/kg
Explanation:
Given data
Velocity v₁=0 m/s
Velocity v₂=34 m/s
to find
Specific energy change Δe
Solution
The specific energy change is simply determined from change in velocity
Δe=(v₂²-v₁²)/2
Put the given values to find the specific energy change

Δe=0.578 kJ/kg
At the tip of either of the magnets poles
Answer:
Set for what? did U finish the question?
Answer:
The average velocity from 1 to 3 seconds is 20 m/s l. X/T = 0 m/1s = zero.
Explanation:
I hope that helped!