Answer:
A larger impulse. A 1-kg ball has twice as much speed as a 10-kg ball.
Explanation:
Like windmills they use the winds to generate their power.
Answer:
Angular acceleration of the disk will be 
Explanation:
We have given mass of the disk m = 5 kg
Diameter of the disk d = 30 cm = 0.3 m
So radius 
Moment of inertia of disk is given by 
Force is given by F=4 N
Torque is given as 
We also know that torque is given by 


Answer:

Explanation:
The electrostatic potential energy is given by the following formula

Now, we will apply this formula to both cases:

So, the change in the potential energy is

My science teacher said that there is no equivalent unit to a newton