Answer:
Explanation:
Given
acceleration is given by

where 

Also acceleration is given by








at 





when air drag is neglected maximum height reached is


Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is 
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
- the peripheral velocity that is directed downward
along the y-axis
- the linear velocity
that is directed along the x-axis
Now;


Also,

where
(angular velocity) = 

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is 
P.E = mgh
This is the formula for potential energy.
This is where m is mass, g is the acceleration due to gravity, and h is height.
All you have to do is multiply all these numbers together.
Answer:
D. ) The force would be zero newtons
Explanation:
Because
If you are at the center of the earth, gravity is zero because all the mass around you is pulling "up" (every direction there is up!)
So F=mg so if g is zero F is also zero
Answer:
Explanation:
Explanation: total displacement =3√2m. and total distance covered=14m. I hope this is right and helps u.