Answer:
h = 16.67m
Explanation:
If the kinetic energy of the cylinder is 510J:


Where the inertia is given by:

Replacing this value:

Speed of the block will therefore be:

By conservation of energy:
Eo = Ef
Eo = 0

So,

Solving for h we get:
h=16.67m
Fg=m•g || IE: Weight = mass x gravity
Therefore, the relationship are as follows:
mass and gravity are inversely proportional
mass and weight are directly proportional
weight and gravity are directly proportional
On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons
BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N
Explanation:
The given data is as follows.
m = 5000 kg, h = 800 km = 
, r = R + h = 
kg, G = 
As we know that,

v = 
And, it is known that formula to calculate angular velocity is as follows.

v = 
= 
= 
Thus, we can conclude that speed of the satellite is
.
I believe this would be an example of Mary's velocity. We have her speed and direction which is all you need to find velocity.