Answer:
a. The magnitude of the tension in the string is greater than the magnitude of the weight of the rock.
Explanation:
During the motion of the rock while it is in downward motion we can say

since it is performing circular motion so we will have its acceleration towards its center



So at the lowest point of the path we can say

so correct answer is
a. The magnitude of the tension in the string is greater than the magnitude of the weight of the rock.
Answer:
The kinetic energy is 
Explanation:
From the question we are told that
The radius of the orbit is 
The gravitational force is 
The kinetic energy of the satellite is mathematically represented as

where v is the speed of the satellite which is mathematically represented as

=> 
substituting this into the equation

Now the gravitational force of the planet is mathematically represented as

Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
![KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A%5B%5Cfrac%7BGMm%7D%7Br%5E2%7D%5D%20%2A%20r)
=> 
substituting values


Answer:
Hey
I am sure from what i know that your answer should be the following:
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Answer:
I need this for may schooling