Answer:v=3.28 m/s
Explanation:
Given
mass of rock 
diameter of circle 
radius 
At highest Point

At highest Point N=0 because mass is just balanced by centripetal Force
thus 




Answer:
The change in internal energy of the system is -17746.78 J
Explanation:
Given that,
Pressure 
Remove heat 
Radius = 0.272 m
Distance d = 0.163 m
We need to calculate the internal energy
Using thermodynamics first equation
...(I)
Where, dU = internal energy
Q = heat
W = work done
Put the value of W in equation (I)

Where, W = PdV
Put the value in the equation


Hence, The change in internal energy of the system is -17746.78 J
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Answer:
there is friction between the two things
Explanation: