The complete question is missing, so i have attached the complete question.
Answer:
A) FBD is attached.
B) The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
Explanation:
A) I've attached the image of the free body diagram.
B) The formula for the net force is given as;
F_net = mv²/r
We know that angular velocity;ω = v/r
Thus;
F_net = mω²r
Now, the minimum downward force is the weight and so;
mg = m(ω_min)²r
m will cancel out to give;
g = (ω_min)²r
(ω_min)² = g/r
ω_min = √(g/r)
The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
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Explanation:
Let acceleration due to Gravity for a planet is given by:
Here,
Escape velocity is given by:
Here, 
and g_X = 2g
Therefore,
I believe that the answer is B. 133 N
Answer:
The magnitude of the electric field is 5.75 N/C towards positive x- axis.
Explanation:
Given that,
Point charge at origin = 2 nC
Second charge = 5 nC
Distance at x axis = 8 m
We need to calculate the electric field at the point x = 2 m
Using formula of electric field
Put the value into the formula
The direction is toward positive x- axis.
Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.
