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).
Given:-
- Mass of the cart (m) = 35 kg
- Speed (consider Velocity) = 1.2 m/s
To Find: Momentum of the cart.
We know,
p = mv
where,
- p = Momentum,
- m = Mass &
- v = Velocity.
Thus,
p = (35 kg)(1.2 m/s)
→ p = 42 kg m/s (Ans.)
Conclusion:-
A. ☑️ 42 kilogram - metre per second.
When you touch<span> a doorknob (or something else made of metal), which has a positive charge with few electrons.</span>
Given:
F = 39 N, the force applied
t = 2 s, the time interval in which the force is applied.
By definition, the impulse is

Answer: 78 N-s