An object dropped from Cliff falls with a constant acceleration of 10 metre per second square find its Speed of two seconds afte
2 answers:
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Answer:
The torque on the pulley, when the system is motionless is approximately 9.81 N·m
Explanation:
The given parameters are;
The mass of the object = 2 kg
The friction between the rope and the pulley = 0
The mass of the rope, = 0.5 kg
The mass of the pulley, = 0.01 kg
The radius of the pulley, r = 0.25 m
The torque on the pulley, τ = I·α = F × D
The torque on the pulley, when the system is motionless, τ = F × D
Where;
F = The force acting on the pulley rope = The weight of the mass ≈ 2 kg × 9.81 m/s² = 19.62 N
D = The diameter of the pulley = 2×r = 2 × 0.25 m = 0.5 m
Therefore;
τ = 19.62 N × 0.5 m = 9.81 N·m
The torque on the pulley, when the system is motionless, τ ≈ 9.81 N·m.
Answer:
Explanation:
Given
radius of electron
centripetal acceleration
we know
(b)For n=10