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:
For a linear elastic material Young's modulus is a constant that is given by:
Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, is the amount by which the length of the object changes and
is the original length of the object. In this case the force is the weight of the mass:
Replacing the given values in Young's modulus formula:
The correct answer is:
Work is negative, the environment did work on the object, and the energy of the system decreases.
In fact, the work-energy theorem states that the work done by the system is equal to its variation of kinetic energy:
In this problem, the variation of kinetic energy is negative (because the final velocity is less than the initial velocity), so the work is negative, and this means that the environment did work on the object, and its energy decreased.
Answer:
s
Explanation:
From the question we are told that
The outer ring with a radius of 30 m
inner Gravity Approximately 9.80 m/s'
Outer Gravity Approximately 5.35 m/s.
Generally the equation for centripetal force is given mathematically as
Centripetal acceleration enables Rotation therefore?
Considering the outer ring,
Therefore solving for Period T
Generally the equation for solving Period T is mathematically given as
s