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alexdok [17]
3 years ago
13

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a ra

dius of 0.25 m. The other end of the rope is attached to a massless suspended platform, upon which 0.5 kg weights may be placed. While the system is initially in equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling the rope. What is the torque on the pulley when the system is motionless?
Physics
1 answer:
worty [1.4K]3 years ago
4 0

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, m_r = 0.5 kg

The mass of the pulley, m_p = 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.

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What is the correct order for the Lunar eclipse
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Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water. If the cylinder
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Answer:

0.10839 m

Explanation:

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P = Total pressure at bottom of mecury = 1.2 atm

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h = d = Depth of mercury

\rho_m = Density of mercury = 1.36\times 10^4\ kg/m^3

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Pressure at the bottom is of the cylinder is given by

P_2=P_1+\rho_wgh\\\Rightarrow P_2=101325+1000\times 9.81(0.7-d)

Pressure at the bottom of mercury is

P=P_2+\rho_mgh\\\Rightarrow 1.2\times 101325=101325+1000\times 9.81(0.7-d)+1.36\times 10^4\times 9.81\times d\\\Rightarrow 1.2\times 101325=123606d+108192\\\Rightarrow d=\dfrac{1.2\times 101325-108192}{123606}\\\Rightarrow d=0.10839\ m

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It is given that,

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Magnetic field, B=8 \times 10^{-5}\ T ( in north )

Using the relation as :

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