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kenny6666 [7]
3 years ago
11

A uniform disk of mass m = 2.4 kg and radius R = 25 cm can rotate about an axle through its center. Four forces are acting on it

as shown in the figure. Their magnitudes are F1 = 8.5 N, F2 = 1.5 N, F3 = 6.5 N and F4 = 6.5 N. F2 and F4 act a distance d = 3.5 cm from the center of mass. These forces are all in the plane of the disk. Write an expression of the magnitude of the torque due to the force F3

Physics
1 answer:
Zinaida [17]3 years ago
6 0

Answer:

0 Nm

Explanation:

Torque is the cross product of the radius vector and the force vector.

τ = r × F

In other words, the magnitude of the torque is equal to the magnitude of the radius times the magnitude of the force times the sine of the angle between them.

τ = rF sin θ

Since F₃ is parallel to the radius vector, θ = 0, so τ = 0.

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Answer:

Hi

Final temperature = 250.11 °C

Final volume = 0,1 m3.

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Explanation:

The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.

This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:

v=0.05 m3/kg

v1=0.057061 m3/kg

T1=242.56°C

v2=0.049779 m3/kg

T2=250.35°C

T=\frac{T2-T1}{v2-v1} x(v-v1)+T1=\frac{250.35°C-242.56°C}{0.049779m3/kg-0.057061m3/kg}x(0.05m3/kg-0.057061m3/kg)+242.56°C=250.11°C

The process work is zero because there is no change in volume during heating:

W=PxΔv=Px0=0

where

W=process work

P=pressure

Δv=change of volume, is zero because the piston was blocked so the volume remains constant.

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517.5Ns

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Statement 1 and 3 are correct.

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