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Anna71 [15]
3 years ago
12

A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

ing contact on the rim of the disk. A uniform magnetic field B= 1.00 T is parallel to the axle of the disk. When the current is 3.0 A, the disk rotates with constant angular velocity. What's the frictional force at the rim between the stationary electrical contact and the rotating rim?
Physics
1 answer:
Galina-37 [17]3 years ago
7 0

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm=\frac{6}{100}=0.06 m

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

dF=IBdr

dF=IBdr

\tau=rdF=IBrdr

\tau=\int_{0}^{R}IBr dr

\tau=IB(\frac{R^2}{2}

Torque due to friction

\tau=R\times F

Where friction force=F

R\times F=\frac{IBR^2}{2}

F=\frac{IBR}{2}

Substitute the values

F=\frac{3\times 1\times 0.06}{2}

F=0.09 N

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Req = R1 + R2 + R3 = 10.0Ω + 10.0Ω + 10.0Ω = 30.0Ω

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Answer: -1038.8 kJ

Explanation:

From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.

from the question we were given

Mass of nitrogen, m = 7kg

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Polytropic index, n = 1.4

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