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chubhunter [2.5K]
3 years ago
14

A metallic circular plate with radius rr is fixed to a tabletop. An identical circular plate supported from above by a cable is

fixed in place a distance dd above the first plate. Assume that dd is much smaller than r.r. The two plates are attached by wires to a battery that supplies voltage V.
Required:
a. What is the tension in the cable?
b. The upper plate is slowly raised to a new height 2d. Determine the work done by the cable by integrating ∫d to 2d F(z)dz, where F(z) is the cable tension when the plates are separated by a distance z.
Express your answer in terms of the variables d, r, V, and constants ϵ0,pi.
W=

c. Compute the energy stored in the electric field before the top plate was raised.
d. Compute the energy stored in the electric field after the top plate was raised.
e. Is the work done by the cable equal to the change in the stored electrical energy? If not, why not?
Physics
1 answer:
vova2212 [387]3 years ago
7 0
The tension cable is in relation to the amount of pressure placed on the upper plate
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Answer:

1) \theta=120^{\circ} from the positive x-axis.

2) t=20\ min

Explanation:

Given:

speed of rowing in still water, v=4\ mph

1)

speed of water stream, v_s=2\ mph

we know that the direction of resultant of the two vectors is given by:

tan\ \beta=\frac{v.sin\ \theta}{v_s+v.cos\ \theta}

where:

\beta=the angle of resultant vector from the positive x-axis.

\theta = angle between the given vectors

When the rower wants to reach at the opposite end then:

\beta =90^{\circ}

so,

tan\ 90^{\circ}=\frac{v.sin\ \theta}{v_s+v.cos\ \theta}

\Rightarrow v_s+v.cos\ \theta=0

2+4\times cos\ \theta=0

cos\ \theta=-\frac{1}{2}

\theta=120^{\circ} from the positive x-axis.

2)

Now the resultant velocity of rowing in the stream:

v_r=\sqrt{v^2+v_s^2+2\times v.v_s.cos\ \theta}

v_r=\sqrt{4^2+2^2+2\times 4\times 2\times cos\ 120}

v_r=12\ mph

Therefore time taken to cross a 4 miles wide river:

t=\frac{4}{12}

t=\frac{1}{3}\ hr

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An orbiting meteor has a mass of 5,123,788 g. What is the mass of the meteor in kg?
gavmur [86]

Answer:

5123.79 kg

Explaination:

I did 5,123,788 divided by 1000 and got 5123.788

Then I rounded to the nearest 10th place.

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The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a vel
Brilliant_brown [7]

Answer:

Explanation:

We shall write the velocities given in vector form to make the solution easy.

The velocity of water with respect to earth that is waV(e) makes 30 degree with north or 60 degree with east so in vector form

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Similarly , velocity of wind with respect to earth that is wiV(e) , is making 50 degree with west or - ve of x axes so we cal write it in vector form as follows

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Now we have to calculate velocity of wind with respect to water that is

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Magnitude of this relative velocity

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