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LUCKY_DIMON [66]
3 years ago
10

You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe

re, there is a strong electric field surrounding your device. Other researchers in your laboratory are complaining that your electric field is affecting their equipment. You think about how you can obtain the large electric field that you need close to the sphere but prohibit the field from reaching your colleagues. You decide to surround your device with a spherical transparent plastic shell. The nonconducting shell is given a uniform charge distribution. Required:a. The shell is placed so that the small sphere is at the exact center of the shell. Determine the charge that must be placed on the shell to completely eliminate the electric field outside of the shell. b. What if the shell moves? Does the small sphere have to be at the center of the shell for this scheme to work?
Physics
1 answer:
madam [21]3 years ago
5 0

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

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Answer:

  \sqrt{\dfrac{2g\lambda^3}{m}}

Explanation:

We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.

  \boxed{\sqrt{\dfrac{2g\lambda^3}{m}}}

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3 years ago
A 25 kg mass is hanging from two cables, each with their own tension. Cable 1 is connected to the
Georgia [21]

Answer:

a. one line down one line to the right one live to the northwest from the object

b. t1=190 t2=310

Explanation:

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2 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a
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Answer:

The x component of the resultant force is -7.27N.

Explanation:

To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:

|F_{1x}|=|F_1|\cos\theta_1=9.20N\cos62.0\°=4.31N \\|F_{2x}|=|F_2|\cos\theta_2=5.00N\cos53.6\°=2.96N

Since both vectors are in the left side of the y-axis, they have a negative x component. So:

F_{1x}=-4.31N;\\F_{2x}=-2.96N

Finally, we sum both components to obtain the component of the resultant force:

F_{Rx}=-4.31N-2.96N=-7.27N

In words, the x component of the resultant force is -7.27N.

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3 years ago
What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a
viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

 COP = 8.49

b

  COP_1 = 9.49  

Explanation:

From the question we are told that

     The lower operation temperature of refrigerator is  T_1 =  -8.00^oC =  265 \  K

     The upper operation temperature of the refrigerator is   T_2 =  23.2 ^oC =  296.2 \  K

Generally the refrigerators coefficient of performance is mathematically represented as

        COP =  \frac{T_1}{T_2 - T_1  }

=>     COP =  \frac{265}{296.2 - 265  }

=>     COP = 8.49

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

            COP_1 =  \frac{T_2}{ T_2 - T_1}  

=>         COP_1 =  \frac{296.2}{ 296.2 - 265 }  

=>         COP_1 = 9.49  

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qaws [65]
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