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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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The wavelength is 1. 87 x 10-7 m. What is the frequency?.
jeka57 [31]

Answer:

11.7 im pretty sure. i did the work so it should be correct

3 0
2 years ago
A stretched string is 2.11 m long and has a mass of 19.5 g. When the string oscillates at 440 Hz , which is the frequency of the
blsea [12.9K]

Answer:

The tension of the string is 41.876 N

Explanation:

Given;

length of the string, L = 2.11 m

mass of the string, m = 19.5 g = 0.0195 kg

frequency of the wave, f = 440 Hz

wavelength, λ = 15.3 cm = 0.153 m

The velocity of the wave is given by;

v = fλ

v = 440 x 0.153

v = 67.32 m/s

Also the velocity of the wave is given by

v = \sqrt{\frac{T}{\mu} }

where;

μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m

T is the tension of the string

T = v²μ

T = (67.32)²(0.00924)

T = 41.876 N

Therefore, the tension of the string is 41.876 N

5 0
4 years ago
Could someone help me on #7?
ozzi

i would sat C then the impact is weaker

8 0
3 years ago
An 18-cm-long bicycle crank arm, with a pedal at one end, is attached to a 20-cm-diameter sprocket, the toothed disk around whic
Tpy6a [65]

Answer:

Part a)

a = 0.056 m/s^2

Part b)

L = 7.85 m

Explanation:

Part a)

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so the angular acceleration is given as

\alpha = \frac{\omega_2 - \omega_1}{\Delta t}

so we will have

\alpha = \frac{2\pi(\frac{90}{60}) - 2\pi(\frac{60}{60})}{10}

\alpha = 0.314 rad/s^2

now the tangential acceleration of the pedal is given as

a = r \alpha

a = 0.18 \times 0.314

a = 0.056 m/s^2

Part b)

Total angular displacement made by the sprocket in the interval of 10 s is given as

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{2\pi (\frac{90}{60}) + 2\pi (\frac{60}{60})}{2}(10)

\theta = 78.5 radian

now length of the chain passing over it is given as

L = R\theta

L = 0.10 \times 78.5

L = 7.85 m

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4 years ago
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