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ELEN [110]
3 years ago
5

A hollow sphere of radius 1.88 m is in a region where the electric field is radial and directed toward the center of the sphere.

If the magnitude of the field at the surface of the sphere is 22.6 N/C, what is the net electric flux through the spherical surface?
Physics
1 answer:
Marta_Voda [28]3 years ago
8 0

Answer:

1003.9 Nm²/C

Explanation:

Parameters given:

Radius of sphere, r = 1.88 m

Electric field at surface of sphere, E = 22.6 N/C

Electric Flux, Φ, is given as:

Φ = E * A

Where A = Surface area

Surface Area of sphere = 4 * pi * r²

= 4 * pi * 1.88²

= 44.4 m²

Φ = 22.6 * 44.4

Φ = 1003.9 Nm²/C

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  • Resistance=R=210Ohm
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Current=I

\boxed{\sf P=I^2R}

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\\ \sf\longmapsto I^2=\dfrac{9.28}{210}

\\ \sf\longmapsto I^2\approx0.04

\\ \sf\longmapsto I\approx\sqrt{0.04}

\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}

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\\ \sf\longmapsto I\approx\dfrac{2}{10}

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3 years ago
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Answer:

271.862 N/m

Explanation:

From Hook's Law,

mgh = 1/2ke²............... Equation 1

Where

m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.

Making k the subject of the equation,

k =2mgh/k²....................... Equation 2

Note: The potential energy of the ball is equal to the elastic potential energy of the spring.

Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m

Substitute into equation 2

k = 2(0.0603)(9.8)(0.537)/0.048317²

k = 0.6346696/0.0023345

k = 271.862 N/m

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Answer:

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Answer:

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Hope this helps! brainliest welcomed! :)

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3 years ago
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