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zheka24 [161]
3 years ago
5

A point charge of magnitude q is at the center of a cube with sides of length L.a) What is the electric flux through each of the

six faces of the cube?b) What would be the flux through a face of the cube if its sides were of length L1?
Physics
1 answer:
agasfer [191]3 years ago
3 0
<h2>The flux through each face is q/6ε₀ .</h2>

Explanation:

The charge q is placed at the center of the cube of side L

According to Gauss's law the flux through any closed surface is q/ε₀

here q is the charge enclosed .

In this case cube has the six faces . The flux through each face = q/6ε₀

In the second case The cube has the face with length L₁

The flux through each face = q/6ε₀

Thus flux through the cube does not depend upon the size of the cube .

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A planet similar to the Earth has a radius 7.4 × 106 m and has an acceleration of gravity of 10 m/s 2 on the planet’s surface. T
Lubov Fominskaja [6]

Answer:

1.5 hr

16.7

Explanation:

Zero apparent weight means there's no normal force.

Sum the forces in the centripetal direction.

∑F = ma

mg = mv²/r

v = √(gr)

v = √(7.4×10⁶ m × 10 m/s²)

v = 8602 m/s

The circumference of the equator is:

C = 2πr

C = 2π (7.4×10⁶ m)

C = 4.65×10⁷ m

So the period is:

T = C / v

T = (4.65×10⁷ m) / (8602 m/s)

T = 5405 s

T = 1.5 hr

The initial speed is:

v = C / T

v = (4.65×10⁷ m) / (25 h × 3600 s/h)

v = 517 m/s

The speed increases by a factor of:

8602 m/s / 517 m/s

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3 0
3 years ago
A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the
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Answer:

47.8rad/s

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For energy to be conserved.

The potential energy sustain by the object would be equal to K.E

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Now K.E = 1/2 × I × (w1^2 - w0^2)

I = 2/3 × M × R2

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W1 = final angular velocity

Wo = initial angular velocity

From P.E = K.E we have;

68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)

(w1^2 - w0^2) = 1948.09

W1^2 = 1948.09 + (18.3^2)

W1^2=2282.98

W1 = √2282.98

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= 47.8rad/s to 1 decimal place.

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326.85 degrees Celsius
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8 0
3 years ago
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