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marta [7]
3 years ago
8

An uncharged, nonconducting, hollow sphere of

Physics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

Explanation:

The whole surface of hollow sphere = 4π r²

= 4 x 3.14 x (10 x 10⁻²)²

= 12.56 x 10⁻² m²

Area of the hole ( both side ) = 2 x π r²

= 2 x 3.14 x (10⁻³)²

= 6.28 x 10⁻⁶ m²

flux coming out of given charge at the centre as per Gauss's theorem

= q / ε₀ where q is charge at the centre and  ε₀ is permittivity of the medium .

= 10 x 10⁻⁶ / 8.85 x 10⁻¹²

= 1.13 x 10⁶

This flux will pass through the surface of sphere so flux passing through per unit area

= 1.13 x 10⁶ / 12.56 x 10⁻²

= 8.99 x 10⁶ weber per m²

flux through area of hole

=  8.99 x 10⁶ x 6.28 x 10⁻⁶

= 56.45 weber .

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Luba_88 [7]

Hey girl

the answer is A

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3 years ago
a sample of iron has the dimensions of 2cm times 3cm times 2cm. if the mass of this rectangular-shaped object is 94g, what is th
Anarel [89]
The answer is 7.83. The three is repeating.

5 0
3 years ago
What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co
MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

4 0
3 years ago
Jenna rides her scooter at a constant speed for 6 km. That part of her ride takes her 1 h. She then
AfilCa [17]

Answer:

5km/h

Explanation:

The equation for average speed is

Vavg = Δd/Δt

where Δd is distance and Δt is time

from what's given, we know she travelled 15km in total and the trip took 3 hours in total. We can now plug in the values into the equation

Vavg = 15/3

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4 0
3 years ago
If a galaxy has an apparent radial velocity of 2000 km/s and the Hubble constant is 70 km/s/Mpc, how far away is the galaxy
ddd [48]

Answer:

28.57 Mpc

Explanation:

This question is going to be solved by applying Hubble's Law.

This Hubble's Law is actually an observation in physical cosmology. This observation makes it clear that galaxies are moving away from the Earth, and are doing so at speeds proportional to their distance. This essentially means that the farther they are from the Earth, the faster they are moving away from Earth.

It is represented by this formula

v = H(0)D, where

v = speed

H(0) = Constant of proportionality, or otherwise, Hubble's constant.

D = Distance to a galaxy

Applying the given parameters to the formula, we have

v = H(0).D

D = v / H(0)

D = 2000 / 70

D = 28.57 Mpc

3 0
3 years ago
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