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OverLord2011 [107]
1 year ago
15

10 C of charge are placed on a spherical conducting shell. A point particle with a charge of –3C is placed at the center of the

cavity. The net charges, in coulombs, on the inner and outer surfaces of the shell, is:
a. -7
b. +7
c. 0
d. +13
e. +6.5
Physics
1 answer:
fenix001 [56]1 year ago
6 0

Answer:

The net charges on inner surface is +3 C

The net charges on outer surface is +7 C

Explanation:

Given that,

Charge on spherical conducting shell = 10 C

Charge at center = -3 C

We know that,

The net electric field inside the conducting shell is zero. The Gaussian surface inside the conductor must have zero net charge.

We need to calculate the net charges on inner surface

The charge on the inner surface of the conductor must be equal and opposite to the present charge of the  center.

So, The net charges on inner surface is +3 C.

We need to calculate the net charges on outer surface

Using formula of net charge

Q'=Q+q

Put the value into the formula

Q'=10-3

Q'=+7C

Hence, The net charges on inner surface is +3 C

The net charges on outer surface is +7 C

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Answer:

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Explanation:

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m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R (5.32\,10^5\,km=5.32\,10^8\,m, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference (2\,\pi\,R) in 388800 seconds, therefore its tangential velocity is:

v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

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Answer

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refractive index of glass, n_g = 1.52

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1 year ago
In the system shown above, the pulley is a uniform disk with a mass of .75 kg and a radius of 6.5 cm. The coefficient of frictio
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Answer:

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A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
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This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
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Fh = Fph + Fgh 
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Answer:

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