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Olenka [21]
2 years ago
12

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

ere is then grounded. Then the ground wire is removed. A is the inner surface and B is the outer surface.
Which statement is correct?

A. The charge is –q/2 on both A and on B.
B. The charge on A is –q; there is no charge on B.
C. There is no charge on either A or B
D. The charge on A is –q; that on B is +q.
E. The charge on B is –q; that on A is +q.
Physics
1 answer:
Olenka [21]2 years ago
4 0

Answer:

B.The charge on A is -q; there is no charge on B.

Explanation:

We are given that

Charge=+q

We have to find the correct statement.

When positive charge is placed at center of uncharged metal sphere

insulated from the ground then negative charge(-q) induced on inner

surface  A  of sphere  and the outer  surface B  is grounded then the surface is neutral .

It means there is no charge on surface B.

Hence, option B is true .

B.The charge on A is -q; there is no charge on B.

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Answer:

y = -8.37 cm

Explanation:

As we know that the equation of SHM is given as

y = A cos(\omega t)

here we know that

\omega = \frac{2\pi}{T}

here we have

T = 0.66 s

now we have

\omega = \frac{2\pi}{0.66}

\omega = 3\pi

now we have

y = (8.8 cm) cos(3\pi t)

now at t = 2.3 s we have

y = (8.8 cm) cos(3\pi \times 2.3)

y = -8.37 cm

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2 years ago
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A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
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W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

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So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

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The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec
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The static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta.

The given parameters;

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<em />

The normal force on the block is calculated as follows;

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The static friction exerted on the block by the incline is calculated as follows;

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Answer:

F = - 50 N

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