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Basile [38]
4 years ago
14

What is the electric field at the center of the metal ball due only to the charges on the surface of the metal ball? (Express yo

ur answer in vector form.) < Entry field with incorrect answer -244800 , Entry field with correct answer 0 , Entry field with correct answer 0 > N/C
Physics
1 answer:
FrozenT [24]4 years ago
8 0

Answer:

The electric field is zero inside the metal sphere

Explanation:

Gauss's law says: the net flow on any enclosed surface is equal to the value of the charge inside divided by the electrical permittivity

     E = Q / ε

Where Q is thec charge into the surface.

If we apply this to the metal sphere, all the charge is on the surface, so there is no charge inside, therefore, the electric field is zero

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What are sound waves
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Answer:  Sound waves are a waves of compression and rarefaction, by which sound is propagated in an elastic medium such as air.

Explanation:

In physics, sound is a vibration that propagates as an acoustic wave, through a transmission medium such as a gas, liquid or solid.

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soil additive

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5. SEP Evaluate Information A classmate states that animals that result from
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A 2.5 kg , 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turnt
emmainna [20.7K]

The concepts required to solve this problem are those related to the conservation of the angular momentum and the moment of inertia of the disk. We will begin by calculating the moment of inertia of the disc, then the moment of inertia of the disc after the two two blocks hits and sticks to the edges of the turn table. In the end we will apply the conservation theorem.

The radius is given as,

R = \frac{20cm}{2} = 10cm = 0.1m

When a block falls from above and sticks to the turn table, the moment inertia of the turntable increases.

Since two blocks are stick to the turn table, the total final moment of inertia of the turntable is the sum moment of inertias of individual turntable, and two blocks.

I_1 = \frac{1}{2} MR^2

I_1 = \frac{1}{2} (2.5)(0.1)^2

I_1 = 0.0125kg \cdot m^2

The moment of inertia of each block is

I_0 = mR^2

Total moment of inertia of two block is

I_0' = 2mR^2

The final moment of inertia of the turn table is

I_2 = I_1 +I'0

I_2 = I_1 +2mR^2

I_2 = 0.01kg\cdot m^2 + 2(500*10^{-3}kg)(0.1m)^2

I_2 = 0.0225kg\cdot m^2

From the conservation of the angular momentum, the initial angular momentum of the system is equal to final angular momentum of the system,

Rearrange the equation we have that

I_1\omega_2 = I_2\omega_2

\omega_2 = \frac{I_1\omega_2}{I_2}

\omega_2 = \frac{0.01*150rpm}{0.0225}

\omega_2 = 66.67rpm

The magnitude of the turntable's angular velocity is 66.67rpm

3 0
3 years ago
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