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quester [9]
3 years ago
14

The surface of an insulating sphere (A) is charged up uniformly with positive charge, and brought very close to an identical–siz

e conducting sphere (B) that has no net charge on it. The spheres do not make contact.
A) Sketch the distribution of charge on each sphere.
B) Will the spheres attract, repel, or not interact with each other? Explain.
C) When the spheres make contact, they repel each other. Explain this behavior.
Physics
1 answer:
Ilya [14]3 years ago
8 0

Answer:

A) A negative charge of value Q is induced on sphere B

B) there is an attraction between sphere

C) The charge of sphere A is distributed between the two spheres,

Explanation:

This is an electrostatic problem, in general charges of the same sign attract and repel each other.

with this principle let's analyze the different situations

A) The sphere A that is insulating has a charge on its surface and zero charge is its interior

   The conducting sphere B has zero charge, but the sphere A creates an attraction in the electrons, therefore a negative charge of the same value as the charge of the sphere A is induced in the part closest and in the part farther away than one that a positive charge.

A negative charge of value Q is induced on sphere B

B) In this case there is an attraction between sphere A with positive charge and sphere B with negative induced charge

C) When the two spheres come into contact, the charge of sphere A is distributed between the two spheres, therefore each one has a positive charge of value half of the initial charge, as now we have net positive charges in the two spheres charges of the same sign repel each other so the spheres separate

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Explanation:

The relationship between frequency and wavelength of a wave is given by the wave equation:

v=f\lambda

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v is the speed of the wave

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For the wave in this problem,

f = 15,500 Hz

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Therefore, the wave speed is

v=(15500)(0.20)=3100 m/s

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Ostrovityanka [42]

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0.358Kg

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The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system

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Using conservation of momentum between the bullet and the block

0.0115(265) = (M + 0.0115)v

3.0475 = (M + 0.0115)v

v = 3.0475/(M + 0.0115)

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12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2

12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )

12.56 = 0.5 × 9.2872/ M + 0.0115

12.56 = 4.6436/ M + 0.0115

12.56 ( M + 0.0115 ) = 4.6436

12.56M + 0.1444 = 4.6436

12.56M = 4.6436 - 0.1444

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4 0
3 years ago
Calculate the height of a cliff if it takes 2.35s for a rock to hit the ground when it is thrown straight up from the cliff with
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Answer:

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in the exercise they indicate the initial velocity v₀ = 8 m / s.

when the rock reaches the ground its height is zero

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        y₀i = -v₀ t + ½ g t²

let's calculate

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Answer:

The answer would be 420 m/s

Explanation:

Look in attachment ⬇

I Hope this Helps!!!

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