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valina [46]
3 years ago
11

Positive charge is distributed uniformly throughout a non-conducting sphere. The highest electric potential occurs: A. Far from

the sphere B. Just outside the surface C. At the surface D. Halfway between the center and surface E. At the center
Physics
1 answer:
Ilya [14]3 years ago
6 0

Answer:E

Explanation:

At the center

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Answer:

a)   E_{L} = -360 V , b)      t = 3 s

Explanation:

The electromotive force in an inductor is

           E_{L} = - L dI/ dt

in the exercise they give us the relation of i (t)

            i (t) = 1.00 t² -6.00t

we carry out the derivative and substitute

          E_{L} = - L (2.00 2t - 6.00 1)

a) the electromotive force at t = 1.00 s

          E_{L} = - 90.0 (4.00 1 - 6.00)

         E_{L} = -360 V

b) for t = 4 s

         E_{L}= - 90 (2 4 2 - 6 4)

         E_{L} = - 720 V

c) for the induced electromotive force to zero, the amount between paracentesis must be zero

           (2.00 t2 - 6.00t) = 0

            t (2.0 t-6.00) = 0

the solutions of this equation are

            t = 0

            2 t -6 = 0

            t = 3 s

to have a different solution the trivial (all zero) we must total t = 3 s

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How can you increase the momentum of an object?
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Hope this helps, have a great day ahead!

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Jorge wants to know if yearly income and amount of tax returns are related in any way. In looking for an association between
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The answer will be b
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Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. A certain gun has an elec
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Answer: v = 2.75 * 10^7 m/s

Explanation: Since the electric field is a uniform one and the distance is small, the motion of the electron is of a constant acceleration, hence newton's laws of motion is applicable.

From the question

E=strength of electric field = 214N/c

q=magnitude of an electronic charge = 1.609* 10^-16c

m= mass of an electronic charge = 9.11*10^-31kg

v = velocity of electron.

S= distance covered = 1cm = 0.01m

a = acceleration of electron.

F = ma but F=Eq

Eq = ma.

a = Eq/m

a = 214 * 1.609*10^-16/ 9.11 * 10^-31

a = 344.32 * 10^-16/ 9.11 * 10 ^-31

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Before the electron is accelerated, they are always not moving, hence initial velocity (u) = 0.

By using the equation of motion, we have

v² = u² + 2aS

But u = 0

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