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melomori [17]
4 years ago
13

A conducting sphere with radius R is charged until the magnitude of the electric field just outside its surface is E. The electr

ic potential of the sphere, relative to the potential for away, is: Group of answer choices 0 E/R E/R2 ER ER2
Physics
1 answer:
dusya [7]4 years ago
8 0

Answer:

he correct answer is V = ER

Explanation:

In this exercise they give us the electric field on the surface of the sphere and ask us about the electric potential, the two quantities are related

                ΔV = ∫ E.ds

where E is the elective field and normal displacement vector.

Since E is radial in a spray the displacement vector is also radial, the dot product e reduces to the algebraic product.

                 ΔV = ∫ E ds

                 ΔV = E s

                 

since s is in the direction of the radii its value on the surface of the spheres s = R

                  ΔV = E R

checking the correct answer is V = ER

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Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

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where

p_p is the initial momentum of the ping-poll ball

p_b is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

p'_p is the final momentum of the ping-poll ball

p'_f is the final momentum of the bowling ball

We can re-arrange the equation as follows

p_p - p'_p = p_b'-p_b

or

-\Delta p_p = \Delta p_b

which means

|\Delta p_p | = |\Delta p_b| (1)

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However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

I=\Delta p (2)

Therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

|I_p| = |I_b|

4 0
4 years ago
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4 years ago
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Answer:

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Some reactions only have one reactant and such a chemical process is often termed decomposition reaction.

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Answer: A

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