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prisoha [69]
3 years ago
12

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics
2 answers:
larisa86 [58]3 years ago
7 0

Answer:

TRUE

Explanation:

Grounding of an object means we connect the given object to the earth by a conducting wire.

here we know that Earth is considered as a large conductor with ZERO potential which means if we connect an object to the earth then it will come to zero potential.

Now if an object is initially charge and then it is connected to Earth for grounding then in that case the final potential of the object must be zero.

In order to make the final potential to be zero we need to transfer all its charge to the earth so in order to make it neutral either electrons are flowing into the object or it will flow out of the object.

so here this is a TRUE statement

Serhud [2]3 years ago
4 0
True,
Explanation: because it’s the same reason that lighting touches the ground/ goes up to the sky, because it is trying to balance out its charges
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The amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is________
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d) 2Fr

Explanation:

We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell  -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².

We now evaluate the integral from r = +r to -r

W = ∫Fdr

= ∫(-e²/4πε₀r²)dr

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W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.

Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.

So W = -2e²/4πε₀r = 2Fr.

So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr

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3 years ago
I hope you are able to read this question?? Help ASAP this question is on the quiz tommorow
Masteriza [31]

You would be correct.

Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.

Hope this helps!

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