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lubasha [3.4K]
3 years ago
14

A small object has charge Q. Charge q is removed from it and placed on a second small object The two objects are placed I m apar

t. For the force that each object exerts on the other to be a maximum, what should q be? a) q 20 c) q Q/2 d) q /4
Physics
1 answer:
Whitepunk [10]3 years ago
4 0

Answer:

q = Q/2

option c is correct

Explanation:

given data

charge = Q

distance r = 1 m

charge = q

to find out

what should q be

solution

we have given charge q is remove so

q1 = Q-q

q2 = q

we know by coulombs law force

force = kq1q2 / r²   .............1

put here value and we know k is constant

F = k (Q-q) q / 1

now take derivative charge q and put = 0

dF/dq = k d(Q-q)q / dq

so  k (Q-q)q =0

Q = 2q

q = Q/2

so option c is correct

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Answer:

a = -8.912 m/s²

Explanation:

Given,

The initial velocity of the car, u = 28 m/s

The final velocity of the car, v = 0

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The velocity displacement relation is given by the formula

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Substituting in the above values in the given equation

                                           t = 88/28

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The acceleration is given by the formula

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                                            = (0 - 28)/3.142

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7 0
4 years ago
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Answer:

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