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pishuonlain [190]
3 years ago
9

Two small objects each with a net charge of Q (positive) exert a force of magnitude F on each other. We replace one of the objec

ts with another whose net charge is 4Q. The original magnitude of the force on the Q charge was F; what is the magnitude of the force on the Q now?
A. 16F
B. 4F
C. F
D. F/4
Physics
1 answer:
erik [133]3 years ago
5 0

Answer:

option B

Explanation:

we know,

Force between two charge is calculated by

F = \dfrac{kq_1q_2}{r^2}

r is the distance between the two charges

k is Coulomb constant.

give two small object with charge Q

now, Force

F = \dfrac{kQQ}{r^2}......(1)

now,

New force if another charge is equal to 4 Q

F' = \dfrac{kQ(4Q)}{r^2}

F' = 4\dfrac{kQQ}{r^2}

now from equation 1

F' = 4 F

Hence, the correct answer is option B

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Evgesh-ka [11]

The average density of the material from which the coin is made is 9.67 g/cm³.

<h3>Volume of the coin</h3>

The volume of the coin at the given diameter is calculated as follows;

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V = πd²/4 x h

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The average density of the material from which the coin is made is calculated as follows;

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Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
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Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

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r is the radius of the cone

h is the height of the cone

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Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

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\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given: \frac {dV}{dt} = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

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-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

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\frac {dh}{dt}=-0.08603\ cm/s

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

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