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Ronch [10]
3 years ago
7

Suppose that two point charges, each with a charge of +3.00 Coulomb are separated by a distance of 3.00 meters. Determine the ma

gnitude of the electrical force of repulsion between them
Physics
1 answer:
Sveta_85 [38]3 years ago
3 0

Answer:

9\cdot 10^9 N

Explanation:

The magnitude of the electrical force between the two point charges is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q_1 =q_2 = +3.0 C is the magnitude of each charge

r = 3.00 m is the separation between the two charges

Substituting the numbers into the formula, we find

F=(9\cdot 10^9 Nm^2C^{-2})\frac{(+3 C)^2}{(3.00 m)^2}=9\cdot 10^9 N

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Answer:

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Explanation:

Given that:

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8 0
3 years ago
A repelling force occurs between two charged objects when​
fgiga [73]
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ It would be 'when they have like charges.'

'Like charges' means the same charge. For example, two positive charged objects have like charges.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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Answer:

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Answer:

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