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ladessa [460]
2 years ago
15

The force between two interacting charges is 9.0 × 10-5 newtons. They are kept 1 meter apart. If the magnitude of one charge is

1.0 × 10-6 coulombs, what is the magnitude of the other charge?
Pluto users ??
Physics
1 answer:
Oliga [24]2 years ago
8 0

The magnitude of other charge will be 1 × 10⁻² coulomb

The formula of electrostatic force is

Electrostatic force = K q1 q1 / r²

where k is the coulomb's constant whose value is 9 × 10⁹

q1 and a2 are the magnitude of charges

and r is the distance between them

magnitude of the force given to us is 9.0 × 10⁻⁵ newtons

magnitude of one charge = 1.0 × 10⁻⁶ coulomb

Force = K q1 q2 / r²

9.0 × 10⁻⁵ = ( ( 9 × 10⁹ ) × ( 1.0 × 10⁻⁶ ) × q2 ) / 1

9.0 × 10⁻⁵ =  9 × 10³ × q2  

10⁻² = q2  

Charge on q2 is 1 × 10⁻² coulomb

So the magnitude of the second charge is came out to be 1 × 10⁻² coulomb after applying the formula of electrostatic force.

Learn more about electrostatic force here:

brainly.com/question/17692887

#SPJ10

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Explanation:

5 0
3 years ago
You are given four resistors, 2 ohms, 3 ohms, 5 ohms, and 10 ohms. Your friend say you can connect them so you obtain an equival
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If we will connect the resistors 2ohms, 3ohms, 5ohms in series and the 10ohms resistance parallel then we get equivalent resistance of 5 ohms.

The equivalent circuit is,

R equivalent for the series connection is,

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The equivalent resistance is 5 ohms.

So your friend is saying true.

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