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Leya [2.2K]
3 years ago
12

What would the electrostatic force be for two 0.005C charges 50m apart?

Physics
1 answer:
irakobra [83]3 years ago
3 0

Answer:

90 N

Explanation:

The electrostatic force between two charges is given by:

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

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem we have

q1 = q2 = 0.005 C

r = 50 m

So the electrostatic force is

F=(9\cdot 10^9 N m^2 C^{-2})\frac{(0.005 C)^2}{(50 m)^2}=90 N

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3 years ago
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
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Answer:

Explanation:

a )

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= 1 / 2 k X²

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4 cm .

=

7 0
3 years ago
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