What must be the distance between point charge q1 = 28.0 μc and point charge q2 = −57.0 μc for the electrostatic force between them to have a magnitude of 5.05 n?

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² Equation 1

Where F = electrostatic force  between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F) Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

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explanation:

answer: the metamorphic rocks are formed when they are exposed to high pressure and high temperatures. this happens when one of the earth's plates comes in contact with the other plate by traveling in opposite directions, straight into each other or slipping under one another. the friction that is caused by this activity results in   melting of rocks and when it cools they are known as metamorphic rocks. this process needs a controlled amount of pressure and temperature otherwise it may turn into magma.