In a vacuum, two particles have charges of q1 and q2, where q1 = +4.2c. they are separated by a distance of 0.31 m, and particle 1 experiences an attractive force of 4.5 n. what is the value of q2, with its sign?

11.44pC.

Explanation:

F = kq1q2/r^2

q2 = Fr^2/kq1

     = 4.5 x 0.31^2/4.2x 9 x 10^9

     = 1.144 x 10^-11C

     = 11.44pC

The sign is minus since they attracted to each other

there is no correct list among the choices you've provided.

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-- baby human

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wavelength divides in half and speed stays the same.