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valentina_108 [34]
3 years ago
15

At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t

hem to be 83.4 N?
Physics
2 answers:
nataly862011 [7]3 years ago
7 0
F = q₁*q₂* C / r²

F =83.4
q₁ = q₂ = 3.2 * 10⁻⁹
C couloumb constant 8.9 * 10⁹

solve for r
STatiana [176]3 years ago
4 0
I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!
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