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please help with physics problem! see attached photo!?
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- ?Lv 76 years agoFavorite Answer
F = kQq / d²
Horizontally: F = 8.99e9N·m²/C² * 7e-9C * 3e-9C / (7m)² = 3.85e-9 N, right
Vertically: F = 8.99e9N·m²/C² * 7e-9C * 2e-9C / (8m)² = 1.97e-9 N, down
Θ = arctan(1.97/3.85) = 27º below the +x axis,
which can also be expressed as Θ = 333º ccw from the +x axis.
Hope this helps!
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