The force of attraction between two point masses is 10^-4N when the distance between them is 0.18m. If the distance is reduced to 0.06m, calculate the force.

1.1 x 10^-5N

3.3 x 10^-5N

3.0 x 10^-4N

9.0 x 10^-4N

Correct answer is D

F \(\alpha \frac{1}{r^2}\)

therefore, \(\frac{F_1}{F_2}\) = \(\frac{r_1}{(r_2)^2}\)

\(\frac{10^{-4}}{F_2}\) = \(\frac{(0.06)^2}{(0.18)^2}\)

F = 9 x 10^-4N

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The force of attraction between two point masses is 10-4N when the distance between them is 0.18m. If the distance is reduced to 0.06m, calculate the force.

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The force of attraction between two point masses is 10^-4N when the distance between them is 0.18m. If the distance is reduced to 0.06m, calculate the force.