Question

The force of gravity between two students is 3.5E-8 Newtons. If

Answer 1

The distance between the two student having a mass of 73.5 Kg and 79.9 Kg with a force of attraction of 3.5×10¯⁸ N is 3.35 m  

From the question given above, the following data were obtained:

Force of attraction (F) = 3.5×10¯⁸ N

Mass of 1st student (M₁) = 73.5 kg

Mass of 2n student (M₂) = 79.9 kg

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Distance apart (r) =?

The distance apart of the two student can be obtained as illustrated below:

$F = \frac{GM_{1}M_{2}}{r^{2}} \\\\3.5*10^{-8} = \frac{6.67*10^{-11} * 73.5 * 79.9 }{r^{2}} \\\\3.5*10^{-8} = \frac{3.9171*10^{-7}}{r^{2}}$

Cross multiply

3.5×10¯⁸ × r² = 3.9171×10¯⁷

Divide both side by 3.5×10¯⁸

$r^{2} = \frac{3.9171*10^{-7} }{3.5*10^{-8}}$

Take the square root of both side

$r = \sqrt{\frac{3.9171*10^{-7} }{3.5*10^{-8}} }$

r = 3.35 m

Therefore, the distance between the two student is 3.35 m  

Learn more: brainly.com/question/20856669