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
The distance apart of the two student can be obtained as illustrated below:
[tex]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}}\\\\[/tex]
Cross multiply
3.5×10¯⁸ × r² = 3.9171×10¯⁷
Divide both side by 3.5×10¯⁸
[tex]r^{2} = \frac{3.9171*10^{-7} }{3.5*10^{-8}}\\\\[/tex]
Take the square root of both side
[tex]r = \sqrt{\frac{3.9171*10^{-7} }{3.5*10^{-8}} }\\\\[/tex]
Therefore, the distance between the two student is 3.35 m
Learn more: https://brainly.com/question/20856669
