Answer: [tex]3.52(10)^{-10} N[/tex]
Explanation:
According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:
[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex]
Where:
[tex]F[/tex] is the module of the force exerted between both bodies
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the universal gravitation constant.
[tex]m_{1}=0.006 kg[/tex] and [tex]m_{2}=55 kg[/tex] are the masses of the pencil and the teenager, respectively.
[tex]r=0.25 m[/tex] is the distance between both bodies
Solving for [tex]F[/tex]:
[tex]F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(0.006 kg)(55 kg)}{(0.25 m)^2}[/tex]
[tex]F=3.52(10)^{-10} N[/tex] This is the force of gravitational attraction between the pencil and the teenager, which is very small in this case
