Question

doua sfere identice din plumb,avind raze egale cu 25 cm se ating.Calculati forta de atractie dintre sfere,daca densitatea plumbului este de 11300kg/m³.

Answer 1

$\displaystyle F=\dfrac{m_1m_2k}{d^2}=\dfrac{m^2k}{4r^2} \\ \\ k=6,674*10^{-11}*m^3*kg^{-1}*s^{-2} \\ \\ r=0,25m \\ \\ m=\dfrac{4\rho_{Pb}\pi r^3}{3}=739,2kg \\ \\ F=\dfrac{739,2^2*6,674*10^{-11}}{4*0,25^2}\approx1,458*10^{-4}N$