sphere of mass 40 kg is attracted by second saphere of mass 80 kg when the centre of 30 cm apart with a force equal to the weight of 1by 4 Mg calculate the constant of gravitation small g is equal to 9.81 M per second square
Answers
Answered by
0
By Newton law of gravitation
F =G mM/r^2
F = 6.626*10^11 * 40 * 80 /( 30)^2
F = 2120.3/900
F = 2 . 355N
F =G mM/r^2
F = 6.626*10^11 * 40 * 80 /( 30)^2
F = 2120.3/900
F = 2 . 355N
Similar questions