Answer:
How do you find the net force?
Explanation:
i think this is want you asked for?
Density <em>ρ</em> is mass <em>m</em> per unit volume <em>v</em>, or
<em>ρ</em> = <em>m</em> / <em>v</em>
Solving for <em>v</em> gives
<em>v</em> = <em>m</em> / <em>ρ</em>
So the given object has a volume of
<em>v</em> = (130 g) / (65 g/cm³) = 2 cm³
Answer:Twice of given mass
Explanation:
Given
Two Particles of Equal mass placed at the base of an equilateral Triangle
let mass of two equal masses be m and third mass be m'
Taking one of the masses at origin
Therefore co-ordinates of first mass be (0,0)
Co-ordinates of other equal mass is (a,0)
if a is the length of triangle
co-ordinates of final mass 
Given its center of mass is at midway between base and third vertex therefore





