Three identical 50-kg masses are held at the corners of an equilateral triangle, 30 cm on each side. if one of the masses is released, what is its initial acceleration, if the only forces acting on it are the gravitational forces due to the other two masses?

a)1.9 x 10-8 m/s^2
b)6.4 x 10-8 m/s^2
c)4.2 x 10-8 m/s^2
d)2.5 x 10-8 m/s^2
e)3.7 x 10-8 m/s^2

a = 3.21 10⁻⁸ m / s² ,  The correct answer is E

Explanation:

For this exercise we will use Newton's second law, where the force is the gravitational attraction force, the attached diagram shows the forces and their components in a horizontal and vertical coordinate system, due to the symmetry of the equilateral triangle the horizontal components be annulled and the two vertical components add up; let's write Newton's education for the vertical axis (Y)

        2 = m a

Let's use trigonometry to find the components

       sin 60 =  / F

         = F sin60

      2 F sin60 = m a

The gravitational force equation is

     F = G m M / r

The mass of the entire sphere is equal to m = M and the distance between    the spheres is the side of the equilateral triangle

    F = G m² / L²

We replace

     2 (G m² / L²) sin60 = m a

     a = 2 G m / L² sin 60

Let's calculate

      a = 6.67 10⁻¹¹  50 / 0.30²  sin 60

      a = 3.21 10⁻⁸ m / s²

The correct answer is E

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.