Four spheres form the corners of a square whose side is 2.0
cmlong. what are the magnitutde and direction of the
netgravitational force from them on a central sphere with
massm5 = 250kg?

  F = 0 N

Explanation:

For this exercise the force is gravitational that is given by

          F = G m₁ m₂ / r₁₂²

We use Newton's second law to find the total force

         F = F₁ₐ + f₂ₐ + F₃ₐ + F₄ₐ

Where

        F₁ₐ is the force between sphere 1 or the central sphere, the distance is

        r₁ₐ = √ (L /)²+ (L / 2)² = L / 2 √ 2

This distance is the same for all spheres, because the spheres are in the corner of a cube,

Let's explicitly write the force

     F₁ₐ = G m₁ mₐ / L √2 / 2

[We can see that the modulus of each force is the same because the masses have the same value and with or the line of action is in the diagonals of the cube the force is annulled two by two

        F₁ₐ = F₃ₐ

        F₂ₐ = F₄ₐ

Therefore the total force on the central sphere is zero

         F = 0 N

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2