Two uniform spheres, each of mass 0.260kg are fixed at points a and b
a.)find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point p and acted on only by forces of gravitational attraction of the spheres at a and b.
b.) find the direction of the initial acceleration of a uniform sphere with mass 0.010 kg.

a) 2.76x10^-9 m/s^2

b) downward

Explanation:

a) In order to do this, we need to use the newton's law of gravitation which is:

F = G.m/r²

Where:

G: Constant of gravitation (6.67x10^-11 N/m)

m: mass of the object

r: distance or radius of the object.

In this case, we have two spheres, so this law becomes:

F = G*m1*m2 / r²

Now, you are not providing a picture with this, however, this problem is very similar to one I solved before, so, I'm gonna use the same values to solve this. If, you have different values for the distance, just replace them in this procedure and should get the correct and more accurate answer.

In this case, the distance that separates sphere in P from sphere A and B is 10 cm. As both spheres have the same mass and they are at the same distance from P, we can assume that F1 = F2 so:

F1 = (6.67x10^-11) * (0.01) * (0.26) / (0.1)² = 1.73x10^-11 N

sphere at P, will roll downward to the middle of sphere A and B. The distance between these two points it's 8 cm (See picture attached)

Now, according to my data, they are exerting a force in the y axis, so:

F1y = Fy * senФ

senФ using trigonometry is:

senФ = opposite lenght / hipotenuse

senФ = 8/10 = 0.8

Replacing:

F1y = 1.73x10^-11 * 0.8 = 1.38x10^-11 N

This force it's exactly the same as F2y, so the total force is:

F = 1.38x10^-11 * 2 = 2.76x10^-11 N

Finally, using the second law of Newton:

F = m*a

Solving for a:

a = F/m

replacing:

a = 2.76x10^-11 / 0.01

a = 2.76x10^-9 m/s²

According to the picture and the value of this magnitude, it's going downward.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6