Four identical masses of mass 800kg each are placed at the corners of a square whose side lengths are 15.0cm .

What is the magnitude of the net gravitational force on one of the masses, due to the other three?

why the hell can't i just look at answers

Sorry for putting what I put above. It was on accident. I don't spell all my words right all the time.

cant see answer.

To find the magnitude of the net gravitational force on one of the masses due to the other three, we can use the formula for gravitational force:

F = G * (m1 * m2) / r^2

where:
F is the gravitational force
G is the universal gravitational constant (approximately 6.674 * 10^-11 Nm^2/kg^2)
m1 and m2 are the masses involved
r is the distance between the center of masses

In this case, the masses are identical (m1 = m2 = 800 kg) and form a square. The distance between the center of masses is the length of a side of the square.

Step 1: Calculate the distance between the center of masses
Since the masses are at the corners of a square, the distance between the centers can be found using the Pythagorean theorem. The side length of the square is given as 15.0 cm, which we can convert to meters:

15.0 cm = 0.15 m

The distance between the centers (diagonal of the square) is √(2) times the side length:

r = √(2) * 0.15 m ≈ 0.212 m

Step 2: Calculate the net gravitational force
Now we can substitute the values into the formula:

F = G * (m1 * m2) / r^2
= (6.674 * 10^-11 Nm^2/kg^2) * (800 kg * 800 kg) / (0.212 m)^2

Calculating this expression will give us the magnitude of the net gravitational force on one of the masses due to the other three.