A 5.2 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. How far to the left of the pivot must a 4.0 kg cat stand to keep the seesaw balanced

The cat must stand 0.96 m from the left side of the pivot to keep the seesaw balanced.

Explanation:

Given;

mass of the cat, m₁ = 5.2 kg

mass of the tuna, m₂ = 2.5 kg

length of the seesaw, L = 4.0 m

Apply the following approach;

                  x m                      <2m >              

        0Δ4m

                  ↓                         2m                                 ↓

                 5.2 kg                                                      2.5 kg

Apply the principle of moment; anticlockwise moment = clockwise moment.

5.2(x ) = 2.5(2)

5.2x = 5

x = 5/5.2

x = 0.96 m

Therefore, the cat must stand 0.96 m from the left side of the pivot to keep the seesaw balanced.

the answer is neither ball has a charge.

because if they were both positively charged, they will repel each other.

if they were positively and negatively charged, they will seek.