<span>It’s 2/5 MR^2 where M is mass and R is the radius of the bas</span>
Answer:
1/3 the distance from the fulcrum
Explanation:
On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

where
W1 is the weight of the boy
d1 is its distance from the fulcrum
W2 is the weight of his partner
d2 is the distance of the partner from the fulcrum
In this problem, we know that the boy is three times as heavy as his partner, so

If we substitute this into the equation, we find:

and by simplifying:

which means that the boy sits at 1/3 the distance from the fulcrum.
Answer:
346.70015 m/s
Explanation:
In the x axis speed is

In the y axis

The resultant velocity is given by

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s
Answer:
A
Explanation:
The next step would be precipitation falling