Amerry-go-round is a common piece of playground equipment. a 3.0-m-diameter merry-go-round with a mass of 300 kg is spinning at 16 rpm . john runs tangent to the merry-go-round at 5.4 m/s , in the same direction that it is turning, and jumps onto the outer edge. john's mass is 30 kg. what is. what is the merry-go-round's angular velocity, in rpm, after john jumps on?

Explanation:

For answer this we will use the law of the conservation of the angular momentum L where:

First, the important data is:

 (mass of the merry-go-round)

    (radius of the merry-go-round)

(angular velocity of the merry-go-round)

(velocity od jhon)

 (mass of jhon)

Then,

where is the moment of inerta after jhon jumps on,  is the moment of inertia of the merry-go-round and is the angular velocity of the merry-go-round after jhon jumps on.

So, we have to find the moment of inertia of the merry-go-round as:

Now we have to find as:

Then, we replace the data on the initial equation and we get:

Solving for :

Finally, we change it to rpm as:

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.