A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 245 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 24.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?

11.65rpm

Explanation:

Moment of inertia of the child = mr²

Inertia = 24 * 1.4 * 1.4 = 47.04kgm²

Converting rpm to rad/s we have

12rpm = 12 * π/30

12rpm = 37.7/30

12rpm = 1.257 rad/s

Angular momentum = Iω

(245 * 1.257) + (47.04 * 0) = 355kgm²

355 = (245 + 47.04) * ω

355 = 292.04ω

ω = 355/292.04

ω = 1.22 rad/s

1.22 rad/s = 1.22 * 9.55 = 11.65 rpm

Thus, the new angular speed of the merry go round is 11.65rpm

there is no correct list among the choices you've provided.

the list might look something like this:

-- baby human

-- piccolo

-- flute

-- soprano singer

-- trumpet

-- tenor singer

-- trombone

-- baritone singer

-- baritone horn

-- bass singer

-- tuba

since i had to make up the list myself, i made it easy on myself, and i didn't try to shuffle the string instruments and woodwinds into it too.

overcharge is the right answer system