Aplayground merry-go-round of radius r = 1.80 m has a moment of inertia i = 255 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?

7 rpm = 0.73 rad/s

Step-by-step explanation:

R = Radius of merry-go-round = 1.8 m

= Moment of inertia of merry-go-round = 255 kg m²

= Moment of inertia of child

ω = 9 rev/min

m = Mass of child = 24 kg

From the conservation of angular momentum

∴ New angular speed of the merry-go-round is 7 rpm =

you should at least use a calculater

step-by-step explanation:

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it