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 =

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

a

step-by-step explanation: