Bob and lily are riding on a merry-go-round. bob rides a horse on the outer edge of the circular platform and lily rides a horse near the center. when the merry-go-round is rotating at a constant angular speed w, bob's speed isa) exactly half as much as lily'sb) exactly twice as much as lily'sc) smaller than lily'sd) larger than lily'se) same as lily's

d) larger than Lily's

Explanation:

If the merry-go-round is rotating at a constant angular speed, this means that any two points, located at different positions along a radius, rotate at the same angular speed, which means that they sweep the same angle at a given time interval.

In order to both points keep aligned along the same radius, we have a single choice (assuming that we are talking about a rigid  body) to meet this premise:

The point farther of the center (Bob) must have a linear speed greater than a point closer to the center (Lily).

Mathematically, we can explain this result as follows:

ω = Δθ / Δt (by definition of angular velocity) (1)

but, by definition of angle, we can say the following:

θ = s/r , where s is the arc along the circumference, and r, the radius.

⇒Δθ = Δs /r

Replacing in (1) we have:

ω = (Δs /Δt) / r

By definition, Δs/Δt = v, so, arranging terms, we get:

v = ω*r

If ω=constant, if r increases, v increases.

So, as Bob is at a distance r from the center larger than Lily's, Bob's speed must be larger than Lily's.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

snell's law

refractve index = sin incidence / sin refraction