A 50.0-kg child stands at the rim of a merry-go-round of radius 2.00 m, rotating in such a way that it makes one revolution in 2.09 s. What minimum coefficient of static friction between her feet and the floor of the merry-go-round is required to keep her in the circular path

2.01

Explanation:

First, we need to find the centripetal acceleration.

We're given that the merry go round rotates 1 revolution in 2.09 seconds. Converting to rpm, we know that it rotates 30 revolution per minute

Now this speed gotten in rpm will be converted to m/s, to ease the calculation

30 rpm = πdN/60 m/s

30 rpm = (3.142 * 4 * 30)/60

30 rpm = 377.04/60

30 rpm = 6.284 m/s

a(c) = v²/r

a(c) = 6.284²/2

a(c) = 39.49 / 2

a(c) = 19.74 m/s²

F = ma

F = 50 * 19.74

F = 987 N

Also, Normal Force, F(n) =

F(n) = mg

F(n) = 50 * 9.81

F(n) = 490.5

We then use this to find the coefficient of static friction, μ

μ = F/F(n)

μ = 987 / 490.5

μ = 2.01

answer: the metamorphic rocks are formed when they are exposed to high pressure and high temperatures. this happens when one of the earth's plates comes in contact with the other plate by traveling in opposite directions, straight into each other or slipping under one another. the friction that is caused by this activity results in   melting of rocks and when it cools they are known as metamorphic rocks. this process needs a controlled amount of pressure and temperature otherwise it may turn into magma.

did yall get answers for this yet