A2.44 gram penny is on a turn table. it starts from rest at a radius of 0.136 meters and the turn table speeds up until the penny flies off with a tangential speed of 0.574 m/s. a. what coefficient of static friction exists between the penny and turntable if the penny doesn't fly off until it reaches this speed? b. suppose a 5.7-gram quarter has the same coefficient of static friction between it and the turntable as does the penny? in what way will the speed at which it les off differ from that of the penny?

a)  μ = 0.247, b) 0.574 m/s

Explanation:

a) Let's use Newton's second law

      F = m a

As the table is spinning the acceleration is centripetal

      a = v² / r

The force that holds the penny is the force of rubbing

      F = fr

      fr = μ N

On the vertical axis

      N-W = 0

     N = mg

Let' replace

     μ mg = m v² / r

  μi g = v² / r

    Mou = v² / gr

Let's calculate

     μ = 0.574² / (9.8 0.136)

     μ = 0.247

b) for this new mass

     fr = m a

     μ mg = m v² / r

     μ g= v² / r

We can see that the speed does not depend on the mass, so the speed of the coin will be the same

lower, it's not as dense as chocolate syrup.

choco syrup is more like a solid than a liquid and even turns solid if it gets slightly cold,

orange juice however is a 95% liquid and has a very low viscosity

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong