Mischievous Joey likes to play with his family's lazy susan (this drives Mom crazy because it is an antique). He puts the salt shaker near the edge and tries to spin the tray at a speed so that the shaker just barely goes around without slipping off. Joey finds that the shaker just barely stays on when the turntable is making one complete turn every two seconds. Joey's older sister measures the mass of the shaker to be 79 grams. She also measures the radius of the turntable to be 0.23 m, and she is able to calculate that the speed of the shaker as it successfully goes around in a circle is 0.7222 m/s. Required:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?

0.179 N

Explanation:

What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?

The horizontal part of the constant force of the turntable on the shaker is the centripetal force of the turntable on the shaker, F.

So, F = mv²/r where m  = mass of shaker = 79 g = 0.079 kg, v = speed of shaker = 0.7222 m/s and r = radius of turntable = 0.23 m

So, substituting the values of the variables into the equation, we have

F = mv²/r

F = 0.079 kg (0.7222 m/s)²/0.23 m

F = 0.0412 kgm/s² ÷ 0.23 m

F = 0.179 kgm/s²

F = 0.179 N

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 ! ~