Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
Answer:
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Explanation:
Answer:

Explanation:
Two identical bodies are sliding toward each other on a frictionless surface.
Initial speed of body 1, m₁ = 1 m/s
Initial speed of body 2, m₂ = 2 m/s
They collide and stick.
We need to find the speed of the combined mass. Let V is the speed of the combined mass.
Using the conservation of momentum.

We have, m₁ = m₂ = m

So, the speed of the combined mass is
.
Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.
Answer:
30 m
Explanation:
The wavelength of a wave is found by the velocity divided by the frequency. Therefore, the wavelength is (300 m/s)/(10 Hz) = 30 m
I hope this helps! :)