Thomas Edison is the answer im 100% sure of it.
Answer:
Because space is a void with no air flow
Explanation:
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:
about 14.7°
Explanation:
The formula for the angle of the first minimum is ...
sin(θ) = λ/a
where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.
The wavelength of the sound is the speed of sound divided by the frequency:
λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m
Then the angle of interest is ...
θ = arcsin(0.261538/1.03) ≈ 14.7°
At an angle of about 14.7°, someone outside the room will hear no sound.