This question is in complete.The question is
A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.
Answer:
distance=0.124 m
Explanation:

-- The train starts at 23 m/s and slows down by 0.25 m/s every second.
So it'll take (23/0.25) = 92 seconds to stop.
-- Its average speed during that time will be (1/2)(23+0) = 11.5 m/s
-- Moving at an average speed of 11.5 m/s for 92 sec, the train will cover
(11.5 m/s) x (92 sec) = <em>1,058 meters</em> .
Answer:
hello your question is not properly arranged attached below is the arranged table and solution
answer : attached table below
Explanation:
Given data:
02 molecules size = 10^-10m
smoke particles size = 0.3 mm
cloud droplets size = 20 mm
Rain droplets size = 3 mm
Attached below is a table showing the kind of scattering that is expected to occur at various wave lengths
Note : For Rayleigh scattering the wave particle is smaller than the wave length while for Non-selective scattering the wave particle is greater than the wavelength.
and For Mie scattering the wavelength is the same as the wavelength.
The seasons affect the scheduling of far more events the moon phases do.
4.096. You just have to multiply 1.6 three times. V=s^3