A circular coil of radius 5.0 cm and resistance 0.20 is placed in a uniform magnetic field perpendicular to the plane of the coi
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First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,
. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
Now we can use the following relationship to find the distance covered by the skier before stopping, S:
where
is the final speed of the skier and 
is the initial speed. Substituting numbers, we find:
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
Now we can use the following relationship to find the distance covered by the skier before stopping, S:
where
The relationship between frequency and wavelength for an electromagnetic wave is
where
f is the frequency
is the wavelength
is the speed of light.
For the light in our problem, the frequency is
, so its wavelength is (re-arranging the previous formula)
where
f is the frequency
For the light in our problem, the frequency is