Aski starts from rest and slides down a 22 incline 75 m long. if the coefficient of friction is 0.090, what is the ski's speed at the base of the incline?

L = length of the incline = 75 m

θ = angle of incline = 22 deg

h = height of skier at the top of incline = L Sinθ = (75) Sin22 = 28.1 m

μ = Coefficient of friction = 0.090

N = normal force by the surface of incline

mg Cosθ = Component of weight of skier normal to the surface of incline opposite to normal force N

normal force "N" balances the component of weight opposite to it hence we get

N = mg Cosθ

frictional force acting on the skier is given as

f = μN

f = μmg Cosθ

v = speed of skier at the bottom of incline

Using conservation of energy

potential energy at the top of incline = kinetic energy at the bottom + work done by frictional force

mgh = f L + (0.5) m v²

mgh = μmg Cosθ L + (0.5) m v²

gh = μg Cosθ L + (0.5) v²

(9.8 x 28.1) = (0.09 x 9.8 x 75) Cos22 + (0.5) v²

v = 20.7 m/s

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

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