need answers to a 2 part question
Physics

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θ = 9.5 °
Calculate the acceleration of the skier in m/s^{2}.
Find the angle of the slope down which this skier could coast at a constant velocity in degrees.
Answers: 0.65 m/s2; 5.7°.
Choose the coordinate system with xaxis directed to the right and the yaxis directed downwards.
The gravity force acting of the skier is W = (0, mg). It can be decomposed into the sum of two forces, F_{N} which is perpendicular to the slope and F_{T} directed along the slope. Their magnitudes are
F_{N} = mg cos θ and F_{T} = mg sin θ, where θ = 9.5°.
The remaining forces acting on the skier are the reaction force F_{N} and the friction force F_{k} which is directed along the slope but opposite to the force F_{T}. The magnitude of the friction force is
Fk = μ_{k} F_{N} = μ_{k} mg cos θ. The resultant of all forces has the magnitude F = F_{T} – F_{k} =
mg(sin θ – μ_{k}cos θ) and creates the acceleration a = F/m = g(sin θ – μ_{k}cos θ) =
9.81 (sin 9.5°  0.10 * cos 9.5°) = 0.65 m/s^{2}.
To answer the second question we need to solve an equation a = 0.
Then sin θ – μ_{k}cos θ = 0, tan θ = μ_{k}, and θ = tan^{–1} μ_{k }= tan^{–1} 0.10 = 5.7°.
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