Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch is of width 62.0 m and the average force of air resistance on the skier is 160 N , how fast is she going after crossing the patch?

14.1 m/s

Explanation:

From the question,

μk = a/g Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g) Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m) Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s

a.

explanation:

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.