A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of 0.390.
How far along the rough ice does she go before stopping?

18.84m

Explanation:

We are given;

Mass; m = 90 Kg

Initial velocity; u = 12 m/s

Coefficient of friction; μ = 0.390

Let,the combined mass of the ice skater and the skate be M

Thus, So,if the coefficient of frictional force is μ,then frictional force acting is; μN. N= Mg. Thus F_friction = μMg

Now, the deceleration due to friction will be, F_friction/M

Thus, deceleration = μMg/M

M will cancel out and we have; μg

Now, deceleration means negative acceleration. Thus acceleration (a) =

-μg

Now, to find the distance, let's use equation of motion which is;

v² = u² + 2as

Putting -μg for a, we have;

v² = u² + 2(-μg)s

v² = u² — 2μgs

We want to know the distance covered before coming to rest. Thus, v = 0m/s

So,

0² = 12² - (2 x 0.39 x 9.8 x s)

0 = 144 - 7.644s

Thus,

7.644s = 144

Thus, s = 144/7.644 = 18.84m

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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4900=490xh

h=10