Apenguin slides at a constant velocity of 1.43 m/s down an icy incline. the incline slopes above the horizontal at an angle of 6.47°. at the bottom of the incline, the penguin slides onto a horizontal patch of ice. the coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal patch. how much time is required for the penguin to slide to a halt after entering the horizontal patch of ice?

t =1.285  s

Explanation:

u=1.43 m/s

∅=6.47°

We know that  μ(k)=tan∅

F(net)=-f(k)

ma =μ(k)*m*g

a =μ(k)*g   equation 1

v=u+at

0=u+(μ(k)*g*t

Putting values

t=u/(g*tan∅)

t=

t=1.285 s

answer: i don't know the answer good luck

explanation:

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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