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st3dent
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1. A bicylist of mass 80 kg (including the bike) can coast down a 4 degree hill at a steady speed of 6.0 km/h. Pumping hard, the bicylist can descend the hill at 30 km/h. Assume the force of friction is proportional to the square of the speed; that is [tex]f_{fr} = bv^2[/tex], where [tex]b[/tex] is a constant.
2. Two mountains are at elevations of 850m and 750m above the valley between them. A ski run extends from the top of the higher peak to the top of the lower one, with a total length of 3.2km and an average slope of 30 degrees,
a) A skier starts from rest on the higher peak. At what speed will be arrive at the top of the lower peak if he just coasts without using poles?
b)How large a coefficient of friction would be tolerable without preventing him from reaching the lower peak?
Sol:
1. no clue how to even start this one
2a) (m)(9.8)(850) = (m)(9.8)(750) + (0.5)(m)([tex]vf^2[/tex])
[tex]vf^2[/tex] = 1960
[tex]vf[/tex] = 44.27 m/s
b)I tried to solve this one...but it does not seem to be working.
Thanks for your help. Much appreciated.
*Ignore the _ that seems to be appearing after the variables. I don't know how to get rid of it.
2. Two mountains are at elevations of 850m and 750m above the valley between them. A ski run extends from the top of the higher peak to the top of the lower one, with a total length of 3.2km and an average slope of 30 degrees,
a) A skier starts from rest on the higher peak. At what speed will be arrive at the top of the lower peak if he just coasts without using poles?
b)How large a coefficient of friction would be tolerable without preventing him from reaching the lower peak?
Sol:
1. no clue how to even start this one
2a) (m)(9.8)(850) = (m)(9.8)(750) + (0.5)(m)([tex]vf^2[/tex])
[tex]vf^2[/tex] = 1960
[tex]vf[/tex] = 44.27 m/s
b)I tried to solve this one...but it does not seem to be working.
Thanks for your help. Much appreciated.
*Ignore the _ that seems to be appearing after the variables. I don't know how to get rid of it.
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