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unimatix
 one year ago
Sam, whose mass is 71.0 kg , takes off down a 51.0 m high, 11.0 ∘ slope on his jetpowered skis. The skis have a thrust of 210 N . Sam's speed at the bottom is 41.0 m/s . What is the coefficient of kinetic friction of his skis on the snow?
unimatix
 one year ago
Sam, whose mass is 71.0 kg , takes off down a 51.0 m high, 11.0 ∘ slope on his jetpowered skis. The skis have a thrust of 210 N . Sam's speed at the bottom is 41.0 m/s . What is the coefficient of kinetic friction of his skis on the snow?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0accelaration of sam=[(45^2).sin12]/(2*54)=A mass=65=M kinetic fric.=K gravity=g MAsin12K(Mgcos12)=150 solve this then ull get K

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0Where did you get those numbers from?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443794291333:dw resolving forces down the plane we have \(\large 210 + mg \sin 11  \mu mg \cos 11 = ma\) where \(\mu \) is the coefficient of friction, m = 71kg, g is gravity, a = acceleration we know that acceleration a from \(\large v^2 = u^2 + 2ax\) ie \(\large 41^2 = 0^2 + 2 \;a (267.3)\). 267.3 is the length of the slope, you have all you need to get its dimensions from trig. put that together and solve for \(\mu\)
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