posted by QUESTION .
this one is kind of hard
A bicyclist can coast down a 7.0 degree hill at a steady 9.5 h^-1 km. If the drag force is proportional to the square fo the speed v, so that Fd = cv^2, calculate
(a) the value fo the constant c and
(b) the average force that must be applied in order to descend the hill at 25 h^-1 km. The mass of the cyclist plus bicycle is 80.0 kg. Ignore other types of friction.
ok for (a) I got 6.7 x 10^-13 m^-1 kg
for (b) I am completly lost the way it's put you would assume there to be zero acceleration so how are you suppose to solve (b)???
wait I think i got it lol
can you tell me if my answer for (a) is right???
ok fixing answer for (a)
ok how is this for (a)
8.2 x 10^-14 m^-1 kg
ok for (a)
13.719 m^-1 kg
ok with correct sig figs (a)
13.7 m^-1 kg
ok i'm stuck on part (b)
part b ...
a velocity of 25 h^-1 km
is it constant velocity were there would be no acceleration if so then there would be no net force...
so i'm lost
ok for (a)
13.7 s^-1 kg
for (b) i got
-.4172 N ;[
ok I know this is right I redid it
(a) 13.7 m^-1 kg
ok for b i got 565 N which was correct with book answer in back
constant speed....so force drag=force gravity
c= mgsin7deg/(2.64m/s)^2 So I don't get your answer for a.
ok for drag force
c = v^-2 Fgx = (2.639 s^-1 m)^-2 95.5 N
Fgx = mg sin theta