An electric scooter has a battery capable of supplying 120 Wh of energy. If friction forces and other losses account for 60.0% of the energy usage, what altitude change can a rider achieve when driving in hilly terrain, if the rider and scooter have a combined weight of 890 N? @Physics

if 60% of battery energy is wasted on friction, 40% of the remaining energy is used to climb on inclined road. 120(0.4)=48 Wh relate energy to force. \[N.m = W.s\] Where J is joules, N is newtons, m is meter, W is watt and s is seconds 48 Wh = 48(60 x 60 x 24) = 48(3600) = 172800 W.s if we say that the road was horizontal, then m = w.s/n where n would be very small and the scooter will go longer distance. remember the vertical forces cancel, the horizontal force of the scooter that climbs it up must be equal or greater then horizontal force of gravity on the scooter which drags it down hill. \[Ncos\theta.m=W.s\] \[m = W.s/Ncos\theta\] as you can see from the last equation, it all depends on the angel of the inclination, if the angel approaches 90 degree, the denominator gets bigger and it will go less distance, if angle is closer to 0, cosine approaches 1 and distance is then more. references: http://en.wikipedia.org/wiki/Joule http://en.wikipedia.org/wiki/Watt#Confusion_of_watts.2C_watt-hours.2C_and_watts_per_hour

plz double check my answer, as i am not sure, i just made it up :P

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