A 2000 kg car is traveling at 15.0 m/s down a long mountain grade of 2.0%. The mountain road is 5.0 km long. The heat generated at the brakes due to friction heats up the brakes. The mass of the brake system is 20.0 kg and has a specific heat of 0.200 kcal/kg °C. What is the increase in the temperature of the brakes?a) 43°C b) 54°C c) 83°C d) 95°C e) 117°C
No, calipers are rebuilt, not replaced.
It would REALLY help if you could give some more info. Like, what is the make of bike/caliper? How old is it? Do you use the bike all year round, or just in good weather?
take off the caliper from the rotor but leave it assembled then press the brake pedal you should see the pistons pushing the brake pads if not try pushing the pistons in to free them using a c clamp or tool made for that,they may be stuck and need to be loosened then see if they move using the pedal,if you cant free them you will need to rebuild them
exchange in KE is going into elevating the temp of the iron 965 kg is going from 15.5 m/s to 0 what proportion joules of skill is this? convert this to energy. what proportion tiers will this many energy advance the temp of a million gram of iron? now on the grounds which you have 17.7 kg of iron what proportion tiers does it characterize?
Assuming the car maintains a constant speed, there's no air resistance and for whatever reason the brake system stores up all the heat, then the answer is e) 117°C. You can work this out from the fact that the car drops a vertical displacement of 100 m, and all of this gravitational potential energy is dumped into the brake system. GPE 2000 * 9.81 * 100 GPE 1,962,000 J GPE 469 kcal (kcal in a physics problem, seriously??) Q mcΔT 469 20 * 0.2 * ΔT 469 4 * ΔT ΔT 469 / 4 ΔT 117 °C
