A passenger bus in Zurich, Switzerland derived its motive power from the energy stored in a large flywheel. The wheel was brought up to speed periodically, when the bus stopped at a station, by an electric motor, which could then be attached to the electric power lines. The flywheel was a solid cylinder with a mass of 1090kg and a diameter of 1.90m ; its top angular speed was 2950rev/min .

Part A

At this angular speed, what is the kinetic energy of the flywheel?

Part B

If the average power required to operate the bus is 1.85

Explanation:

moment of inertia of flywheel

= 1/2 m R²

= .5 x 1090 x .95²

I = 491.8625 kgm²

angular speed ω = 2πn where n = revolution per second

= 2π x 2950 / 60

= 308.77 rad /s

Rotational kinetic energy

= 1/2 Iω²

= .5 x 491.8625 x 308.77²

= 23.446 x 10⁶ J

B )

power being used by bus = 1.85 x 10⁴ W .

energy provided by flywheel = 23.446 x 10⁶ J

energy used per second =  1.85 x 10⁴ J

Time duration of travel between two station

= 23.446 x 10⁶ / 1.85 x 10⁴

= 1267 s

21.11 minutes.

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).