As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. she runs at 6.0 m> s to catch the bus, but the door closes when she’s still 8.0 m behind the door, and the bus leaves the stop at a constant acceleration of 2.0 m> s2. she has missed her bus, but as a physics exercise she keeps running at 6.0 m> s until she draws even with the bus door. (a) calculate ∆t from the instant the bus leaves the stop to the instant the instructor draws even with the door. (b) at the instant she draws even, how fast is she moving? (c) at the instant she draws even, how fast is the bus moving? (d) plot v(t) for the bus and the instructor. (e) plot x(t) for the bus and the instructor. ⦁⦁⦁

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explanation:

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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).