Astudent stands on the edge of a merry-go-round which rotates five times a minute and has a radius of two meters one evening as the sun is setting. the student produces a shadow on the nearby building. (a) write an equation for the position of the shadow. (b) write an equation for the velocity of the shadow.

To solve this problem it is necessary to apply the concepts related to the simple harmonic movement, to the speed in terms of displacement and the timpo, as well as the angular frequency and the period of frequency.

PART A) According to the description given, 5 revolutions are made in one minute (or 60 seconds) that is to say that the frequency would be given by

Therefore the angular velocity can be found as

The displacement that determines the maximum displacement based on the angular velocity time and the time in the simple harmonic movement, is equal to the radius of the circle, in other words

Where,

A = Amplitude

Angular velocity

t = time

If for our given values the value of the amplitude is 2m and the value of the angular velocity is 0.52rad/s

So the equation for the position of the shadow is of

PART B) The equation for the velocity of the shadow is calculated as a expression of the displacement against the time, if we differenciate the previous value found, we have that,

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).