Exlpain the process of harvesting

Copper has free electrons per cubic meter. a 71.0-cm length of 12-gauge copper wire that is 2.05 mm in diameter carries 4.85 a of current. (a) how much time does it take for an electron to travel the length of the wire? (b) repeat part (a) for 6-gauge copper wire (diameter 4.12 mm) of the same length that carries the same current. (c) generally speaking, how does changing the diameter of a wire that carries a given amount of current affect the drift velocity of the electrons in the wire?

Fft review: linspace, fs, fftshift, nfft 1. generate one second of a cosine of w,-10hz sampled at f, = 100hz and assign it to x. define a tt as your time axis 2. take 64 points fft. 3. as you remember, the dft (which the fft implements) computes n samples of s2t where k-0,1,2, n -1. plot the magnitude of this 64-points fft at range 0 to 63, what do you think of this graph? 4â·to get the x-axis into a hz-frequency form, plot this 64-points fft between-50 to 50 (the 100hz sampling rate) and have n-points between them. 5. according to your figure, what frequency is this cosine wave at? 6. remember that the fft is evaluating from 0 to 2ď€. we are used to viewing graphs from-ď€ to ď€. therefore, you need to shift your graph. 7. now according to your shifted graph. what frequency is this at? 8. note that the spikes have long drop-offs? try a 1024-point dft. note that the peak is closer to 10 and the drop-off is quicker. although, now sidelobes are an issue

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).