In the diagram, q1 = -3.33*10^-9 C and q2 = -5.22*10^-9 C. The electric field at
point P is zero. What is the distance
from P to q2?
91
P
92
?
0.255 m-4
(Remember, E points away from + charges,
and toward - charges.)
(Unit = m)

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

Can anybody me with this? ? (picture included.)

Two horses begin at rest. after a few seconds, horse a is traveling with a velocity of 10 m/s west, while horse b is traveling with a velocity of 13 m/s west. which horse had a greater acceleration? a. horse a because its change in velocity was greater than horse b's in the same time. b. horse b because its change in velocity was greater than horse a's in the same time. c. horse b because its change in velocity was the same as horse a's, but it took less time. d. both horses had the same acceleration.