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# Homework - a-1.8.6.4.2 0.2.4.6.8 1-6-4-2 2 4 6 time(s...

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Homework #3 1. a) b) x = v 0 t + 1/2at 2 x/t = v 0 + at/2 Plotting x/t vs t/2 will result in a linear plot with a slope equal to a(=g*sinθ), and a y intercept equal to v 0 .

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 50 100 150 200 250 300 time/2 (s) position/time (cm/s) position/time vs. time/2 c) m = 480 ± 20 cm/s 2 b = 24 ± 5 cm/s d) a = 480 ± 20 cm/s 2 = 4.8 ± 0.2 m/s 2 a = g(sin θ) g = [4.8 ± 0.2 m/s 2 ] / sin(30) = 9.6 ± 0.4 m/s 2 v 0 = 24 ± 5 cm/s = 0.24 ± 0.05 m/s
e) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 50 100 150 200 250 300 position/time vs time/2 time/2 (s) position/time (cm/s) line of best fit data y = 480x + 24 f) The experiment was successful; the measured value of g is within one standard deviation of the accepted value of 9.8 m/s 2 .

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2.

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Unformatted text preview: a) -1-.8 -.6 -.4 -.2 0 .2 .4 .6 .8 1-6-4-2 2 4 6 time (s) amplitude (V) amplitude vs. time y = 3cos(400*pi*t ) y = 6cos(200*pi*t + pi/4) b) Assuming CH1 = X and CH2 = Y, the Lissajous figure is: -3-2-1 1 2 3-6-4-2 2 4 6 3. a) 5 10 15 20 25 30 35 40 45 50 5 10 15 20 25 30 35 40 45 50 number of returns height (cm) height vs. number of returns This relationship is given by the function H = 50*e-0.05i b) (.409)(mgh) = 2.0041 J [assume mass is 1 kg] the corresponding height for this energy would be 2.0041 J /9.8 m/s 2 = 20.45 cm 20.45 = 50*e-0.05i .409 = e-0.05i ln(.409) = -0.05i i = 18 returns...
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## This note was uploaded on 04/07/2008 for the course PHYS 122L taught by Professor Gougousi during the Spring '07 term at UMBC.

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Homework - a-1.8.6.4.2 0.2.4.6.8 1-6-4-2 2 4 6 time(s...

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