ENGR_1181_EXAM_2_Practice_Problem_Solutions_Rev_2015_11_19

# ENGR_1181_EXAM_2_Practice_Problem_Solutions_Rev_2015_11_19...

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ENGR 1181 | Midterm Exam 2: Practice Problem Solutions Problem 1: Solution clear close all clc disp( 'Urban Meyer, Seat #01' ); disp( 'ENGR 1181' ); disp( 'Practice Problem 1' ); disp( 'AU 14 Prof: Woody Hayes' ); disp( '11/4/14' ); g = 9.81; for i = 1:3 v0 = input( 'Enter the initial speed for this launch (m/s): ' ); theta = input( 'Enter the angle of launch (degrees): ' ); D(i) = v0^2*sin(2*theta/180*pi)/g; fprintf( 'For Launch %i the distance traveled was %.2f meters\n' ,i,D(i)); end MATLAB Command Window Page 1 Urban Meyer, Seat #01 ENGR 1181 Practice Problem 1 AU 14 Prof: Woody Hayes 11/4/14 Enter the initial speed for this launch (m/s): 50 Enter the angle of launch (degrees): 30 For Launch 1 the distance traveled was 220.70 meters Enter the initial speed for this launch (m/s): 25 Enter the angle of launch (degrees): 60 For Launch 2 the distance traveled was 55.17 meters Enter the initial speed for this launch (m/s): 36 Enter the angle of launch (degrees): 45 For Launch 3 the distance traveled was 132.11 meters >> Problem 2: Solution clear close all clc disp( 'Urban Meyer, Seat #01' ); disp( 'ENGR 1181' ); disp( 'Practice Problem 2' ); disp( 'AU 14 Prof: Woody Hayes' ); disp( '11/4/14' ); % Req = R1 + (1/R2 + 1/R3)^-1; % This is because R2 and R3 are in parallel % That combination is in series with R1 % % Vpower = I1*Req

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N ame : Se at N u mb e r: __________ % I1 = Vpower/Req % This is because I1 is the current through the power supply traveling % through the equivalent resistance % % V23 = Vpower - I1*R1 % This is because the voltage across the combination in parallel is the % difference between the supply voltage and the voltage across R1 % V23 = I2*R2 % I2 = V23/R2 % V23 = I3*R3 % I3 = V23/R3
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• Fall '14
• SEPTA Regional Rail, power supply, Drag equation, Woody Hayes, Urban Meyer

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