N End_time_3) fprintf n fprintf n fprintf n fprintf n fprintf n fprintf n fprintf n the the the the the the the value value value value value value value for for for for for for for x1 x2 x3 x4 x5 x6. 1) build a padded character array of the metals called metal metal char Aluminum 'copper 'iron 'molybdenum 'cobalt 2) build a padded character array of the symbols called symbol symbol char Al 'cu 'fe 'mo 'co 3) build an integer array of the atomic numbers. If we look at the metal, the first metal is in the first row, however its charachers are in indivual collons, so that's why we use it is a symbol that allows to reach the max vale, without hardcoding. For k 1:length(A_num element_struct(k).aw aw(k element_struct(k).density density(k element_struct(k).structure structure(k end part Three: Find element with the highest density 1 first we will extract the element with the highest densisty density, vertical_location max(element_nsity now we will use something like the index method to find the vertical. After choosing the element, the user will be prompted to give some information of the element, it's atomic mass, density, structure and such. simply showing that we are going to keep with the index value in the previous created for loop. Kk; however because we don't want to overwrite any of the previous fields we must introduce a new lenght, 1 through 5 is already taken, so the next value would.
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R_inches -10;5;0; convert length into feet R_feet r_inches./12; your Answer Final_Answer cross(Vector, R_feet display fprintf For the x axis given bracket.6f fprintf For the y axis given bracket.6f fprintf For the z axis given bracket.6f For the for the. we are given lines of equations, but we are given an array of unknown variables, so before we can tackle the problem, we must first figure out the unknown variables. we can tackle this problem, and in fact we can use inverse method, or left division. so we are going to use both methods to tackle this problem, and time the process to see which method is more time efficient for Matlab first Problem 1st Matrix: Contains all the amounts of the unknown variables. A 3,4,2,-1,1,7,1; 2,-2,3,-4,5,2,8; 1,2,3,1,2,4,6; 5,10,4,3,9,-2,1; 3,2,-2,-4,-5,-6,7; -2,9,1,3,-3,5,1; 1,-2,-8,4,2,4,5; 2nd Matrix: Contain the answers of all equations. B42,32,12,-5,10,18,17; you have want to have the inner Dimensions of the matrix to agree, so you have to transpose one of the matrices. By transposing b is now 7x1, while a is 7x7, so now the inner values matrices equal bb inverse method we can use tic toc set up to time, we will first use mathematics to solve problem. We are using -1 tic X1 (A-1 b end_time_1 toc; Inverse method we can use tic toc set up to time, and we can to inv because the matrix is a square report (2x2 3x3 4x4.) tic 1 X2 inv(a b end_time_2 toc; left division. Tic X3 ab end_time_3 toc; re-formate the three values to have defined decimal count X1single(X1) X2single(X2) X3single(X3) if X1X2X3 fprintf When we use power to the -1 to perform inverse methodn to solve the problem Matlab takes.9f seconds to run the problem. N End_time_1) fprintf When we use the inv function perform inverse methodn to solve the nproblem Matlab takes.9f seconds to run the problem. N End_time_2) fprintf When we use backward slash to perform left divisionn to solve the problem Matlab takes.9f seconds to run the problem.
For Index 1:length(x y(Index) end disp for loop plus actual value of c toc Matrix math Elapsed time.001486 seconds. For loop square Elapsed time.085082 seconds. For loop plus c elapsed time.085998 seconds. For loop plus actual value of C 2 Elapsed time.110865 seconds. Published with matlab r2015b 3 business beacham Chapter 10 Problem 16 clc clear given force in equation. Force_lb 35; degree of the force Alpha_angle 55; the angle used for trig equations Alpha_angle_theta Alpha_angle/180*pi; creating matrix shooing relation note that we are working with trigonometry, so all values are radians. note that because the force is a downwards force, it will have a negative sign. Vector lenghts gvven of the brsckets Note: that tthe x axis is on the left from the origin, and therefore negative.
Disp for loop square toc.18d clear x y in this case we are going to follow the steops.18b, and add c in addition of squaring. X rand(1,100000 our answer will be known as y, and for the way the for loop works, we much equal it to x yx; new essay addition, from book. C start of timer. Tic for loop ant work. For Index 1:length(x y(Index) sqrt(x(Index)C; end End of timer. Disp for loop plus c toc.18d second our matrix with 100,000 random values, so we used rand function. X rand(1,100000 our answer will be known as y, and for the way the for loop works, we much equal it to x yx; rather than using c, we will add the actual value.
Table degree;radians disp degree to radians disp degree radians fprintf.0f.2f pi n Table) Table columns 1 through.00.00.00.00.00.06.11.17.22.28.00.39.00.44.00.50 100.00.56 130.00.72 140.00.78 150.00.83. And we are going to time how long Matlab will take to complete the loop. our matrix with 100,000 random values, so we used rand function. X rand(1,100000 our main component for the While loop. Tic y sqrt(x disp matrix math toc.18b clear x y essentially we are going to do the same.18a, but we are going to use the for loop to tackle this problem, (it's much easier to use this method for this type. X rand(1,100000 our answer will be known as y, and for the way the for loop works, we much equal it to x yx; Start of timer. Tic for loop going to work for Index 1:length(x y(Index) sqrt(x(Index end End of timer.
Matlab for Linear Algebra
case 5 disp you will need at least 129 credits to graduate. end disp Tnank you for using Matlab, have a nice day. you will need at least 122 credits to graduate. Tnank you for using Matlab, have a nice day. Published with matlab r2015b 1 beacham.3 tiger Example. Clc clear In this problem we will be making a table that will show the relationship of radians and degrees.
we will finding the coronation by using a loop, a while loop to be exact. Degree: This is the side we are starting with. Radians: we are going to convert our initial amount (degree) to get find out the radians (degree/180*pi) Index: name chosen as our index value, main component of While loop, logical component (Index 36) Table: The output. While loop taking place. Index 1; degree 0:10:360; radians degree pi/180 it should've been 36, but remember that Wile loops stop at the first unacceptable value, and because are max value is 360 it repeats, so that's why we did 36-135 to have all the numbers, without any repetitions. While Index 35 degree(Index)Index*10; radians(Index) degree(Index pi/180; Index Index1; end we're going to re-format the radians value to have a better representation.(2pi set up) radians radians/pi; making a visual representation of the comparison between degree and radians.
One_matrix_for_dimensions ones(51 the actuall Max height beam. Graphed_Max_Height this is the ground, anything below means the rocket is going underground. we use the zeros matrix to reach the required dimensions. Ground_level zeros(51 Actual graph. Graph plot(time, graphing info title rocket launched, problem.2 xlabel time, in seconds ylabel height, in meters axis(0,100,-3000,2000) grid on legend The rocket s movements Max height Ground level published with matlab r2015b 4 beacham.10 clc clear notation example: x 5 y 8, this. Disp In this experiment we will see comparisons between x, and y disp Only use nuubers to fill in the values.
x input Please plug in a value for X: y input Please plug in a value for Y: if x y disp The x value is greater than the y vaalue elseif y x disp your x value is less than the y canlue elseif. Error using input Cannot call input from evalc. Error in (line 9) x input Please plug in a value for X: published with matlab r2015b 1 bbeacham.16 clc clear Minimum_required_credits menu Please select a engineering Major civil Engineer Chemical Engineering Computer Engineer Electrical Engineer Mechanical Engineer switch Minimum_required_credits case 1 disp you. case 2 disp you will need at least 130 credits to graduate. case 3 disp you will need at least 122 credits to graduate. case 4 disp you will need at least 126.5 credits to graduate.
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Last_pos_Index max(Positive_Heights_Index) Equation.2 we will be looking for the collision time, and therefore answer gps the question. better projection of answer. Disp.2 problem a disp Time of impact in lined seconds disp(Time_of_collision) Positive_Heights_Index columns 1 through.00.00.00.00.00.00.00.00.00.00.00.00.00 Columns 6 through.00 Columns 11 through.00 Columns 16 through 20.00. Equation.3 we are looking for the highest distance the rocket went, and see the corresponding Index value for the rows (time). Maximum_Height, time_index max(Height Equation.4 we will now convert the Index value to the time frame it's representing. Time_of_Max_Heightt disp.2 problem b disp Time when maximum height was reached, in seconds disp(Time_of_Max_Heightt).2 problem b Time when maximum height was reached, in seconds.00.2c we want to make a graph that shows the relation of distance (height in meters) travelled. 3 This matrix is required to project the maximum height beam.
time_of_Max_Heightt: This will give us the time frames in which the 1 rocket reached its top height. variables required for all three problems. Time0:2:100;.2a we are finding at what time the rocket hit the floor it doesn't stop when it hits the ground though, it just keeps going in a downwards direction). we know that one index value represents two seconds (look at problem). Index_to_seconds 2; we also know that the index value is one less than the values shown pollution (zero has no index value, it starts at 1) so we have to make up for it by subtracing. Equation.1 we first have to find the points in which the height is still positive. calculated variable this is the final positive position given from equation.1.
of Index Values of all the heights that are above 0 (Positive values still in the air, and not underground). last_pos_Index: The final Index Value produced that stayed in accordance of the request. maximum_Height: This will be the highest the rocket has traveled in these conditions. time_index: The Index value that corresponds with the maximum_Height. Graphed_Max_Height: This is the version of the max height that can be disolayed onthe graph. Ground_level: This is the ground (x0) because the rocket goes underground, that's why this should be displayed, for greater emphasis. (Note i could've used the_one_matrix_for_dimensions, but matlab has a build in function to create matrixes completely out of zeros Outputs time_of_collision: This will give you the time when the rocket hit the floor.
Diameter7926, 4218; 2 to find the radians of the planets, you have to divide the dimater. Rad diameter./2; 3 The heights that we wish to find is an array. Hi_feet 0:1000:10000; 4 The conversion factor from feet to miles. Feet_Mile.000189394; 5 Note the heights need to be converted into miles to make it in the equation. Hi hi_feet.*Feet_Mile; The final answer, the distance traveled. Answer Distance(rad, hi disp distance traveled disp earth Mars remote disp(Answer) Distance traveled Earth Mars 0.7450.2648.7943.9734.1099.9583.4927.5334.6405.2077.9111.2421 102.5169.7917 109.5965.9575 116.2460.8097 122.5355.3993 Published with matlab r2015b 1 Table of Contents.17. Clc clear Angle_radians.0*pi:0.1*pi:2.0*pi; Sine_angle sin(Angle_radians Angle_degree angle_radians*180/pi; Table disp degrees to radians cross reference disp radiant, sin, degree fprintf.2f,.2f,.2f)n Table) degrees to radians cross reference radiant, sin, degree (0.00,.00,.00) (0.31,.31,.00) (0.63,.59,.00) (0.94,.81,.00) (1.26. time: a vector that goes from 0 to 100, with increments of 2(in seconds). height: distance traveled in the air (vertical, and in meters).
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Book solution "Matlab for Engineers holly moore - homework assignments chapter with. Beacham - stuDocu out of 46, beacham clc clear format bank rad0*0:0.10*pi:2.0*pi degree180/pi*rad tablerad degree' rad, columns 1 through 220.127.116.11.18.104.22.168.83.46.77.08.40.03.34.65.97.00.00.00.00 108.00 126.00 144.00 162.00. Columns 16 through.71, column.28 degree columns 1 through 5 0 Columns 6 through.00 Columns 11 through 15 180.00 198.00 Columns 16 through 20 270.00 288.00 1 Column 21 360.00 table.22.214.171.124.126.96.36.199. R*T) plot(T,k) title.4 xlabel K dgrees ylabel k summary values k columns 1 through.00.00.00.00.05.16.38 Columns 6 through.01.00.10 Rad0:0.1*pi:2.0*pi sin(Rad) cos(Rad) tan(Rad) Rad columns 1 through 188.8.131.52.26.88. Dat problem.4 a time1; sensor_12; sensor_23; sensor_34; sensor_45; sensor_56; beacham 9/16/13 problem.4c mean_sensor_1 mean(sensor sensor_1) std_sensor_1 std(sensor sensor_1) mean_sensor_2 mean(sensor sensor_2) std_sensor_2 std(sensor sensor_2) mean_sensor_3 mean(sensor sensor_3) std_sensor_3 std(sensor sensor_3) mean_sensor_4 mean(sensor sensor_4) std_sensor_4 std(sensor sensor_4) mean_sensor_5 mean(sensor sensor_5) std_sensor_5 std(sensor sensor_5) Problem. Answer_1 hm_p.*hw_p Or should've i've used meshgrid? Height_Mercury, mercury_Plus_Water height_Mercury.* height_Water; value_sensor_3.96 position_sensor_3.00 time_sensor_3.00 min_value_sensor_3.76 position_sensor_3.00 min_time_sensor_3.00 value_sensor_4.56 2 position_sensor_4.00 time_sensor_4.00 min_value_sensor_4.09 position_sensor_4.00 min_time_sensor_4.00 value_sensor_5.18 position_sensor_5.00 time_sensor_5.00 min_value_sensor_5.37 position_sensor_5.00 min_time_sensor_5.00 3 mean_sensor_1.73 std_sensor_1.