Write a program whose source file is named UnitProgl.java that do.pdf

•

0 likes•2 views

Write a program whose source file is named \"UnitProgl.java\" that does the following: Creates a frame with a GridLayout Manager Creates two panels and adds them to the frame, stacked one over the other. In each panel creates three buttons, side by side. The buttons should be labeled \"Button 1\", \"Button 2\", \"Button 3\", \"Button 4\", \"Button 5\" and \"Button 6\". Displays this GUI when run. The only buttons that actually works will be the buttons on the window title bar: minimize, maximize and the exit button in the upper right hand corner of the programs window. Part 2 - Event Driven Programming Write a program whose source file is named \"UnitProg2.java\". Do the following: Start by copying the program code from part 1 and modifying it. This program will have a GUI that looks just like the previous one. The goal is to make the JButtons work. Add code to handle the events when the user presses any of the six buttons on the GUI. When the user presses a button simply write out to the console \"The user pressed Button X\", where \"X\" is the number of the button. For example, when \"Button 2\" is pressed the message written to the console should be \"The user pressed Button 2.\". Solution UnitProg1.java import java.awt.*; import javax.swing.*; public class UnitProg1 { private JFrame mainFrame; private JPanel controlPanel; private JPanel controlPanel1; public UnitProg1(){ prepareGUI(); } public static void main(String[] args){ UnitProg1 prog1 = new UnitProg1(); } private void prepareGUI(){ mainFrame = new JFrame(\"UnitProg1\"); mainFrame.setSize(400,400); mainFrame.setLayout(new GridLayout(2,1)); controlPanel = new JPanel(); controlPanel.setLayout(new FlowLayout()); controlPanel1=new JPanel(new FlowLayout()); JButton button1=new JButton(\"Button1\"); JButton button2=new JButton(\"Button2\"); JButton button3=new JButton(\"Button3\"); JButton button4=new JButton(\"Button4\"); JButton button5=new JButton(\"Button5\"); JButton button6=new JButton(\"Button6\"); controlPanel.add(button1, controlPanel); controlPanel.add(button2, controlPanel); controlPanel.add(button3, controlPanel); controlPanel1.add(button4, controlPanel1); controlPanel1.add(button5, controlPanel1); controlPanel1.add(button6, controlPanel1); mainFrame.add(controlPanel); mainFrame.add(controlPanel1); mainFrame.setVisible(true); } } UnitProg2.java import java.awt.*; import java.awt.event.ActionEvent; import java.awt.event.ActionListener; import java.awt.event.ItemEvent; import java.awt.event.ItemListener; import javax.swing.*; public class UnitProg2 { private JFrame mainFrame; private JPanel controlPanel; private JPanel controlPanel1; public UnitProg2(){ prepareGUI(); } public static void main(String[] args){ UnitProg2 prog2 = new UnitProg2(); } private void prepareGUI(){ mainFrame = new JFrame(\"UnitProg2\"); mainFrame.setSize(400,400); mainFrame.setLayout(new GridLayout(2, 1)); controlPanel = new JPanel(); controlPanel.setLayout(new FlowLayout()); controlPanel1=new JPanel(new F.

Education

mainFrame.setSize(400,400);
mainFrame.setLayout(new GridLayout(2,1));
controlPanel = new JPanel();
controlPanel.setLayout(new FlowLayout());
controlPanel1=new JPanel(new FlowLayout());
JButton button1=new JButton("Button1");
JButton button2=new JButton("Button2");
JButton button3=new JButton("Button3");
JButton button4=new JButton("Button4");
JButton button5=new JButton("Button5");
JButton button6=new JButton("Button6");
controlPanel.add(button1, controlPanel);
controlPanel.add(button2, controlPanel);
controlPanel.add(button3, controlPanel);
controlPanel1.add(button4, controlPanel1);
controlPanel1.add(button5, controlPanel1);
controlPanel1.add(button6, controlPanel1);
mainFrame.add(controlPanel);
mainFrame.add(controlPanel1);
mainFrame.setVisible(true);
}
}
UnitProg2.java
import java.awt.*;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.awt.event.ItemEvent;
import java.awt.event.ItemListener;
import javax.swing.*;
public class UnitProg2 {
private JFrame mainFrame;
private JPanel controlPanel;

private JPanel controlPanel1;
public UnitProg2(){
prepareGUI();
}
public static void main(String[] args){
UnitProg2 prog2 = new UnitProg2();
}
private void prepareGUI(){
mainFrame = new JFrame("UnitProg2");
mainFrame.setSize(400,400);
mainFrame.setLayout(new GridLayout(2, 1));
controlPanel = new JPanel();
controlPanel.setLayout(new FlowLayout());
controlPanel1=new JPanel(new FlowLayout());
JButton button1=new JButton("Button1");
button1.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
System.out.println("The user pressed Button 1");
}
});
JButton button2=new JButton("Button2");
button2.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
System.out.println("The user pressed Button 2");
}
});
JButton button3=new JButton("Button3");
button3.addActionListener(new ActionListener() {

@Override
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
System.out.println("The user pressed Button 3");
}
});
JButton button4=new JButton("Button4");
button4.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
System.out.println("The user pressed Button 4");
}
});
JButton button5=new JButton("Button5");
button5.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
System.out.println("The user pressed Button 5");
}
});
JButton button6=new JButton("Button6");
button6.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
System.out.println("The user pressed Button 6");
}
});
controlPanel.add(button1, controlPanel);
controlPanel.add(button2, controlPanel);

controlPanel.add(button3, controlPanel);
controlPanel1.add(button4, controlPanel1);
controlPanel1.add(button5, controlPanel1);
controlPanel1.add(button6, controlPanel1);
mainFrame.add(controlPanel);
mainFrame.add(controlPanel1);
mainFrame.setVisible(true);
}
}

Similar to Write a program whose source file is named UnitProgl.java that do.pdf

Getting started with GUI programming in Java_1

Muhammad Shebl Farag

1) Write a shortsnippetofcodethatcreates a J Panel objectcalled p1, and creates a border around it containing the title \"this is a JPanel.\" 2) Write a keyPressed method that behaves as follows. If the user presses the up arrow, the method should output \"You pressed up\"using the System.out.printin method. Ifthe user presses the down arrow, the method should output \"You pressed down\" using the System.out.printin method. 3) Write a short class that represents a panelwith a single radio button that has the option \"Yes\"and the option \"No.\" Bydefault, the Yes buttonshould be checked. 4) Suppose we have created a class called MyGUI, which represents a GUI. Write a programthat creates a JFrame object, adds a MyGUI object to the frame and makes it visible. 5) Write a short class that represents a panelwith a single sliderthat has values from 0 to 250, with largetick marks in incrementsof 50 and smalltick marks in increments of 10. Solution 1) /** * The java Demo1 that cretes a titled border * for jpanel. * */ //Demo1.java import javax.swing.JFrame; import javax.swing.JPanel; import javax.swing.border.TitledBorder; public class Demo1 { public static void main(String[] args) { JFrame frame=new JFrame(); //Create an instance of Jpanel JPanel p1=new JPanel(); //Create an instance of TitledBorder TitledBorder border = new TitledBorder(\"this is a JPanel\"); border.setTitleJustification(TitledBorder.CENTER); //add border to panel ,p1 p1.setBorder(border); //add panle to frame frame.add(p1); frame.setSize(250, 250); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); frame.setVisible(true); } } 2) /** * The java Demo2 that display a message * for arrow up is pressed and arrow down * key is pressed * * */ //Demo1.java import java.awt.event.KeyEvent; import java.awt.event.KeyListener; import javax.swing.JFrame; import javax.swing.JPanel; import javax.swing.border.TitledBorder; public class Demo2 { public static void main(String[] args) { JFrame frame=new JFrame(); //Create an instance of Jpanel JPanel p1=new JPanel(); //Create an instance of TitledBorder TitledBorder border = new TitledBorder(\"this is a JPanel\"); border.setTitleJustification(TitledBorder.CENTER); //add border to panel ,p1 p1.setBorder(border); p1.setFocusable(true); //Add a KeyListener interface to the panel,p1 p1.addKeyListener(new KeyListener() { @Override public void keyTyped(KeyEvent e) { // TODO Auto-generated method stub } @Override public void keyReleased(KeyEvent e) { // TODO Auto-generated method stub } @Override /*Write to display pressed up and pressed down */ public void keyPressed(KeyEvent e) { int keyCode = e.getKeyCode(); switch( keyCode ) { case KeyEvent.VK_UP: System.out.println(\"You pressed up\"); break; case KeyEvent.VK_DOWN: System.out.println(\"You pressed down\"); break; } } }); //add panle to frame frame.add(p1); frame.setSize(250, 250); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); frame.setVisible(true); } } 3) /** * the java program that deomons.

1) Write a shortsnippetofcodethatcreates a J Panel objectcalled p1, .pdf

optokunal1

14a-gui.ppt

DrDGayathriDevi

ch20.pptx

EnriqueMartinezTelle2

Working with Layout Managers. Notes: 1. In part 2, note that the Game class inherits from JPanel. Therefore, the panel you are asked to add to the center of the content pane is the \"game\" object. 2. In part 4, at the end of the function, call validate(). This is not mentioned in the book, but it is mentioned in the framework comments. import javax.swing.*; import java.awt.*; import java.awt.event.*; public class Game extends JPanel { private JButton [][] squares; private TilePuzzle game; public Game( int newSide ) { game = new TilePuzzle( newSide ); setUpGameGUI( ); } public void setUpGame( int newSide ) { game.setUpGame( newSide ); setUpGameGUI( ); } public void setUpGameGUI( ) { removeAll( ); // remove all components setLayout( new GridLayout( game.getSide( ), game.getSide( ) ) ); squares = new JButton[game.getSide( )][game.getSide( )]; ButtonHandler bh = new ButtonHandler( ); // for each button: generate button label, // instantiate button, add to container, // and register listener for ( int i = 0; i < game.getSide( ); i++ ) { for ( int j = 0; j < game.getSide( ); j++ ) { squares[i][j] = new JButton( game.getTiles( )[i][j] ); add( squares[i][j] ); squares[i][j].addActionListener( bh ); } } setSize( 300, 300 ); setVisible( true ); } private void update( int row, int col ) { for ( int i = 0; i < game.getSide( ); i++ ) { for ( int j = 0; j < game.getSide( ); j++ ) { squares[i][j].setText( game.getTiles( )[i][j] ); } } if ( game.won( ) ) { JOptionPane.showMessageDialog( Game.this, \"Congratulations! You won!\ Setting up new game\" ); // int sideOfPuzzle = 3 + (int) ( 4 * Math.random( ) ); // setUpGameGUI( ); } } private class ButtonHandler implements ActionListener { public void actionPerformed( ActionEvent ae ) { for( int i = 0; i < game.getSide( ); i++ ) { for( int j = 0; j < game.getSide( ); j++ ) { if ( ae.getSource( ) == squares[i][j] ) { if ( game.tryToPlay( i, j ) ) update( i, j ); return; } // end if } // end inner for loop } // outer for loop } // end actionPerformed method } // end ButtonHandler class } // end Game class import javax.swing.*; import java.awt.*; import java.awt.event.*; public class NestedLayoutPractice extends JFrame { private Container contents; private Game game; private BorderLayout bl; private JLabel bottom; // ***** Task 1: declare a JPanel named top // also declare three JButton instance variables // that will be added to the JPanel top // these buttons will determine the grid size of the game: // 3-by-3, 4-by-4, or 5-by-5 // Part 1 student code starts here: // Part 1 student code ends here. public NestedLayoutPractice() { super(\"Practicing layout managers\"); contents = getContentPane(); // ***** Task 2: // instantiate the BorderLayout manager bl // Part 2 student code starts here: // set the layout manager of the content pane contents to bl: game = new Game(3); // instantiating the GamePanel object // add panel (game) to the center of the content pane // Part 2 student code ends here. bottom = new JLabel(.

Working with Layout Managers. Notes 1. In part 2, note that the Gam.pdf

udit652068

Swing

Nataraj Dg

intro_gui

filipb2

Programming Without Coding Technology (PWCT) - Simple GUI Application

Mahmoud Samir Fayed

03_GUI.ppt

DrDGayathriDevi

Introduction

شكسبير ويليام

I need help creating a java gui that draws lines, shapes, characters in response to mouseDragged, mousePressed, keyTyped. The user should be able to control the size using a Jslider, and control the color of all the shapes, lines, and letters using JcolorChooser and be able to control the Font of the letters. Solution import java.awt.*; import java.awt.event.*; import javax.swing.*; public class DPanel extends JFrame { private DObjects panel = new DObjects(); private JPanel BPanel = new JPanel(); private JFrame window = new JFrame(); DPanel(){ buildGUI(); } void buildGUI(){ window.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); window.setLayout(new GridLayout(2,2)); window.add(panel); window.add(BPanel); BPanel.setBackground(Color.blue); //define buttons and add to panel JButton rect = new JButton(\"Rect\"); JButton oval = new JButton(\"Oval\"); BPanel.add(rect); BPanel.add(oval); rect.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { panel.setType(1); } }); oval.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { panel.setType(2); } }); window.setVisible(true); window.setSize(1024, 800); } public static void main(String[] args) { new DPanel(); } } import java.awt.*; import java.awt.event.*; import javax.swing.*; class DObjects extends JPanel { public int x1,x2,y1,y2; public int type = 1; public DObjects() { init(); } public void init(){ setBackground(Color.WHITE); addMouseListener(new MouseAdapter() { public void mousePressed(MouseEvent m) { x1 = m.getX(); y1 = m.getY(); repaint(); } public void mouseReleased(MouseEvent m) { x2 = m.getX(); y2 = m.getY(); repaint(); } }); addMouseMotionListener(new MouseMotionAdapter() { public void mouseDragged(MouseEvent m) { x2 = m.getX(); y2 = m.getY(); repaint(); } }); } public void setType(int arg){ if(arg == 1){ type = 1; }else if(arg == 2){ type = 2; } } public void paintComponent(Graphics g) { super.paintComponent(g); if(type == 1) { g.drawRect(x1,y1,x2,y2); } else if (type == 2) { g.drawOval(x1,y1,x2,y2); } } }.

I need help creating a java gui that draws lines, shapes, characters.pdf

ezhilvizhiyan

step by step to write a gnome-shell extension

Yuren Ju

Program klik sederhana

Henfry Kai

Ingles 2do parcial

Harry Ostaiza

The Ring programming language version 1.7 book - Part 76 of 196

Mahmoud Samir Fayed

Programming Without Coding Technology (PWCT) - Spinner control

Mahmoud Samir Fayed

Programming Without Coding Technology (PWCT) - Image control

Mahmoud Samir Fayed

Programming Without Coding Technology (PWCT) - Progressbar control

Mahmoud Samir Fayed

I hope the below code is helpful to you, please give me rewards.im.pdf

apexelectronices01

Programming Without Coding Technology (PWCT) - Grid control

Mahmoud Samir Fayed

Similar to Write a program whose source file is named UnitProgl.java that do.pdf (20)

Getting started with GUI programming in Java_1

1) Write a shortsnippetofcodethatcreates a J Panel objectcalled p1, .pdf

14a-gui.ppt

ch20.pptx

Working with Layout Managers. Notes 1. In part 2, note that the Gam.pdf

Swing

intro_gui

Programming Without Coding Technology (PWCT) - Simple GUI Application

03_GUI.ppt

Introduction

I need help creating a java gui that draws lines, shapes, characters.pdf

step by step to write a gnome-shell extension

Program klik sederhana

Ingles 2do parcial

The Ring programming language version 1.7 book - Part 76 of 196

Programming Without Coding Technology (PWCT) - Spinner control

Programming Without Coding Technology (PWCT) - Image control

Programming Without Coding Technology (PWCT) - Progressbar control

I hope the below code is helpful to you, please give me rewards.im.pdf

Programming Without Coding Technology (PWCT) - Grid control

More from arihantcomputersddn

write an assemblyprogram that uses indirect addressing to switch the 1st and 2nd elements of the dword array. Then switch the 3rd and 4th elements, and so on. Then output the final contents of the array dwarraay. Use this template INCLUDE Irvine32.inc .data dwarray dword 0,2,5,9,10,12 .code main proc all waitMsg exit mainENDP END main Solution section .text global _start ;must be declared for linker (ld) _start: ;tells linker entry point mov edx,len ;message length mov ecx,msg ;message to write mov ebx,1 ;file descriptor (stdout) mov eax,4 ;system call number (sys_write) int 0x80 ;call kernel mov eax,1 ;system call number (sys_exit) int 0x80 ;call kernel section .data.

write an assemblyprogram that uses indirect addressing to switch the.pdf

arihantcomputersddn

Why is there no blood supply that extends into the epidermis? Thanks. Why is there no blood supply that extends into the epidermis? Thanks. Solution Answer: The epidermis has a stratified squamous epithelium, which normally undergoes ecdysis or moulting & do not require nutritious supply so that there no blood supply that extends into the epidermis. Epidermis is the outermost layer of the skin made of keratinized stratified squamous epithelial tissue in which outer or apical surface cells are hardened or dead. It does not contain blood vessels (not vascularized), and the dermal layer supplies oxygen and nutrients by means of diffusion. The layer of the epidermis that contains transulacent cells filled keratin are often referred as “thick stratum lucidum”. The other layers often sloughed off from the surface composed of dead cells that are flat and filled with keratin often referred as “stratum corneum”.

Why is there no blood supply that extends into the epidermis Than.pdf

arihantcomputersddn

Which of the following protocols was developed in the mid-1970s for use in Systems Network Architecture (SNA) environments? A. SDLC B. ISDN C. LAP-B D. X.25 Solution A. The Synchronous Data Link Control (SDLC) protocol was developed in the mid-1970s for use in Systems Network Architecture (SNA) environments. SDLC is unique in that it was the first synchronous, link layer, bit-oriented protocol. The ISO modified SDLC to create the High- Level Data Link Control (HDLC) protocol and release it as a standard. Answer B is incorrect because ISDN is an end-to-end telephone service that is digital in nature. Answer C is incorrect because Link Access Procedure–Balanced (LAP-B) is a subset of HDLC and is not used by SNA. Answer D is incorrect because X.25 is an efficient protocol developed in the 1970s for packet-switched networks..

Which of the following protocols was developed in the mid-1970s for .pdf

arihantcomputersddn

Which of the following is an weak implementation of EAPA. EAP-FAS.pdf

arihantcomputersddn

Where are chaperones found? periplasm, cytoplasm, or cell membrane (can be any combination of those) Solution Chaperones are found in all types of cells and in most of the cell organelles. They are found in both periplasm and cytoplasm as the chaperones are involved in proper folding of protein molecules, once they are translated. There are some chaperones which are also found in extracellular spaces. These chaperones are involved in functional folding of those protein molecules which are to be released extracellularly..

Where are chaperones found periplasm, cytoplasm, or cell membrane (.pdf

arihantcomputersddn

Talks about McDonaldization What do you think of this concept? Do you see a standardization of the world around us? What do you like and dislike about the standardization of society? Solution McDonaldization is the process of rationalization, albiet taken to extreme levels Rationalization is a sociological term that simply means the substitution of logically consistent rules for traditional (or illogical) rules One of the fundamental aspects of McDonaldization is that almost any task can (and should) be rationalized example: the premise of fast food often turns out to be just the opposite - long waits in lines. Fast food is not necessarily good food - in fact, McDonald\'s food is extremely unhealthy and the taste is average and bland. The system of efficiently producing and distributing their food has some other consequences, namely millions of tons of trash each year (disposability) and a food cultivation system of questionable ethics the four main dimensions of McDonaldization are: Dislike of standardization---- 1. Irrationality of Rationality 2.De-McDonaldization 3.cultural hybridization.

Talks about McDonaldization What do you think of this concept Do yo.pdf

arihantcomputersddn

What kind of state information is maintained by an FTP server? Solution In a FTP session, the user should be sitting in front of one host machine and wants to transfer files to or from a remote host. In order for the user to access the remote account, the user must provide user id and a password. After the authorization information is provided and correct, the user can transfer files from the local file system to the remote file system and vice versa. The FTP server sholud maintain state about the user. Server must associate the control connection with a specific user , and the server must keep track of the user\'s current directory as the user knows about the remote directory. Keeping track of this state information of each user and every session, the total number of session that FTP can maintain simultaneously..

What kind of state information is maintained by an FTP serverSo.pdf

arihantcomputersddn

When H_p is equal to r2, the stable equilibrium is K2 and referred .pdf

arihantcomputersddn

The standard deviation of all possible values is called thea. stan.pdf

arihantcomputersddn

The difference between bipolar I disorder and bipolar II disorder is: Select one: a. the seasonal variation in the episodes. b. the number of depressive and manic episodes. c. the number of depressive episodes. d. the severity of the manic episodes. Solution biolar disorders are mental health problems mainly affects your mood. these disorders mainly characterzied by mania and depression. bipolar disorders are classified as type I and II. in type I is characterized by one or more manic episodes or mixed episodes (mania and depression) but is marked by extreme manic episodes. in bipolar disorders II is characterized by one or more major depressive episodes and at least one hypomania. as see above both types I and II there is occurence of depressive episodes but main difference is the severity of the manic episodes. type I -mania but in type II hypomania.

The difference between bipolar I disorder and bipolar II disorder is.pdf

arihantcomputersddn

solve 6 , quiz 2 link of book :http://www.irccyn.ec- nantes.fr/~martinet/Mobrob/FundamentalsofVehicleDynamics.pdf 122% Document Elements Tables charts Smart Art Review A Home Layout Paragraph Style Cambria (Body) 12 Aa A AaBbc AaBbCcDdE Normal No spacing Heading 1 Heading 2 Tit Problem 6 Propose a currently non-commercially available car mechatronics system. Describe the sensors, ac uators and control system you ntend to use. Explain the purpose of this innovation and specify what the added value is for the user [i.e why would anyone want to buy it?]. Print Layout View Sec 1 Pages: rds: 32 of 45 Search in Document Aa Text Box Shape Picture Themes 22% O Solution COMPUTER AIDED DESIGN OF MECHATRONIC SYSTEMS I Use of advanced computer tools in the product development process is necessary to meet the market needs for fast product development. I These tools allows for the verification of the design in all phases of the development process. I Virtual prototypes, I The main drawback of using the computer tools is that they are mostly limited to specific domains. The main requirement to the computer tools for mechatronic system design I to apply to the systems from various domains. I to simultaneously use different system representations (block diagram, physical model, bond graph) Matuško&Koloni 2 of 43 MECHATRONICS SYSTEMS 2014/2015 CLASSICAL (NON- MECHATRONIC) OPTIMIZATION I Classical approach to the optimization during the design process assumes that the parameters of the process, sensors, actuators are known and constant and only controller parameters are optimized. CONTROLLER PROCESS INTERFACE SENSOR OPTIMIZATION + - yR(t) y(t) Matuško&Koloni 3 of 43 MECHATRONICS SYSTEMS 2014/2015 OPTIMIZATION IN MECHATRONIC SYSTEM I Since the main characteristic of the mechatronic design approach is simultaneous design of all system components, the optimization process is performed, not only on the controller parameters but also on the parameters of the other parts of the system. CONTROLLER PROCESS INTERFACE SENSOR OPTIMIZATION + - yR(t) y(t) Matuško&Koloni 4 of 43 MECHATRONICS SYSTEMS 2014/2015 MODELS OF MECHATRONIC SYSTEMS I Physical models, I Bond graphs I Control oriented models I Block diagrams, I Bode diagrams, I Nyquist diagram, I Sate space models. I Time domain models I Animation I C-code Matuško&Koloni 5 of 43 MECHATRONICS SYSTEMS 2014/2015 PHYSICAL MODELS I Physical models describe the system behavior both on functional and structural level, I Model parameters are real physical parameters, I Generally, physical models are not suitable for the control system design, I Building such models requires a deep insight into a physical behavior of the system. Matuško&Koloni 6 of 43 MECHATRONICS SYSTEMS 2014/2015 BOND GRAPHS I Universal representation of the system behavior, independent of the system nature. I Based on power flow analysis of the system. I Two quantities that define the system behavior and power flow within the system (and with its environment).

solve 6 , quiz 2link of book httpwww.irccyn.ec-nantes.fr~mart.pdf

arihantcomputersddn

Question 6 A wastewater sample was collected after screening and grit removal step from Waterloo WWTP. The sample was analyzed in the lab. The sample was found to have the following properties: Total COD 350 mg/L Soluble COD 200 mg/L BOD, 220 mg/L VSS/TSS ratio is 0.7 Removal efficiency of the primary clarifier in the WWTP is 60% for both VSS (both nbVSS and bVSS) and FSS. Assume that biological activity is restricted in primary clarifier, and all soluble compounds pass through. The effluent from primary clarifier is further treated in active sludge Solution To calculate rb COD we use formula sCOD = rbCOD + nbsCOD Where sCOD is soluble COD = 200mg/L, so rb COD 200 = rbCOD + 50; rbCOD = 150 mg/L Now for calculating sb COD we use formula bCOD = rbCOD + sbCOD However, we don’t have bCOD [Biodegradable COD], to calculate that we use bCOD (1.7) of (BOD) bCOD = 1.6 x 220 = 352 mg/L Now bCOD = rbCOD + sbCOD 352 = 150 +sbCOD; sbCOD = 352-150 = 198 mg/L.

Question 6 A wastewater sample was collected after screening and grit.pdf

arihantcomputersddn

1. Cathy has just given birth to a little girl. When the nurse tried to feed her or bather her, the baby become cyanotic. Physical examination indicates no structural deformities. Blood gas levels show only 60% saturation with oxygen. What may be causing this problem? A. The infant has malformed digestive system B. the lungs are hyperventilating C. the lungs are hypoventilating D. the cardiovascular supply of the lungs is impaired E. the infant is born with only one lungs 2. The most effective long-term protection against pH changes of body fluid is: A. buffer system of plasma B. Renal control of H+/HCO3- C. Presence of dissolved proteins in the plasma D. Presence of electrolytes in the plasma E. None of the above 3. Jean has been vomiting off & on for a few days due to viral enteritis. His doctor orders SMA7 (seven blood chemistries including electrolytes) the serum will show: A. Low chloride level B. High chloride leve C. Ca++ will be high D. High H+ ions E. Plasma pH will be low 4. The process of invasion by malignant cells is dependent on? A. activation of TNM system B. Release of enzymes by invading cells C. Decreasing anaplasia D. Tissue damage by chemotherapy E. Stromal adesion 5. T- Cells classified as CD8 cells are: A. Cytotoxic T-Cells B. Killers T-cells C. Helper T- cells D. Memory T-cells E. Regulator T-cells 6. Herpes viruses are transmitted to humans by: A. Birds B. Fleas C. Rodents D.Rodents D. Carnivories E. None Solution 1. The correct option is d because the baby becomes cyanotic is mainly because of the insufficient supply of oxygen which is due to the impaired cardiovascular supply of the lungs. 2. the correct option is a because the blood plasma is responsible for the pH maintenance and works with lungs and kidney to maintain the pH of the body fluids by maintaining the acid and base ratio. 3. The correct option is a because the low levels of chloride level suggest that the body is losing the ions away during vomiting along wth the food particles. 4. The correct option is b because the level of enzymes that are released by the invading cells for the destruction of the proteins of the extracellular matrix which separates the cancer cells and the adjoining tissue. The level of degrading enzymes is responsible for the proliferation and escape of the cancer cell into the surrounding tissue. 5. The option is b killer t cells. 6. The correct option is e. because the Herpes viruses are transmitted through skin to skin contact in humans. mainly when in contact with infected person..

1. Cathy has just given birth to a little girl. When the nurse tried.pdf

arihantcomputersddn

QUESTION 1-construction financeWhich one has not correctly describ.pdf

arihantcomputersddn

Problem #4 (20 points) a.A bus route having seven scheduled stops has the following number of passengers alighting and boarding the bus Sto Alighting 0 Boarding 20 0 per passenger 1.8 seconds and boarding Passengers board and alight through the front door. Alighting time p time per passenger 3 seconds. Compute the dwell time at each stop and the total dwell time. Assume the clearance ime 5 seconds. Briefly describe the four categories of transit system characteristics. c. Briefly describe the various transit modes of transportation End of Problem #4 Problemi #5 (20 points) Transportation Engineering is a multidisciplinary area of study. It is a profession that carries distinct societal responsibilities. What would a comprehensive training include? Provide examples of each training tem a) b) c) d) Explain the difference between direct elasticity and cross elasticity. Give one example of each What does CBD mean? Give an example of a CBD A transportation network having the following characteristics Travel time (minutes) between nodes Nodes From 10 10 10 0 16 12 Each node represents an activity center, and each link represents travel time in minutes. Draw a link/node diagrams for the travel times and distances, if the average travel speed on all links is 30 miles per hour End of Problem #5 Solution 4. (a). There are 2 ways in general to model the dwell time at a particular stop which depends on whether the boarding and alighting processes are sequential or simultaneous. Assuming in the question given that sequential boarding and alighting takes place since only one front door is available for flow of passengers: Dw = a * A + b*B + td ; where A and B are the number of boarding and alighting passengers respectively a and b are time intervals for boarding and alighting per passenger. td is the average clearance time for opening and closing the doors. Here a= 3 sec, b=1.8sec Therefore, at Stop 1: Dw1 = 3*20 + 1.8*0 + 5 = 65 sec Stop 2: Dw2 = 3*5 + 1.8*3 + 5 = 25.4 sec Stop 3: Dw3 = 3*7 + 1.8*5 + 5 = 35 sec Stop 4: Dw4 = 3*3 + 1.8*6 + 5 = 24.8 sec Stop 5: Dw5 = 3*4 + 1.8*8 + 5 = 31.4 sec Stop 6: Dw6 = 3*2 + 1.8*7 + 5 = 23.6 sec Stop 7: Dw7 = 3*0 + 1.8*12 + 5 = 26.6 sec Total dwell time= 231.8 seconds (b). Transit system characteristics can be classified into four main categories: i. System performance elements: Contains entire set of physical performance related elements which decides the raw efficiency of the central system employed as the primary functional core unit of the transit model. This includes Service frequency, operating speed, reliability and safety measures and line and productive capacities. ii. Level of Service (LOS) is an important service parameter that affects the users drastically due to differential change in its delivery. It is a simple basic most powerful characteristic that can attract the users or provide bad experiences to them if miniscule error occurs in provider’s implementation. The factors that affect LOS are performance.

Problem #4 (20 points) a.A bus route having seven scheduled stops h.pdf

arihantcomputersddn

please read carefully before answering. Thank you. Choose at least two concepts in financial reporting (Intermediate Accounting: Reporting and Analysis) you have learned and explain how you may use them in your present or future accounting position. Make sure you provide details to include how each concept will help you support the financial goals of the company you currently work for or will work for in the future. Solution CONCEPT OF MATERIALITY-THE MOST IMPORTANT CONCEPT IN FINANCIAL REPORTING IS RELATED TO MATERIALITY BECAUSE IT EMPHASISES ON RECORDING OF TRANSACTIONS BASED ON TRANSACTIONS MATERIALITY OR IT\'S IMPORTANCE WITH RESPECT TO THE COMPANY CONCERNED.THE CONCEPT OF MATERIALITY STATES THAT ONLY THOSE TRANSACTION SHOULD BE RECORDED WHICH ARE MATERIAL FINANCIALLY TO THE ORGANISATION OR ELSE SUCH TRANSACTONS SHOULD BE IGNORED.SO IT BECOMES VERY IMPORTANT TO DETERMINE WHAT IS MATERIAL AND WHAT IS NOT FOR THE ORGANISATION FOR A FUTURE FINANCIAL MANAGER AND HENCE WILL REQUIRE PERFECT PROFESSIONAL JUDGEMENT. CONCEPT OF PRUDENCE-IT STATES THAT A TRANSACTION SHOULD BE RECORDED BASED ON IT\'S SUBATANCE FORM RATHER THAN IT\'S LEGAL FORM WHICH MEANS THST IF A TRANSACTION HAS BEEN OCCURED IN SUBSTANCE AND ONLY A LEGAL DOCUMENTATION OR FORMALITIES ARE REMAINING FOR SUCH TRANSACTION THEN SUCH TRANSACTIONS SHOULD BE RECORDED IN BOOKS SO THAT A TRUE AND FAIR VIEW OF ORGANISATION IS GIVEN TO IT\'S USERS.IT IS ALAO A IMPORTANT CONCEPT AS IT WILL LEAD TO RELIABILTY OF ORGANISATION FINANCIAL PERFORMANCE AND THUS WILL CONTRIBUTE TO IT\'S FUTURE GOAL..

please read carefully before answering. Thank you. Choose at least.pdf

arihantcomputersddn

As your reward for saving the Kingdom of Bigfunnia from the evil monster “Exponential Asymptotic,” the king has given you the opportunity to earn a big reward. Behind the castle there is a maze, and along each corridor of the maze there is a bag of gold coins. The amount of gold in each bag varies. You will be given the opportunity to walk through the maze, picking up bags of gold. You may enter only through the door marked “ENTER” and exit through the door marked “EXIT.” (These are distinct doors.) While in the maze you may not retrace your steps. Each corridor of the maze has an arrow painted on the wall. You may only go down the corridor in the direction of the arrow. There is no way to traverse a “loop” in the maze. You will receive a map of the maze, including the amount of gold in and the direction of each corridor. Describe and analyze an efficient algorithm to help you pick up the most gold in this maze while traversing a path from the start to the finish. Solution #include using namespace std; // Returns count of possible paths to reach cell at row number m and column // number n from the topmost leftmost cell (cell at 1, 1) int numberOfPaths(int m, int n) { // Create a 2D table to store results of subproblems int count[m][n]; // Count of paths to reach any cell in first column is 1 for (int i = 0; i < m; i++) count[i][0] = 1; // Count of paths to reach any cell in first column is 1 for (int j = 0; j < n; j++) count[0][j] = 1; // Calculate count of paths for other cells in bottom-up manner using // the recursive solution for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) // By uncommenting the last part the code calculatest he total // possible paths if the diagonal Movements are allowed count[i][j] = count[i-1][j] + count[i][j-1]; //+ count[i-1][j-1]; } return count[m-1][n-1]; } #include using namespace std; // Returns count of possible paths to reach cell at row number m and column // number n from the topmost leftmost cell (cell at 1, 1) int numberOfPaths(int m, int n) { // Create a 2D table to store results of subproblems int count[m][n]; // Count of paths to reach any cell in first column is 1 for (int i = 0; i < m; i++) count[i][0] = 1; // Count of paths to reach any cell in first column is 1 for (int j = 0; j < n; j++) count[0][j] = 1; // Calculate count of paths for other cells in bottom-up manner using // the recursive solution for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) // By uncommenting the last part the code calculatest he total // possible paths if the diagonal Movements are allowed count[i][j] = count[i-1][j] + count[i][j-1]; //+ count[i-1][j-1]; } return count[m-1][n-1]; }.

As your reward for saving the Kingdom of Bigfunnia from the evil mon.pdf

arihantcomputersddn

Name 3 openings that are transmitted through the levator ani muscle? Name 3 openings that are transmitted through the levator ani muscle? Solution Levator ani muscle extends through and makes internal floor portion of the pelvis (hip portion). The muscle is arranged such that visceral organs. Levator ani allows horizontal openings on the lower side for passage of the following. 1. Anal opening 2. Urinal opening 3. Genital opening.

Name 3 openings that are transmitted through the levator ani musc.pdf

arihantcomputersddn

Answer the following questions about molecule movement across a membrane. a. How is osmosis similar to simple diffusion of CO_2 across a membrane? How is it different? If an aquaporin is involved, how does this change your answer? b. Compare and contrast carrier proteins and channel proteins. Include information about both structure and mechanism of transport. c. Compare and contrast indirect and direct active transport. Give a specific example for each type of transport. Solution 5a: Osmosis is the movement of water molecules down the concentration gradiant through a semipermeable membrane. Both osmosis and simple diffusion of CO2 across a membrane occur down the concentration gradiant and are passive transport. However, osmosis transports hydrophilic water molecules while simple diffusion transports hydrophobic non polar CO2 molecules. If an aquaporin is involved, the proces would be called as facililated diffusion which refers to transport of polar, charged, hydrophilic molecules down the concentration gradiant with the help of membrane transport proteins. 5b: Carrier proteins are the transporter proteins that are actually the integral membrane proteins serving in transport of ions across the membrane. These proteins have ion binding sites and exhibit changes in shape to transport the ions. Channel proteins are the integral membrane proteins that form the hydrophilic channels for transport of polar, charged and hydrophilic substances across the membrane. They do not undergo the shape change. 5c: Direct active transport couples the exergonic chemical reaction for increasing the solute/ion concentration on one side of the membrane. The indirect active transport includes coupling of favorable downhill movement of one solute to the unfavorable uphill movement of other substance. For ex: sodium potassium pump couples ATP hydrolysis with uphill movement of sodium (direct active transport). Downhill movement of sodium coupled with uphill movement of glucose (indirect active transport)..

Answer the following questions about molecule movement across a membr.pdf

arihantcomputersddn

In how many ways, three indistinguishable Owls, one Parrot and one Sparrow can sit together in a line on a string? (b) How many permutations of the letters are possible from the letters in the word: OOOPS!? (a) How many different letter arrangements can be made from the word \"DINOSAUR\"? (b) How many different letter arrangements can be made from the word \"DRAGON\"? Solution S.7 (a) Let us denote Owl as O, Parrot as P and Sparrow as S. The different ways they can sit together in a line on a string are given below: Case I: All three owls sit next to each other O O O P S O O O S P P S O O O S P O O O P O O O S S O O O P Case II: Only two owls sit next to each other O O P S O O O S P O O O P O S O O S O P O P S O O O S P O O P O S O O S O P O O O P O O S O S O O P P O O S O S O O P O Case III: None of the owls sit next to each other O S O P O O P O S O Therefore, there are (6 + 12 + 2) = 20 different ways they can sit together in a line on a string. (Answer) (b) The given word is OOOPS The objective is to determine the number of permutations of the letters possible from the letters of the word OOOPS. The problem is similar to the earlier problem (a) as we have denoted Owl as O, Parrot as P and Sparrow as S. Therefore the answer is 20..

In how many ways, three indistinguishable Owls, one Parrot and one Sp.pdf

arihantcomputersddn

More from arihantcomputersddn (20)