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2 years ago

Which of the following is defined as the progression of Ideas leading to the final delivery of a product?

Engineering

1 answer:

Rama09 [41]2 years ago

6 0

The progression of ideas leading to the final delivery of a product is in option D optimization.

<h3>What is optimization?</h3>

Optimization means upgrading something or making it best. The process of making the best effective use of a situation or resource.

Hence, if the product is going to shape to its final form, then it will go through the process of optimization.

Hence, option D is true.

Learn more about optimization here,

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A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c

Eduardwww [97]

Answer:

$h=1.99998\ W/m^2.C$

$k=33.333\ W/m.C$

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

$\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0$ Eq (1)

Where:

k is thermal Conductivity

$\dot e_{gen}$ is uniform thermal generation

$T(x) = a(L^2-x^2)+b$

$\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a$

Putt in Eq (1):

$-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C$

Energy balance is given by:

$Q_{convection}=Q_{conduction}$

$h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L$ Eq (2)

$T(x) = a(L^2-x^2)+b$

Putting x=L

$T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC$

$\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2$

From Eq (2)

$h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C$

7 0

3 years ago

This assignment is designed to test your understanding of graphical user interface components and Event Driven programming in Ja

lys-0071 [83]

Answer:

Java program is given below

Explanation:

JavaPadGUI.java

package javapad.gui;

import java.awt.BorderLayout;

import java.awt.Container;

import java.awt.Dimension;

import java.awt.event.ActionEvent;

import java.awt.event.ActionListener;

import java.awt.event.WindowAdapter;

import java.awt.event.WindowEvent;

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

import javax.swing.JButton;

import javax.swing.JFrame;

import javax.swing.JLabel;

import javax.swing.JOptionPane;

import javax.swing.JPanel;

import javax.swing.JScrollPane;

import javax.swing.JTextArea;

public class JavaPadGUI extends JFrame implements ActionListener{

//constants

private static final String TITLE = "Macrosoft JavaPad XP";

private static final String SLOGAN = "Macrosoft: Resistance is futile";

//button names

private static final String NEW = "New";

private static final String LOAD = "Load";

private static final String SAVE = "Save";

private static final String QUIT = "Quit";

private static final String FILENAME = "hardcode.txt";

//messages

private static final String QUIT_MSG = "Quitting, Save?";

private static final String LOAD_ERR = "Could not access file " + FILENAME;

//fields

private JTextArea txtContent;

private JButton butNew, butLoad, butSave, butQuit;

public JavaPadGUI()

{

super(TITLE);

createUI();

setSize(new Dimension(400, 300));

setDefaultCloseOperation(EXIT_ON_CLOSE);

}

private void createUI()

{

Container c = getContentPane();

c.setLayout(new BorderLayout(30,30));

//create the buttons panel

JPanel butPanel = new JPanel();

butPanel.add(butNew = new JButton(NEW));

butPanel.add(butLoad = new JButton(LOAD));

butPanel.add(butSave = new JButton(SAVE));

butPanel.add(butQuit = new JButton(QUIT));

//set command names for the buttons, these will be when button is clicked

butNew.setActionCommand(NEW);

butLoad.setActionCommand(LOAD);

butSave.setActionCommand(SAVE);

butQuit.setActionCommand(QUIT);

JPanel sloganPanel = new JPanel();

sloganPanel.add(new JLabel(SLOGAN));

//create textarea with scrollbar

txtContent = new JTextArea(15, 25);

txtContent.setLineWrap(true);

JScrollPane scroll = new JScrollPane(txtContent);

//now add all components

c.add(butPanel, BorderLayout.NORTH);

c.add(scroll, BorderLayout.CENTER);

c.add(sloganPanel, BorderLayout.SOUTH);

//set the actionlistner to buttons to handle button click event

butNew.addActionListener(this);

butLoad.addActionListener(this);

butSave.addActionListener(this);

butQuit.addActionListener(this);

}

public void actionPerformed(ActionEvent e)

{

String cmd = e.getActionCommand();

if(cmd.equals(NEW))

txtContent.setText("");

else if (cmd.equals(LOAD))

load();

else if(cmd.equals(SAVE))

save();

else if (cmd.equals(QUIT))

quit();

}

private void quit()

{

int option = JOptionPane.showConfirmDialog(this, QUIT_MSG);

if(option == JOptionPane.YES_OPTION)

{

save();

}

System.exit(0);

}

private void save()

{

try {

PrintWriter w = new PrintWriter(new File(FILENAME));

w.write(txtContent.getText());

w.close();

} catch (FileNotFoundException e) {

JOptionPane.showMessageDialog(this,"I/O Error", LOAD_ERR, JOptionPane.ERROR_MESSAGE);

}

private void load()

{

try {

Scanner s = new Scanner(new File(FILENAME));

String str = "";

while(s.hasNextLine())

str += s.nextLine();

txtContent.setText(str);

s.close();

} catch (FileNotFoundException e) {

JOptionPane.showMessageDialog(this, LOAD_ERR, "I/O Error",JOptionPane.ERROR_MESSAGE);

}

RunJavaPad.java

package javapad;

import javapad.gui.JavaPadGUI;

public class RunJavaPad {

public static void main(String[] args) {

JavaPadGUI gui = new JavaPadGUI();

gui.setVisible(true);

}

7 0

3 years ago

Fully developed conditions are known to exist for water flowing through a 25-mm-diameter tube at 0.01 kg/s and 27 C. What is the

Irina18 [472]

Answer:

0.0406 m/s

Explanation:

Given:

Diameter of the tube, D = 25 mm = 0.025 m

cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²

Mass flow rate = 0.01 kg/s

Now,

the mass flow rate is given as:

mass flow rate = ρAV

where,

ρ is the density of the water = 1000 kg/m³

A is the area of cross-section of the pipe

V is the average velocity through the pipe

thus,

0.01 = 1000 × 4.9 × 10⁻⁴ × V

V = 0.0203 m/s

also,

Reynold's number, Re = $\frac{VD}{\nu}$

where,

ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s

thus,

Re = $\frac{0.0203\times0.025}{0.833\times10^{-6}}$

Re = 611.39 < 2000

thus,

the flow is laminar

hence,

the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s

maximum velocity = 0.0406 m/s

5 0

3 years ago

how are the construction steps for bisecting an angle similar to the steps you would have taken to construct and bisect a line s

GREYUIT [131]

Answer:

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Explanation:

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3 years ago

To check for ripple voltage from the alternator, connect a digital multimeter and select

Roman55 [17]

Answer:

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3 years ago