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3 years ago

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A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b

oth and phases. If the composition of the phase is 87 wt% B-13 wt% A, what is the composition of the phase

1 answer:

Dennis_Churaev [7]3 years ago

8 0

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition Cβ = 87

we use here lever rule that is

Wα = Wβ ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

$\frac{C_\beta - Co}{C_\beta - Ca}$ = 0.5

$\frac{87 - 57}{87 - Ca}$ = 0.5

solve it we get

Ca = 27

so composition of alpha phase is 27% B

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A rectangular workpiece has the following original dimensions: 2a=100mm, h=25mm, and width=20mm. The metal has a strengh coeffic

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Answer:

See attachment for detailed answer.

Explanation:

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3 years ago

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 1.7. If, afte

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Answer:

It would take approximately 305 s to go to 99% completion

Explanation:

Given that:

y = 50% = 0.5

n = 1.7

t = 100 s

We need to first find the parameter k from the equation below.

$exp(-kt^n)=1-y$

taking the natural logarithm of both sides:

$-kt^n=ln(1-y)\\kt^n=-ln(1-y)\\k=-\frac{ln(1-y)}{t^n}$

Substituting values:

$k=-\frac{ln(1-y)}{t^n}= -\frac{ln(1-0.5)}{100^1.7} = 2.76*10^{-4}$

Also

$t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}$

Substituting values and y = 99% = 0.99

$t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s$

∴ t ≅ 305 s

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

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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);

}

}

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3 years ago

Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)

IceJOKER [234]

Answer:

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa) Number- Average Molecular Weight (g/mol)

82 12,700

156 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

where;

$T_S$ = Tensile Strength

$T_{S \infty}$ = Tensile Strength (Infinity)

$M_n$ = Number- Average Molecular Weight (g/mol)

SO;

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

From equation (1) ; collecting the like terms; we have :

$T_{S \infty} =82+ \dfrac{A}{12700}$

From equation (2) ; we have:

$T_{S \infty} =156+ \dfrac{A}{28500}$

So; $T_{S \infty} = T_{S \infty}$

Then;

$T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}$

Solving by L.C.M

$\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}$

$\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}$

By cross multiplying ; we have:

$({4446000 + A})* {12700} ={28500} *({1041400 + A})$

$(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)$

Collecting like terms ; we have

$(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)$

$2.67843*10^{10} = 15800 \ A$

Dividing both sides by 15800:

$\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}$

A = 1695208.861

From equation (1);

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

Replacing A = 1695208.861 in the above equation; we have:

$82= T_{S \infty} - \dfrac{1695208.861}{12700}$

$T_{S \infty}= 82 + \dfrac{1695208.861}{12700}$

$T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{2736608.861 }{12700}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

From equation(2);

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

Replacing A = 1695208.861 in the above equation; we have:

$156= T_{S \infty} - \dfrac{1695208.861}{28500}$

$T_{S \infty}= 156 + \dfrac{1695208.861}{28500}$

$T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{6141208.861}{28500}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

Then:

$181 = 215.481- \dfrac{1695208.861 }{M_n}$

Collecting like terms ; we have:

$\dfrac{1695208.861 }{M_n} = 215.481- 181$

$\dfrac{1695208.861 }{M_n} =34.481$

1695208.861= 34.481 $M_n$

Dividing both sides by 34.481; we have:

$M_n$ = $\dfrac{1695208.861}{34.481}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

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3 years ago

How many people made machines

olganol [36]

Answer:

The total number of people whom have made machines is not a recorded figure? Need to be more specific :/

Explanation:

Sorry not very helpful, your question is REALLY broad

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3 years ago

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