How can you do this 5.2.4: Rating?

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

evablogger [386]

Answer:

Compute the number of gold atoms per cubic centimeter = 9.052 x 10^21 atoms/cm3

Explanation:

The step by step and appropriate substitution is as shown in the attachment.

From number of moles = Concentration x volume

number of moles = number of particles/ Avogadro's number

Volume = mass/density, the appropriate derivation to get the number of moles of atoms

5 0

2 years ago

What would be the structure for the body points for a persuasive presentation?.

Nataliya [291]

A persuasive speech is structured like an informative speech. It has an introduction with an attention-getter and a clear thesis statement. It also has a body where the speaker presents their main points and it ends with a conclusion that sums up the main point of the speech.

5 0

2 years ago

1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-

Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * $\frac{79.25}{21.2}$

= 0.0214 * 3.738

= 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 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

A piston-cylinder device contains 0.1 m3 of liquid water and 0.9 m² of water vapor in equilibrium at 800 kPa. Heat is transferre

docker41 [41]

Answer:

Initial temperature = 170. 414 °C

Total mass = 94.478 Kg

Final volumen = 33.1181 m^3

Diagram = see picture.

Explanation:

We can consider this system as a close system, because there is not information about any output or input of water, so the mass in the system is constant.

The information tells us that the system is in equilibrium with two phases: liquid and steam. When a system is a two phases region (equilibrium) the temperature and pressure keep constant until the change is completed (either condensation or evaporation). Since we know that we are in a two-phase region and we know the pressure of the system, we can check the thermodynamics tables to know the temperature, because there is a unique temperature in which with this pressure (800 kPa) the system can be in two-phases region (reach the equilibrium condition).

For water in equilibrium at 800 kPa the temperature of saturation is 170.414 °C which is the initial temperature of the system.

to calculate the total mass of the system, we need to estimate the mass of steam and liquid water and add them. To get these values we use the specific volume for both, liquid and steam for the initial condition. We can get them from the thermodynamics tables.

For the condition of 800 kPa and 170.414 °C using the thermodynamics tables we get:

Vg (Specific Volume of Saturated Steam) = 0.240328 m^3/kg

Vf (Specific Volume of Saturated Liquid) = 0.00111479 m^3/kg

if you divide the volume of liquid and steam provided in the statement by the specific volume of saturated liquid and steam, we can obtain the value of mass of vapor and liquid in the system.

Steam mass = *0.9 m^3 / 0.240328 m^3/kg = 3.74488 Kg

Liquid mass = 0.1 m^3 /0.00111479 m^3/kg = 89.70299 Kg

Total mass of the system = 3.74488 Kg + 89.70299 Kg = 93,4478 Kg

If we keep the pressure constant increasing the temperature the system will experience a phase-change (see the diagram) going from two-phase region to superheated steam. When we check for properties for the condition of P= 800 kPa and T= 350°C we see that is in the region of superheated steam, so we don’t have liquid water in this condition.

If we want to get the final volume of the water (steam) in the system, we need to get the specific volume for this condition from the thermodynamics tables.

Specific Volume of Superheated Steam at 800 kPa and 350°C = 0.354411 m^3/kg

We already know that this a close system so the mass in it keeps constant during the process.

If we multiply the mass of the system by the specific volume in the final condition, we can get the final volume for the system.

Final volume = 93.4478 Kg * 0.354411 m^3/kg = 33.1189 m^3

You can the P-v diagram for this system in the picture.

For the initial condition you can calculate the quality of the steam (measure of the proportion of steam on the mixture) to see how far the point is from for the condition on all the mix is steam. Is a value between 0 and 1, where 0 is saturated liquid and 1 is saturated steam.

Quality of steam = mass of steam / total mass of the system

Quality of steam = 3.74488 Kg /93.4478 Kg = 0,040 this value is usually present as a percentage so is 4%.

Since this a low value we can say that we are very close the saturated liquid point in the diagram.

6 0

3 years ago