Low relative humidity can cause wood furniture joints to swell

Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f

natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of and to the argument of the decorator. The return value of the wrapper should look like: return “” + func() + “”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of and to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “” + func() + “.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "{0}".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "{0}".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0

3 years ago

I will definitely rate 5 stars/brainliest!!! HELP PLEASE!!! State University must purchase 1,100 computers from three vendors. V

romanna [79]

Why 1+12+ Y3 < 1100
Says the state of university Need to purchase 1100 computers in total, we have the following answer on the way top

3 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

Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed

Vilka [71]

Answer:

The direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube.

Explanation:

The question is incomplete. So the picture, which is missing in question, is attached for your review.

As it can be seen in the picture, the ball coming out of the tube will have two components of velocity. One is along the length of tube (because ball is moving in that direction and is coming out from the hole), other is velocity component will be perpendicular to the tube (because the ball is made to move in that direction as the tube is rolling on the surface).

So, taking the resultant of two vectors of velocity, the resultant direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube.

6 0

3 years ago

Determine the flow velocities at the inlet and exit sections of an

VikaD [51]

Answer:

The inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

Explanation:

Using Bernoulli's equation

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²

P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

ΔP + ρgΔh = 1/2ρ(v₂² - v₁²) (1)

where ΔP = pressure difference = 12.35 kPa = 12350 Pa

Δh = height difference = 1.35 m

From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.

Substituting v₁ into (1), we have

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂²v₂/d₁² )²)

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂/d₁)⁴v₂²)

v₂ = √[2(ΔP + ρgΔh)/ρ(1 - (d₂/d₁)⁴)}

substituting the values of the variables, we have

v₂ = √[2(12350 Pa + 1000 kg/m³ × 9.8 m/s² × 1.35 m)/(1000 kg/m³ (1 - (0.44 m/0.95 m)⁴))}

= √[2(12350 Pa + 13230 Pa)/(1000 kg/m³ × 0.954)]

= √[2(25580 Pa)/954 kg/m³]

= √[51160 Pa/954 kg/m³]

= √53.627

= 7.32 m/s

v₁ = d₂²v₂/d₁²

= (0.44 m/0.95 m)² × 7.32 m/s

= (0.954)² × 7.32 m/s

= 6.66 m/s

The volume flow rate Q = A₁v₁

= πd₁²v₁/4

= π(0.95 m)² × 6.66 m/s ÷ 4

= 18.883 m³/s ÷ 4

= 4.72 m³/s

So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

7 0

3 years ago

Low relative humidity can cause wood furniture joints to swell​

Low relative humidity can cause wood furniture joints to swell