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AleksandrR [38]
3 years ago
7

You are driving on a road where the speed limit is 35 mph. If you want to make a turn, you must start to signal at least _______

_ before you turn.
The options are as follows:
A. 10 feet
B. 100 feet
C. 75 feet
D. 20 feet
Engineering
1 answer:
stich3 [128]3 years ago
3 0
I believe it’s D) 20 feet
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Hi I don't know of yall remeber me, but I'm Jadin aka J. I am looking for my friend group, that I have missed but can't find cau
kirill [66]

Answer:

The young lady was his daughter.The shoemaker was frightened when he saw that she wants to sit near him and took his knife to frighten her and leave him alone to do his work

Explanation:

could uh name them since if i know any i would surely tryin help

7 0
2 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
Lyrx [107]

Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

5 0
3 years ago
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0
3 years ago
Write a GUI-based program that plays a guess-the-number game in which the roles of the computer and the user are the reverse of
AURORKA [14]

Answer:

import javax.swing.*;

import java.awt.*;

import java.util.Random;

import java.awt.event.*;

public class Guess extends JFrame

{

   private static final long serialVersionUID = 1L;

   private JButton newGame;

   private JButton enter;

   private JButton exit;

   private JTextField guess;

   private JLabel initialTextLabel;

   private JLabel enterLabel;

   private JLabel userMessageLabel;

   private int randNum;

   private int userInput;

   private int maxtries = 0;

   public Guess()

   {

       super("Guessing Game");

       newGame = new JButton("New Game");

       exit = new JButton("Exit Game");

       enter = new JButton("Enter");

       guess = new JTextField(4);

       initialTextLabel = new JLabel("I'm thinking of a number between 1 and 100. Guess it!");

       enterLabel = new JLabel("Enter your guess.");

       userMessageLabel = new JLabel("");

       randNum = new Random().nextInt(100) + 1;

       setLayout(new FlowLayout());

       add(initialTextLabel);

       add(enterLabel);

       add(guess);

       add(newGame);

       add(enter);

       add(exit);

       add(userMessageLabel);

   

       setSize(500, 300);

       addWindowListener(new WindowAdapter()

       {

           public void windowClosing(WindowEvent e)

           {

               System.exit(0);

           }

       });

       newGameButtonHandler nghandler = new newGameButtonHandler();

       newGame.addActionListener(nghandler);

       ExitButtonHandler exithandler = new ExitButtonHandler();

       exit.addActionListener(exithandler);

       enterButtonHandler enterhandler = new enterButtonHandler();

       enter.addActionListener(enterhandler);

   }

   class newGameButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           setBackground(Color.ORANGE);

           guess.setEnabled(true);

           guess.setText("");

           enter.setEnabled(true);

           maxtries = 0;

           userMessageLabel.setText("");

           randNum = new Random().nextInt(100) + 1;

       }

   }

   class ExitButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           System.exit(0);

       }

   }

   class enterButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           userInput = Integer.parseInt(guess.getText());

           checkGuess(randNum);

      if(userInput > 100 )

          {

                               userMessageLabel.setText("invalid entry");

          }

       }

   }

   public void checkGuess(int randomNumber)

   {

       maxtries++;

     if(maxtries==10){

           userMessageLabel.setText("You Lose!!");

           guess.setEnabled(false);

           enter.setEnabled(false);

         

       }else if (userInput == randomNumber)

           {

               userMessageLabel.setText("Correct !");

           }

       else if (userInput > randomNumber)

           {

               userMessageLabel.setText("Too high");

           }

       else if (userInput < randomNumber)

           {

               userMessageLabel.setText("Too Low");

           }

   }

   public static void main(String[] args)

   {

       Guess game = new Guess();

       game.setVisible(true);

   }

}

8 0
3 years ago
The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort &lt; 0. Fort 2 0 they areV =75 ~75e-10
masya89 [10]

Answer:

maximum value of the power delivered to the circuit =3.75W

energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

Explanation:

V =75 - 75e-1000t V

l = 50e -IOOOt mA

power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)

=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)

=

maximum value of the power delivered to the circuit =3.75W

the total energy delivered to the element = \int\limits^t_0  {3.75(e^{ -IOOOt} - e ^{-2OOOt} )} , dx \\\\

3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

5 0
3 years ago
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