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zimovet [89]
3 years ago
6

A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

and downstream of the segment midpoint. It is on rolling terrain with 10% heavy vehicles and is operating at capacity with a peak-hour factor of 0.900. If the road is expanded to four 11-ft lanes with a 2-ft right shoulder, and traffic after the expansion is projected to increase by 10% with the same heavy vehicle percentage and peak hour factor, what is the new LOS and estimated density?
Engineering
1 answer:
alisha [4.7K]3 years ago
6 0

Answer:

The level of the service is loss and the density is 34.2248 pc/mi/ln

Explanation:

the solution is attached in the Word file

Download docx
You might be interested in
A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of
Burka [1]

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end  + area of the curve surface

=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}

= 2.136 *10^-4 m^{2}

Heat flux =\frac{0.6}{2.136 * 10^{-4} } = 2808.99 W/m^{2}

fraction of heat dissipated from the top and bottom surface

=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100}  )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175

=11.75%

8 0
3 years ago
Read 2 more answers
Two identical 3 in. major-diameter power screws (single-threaded) with modified square threads are used to raise and lower a 50-
sp2606 [1]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

6 0
3 years ago
Ext
Galina-37 [17]
Engineering is the technical
8 0
2 years ago
The second programming project involves writing a program that accepts an arithmetic expression of unsigned integers in postfix
Tpy6a [65]

Answer:

Explanation:

Note: In case of any queries, just comment in box I would be very happy to assist all your queries

SourceCode:

// MyGUI.java:

// Import packages

import java.awt.FlowLayout;

import java.awt.GridLayout;

import java.awt.event.ActionEvent;

import java.awt.event.ActionListener;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.EmptyStackException;

import java.util.Stack;

import javax.swing.JButton;

import javax.swing.JFrame;

import javax.swing.JLabel;

import javax.swing.JOptionPane;

import javax.swing.JPanel;

import javax.swing.JTextField;

import javax.swing.SwingConstants;

// Declaare and define the class MyGUI

abstract class MyGUI extends JFrame implements ActionListener {

JTextField userInput;

JLabel inputDescLbl, resultLbl;

JPanel inputPanel, resultPanel;

JButton evlBtn;

Stack<Object> stk;

// Define the constructor MyGUI

MyGUI() {

super("Tree Address Generator");

inputPanel = new JPanel(new FlowLayout());

resultPanel = new JPanel(new FlowLayout());

setLayout(new GridLayout(2, 1));

userInput = new JTextField(20);

inputDescLbl = new JLabel("Enter Postfix Expression:");

evlBtn = new JButton("Construct Tree");

evlBtn.addActionListener(this);

resultLbl = new JLabel("Infix Expression:", SwingConstants.LEFT);

add(inputPanel);

add(resultPanel);

inputPanel.add(inputDescLbl);

inputPanel.add(userInput);

inputPanel.add(evlBtn);

resultPanel.add(resultLbl);

stk = new Stack<Object>();

}

}

//Stack.java:

// Declare and define the class Stack

class Stack {

private int[] a;

private int top, m;

public Stack(int max) {

m = max;

a = new int[m];

top = -1; }

public void push(int key) {

a[++top] = key; }

public int pop() {

return (a[top--]); }

}

// Declare and define the class Evaluation()

class Evaluation {

public int calculate(String s) {

int n, r = 0;

n = s.length();

Stack a = new Stack(n);

for (int i = 0; i < n; i++) {

char ch = s.charAt(i);

if (ch >= '0' && ch <= '9')

a.push((int) (ch - '0'));

else if (ch == ' ')

continue;

else {

int x = a.pop();

int y = a.pop();

switch (ch) {

case '+':

r = x + y;

break;

case '-':

r = y - x;

break;

case '*':

r = x * y;

break;

case '/':

r = y / x;

break;

default:

r = 0;

}

a.push(r);

}

}

r = a.pop();

return (r);

}

}

// PostfixToInfix.java:

// Import packages

import java.util.Scanner;

import java.util.Stack;

// Declare and define the class PostfixToInfix

class PostfixToInfix {

// Determine whether the character entered is an operator or not

private boolean isOperator(char c) {

if (c == '+' || c == '-' || c == '*' || c == '/' || c == '^')

return true;

return false;

}

// Declare and define the convert()

public String convert(String postfix) {

Stack<String> s = new Stack<>();

for (int i = 0; i < postfix.length(); i++) {

char c = postfix.charAt(i);

if (isOperator(c)) {

String b = s.pop();

String a = s.pop();

s.push("(" + a + c + b + ")");

} else

s.push("" + c);

}

return s.pop();

}

// Program starts from main()

public static void main(String[] args) {

PostfixToInfix obj = new PostfixToInfix();

Scanner sc = new Scanner(System.in);

// Prompt the user to enter the postfix expression

System.out.print("Postfix : ");

String postfix = sc.next();

// Display the expression in infix expression

System.out.println("Infix : " + obj.convert(postfix));

}

}

Output:

e Console X terminated PostfixTolnfix [Java Application] C:\Program Files\Java\jrel.8.0_121\bin\javaw.exe Postfix : ABD++C-D/ .

3 0
3 years ago
A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons
antoniya [11.8K]

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given  :

PV^{1.3}=C

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}

where r_{c} is compression ratio = \frac{v_{c}+v_{s}}{v_{c}}

           r_{c} = 1+\frac{v_{s}}{v_{c}}

where v_{c} is compression volume

           v_{s} is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, v_{30}=v_{c}+0.7v_{s}

and at 70% compression, 30% of the swept volume remains

    ∴    v_{70}=v_{c}+0.3v_{s}  

We know,

\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}

\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}

\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\

1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}

v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}

0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}

0.2218v_{s} = 0.594v_{c}

v_{c}=0.3734 v_{s}

∴   r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}

Therefore, compression ratio is r_{c} = 3.678

Now efficiency, \eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}

 \eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}

 \eta =0.32342 , this is the ideal efficiency

Therefore actual efficiency, \eta_{act} =0.5\times \eta _{ideal}

           \eta_{act} =0.5\times 0.32342

           \eta_{act} =0.1617

Therefore total power required = 1 kW x 3600 J

                                                    = 3600 kJ

∴ we know efficiency, \eta=\frac{W_{net}}{Q_{supply}}

Q_{supply}=\frac{W_{net}}{\eta _{act}}

Q_{supply}=\frac{3600}{0.1617}

Q_{supply}=22261.78 kJ

Therefore fuel required = \frac{22261.78}{45000}

                                        = 0.4947 kg/hr      

5 0
4 years ago
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