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Sloan [31]
3 years ago
10

The density of a certain material is such that it weighs 9 pounds per cubic foot of

Engineering
1 answer:
My name is Ann [436]3 years ago
8 0

Answer:

The answer is "101"

Explanation:

Given value:

We must convert density into pounds per cubic foot to 6900 grams per 4.5 quarter of the length.

1\  \text{Liquid quart} = 0.03342 \ \text{cubic foot} \\\\1 \ \text{gram} = 0.0022 \ \text{pound}\\

Formula:

\bold{density = \frac{6900}{4.5} \frac{gram}{quart}}\\

=\frac{6900 \times 0.0022 \ pound }{4.5 \times 0.03342 \cubic \ foot}\\\\= \frac{15.2119}{0.15039} \\\\= 101.1497 \ or \ 101 \ \text{pound per cubic feet}\\

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Is it better to do blue prints with paper and pencil or a computer program if you’re going to design a house? Why?
Hatshy [7]

Answer:

Computer program

Explanation:

I use Revit and its way better to do all that you can see 2D 3D the measurements and its super easy to use hope this helps

6 0
3 years ago
Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e −2t + 3t
IrinaK [193]

Answer:

V=L(di/dt) where i is current, V=0.208

Explanation:

using expression iL(t)=5e-2t+3te-2t-2 and L=0.05H(50/1000)

V=0.05*d(5e-2t+3te-2t-2)/dt

since there is no power of e, I'll assume the power to be 1

V=0.05*(-2+3e-2)

at t=0.25

V=0.15e-0.2

V=0.208

3 0
3 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
1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. T
Elenna [48]

Answer:

Concentration = 10.33 kg/m³

Explanation:

We are given;

Mass of solids; 10,000 kg

Volume; V = 440,000 L = 440 m³

Rate at which water is pumped out = 40,000 liter/h

Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³

Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;

200m³/440m³ = 5/11

Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;

Thus,

Amount left; = 10000 x (5/11) = 4545 kg

The concentration would be calculated by:

Concentration = amount left/initial volume

Thus,

Concentration = 4545/440 = 10.3 kg/m³

8 0
3 years ago
An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at
polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end  = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos  = 0.173 W/mK

solution

first we will get here side of cross section that is express as

S = S1 + \frac{S2-S1}{L} x     ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + \frac{10-5}{15} x

S = \frac{0.15 + x}{3}    m

and  

now we get here Area of section at distance x is

area A = S²    ...............2

area A = (\frac{0.15 + x}{3})^2    m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = \frac{-k\times A\  dt}{dx}      ...............3

put here value and we get

heat conduction q = -k\times (\frac{0.15 + x}{3})^2 \   \frac{dt}{dx}  

it will be express as

q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)  

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W

5 0
3 years ago
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