Answer:
0.028 kg per Seconds; 8.4
Explanation:
N-S, 4 x 100 = 400 m^2, concentration 5/400= 0.0125 kg/s
E-W 4 x 100 =400 m^2 , => 10/400 =0.025 kg/s
(0.0125^2 + 0.025^2)^(1/2) =0.028 kg/s
in 5 minutes = 5 x 60 x 0.028 =8.4
You are designing a spherical tank to hold water for a small village in a developing country. The volume of liquid it can hold can be computed as V = πh2[3R − h]3 where V = volume (m3), h = depth of water in tank (m), and R = the tank radius (m).
Answer:
A. [tex]9\pi h^2 - \pi h^3 -90 = 0\\\\[/tex]
B. h1 = 2.0813772719
h2 = 2.0272465228
h3 = 2.0269057423
Explanation:
V = πh^2 x [3R − h]/3
Given v=30, R=3
[tex]30 = \pi h^2 [\frac{9-h}{3} ]\\\\9\pi h^2 - \pi h^3 -90 = 0\\\\[/tex]
B. An initial guess within the interval [0; 6] is selected,
initial guess h0 = 1.5
Newton method x₂ = x₁ - f'(x₁)/f(x₁)
h₂ = h₁ - f'(x₁)/f(x₁)
[tex]h_2 = h_1 - f'(x_1)/f(x_1)\\\\h_2 = h_1 - \frac{18\pi h-3\pi h^2}{9\pi h^2 - \pi h^3 -90}[/tex]
h1 = 2.0813772719
h2 = 2.0272465228
h3 = 2.0269057423
To accomplish the design project's Outback Challenge mission, your third design topic to research is a Flight Controller (i.e. manual RC transmitter) and GPS Module. The Flight Controller is not an autopilot, but rather a manual handheld flight control transmitter. If the autonomous operations fail, the Ground Base Control will take control of the UAS to manually release the rescue package (water bottle) to Outback Joe. Using the criteria found in the Outback Challenge, research and select the type of flight controller (i.e. manual handheld RC transmitter) and GPS module you feel would best accomplish the two-fold mission of searching for and delivering a rescue package to Outback Joe.
Answer:
Explanation: see attachment below
Calculate the diffusion coefficient for magnesium in aluminum at 500oC. Pre-exponential and activation energy values for this system are 1.2 x 10-4m2/s and 144,000 J/mole respectively. R = 8.314 J/mol K
Answer:
The diffusion coefficient for magnesium in aluminum at the given conditions is 2.22x10⁻¹⁴m²/s.
Explanation:
The diffusion coefficient represents how easily a substance (solute, e.g.: magnesium) can move through another substance (solvent, e.g.: aluminum).
The formula to calculate the diffusion coefficient is:
D=D₀.exp(- EA/R.T)
Each term of the formula means:
D: diffusion coefficient
D₀: Pre-exponential factor (it depends on each particular system, in this case the magnesium-aluminium system)
EA: activation energy
R: gas constant
T: temperature
1st) It is necessary to make a unit conversion for the temperature from °C to K, because we need to solve the problem according to the units of the gas constant (R) to assure the units consistency, so:
Temperature conversion from °C to K: 500 + 273 = 773 K
2nd) Replace the terms in the formula and solve:
D=D₀.exp(- EA/R.T)
D=1.2x10⁻⁴m²/s . exp[- 144,000 J/mole / (8.314 J/mol K . 773K)]
D=1.2x10⁻⁴m²/s . exp(-22.41)
D=1.2x10⁻⁴m²/s . 1.85x10⁻¹⁰
D= 2.22x10⁻¹⁴m²/s
Finally, the result for the diffusion coefficient is 2.22x10⁻¹⁴m²/s.
2 samples of water of equal volume are put into dishes and kept at room temp for several days. the water in the first dish is completely vaporized after 2.8 days while the water in the second dish takes 8.3 days to completely evaporate. What can you conclude about the two dishes
Answer:
Vaporization is the process by which a substance changes from its solid or liquid state to a gaseous state.
Since both liquids are of the same volume and are placed under the same temperature condition, for them to not to vaporize at the same time, they must have been in different containers.
For vaporization to take place, the volume of liquid, amount of air exposure and area of the surface must be considered.
Maybe the first liquid was in a dish which has a large opening, thereby exposing a large amount which can make water to evaporate faster, whereas the second liquid was somehow enclosed (in a deeper dish).
Define the Artist class in Artist.py with a constructor to initialize an artist's information. The constructor should by default initialize the artist's name to "None" and the years of birth and death to 0. Define the Artwork class in Artwork.py with a constructor to initialize an artwork's information. The constructor should by default initialize the title to "None", the year created to 0, and the artist to use the Artist default constructor parameter values. Add an import statement to import the Artist class. Add import statements to main.py to import the Artist and Artwork classes.
The codes for the files are:
File: Artist.py
This file will contain the Artist class with a constructor initializing the artist's name, birth year, and death year.
# Artist. py
class Artist:
def __init__(self, name="None", birth_year=0, death_year=0):
self.name = name
self. birth_year = birth_year
self. death_year = death_year
File: Artwork.py
This file will define the Artwork class and import the Artist class from Artist.py. It initializes the artwork's title, year of creation, and artist.
# Artwork. py
from Artist import Artist
class Artwork:
def __init__(self, title="None", year_created=0, artist=Artist()):
self. title = title
self. year_created = year_created
self. artist = artist
File: main. py
This file will import both Artist and Artwork classes and create instances of them to demonstrate their usage.
# main. py
from Artist import Artist
from Artwork import Artwork
# Creating an instance of Artist
artist1 = Artist("Vincent van Gogh", 1853, 1890)
# Creating an instance of Artwork using the artist instance
artwork1 = Artwork("Starry Night", 1889, artist1)
# Printing artist and artwork details to verify
print(f"Artist: {artwork1. artist. name}, Born: {artwork1. artist. birth_year}, Died: {artwork1. artist. death_year}")
print(f"Artwork: {artwork1. title}, Created in: {artwork1. year_created}")
Execution and Output
When you run the main.py script, it will create an Artist object for Vincent van Gogh and an Artwork object for one of his most famous pieces, "Starry Night". The script will then print the details of both the artist and the artwork.
Expected Output:
Artist: Vincent van Gogh, Born: 1853, Died: 1890
Artwork: Starry Night, Created in: 1889
The statements in the file main.cpp are in incorrect order.
Rearrange the statements so that they prompt the user to input:
The shape type (rectangle, circle, or cylinder)
The appropriate dimension of the shape.
Note: For grading purposes place the cylinder height statement before the radius statement.
The C++ program then outputs the following information about the shape:
For a rectangle, it outputs the area and perimeter
For a circle, it outputs the area and circumference
For a cylinder, it outputs the volume and surface area.
After rearranging the statements, your program should be properly indented.
Here is the code out-of-order:
using namespace std;
#include
int main()
{
string shape;
double height;
#include
cout << "Enter the shape type: (rectangle, circle, cylinder) ";
cin >> shape;
cout << endl;
if (shape == "rectangle")
{
cout << "Area of the circle = "
<< PI * pow(radius, 2.0) << endl;
cout << "Circumference of the circle: "
<< 2 * PI * radius << endl;
cout << "Enter the height of the cylinder: ";
cin >> height;
cout << endl;
cout << "Enter the width of the rectangle: ";
cin >> width;
cout << endl;
cout << "Perimeter of the rectangle = "
<< 2 * (length + width) << endl;
double width;
}
cout << "Surface area of the cylinder: "
<< 2 * PI * radius * height + 2 * PI * pow(radius, 2.0)
<< endl;
}
else if (shape == "circle")
{
cout << "Enter the radius of the circle: ";
cin >> radius;
cout << endl;
cout << "Volume of the cylinder = "
<< PI * pow(radius, 2.0)* height << endl;
double length;
}
return 0;
else if (shape == "cylinder")
{
double radius;
cout << "Enter the length of the rectangle: ";
cin >> length;
cout << endl;
#include
cout << "Enter the radius of the base of the cylinder: ";
cin >> radius;
cout << endl;
const double PI = 3.1416;
cout << "Area of the rectangle = "
<< length * width << endl;
else
cout << "The program does not handle " << shape << endl;
cout << fixed << showpoint << setprecision(2);
#include
In the rearranged code, the statements are placed in the correct order to prompt the user for input of the shape type and appropriate dimensions.
```cpp
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
string shape;
double height, radius, width, length;
const double PI = 3.1416;
cout << "Enter the shape type: (rectangle, circle, cylinder) ";
cin >> shape;
cout << endl;
if (shape == "rectangle")
{
cout << "Enter the width of the rectangle: ";
cin >> width;
cout << endl;
cout << "Enter the length of the rectangle: ";
cin >> length;
cout << endl;
cout << "Area of the rectangle = "
<< length * width << endl;
cout << "Perimeter of the rectangle = "
<< 2 * (length + width) << endl;
}
else if (shape == "circle")
{
cout << "Enter the radius of the circle: ";
cin >> radius;
cout << endl;
cout << "Area of the circle = "
<< PI * pow(radius, 2.0) << endl;
cout << "Circumference of the circle: "
<< 2 * PI * radius << endl;
}
else if (shape == "cylinder")
{
cout << "Enter the height of the cylinder: ";
cin >> height;
cout << endl;
cout << "Enter the radius of the base of the cylinder: ";
cin >> radius;
cout << endl;
cout << "Volume of the cylinder = "
<< PI * pow(radius, 2.0) * height << endl;
cout << "Surface area of the cylinder: "
<< 2 * PI * radius * height + 2 * PI * pow(radius, 2.0) << endl;
}
else
cout << "The program does not handle " << shape << endl;
cout << fixed << showpoint << setprecision(2);
return 0;
}
```
In the rearranged code, the statements are placed in the correct order to prompt the user for input of the shape type and appropriate dimensions. Depending on the shape selected, the program computes and outputs the corresponding area, perimeter, circumference, volume, and surface area. The code is properly indented for readability.
Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime X (in weeks) has a gamma distribution with mean 28 weeks and standard deviation 14 weeks. (a) What is the probability that a transistor will last between 14 and 28 weeks? (Round your answer to three decimal places.) (b) What is the probability that a transistor will last at most 28 weeks? (Round your answer to three decimal places.) Is the median of the lifetime distribution less than 28? Why or why not? The median is 28, since P(X ≤ ) = .5. (c) What is the 99th percentile of the lifetime distribution? (Round your answer to the nearest whole number.) (d) Suppose the test will actually be terminated after t weeks. What value of t is such that only 0.5% of all transistors would still be operating at termination? (Round your answer to the nearest whole number.) t = weeks
Answer:
a The probability Needed is 0.424
b1 The probability Needed is 0.567
b2 The median of the life time distribution is less than 28 cause its probability is less than the probability of 28
c The Needed life time is 70 weeks
d The needed lifetime is t = 77 weeks
Explanation:
Let A represent the life time of the transistor in weeks and follows a gamma distribution with parameters ([tex]\alpha,\beta[/tex]).
Looking at what we are give we can say that
E(A) = 28 weeks and [tex]\sigma_A[/tex] = 14 weeks
Hence the mean of the gamma distribution is E(A) = [tex]\alpha\beta ------(1)[/tex]
and the variance of the gamma distribution is V(A) = [tex]\alpha\beta^2 ------(2)[/tex]
Looking at Equation 2
[tex]\alpha\beta^2 =V(A)[/tex]
[tex](\alpha\beta)\beta = V(A)[/tex]
[tex]E(A)\beta = V(A)[/tex]
[tex]\beta = \frac{V(A)}{E(A)}[/tex]
[tex]Note:\alpha\beta = E(A)\\\ and \ \sigma_A^2 = V(A)[/tex]
[tex]\beta = \frac{\sigma_A^2}{E(A)}[/tex]
[tex]\beta = \frac{14^2}{28} = \frac{196}{28} = 7[/tex]
Looking at Equation 1
E(A) = [tex]\alpha \beta[/tex]
28 = [tex]\alpha (7)[/tex]
=> [tex]\alpha = \frac{28}{7}[/tex]
=> [tex]\alpha = 4[/tex]
Now considering the first question
i.e to find the probability that a transistor will last between 14 and 28 weeks
P(14 < A < 28) = P(A ≤28) - P(A ≤ 14)
The general formula for probability of gamma distribution
= [tex]F[\frac{a}{\beta},\alpha ] - F[\frac{a}{\beta},\alpha ][/tex]
[tex]= F[\frac{28}{7},4 ] - F[\frac{14}{7},4 ][/tex]
[tex]= F(4,4) - F(2,4)[/tex]
[tex]= 0.567 - 0.143[/tex] (This is gotten from the gamma table)
[tex]= 0.424[/tex]
Now considering the second question
i.e to find the probability that a transistor will last at most 28 weeks
[tex]P(A \le 28)[/tex] = [tex]F[\frac{28}{7},4][/tex]
[tex]=F(4,4)[/tex]
[tex]= 0.567[/tex] (This is obtained from the gamma Table)
we need to determine the median of the life time distribution less than 28
from what we are given [tex]P(X \le \mu ) = 0.5[/tex] we can see that [tex]\mu[/tex]<28 since [tex]P(A \le 28)[/tex] = 0.567 and this is greater than 0.5
Now considering the Third question
i.e to find the [tex]99^{th}[/tex] percentile of the life time distribution
Generally
F(a : [tex]\alpha[/tex]) =99%
F(a : 4) = 0.99
Looking at the gamma table the tabulated values at [tex]\alpha =4[/tex] and the corresponding 0.99 percent value to a is 10
Now , find the product of [tex]\beta =7[/tex] with a = 10 to obtain the [tex]99^{th}[/tex] percentile
[tex]99^{th} \ percentile = a\beta[/tex]
[tex]= 10 * 7[/tex]
[tex]=70[/tex]
Now considering the Fourth question
i.e to determine the number of weeks such that 0.5% of all transistor will still be operating
The first thing to do is to find the value in the incomplete gamma function with [tex]\alpha = 4[/tex] in such a way that
F(a:4) = 1 - 0.5%
F(a: 4) = 1 - 0.005
F(x: 4) = 0.995
So we will now obtain the 99.5 percentile
Looking at the gamma tabulated values at [tex]\alpha = 4[/tex] and the corresponding 0.995 percent value for a (x depending on what you want to denoted it with) is 11
So we would the obtain the product of [tex]\beta =7[/tex] with a = 11 which is the [tex]99.5^{th}[/tex] percentile
[tex]t = a\beta[/tex]
[tex]= 11 *7[/tex]
[tex]= 77[/tex]
The probability will be:
(a) 0.4240
(b) 0.5670
(c) 70
(d) 77
According to the question,
→ [tex]\mu = \alpha \beta[/tex]
[tex]28= \alpha \beta[/tex] ...(equation 1)
→ [tex]\sigma^2 = \alpha \beta^2[/tex]
[tex]14^2 = \alpha \beta^2[/tex] ...(equation 2)
By putting "equation 1" in "equation 2", we get
→ [tex]14^2 = \alpha \beta\times \beta[/tex]
[tex]14^2 = 28\times \beta[/tex]
→ [tex]196 = 28\times \beta[/tex]
[tex]\beta = 7[/tex]
[tex]\alpha = 4[/tex]
(a) P(14 and 28)
→ [tex]P(14< X< 28 ) = F(\frac{x}{\beta}, \alpha )- F (\frac{x}{\beta}, d )[/tex]
[tex]= F(\frac{28}{7}, 4 )- F(\frac{14}{7} ,4)[/tex]
[tex]= F(4,4)-F(2,4)[/tex]
By using gamma tables, we get
[tex]= 0.5670-0.1430[/tex]
[tex]= 0.4240[/tex]
(b) P(almost 28)
→ [tex]P(x \leq 28) = F(\frac{x}{\beta}, \alpha )[/tex]
[tex]= F (\frac{28}{7} ,4)[/tex]
[tex]= F(4,4)[/tex]
[tex]= 0.5670[/tex]
(c) 99 percentile
[tex]F(X \leq x, \alpha) = 0.99[/tex][tex]F(x, \alpha) = 0.99[/tex][tex]F(x:4)=0.99[/tex]at x = 10,
99th percentile,
→ [tex]x \beta = 10\times 7[/tex]
[tex]= 70[/tex]
(d)
[tex]F(x;4) = 1-0.005[/tex][tex]F(x;4) = 0.995[/tex]x = 11
→ [tex]t = x \beta[/tex]
[tex]= 11\times 7[/tex]
[tex]= 77[/tex]
Thus the above answers are correct.
Learn more:
https://brainly.com/question/14281632
A cylindrical specimen of cold-worked steel has a Brinell hardness of 250.(a) Estimate its ductility in percent elongation.(b) If the specimen remained cylindrical during deformation and its original radius was 5 mm (0.20 in.), determine its radius after deformation.
Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm
Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is set by changing the value in the code itself. If you are not sure of how it should work then look at the Sample Runs of the next part. This part handles the beginning where it lists all the cheese types available and their prices. Note: it is a very simple fix that needs to be added to all the statements that have an array access.
Answer:
Code fixed below using Java
Explanation:
Error.java
import java.util.Random;
public class Error {
public static void main(String[] args) {
final int MAXCHEESE = 10;
String[] names = new String[MAXCHEESE];
double[] prices = new double[MAXCHEESE];
double[] amounts = new double[MAXCHEESE];
// Three Special Cheeses
names[0] = "Humboldt Fog";
prices[0] = 25.00;
names[1] = "Red Hawk";
prices[1] = 40.50;
names[2] = "Teleme";
prices[2] = 17.25;
System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");
System.out.println(names[0] + ": $" + prices[0] + " per pound");
System.out.println(names[1] + ": $" + prices[1] + " per pound");
System.out.println(names[2] + ": $" + prices[2] + " per pound");
Random ranGen = new Random(100);
// error at initialising i
// i should be from 0 to MAXCHEESE value
for (int i = 0; i < MAXCHEESE; i++) {
names[i] = "Cheese Type " + (char) ('A' + i);
prices[i] = ranGen.nextInt(1000) / 100.0;
amounts[i] = 0;
System.out.println(names[i] + ": $" + prices[i] + " per pound");
}
}
}
In many problems where the potential energy is considered between two objects, the potential energy is defined as zero when the particles are in infinite distance apart. Using the definition of potential energy, explain why the potential energy would be positive if the force between the particles is repulsive and negative if the force between the particles is attractive for noninfinite distances.
Answer:
potential energy=ε Qq/r. Putting values of charge along with sign and calculating the potential energy will tell whether the resultant force is attractive or repulsive.
Explanation:
potential energy=ε Qq/r
q and Q are charges on each particle
So in case the particles have attractive force between them, one particle will be negative and the other will be positiive. The result will be negative.
In case the particles have repulsive forces, the charge on both particles will be either positive or negative. The result will be positive
A specimen of some metal having a rectangular cross section 10.5 mm x 13.7 mm is pulled in tension with a force of 2650 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.
The calculated resulting strain of the metal is 0.0002329
How to determine the resulting strain
To calculate the resulting strain [tex](\(\epsilon\))[/tex], we can use Hooke's Law
This is represented as
[tex]\[\epsilon = \frac{\sigma}{E}\][/tex]
Where:
-[tex]\(\sigma\)[/tex] is the stress
- E is the elastic modulus
From the question, we have
Force F = 2650 N
The cross-sectional area is calclated as
[tex]\(A\) = 10.5 mm x 13.7 mm[/tex]
This gives
[tex]A = \(143.85 \times 10^{-6}\) m\(^2\)[/tex] (converted to [tex]m\(^2\)[/tex])
Calculating the stress [tex](\(\sigma\))[/tex] using the formula:
[tex]\[\sigma = \frac{F}{A}\][/tex]
Substitute the given values:
[tex]\[\sigma = \frac{2650}{143.85 \times 10^{-6}} = \frac{2650}{143.85 \times 10^{-6}} = 18422118.06 \text{ Pa}\][/tex]
Next, we have
[tex]\[\epsilon = \frac{\sigma}{E} = \frac{18422118.06}{79 \times 10^9} = \frac{18422118.06}{79 \times 10^9} = 0.0002329\][/tex]
Hence, the resulting strain is 0.0002329
Consider the animation code from the end of Chapter 4. Modify the ShapeIcon class so that it aggregates a collection of objects implementing the MovableShape interface and its paintIcon() method takes care of painting all the given MovableShape objects. Modify the animation program (AnimationTester.main) to display 5 cars moving horizontally, each starting from a different vertical coordinate, so that the car with index i moves twice as fast as the car with index i - 1.
Answer:
import java.awt.*;
import java.util.*;
import javax.swing.*;
/**
An icon that contains a moveable shape.
*/
public class ShapeIcon implements Icon
{
private int width;
private int height;
private MoveableShape shape;
private ArrayList <MoveableShape> cars;
public ShapeIcon(MoveableShape shape,
int width, int height)
{
this.shape = shape;
this.width = width;
this.height = height;
cars = new ArrayList<MoveableShape>();
}
public int getIconWidth()
{
return width;
}
public int getIconHeight()
{
return height;
}
public void paintIcon(Component c, Graphics g, int x, int y)
{
Graphics2D g2 = (Graphics2D) g;
shape.draw(g2);
}
}
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
/**
This program implements an animation that moves
a car shape.
*/
public class AnimationTester
{
private static final int ICON_WIDTH = 400;
private static final int ICON_HEIGHT = 100;
private static final int CAR_WIDTH = 100;
public static void main(String[] args)
{
JFrame frame = new JFrame();
final MoveableShape shape
= new CarShape(0, 0, CAR_WIDTH);
ShapeIcon icon = new ShapeIcon(shape,
ICON_WIDTH, ICON_HEIGHT);
final JLabel label = new JLabel(icon);
frame.setLayout(new FlowLayout());
frame.add(label);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.pack();
frame.setVisible(true);
final int DELAY = 10; //100
// Milliseconds between timer ticks
Timer t = new Timer(DELAY, new
ActionListener()
{
public void actionPerformed(ActionEvent event)
{
shape.translate(1, 0);
label.repaint();
}
});
t.start();
}
}
import java.awt.*;
import java.awt.geom.*;
import java.util.*;
/**
A car that can be moved around.
*/
public class CarShape implements MoveableShape
{
private int x;
private int y;
private int width;
/**
Constructs a car item.
@param x the left of the bounding rectangle
@param y the top of the bounding rectangle
@param width the width of the bounding rectangle
*/
public CarShape(int x, int y, int width)
{
this.x = x;
this.y = y;
this.width = width;
}
public void translate(int dx, int dy)
{
// x += dx;
// y += dy;
if (x <= 400)
x += dx;
else
x = 0;
y += dy;
}
public void draw(Graphics2D g2)
{
Rectangle2D.Double body
= new Rectangle2D.Double(x, y + width / 6,
width - 1, width / 6);
Ellipse2D.Double frontTire
= new Ellipse2D.Double(x + width / 6, y + width / 3,
width / 6, width / 6);
Ellipse2D.Double rearTire
= new Ellipse2D.Double(x + width * 2 / 3, y + width / 3,
width / 6, width / 6);
// The bottom of the front windshield
Point2D.Double r1
= new Point2D.Double(x + width / 6, y + width / 6);
// The front of the roof
Point2D.Double r2
= new Point2D.Double(x + width / 3, y);
// The rear of the roof
Point2D.Double r3
= new Point2D.Double(x + width * 2 / 3, y);
// The bottom of the rear windshield
Point2D.Double r4
= new Point2D.Double(x + width * 5 / 6, y + width / 6);
Line2D.Double frontWindshield
= new Line2D.Double(r1, r2);
Line2D.Double roofTop
= new Line2D.Double(r2, r3);
Line2D.Double rearWindshield
= new Line2D.Double(r3, r4);
g2.draw(body);
g2.draw(frontTire);
g2.draw(rearTire);
g2.draw(frontWindshield);
g2.draw(roofTop);
g2.draw(rearWindshield);
Explanation:
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636 and 19.661 mm, respectively, and its final length is 75.9 mm, calculate its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 108 GPa and 37.1 GPa, respectively.
Answer:
The original length of the specimen is found to be 76.093 mm.
Explanation:
From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:
Initial Volume = Final Volume
πd1²L1/4 = πd2²L2/4
d1²L1 = d2²L2
L1 = d2²L2/d1²
where,
d1 = initial diameter = 19.636 mm
d2 = final diameter = 19.661 mm
L1 = Initial Length = Original Length = ?
L2 = Final Length = 75.9 mm
Therefore, using values:
L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²
L1 = 76.093 mm
Two physical properties that have a major influence on the cracking of workpieces, tools, or dies during thermal cycling are thermal conductivity and thermal expansion.Explain why.
Answer:
Explanation:
It is generally known that the thermal stresses developed during thermal cycle results into cracking, and these thermal stresses are due to temperature gradients .
Stresses will be equivalently lower for a particular temperature gradient when the thermal expansion is low.
It also known that there will be a reduction in the temperature gradient if the thermal conductivity is high, as heat is dissipated faster and more equally and with it, as well as when deformation takes place due to thermal stresses, cracking occurs but if the ductility is high, more deformation will be allowed without cracking and thus reduces the tendency for cracking.
Thermal expansion and conductivity influence cracking during thermal cycling because materials expand or contract at different rates causing stress. Differences in these properties between bonded materials or different parts of a structure can lead to cracks. Understanding these properties is important in designing materials to minimize thermal stress.
Explanation:The two physical properties that have a major influence on the cracking of workpieces, tools, or dies during thermal cycling are thermal conductivity and thermal expansion. Thermal expansion occurs due to the tendency of a material to change in volume in response to a change in temperature. When different parts of a workpiece, or different materials bonded together, have different thermal expansion coefficients or varied thermal conductivities, thermal stress can result as the materials expand or contract at different rates. This stress leads to the formation of cracks, especially if the material is rigid and cannot accommodate the stress through deformation.
For example, metal implants in the body may need replacement because of the lack of bonding between metal and bone due to different expansion coefficients. Similarly, the expansion of fillings in teeth can be different from that of tooth enamel, causing discomfort or damage. The understanding of thermal expansion and conductivity is critical in designing materials and structures, such as railroad tracks and roadways with adequate expansion joints to prevent buckling, or using materials like Pyrex for cooking pans to minimize cracking from thermal stress.
A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at the end of a randomly chosen red light. The engineer believes that the probability that X = x is proportional to (x + 1)(8 − x) for x = 0, . . . , 7 (the possible values of X), i.e. P(X = x) = ( C(x + 1)(8 − x) for x = 0, 1, . . . , 7 0 otherwise 1. Determine the value of C. 2. Find the probability that X will be at least 5.
Answer:
A) C = 1/120
B) P( x ≥ 5) = 1/3
Explanation:
I've attached explanations to this.
The value of C is: C = 1/126
The probability that X will be at least 5 is 1/21.
To determine the value of C, we need to make sure that the sum of the probabilities of all possible values of X is equal to 1. This means that the following equation must hold:
\sum_{x=0}^7 P(X = x) = 1
Substituting in the given expression for P(X = x), we get:
\sum_{x=0}^7 C(x + 1)(8 - x) = 1
Expanding the summation and simplifying, we get:
7C + 21C + 35C + 35C + 21C + 7C = 1
126C = 1
Therefore, the value of C is:
C = 1/126
To find the probability that X will be at least 5, we need to calculate the following sum:
P(X \ge 5) = P(X = 5) + P(X = 6) + P(X = 7)
Substituting in the expression for P(X = x), we get:
P(X \ge 5) = C(5 + 1)(8 - 5) + C(6 + 1)(8 - 6) + C(7 + 1)(8 - 7)
P(X \ge 5) = C(6)(3) + C(7)(2) + C(8)(1)
Substituting in the value of C, we get:
P(X \ge 5) = (1/126)(6)(3) + (1/126)(7)(2) + (1/126)(8)(1)
P(X \ge 5) = 1/21
Therefore, the probability that X will be at least 5 is 1/21.
For such more question on probability
https://brainly.com/question/31064097
#SPJ3
) An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of material having a thermal conductivity of 5 W/m·K. Estimate the temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
Answer:
Th = 40.91 C
Explanation:
We must use the equation for heat current in conduction
[tex]H = \frac{kA(Th-Tc)}{L}[/tex]
Where k is the thermal conductivity of the material
A is the cross-sectional area
(Th -Tc) is the temperature difference
and L is the length of the material
Thus
[tex]A = 2\pi r*0.1 = 0.00157079 m^{2}[/tex]
[tex]50 = \frac{5*0.00157079(Th-25)}{0.0025}[/tex]
Isolating Th, we have
[tex]Th = \frac{50*0.0025}{5*0.00157079}+25[/tex]
Th = 40.91 C
The question relates to estimating the temperature of a heater based on heat conduction, but it lacks the necessary details about the heater's material properties to provide a numerical answer.
Explanation:The student is asking about the temperature reached by an electrical heater when dissipating power. To estimate this temperature, one must consider the concept of heat conduction in which the rate of heat transfer (Q/t) is influenced by the thermal conductivity (k) of the material, surface area (A), thickness (d), and temperature difference across the material. In this problem, the heater dissipates 50 W of power, the length of the heater is 100 mm, the diameter is 5 mm, and the thermal conductivity of the surrounding material is 5 W/m·K. To calculate the temperature reached by the heater, we would typically use the equation for steady-state heat transfer. However, the question lacks sufficient details such as the properties of the heater material, which would be necessary to determine the temperature distribution and to estimate the temperature reached by the heater. Therefore, with the information provided, we can only discuss the factors involved in calculating the temperature reached by the heater but cannot provide a numerical answer.
Find the value of R2 required if vs = 18 V , R1 = 50 kΩ , and the desired no-load output voltage is vo = 6 VPart b)Several different loads are going to be used with the voltage divider from Part A. If the load resistances are 300 kΩ , 200 kΩ , and 100 kΩ , what is the output voltage that is the most different from the design output voltage vo = 6 V ?part c)The circuit designer wants to change the values of R1 and R2 so that the design output voltage vo = 6 V is achieved when the load resistance is RL = 200 kΩ rather than at no-load. The actual output voltage must not drop below 5.4 V when RL = 100 kΩ . What is the smallest resistor value that can be used forR1?
Answer:
Explanation:
Check attachment for solution
a. The value of R2 required for the desired no-load output voltage R2 = 10 kΩ
b. The most different output voltage is 5.8 V, which occurs when the load resistance is 300 kΩ.
c. The smallest resistor value that can be used for R1 is 45 kΩ.
Part A:
To find the value of R2 required for the desired no-load output voltage, we can use the following voltage divider equation:
vo = vs * R2 / (R1 + R2)
Substituting in the known values, we get:
6 V = 18 V * R2 / (50 kΩ + R2)
Solving for R2, we get:
R2 = 18 V * 6 V - 6 V * 50 kΩ / 6 V - 18 V
R2 = 10 kΩ
Part B:
To find the output voltage that is the most different from the design output voltage, we need to calculate the output voltage for each load resistance. We can use the following voltage divider equation:
vo = vs * (R2 / (R1 + R2) || RL)
Substituting in the known values for each load resistance, we get:
vo (RL = 300 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 300 kΩ)
vo (RL = 300 kΩ) = 5.8 V
vo (RL = 200 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 200 kΩ)
vo (RL = 200 kΩ) = 7.2 V
vo (RL = 100 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 100 kΩ)
vo (RL = 100 kΩ) = 8.6 V
Therefore, the output voltage that is the most different from the design output voltage is 5.8 V, which occurs when the load resistance is 300 kΩ.
Part C:
To change the values of R1 and R2 so that the design output voltage vo = 6 V is achieved when the load resistance is RL = 200 kΩ rather than at no-load, we can use the following voltage divider equation:
vo = vs * R2 / (R1 + R2)
Substituting in the known values, we get:
6 V = 18 V * R2 / (R1 + 200 kΩ)
Solving for R2, we get:
R2 = 30 kΩ
Next, we need to calculate the value of R1 to ensure that the actual output voltage does not drop below 5.4 V when RL = 100 kΩ. We can use the following voltage divider equation:
vo = vs * (R2 / (R1 + R2) || RL)
Substituting in the known values, we get:
5.4 V = 18 V * (30 kΩ / (R1 + 30 kΩ) || 100 kΩ)
Solving for R1, we get:
R1 = 45 kΩ
Therefore, the smallest resistor value that can be used for R1 is 45 kΩ.
For such more question on voltage
https://brainly.com/question/28632127
#SPJ3
A piston-cylinder assembly contains 0.5 lb of water. The water expands from an initial state where p1 = 40 lbf/in.2 and T1 = 300o F to a final state where p2 = 14.7 lbf/in.2 During the process, the pressure and specific volume are related by the polytropic process pv 1.2 = constant. Determine the energy transfer by work, in Btu.
To find the work done during the polytropic expansion of water in a piston-cylinder assembly, the initial and final volumes are needed to apply the polytropic process work formula. Without this information, it's not possible to calculate the energy transfer by work.
Explanation:The question asks to determine the energy transfer by work during a polytropic process where pressure p and specific volume v are related by the equation pv1.2 = constant. To find the work done by the water when it expands in a piston-cylinder assembly, you would typically use the polytropic process work formula:
W = (P1V1 - P2V2) / (n - 1)
However, to use this formula, you would need the initial and final volumes, V1 and V2, which are not provided in the question. Without these volumes or additional information such as tables or diagrams that could help find these values through the relationship of states, it is impossible to provide a numerical answer to the question asked.
An equal-tangent vertical curve is to be constructed between grades of -2% (initial) and 1% (final). The PVI is at station 110 00 and at elevation 420 ft. Due to a street crossing the roadway, the elevation of the roadway at station 112 00 must be at 424.5 ft. Design the curve, and determine the elevations and stations for PVC and PVI.
Answer:
The curve length (L) will be = 1218 ft
The elevations and stations for PVC and PVI
a. station of PVC = 103 + 91.00
b. station of PVI = 116 + 09.00
c. elevation of PVC = 432.18ft
d. elevation of PVI = 426.09ft
Explanation:
First calculate for the length (L)
To calculate the length, use the formula of "elevation at any point".
where, elevation at any point = 424.5.
and ∴ PVC Elevation = (420 + 0.01L)
Then, calculate for Station of PVC and PVI and elevation of PVC and PVI
The surface energy of a single crystal depends on crystallographic orientation. Does this surface energy increase or decrease with an increase in planar density
Answer:
The surface energy of a single crystal depends on the planar density (i.e., degree of atomic packing) of the exposed surface plane because of the number of unsatisfied bonds. As the planar density increases, the number of nearest atoms in the plane increases, which results in an increase in the number of satisfied atomic bonds in the plane, and a decrease in the number of unsatisfied bonds. Since the number of unsatisfied bonds diminishes, so also does the surface energy decrease. (That is, surface energy decreases with an increase in planar density.)
Explanation:
Water at 158C (r 5 999.1 kg/m3 and m 5 1.138 3 1023 kg/m·s) is flowing steadily in a 30-m-long and 5-cm-diameter horizontal pipe made of stainless steel at a rate of 9 L/s. Determine (a) the pressure drop, (b) the head loss, and (c) the pumping power requirement to overcome this pressure drop.
Complete Question
The complete question is shown on the first uploaded image
Answer:
(a) the pressure drop is [tex]\Delta P_L = 100.185\ kPa[/tex]
(b) the head loss is the [tex]h_L =10.22\ m[/tex]
(c) the pumping power requirement to overcome this pressure drop [tex]W_p = 901.665\ W[/tex]
Explanation:
So the question we are told that the water is at a temperature of 15°C
And also we are told that the density of the water is [tex]\rho=999.1\ kg/m^3[/tex]
We are also told that the dynamic viscosity of the water is [tex]\mu = 1.138 *10^{-3} kg/m \cdot s[/tex]
From the diagram the length of the pipe is [tex]L=[/tex] 30 m
The diameter is given as [tex]D=[/tex] 5 cm [tex]\frac{5}{100} m = 0.05m[/tex]
The volumetric flow rate is given as [tex]Q=[/tex] 9 L/s [tex]= \frac{9}{1000} m^3/ s = 0.009\ m^3/s[/tex]
Now the objective of this solution is to obtain
i the pressure drop, ii the head loss iii the pumping power requirement to overcome this pressure drop
to obtain this we need to get the cross-sectional area of the pipe which will help when looking at the flow analysis
So mathematically the cross-sectional area is
[tex]A_s =\frac{\pi}{4} D^2[/tex]
[tex]= \frac{\pi}{4} (5*10^{-2})[/tex]
[tex]=1.963*10^{-3} m^2[/tex]
Next thing to do is to obtain average flow velocity which is mathematically represented as
[tex]v = \frac{Q}{A_s}[/tex]
[tex]v =\frac{9*10^{-3}}{1.963*10^{-3}}[/tex]
[tex]=4.585 \ m/s[/tex]
Now to determine the type of flow we have i.e to know whether it a laminar flow , a turbulent flow or an intermediary flow
We use the Reynolds number
if it is below [tex]4000[/tex] then it is a laminar flow but if it is higher then it is a turbulent flow ,now when it is exactly the value then it is an intermediary flow
This Reynolds number is mathematically represented as
[tex]Re = \frac{\rho vD}{\mu}[/tex]
[tex]=\frac{(999.1)(4.585)(0.05)}{1.138*10^{-3}}[/tex]
[tex]=2.0127*10^5[/tex]
Now since this number is greater than 4000 the flow is turbulent
So we are going to be analyse the flow using the Colebrook's equation which is mathematically represented as
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{\epsilon/D_h}{3.7} +\frac{2.51}{Re\sqrt{f} } ][/tex]
Where f is the friction factor , [tex]\epsilon[/tex] is the surface roughness ,
Now generally the surface roughness for stainless steel is
[tex]\epsilon = 0.002 mm = 2*10^{-6} m[/tex]
Now substituting the values into the equation we have
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{2*10^{-6}/5*10^{-2}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ][/tex]
So solving to obtain f we have
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{4*10^{-5}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ][/tex]
[tex]= -2.0log[1.0811 *10^{-5} +\frac{1.2421*10^{-5}}{\sqrt{f} } ][/tex]
[tex]f = 0.0159[/tex]
Generally the pressure drop is mathematically represented as
[tex]\Delta P_L =f\ \frac{L}{D} \ \frac{\rho v^2}{2}[/tex]
Now substituting values into the equation
[tex]= 0.0159 [\frac{30}{5*10^{-5}} ][\frac{(999.1)(4.585)}{2} ][/tex]
[tex]=100.185 *10^3 Pa[/tex]
[tex]=100.185 kPa[/tex]
Generally the head loss in the pipe is mathematically represented as
[tex]h_L =\frac{\Delta P_L}{\rho g}[/tex]
[tex]= \frac{100.185 *10^3}{(999.1)(9.81)}[/tex]
[tex]=10.22m[/tex]
Generally the power input required to overcome this pressure drop is mathematically represented as
[tex]W_p = Q \Delta P_L[/tex]
[tex]=(9*10^{-3}(100.185*10^3))[/tex]
[tex]= 901.665\ W[/tex]
Air enters the compressor of an ideal gas refrigeration cycle at 7∘C and 35 kPa and the turbine at 37∘C and 160 kPa. The mass flow rate of air through the cycle is 0.2 kg/s. Assuming variable specific heats for air, determine
(a) the rate of refrigeration, (b) the net power input, and (c) the coefficient of performance.
It appears that your answer contains either a link or inappropriate words. Please correct and submit again! error
Had to screenshot the solution check attached
Let m be an integer in the set {0,1,2,3,4,5,6,7,8, 9}, and consider the following problem: determine m by asking 3-way questions, i.e. questions with at most 3 possible answers. For instance, one could ask which of 3 specific subsets m belongs to.
Give a decision tree argument showing that at least 3 such questions are necessary in worst case. In other words, prove that no correct algorithm can solve this problem by asking only 2 questions in worst case.
Answer:
Take any algorithm if that algorithm solves this problem it can be represented as a ternary decision tree. Therefore each question has at most three answers.
There are ten possible verdicts, the height of such kind of tree should satisfy
ℎ >= ⌈log3(10)⌉ = 3
Hence no such algorithm can ask less than three questions in the worst case.
---
b)
Each and every internal node represents a question asking whether m belongs to one of three possible subset of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} or not
For example 0123|456|789 represented the questionDoes “m: belongs to {0, 1, 2, 3}, to {4, 5, 6}, or to {7, 8, 9}?"
Verdicts are placed in brackets "[ ]"
Explanation:
decision tree is attached below
Which of these statements is true?
1-Gutters are installed against the soffit.
2-Fascia requires openings for ventilation.
3-Drip edges prevent water from running underneath an overhang.
4-A gutter is installed flush with the fascia.
5-A vent spacer is installed underneath the rafter insulation.
Answer:
3-Drip edges prevent water from running underneath an overhang.
Explanation:
The only correct statement is in option 3. Drip edges are structures that are connected to the roof edges of buildings to ensure that the flow of water is properly controlled. They are typically used yo prevent water from getting to other parts of the building. They are made of non-corrosive and non-staining materials to make roofs of buildings beautiful.
Answer:
Drip edges prevent water from running underneath an overhang.
Explanation:
Consider the following matrix transpose routine:
typedef int array[4] [4];
void transpose2(array dst, array src)
{
int i, j;
for (i = 0; i < 4; i++);
for (j = 0; j < 4; j++);
dst [j] [i] = src[i][j];
}
}
}
Assume this code runs on a machine with the following properties:
(a) sizeof (int) = 4
(b) The src array starts at address 0 and the dst array starts at address 64 (decimal).
(c) There is a single LI data cache that is direct-mapped, write-through, write-allocate, with a block size of 16 bytes.
(d) The cache has a total size of 32 data bytes (i.e. not considering tags, etc.), and the cache is initially empty.
(e) Accesses to the src and dst arrays are the only sources of read and write misses, respectively.
For each row and col, indicate whether the access to src [row] [col] and dst [row] [col] is a hit (h) or a miss (m). For example, reading src [0] [0] is a miss and writing dst [0] [0] is also a miss.
The question is about determining cache hits and misses when performing a matrix transposition, taking into account the cache's specifications. Given the setup, each new cache block accessed will result in a miss, followed by hits if subsequent accesses fall within the same block. The non-sequential access pattern of matrix transposition is likely to incur more cache misses compared to sequential access.
Explanation:The student is asking about cache behavior when transposing a matrix, which involves changing the positions of each element in the array so that rows become columns and vice versa. In a cache memory system that is direct-mapped, block size determines how many elements will fit within one block of cache. Because the array starts at a known memory address and each integer occupies 4 bytes (given by the property that sizeof(int) is 4), we can predict how the cache will behave during the execution of the transpose routine.
Given the cache specifics - direct-mapped, write-through, write-allocate, and block size of 16 bytes (which can hold 4 integers), and size of 32 bytes total - the access pattern will largely depend on the congruence of array indices and cache block boundaries. For example, when accessing the source array src at address 0 (src[0][0]), it will be a cache miss as the cache is initially empty. However, subsequent accesses like src[0][1], src[0][2], and src[0][3] may hit as they reside in the same cache block if the block can accommodate them. Similarly, writes to the destination array dst will initially miss and then follow a similar pattern of hits and misses based on cache boundaries and block allocation behavior after write misses (due to write-allocate policy).
Matrix transposition involves swapping elements such that element (i, j) becomes element (j, i). As a result of accessing elements in this non-sequential manner, there could be more cache misses compared to sequential access because the transpose operation accesses elements across different rows and hence, potentially different cache lines.
piston cylinder has 5kg water at 500kPa and 300c. Heated by 100V and 4A that operates for 500 secs. It's cooled to a saturated liquid. what is the boundary work and heat lost?
Answer: boundary work is 1.33kJ and heat lost is 200kJ
Explanation: mass of water m = 5kg
Molar mass of water mm = 18kg/mol
No of moles n = 5/18 = 0.28
Pressure P = 500x10^3
Temperature T = 300°c = 573K
R = 8.314pa.m3/mol/k a constant.
Using PV = nRT
V = nRT/P
V = (0.28x8.314x573)÷(500x10^3)
V = 0.00266m3
Work on boundary w = PV
W = 500x10^3 x 0.00266 = 1330 J
= 1.33kJ
Heat lost on cooling is equal to electrical heat Q used to heat the water initially
Q =Ivt
I = current = 4A
v = voltage = 100
T = time = 500sec
Q = 4x100x500 = 200000 = 200kJ
2. If you have 10 ft3 of an ideal gas stored at 75 F and at 12 psia that is expanded to 15 ft3 , what is the resulting pressure? 3. The atmospheric pressure is 14.3 psi. How many inches of mercury (s=13.6) does the National Weather Service report this pressure as? Pressure = __________________
Answer:
2. The decrease in pressure will be 8psia
3. The NWS will report the pressure to be 0.0657inchHg
Explanation:
2. This is an isothermal system because it was only the volume of the gas that was increased, and temperature was constant. To find the pressure, use the Boyle's law equation below.
P1V1 ÷ P2V2..............(1)
make P2 the subject of the formula
P2 = P1V1 ÷ V2 ..............(2)
P1 = 12psia
V1 = 10ft3
V2 = 15ft3
Using equation 2
P2 = (12 × 10) ÷ 15 = 8psia
Therefore, the resulting pressure of the gas will decrease to 8psia, which obeys Boyle's law that states that pressure is inversely proportion to volume in a given mass of gas. That means, as volume increases pressure should decrease.
3. The National weather Service report, are the national agency that forecast the weather, to report how the climate will change. To report the pressure of a weather, their use a pressure barometer, and reports the reading in the pressure barometer, which is the hight the Mercury in the barometer increased to.
Using a pressure barometer.
Patm = pgh..............(3)
Make h the subject of formula
h = Patm ÷ pg ..............(4)
Patm is the Atmospheric pressure = 4.3psi
p is the density of Mercury = 13.6
g is the force of gravity = 9.8
h is the hight of the Mercury in the barometer, which is the pressure that is measured.
Using equation 4
h = 4.3 ÷ (13.6 × 9.8) = 0.0322629psi
To convert Pound- force per square inch to inch of mecury (that's is psi to inchHg)
1psi = 2.03602inchHg
Therefore
0.0322629psi = 0.065689399inchHg
The NWS will report the pressure to be 0.0657inchHg
Tanya Pierce, President and owner of Florida Now Real Estate is seeking your assistance in designing a database for her business. One of her employees has experience in developing and implementing Access-based systems, but has no experience in conceptual or logical data modeling. So, at this point Tanya only wants you to develop a conceptual data model for her system. You are to use our entity-data diagramming notation - Crows foot symbols.
Tanya has some very specific needs for her system. There are several aspects of the business that need to be represented in the data model. Of central interest are properties that are listed or sold by the company. Note that a separate division of Florida Now handles raw land, so your system only has to deal with developed property. For all types of properties, Tanya wants to keep track of the owner (client), the listing and selling dates, the asking and selling prices, the address of the property, the Multiple Listing Service (MLS) number, and any general comments. In addition, the database should store the client that purchases a property. There are some specialized data that need to be stored, depending on the type of property. For single family houses she wants to store the area (for example UCF or Conway), the size of the house in square feet, the number of bedrooms and baths, the size of the garage (for example, 2 car), and the number of stories. For condominiums, the database should track the name of the complex, the unit number, the size in square feet, the number of bedrooms and baths, and the type of community (coop or condo). For commercial properties, she needs to know the zoning, size in square feet, type (industrial, retail, or office), and the general condition of the property.
Tanya also wants the database to store information about her real estate agents, such as their name, home address, home phone number, mobile phone number, email address, and their real estate license number. Also, the database should track which agent lists and sells each piece of property. Note that she has a separate system that tracks selling agents and listings from outside brokerages, so you don’t have to worry about external agents. However, sometimes a property will be listed and sold by the same Florida Now agent, but other times one Florida Now agent will list a property and another will sell it.
Of course, Tanya thinks that it would be good to have the database track information about Florida Now’s clients, such as their name, phone number, and street and email addresses. Also, she wants to be able to record comments about the client. Florida Now offers referral fees to clients who refer potential customers to the brokerage. The database should store these referral relationships between clients, including the amount and date of the referral payment. Note that only one client can be paid for a referral. In other words, it is not possible for two clients to be paid for referring the same client.
Finally, Tanya wants to be able to use the database to examine the effectiveness of various advertising outlets. For each outlet, the database should store the name of the outlet (for example, realestate.com or the Orlando Sentinel), the main contact person and their phone number. She also wants to know how much it cost to advertise each property on any outlet used, and when a property was advertised on each outlet used for that property. Keep in mind that a property may be advertised on multiple outlets, and that the cost of advertising on an outlet might vary from property to property.
Create an entity relationship diagram that captures Tanya’s database requirements. The ERD should indicate all entities, attributes, and relationships (including maximum and minimum cardinalities). Also be sure to indicate primary key attributes.
You must draw the ERD using computer software such as Visio, PowerPoint or Word. In addition, you must resolve all M:N (many- to-many) relationships and multi-valued attributes. State any assumptions you make. Remember that your assumptions must be reasonable and must not violate any stated requirements. Think through the entities, and attributes for each entity.
Answer:
the answer is attributes for each entity
8.2.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's first two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new
Answer:
The output will be (3, 4) becomes (8, 10)
Explanation:
#include <stdio.h>
//If you send a pointer to a int, you are allowing the contents of that int to change.
void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){
*xNew = (xVal+1)*2;
*yNew = (yVal+1)*2;
}
int main(void) {
int xValNew = 0;
int yValNew = 0;
CoordTransform(3, 4, &xValNew, &yValNew);
printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);
return 0;
}
The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The motor has an efficiency of ε = 0.65. Determine the power that must be supplied to the motor at the instant the load has been hoisted s = 27 ft starting from rest.
The power that must be supplied to the motor is 136 hp
Explanation:
Given-
weight of the elevator, m = 1000 lb
Force on the table, F = 500 lb
Distance, s = 27 ft
Efficiency, ε = 0.65
Power = ?
According to the equation of motion:
F = ma
[tex]3(500) - 1000 = \frac{1000}{32.2} * a[/tex]
a = 16.1 ft/s²
We know,
[tex]v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s[/tex]
To calculate the output power:
Pout = F. v
Pout = 3 (500) * 29.48
Pout = 44220 lb.ft/s
As efficiency is given and output power is known, we can calculate the input power.
ε = Pout / Pin
0.65 = 44220 / Pin
Pin = 68030.8 lb.ft/s
Pin = 68030.8 / 500 hp
= 136 hp
Therefore, the power that must be supplied to the motor is 136 hp
To calculate the power supplied to the motor, we use the work done by the motor and the motor efficiency. However, without the time, it takes to lift the elevator, a crucial piece of information is missing, thus preventing an exact calculation.
Explanation:To determine the power that must be supplied to the motor at the instant the 1000-lb elevator has been hoisted 27 ft from rest, we must first calculate the useful power output. Since the motor has an efficiency of 0.65 and exerts a constant force of 500 lb, we can use the work-energy principle along with the efficiency to find the required power.
The work done by the motor on the elevator is the force applied times the distance moved. The useful work is W = force × distance = 500 lb × 27 ft.
The useful power output, or work done per unit time, can be found by P = W / t. However, the time is not provided directly in this scenario; we assume this process occurs instantaneously, which is not realistic for real-world scenarios.
Since power supplied is equal to the useful power output divided by the efficiency, we have Power supplied = Useful power / ε.
Assumptions in the CalculationWe would normally need time to calculate power, which is work done or energy converted per unit time.The efficiency of the motor is used to calculate the actual power supplied versus the useful power output.The actual lifting process takes time, and an instantaneous lift is not realistic.Due to missing time information, the exact numerical answer cannot be provided without assumptions or additional data concerning how quickly the elevator is lifted.