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lys-0071 [83]
3 years ago
7

Extra Credit: The Linc (parking lot and stadium)In celebration of the upcoming Super Bowl, for a maximum 10 points of extra cred

it, you may try to reproduce the ASCII Art shown below of Lincoln Stadium, home of the Philadelphia Eagles. You should still include a class constant for the SIZE; in Dr. Yates' implementation, the SIZE value that produces the picture below is 4, and works for any size >= 2. You must include loops and nested loops to make this work correctly; you CANNOT simply include a separate println statement for each line of the drawing. You will get the full extra credit points only if you duplicate the drawing EXACTLY. (Note: this is a fairly tricky figure to do right.) The parking lot alone is worth a maximum of 2 points.
Engineering
1 answer:
Sidana [21]3 years ago
8 0

Answer:

Explanation:

// Below is the code to draw parking lot only , i have also put the o/p of code.

import java.util.*;

import java.lang.*;

import java.io.*;

public class Linc

{

static int size=4;

static int numBoxes = 1; // one row.

static int height =(int) Math.pow(size, 2); //Just how many | will it have.

static int width = 2; // two boxes in a row

public static void main (String[] args) throws java.lang.Exception

{

top();

printHeight();

}

  public static void top(){

  for(int i = 0; i <= numBoxes*width;i++)

  {

  if(i%width == 0) {//When this happens we're in a new box.

      System.out.print(" ");

      System.out.print("____________");

  }

  else

  System.out.print(" ");

  }

  System.out.println(); //Move to next line.

 

  }

  public static void printHeight(){

  for(int j = 0; j < height;j++){

  for(int i = 0; i <= numBoxes*width;i++)

  {

  if(i%width == 0) {

  System.out.print("|"); //Whenever this happens we're in a new box.

  System.out.print("____________");

     

  }

  else

  System.out.print(" ");

 

  }

  System.out.print("|");

  System.out.println(); //Move to next line.

  }

  }

}

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Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase
ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite (\% V_{Gr}) is determined by the following expression:

\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%

\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%

Where:

V_{Gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{Fe} - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}

V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}

Where:

m_{Gr}, m_{Fe} - Masses of the graphite and ferrite phases, measured in grams.

\rho_{Gr}, \rho_{Fe} - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%

\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

m_{Gr} = \frac{2.5}{100}\times (100\,g)

m_{Gr} = 2.5\,g

m_{Fe} = 100\,g - 2.5\,g

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If m_{Gr} = 2.5\,g, m_{Fe} = 97.5\,g, \rho_{Fe} = 7.9\,\frac{g}{cm^{3}} and \rho_{Gr} = 2.3\,\frac{g}{cm^{3}}, the volume percent of graphite is:

\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%

\% V_{Gr} = 91.906\,\%

The volume percent of graphite is 91.906 per cent.

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