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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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A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o
victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width= 25 mm = 25\times 10^{-3} m

thickness = 6.5 mm = 6.5\times 10^{-3} m

crack length 2c = 0.5 mm at centre of specimen

\sigma _{applied} =  1000 N/cross sectional area

stress intensity factor  =  k  will be

\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}

                   = 6.154\times 10^{6} Pa

we know that

k =\sigma_{applied} (\sqrt{\pi C})

  =6.154\sqrt{\pi (2.5\times 10^{-04})}          [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

ifK_C = 1.15 Mpa m^{1/2} then load will be

Kc = \sigma _{frac}(\sqrt{\pi C})

1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}

\sigma _{frac} = 41.04 MPa

load = \sigma _{frac}\times Area

load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N

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2 years ago
Why might construction crews want to install pipes before the foundation is poured
Crazy boy [7]

The answer is choice C

Explanation:

As during construction ,the site is cleared for all debris before laying out the foundation. Even the sewer lines are dug out .

So it will be useful for the construction crews to  connect the pipes to the sewer lines before the foundation is poured.

But usually the steps take in construction activity is:- first the site is cleared for the foundation to be poured  and once the foundation wall is set , then all utilities , including plumbing and electrical activities are done.,

After this process is over, the city inspector comes to check whether the foundation has been laid down as per the code of construction.

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A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The
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Answer:

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Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = \frac{60+180}{2}  = 120 kPa

and,

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4 0
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Answer:

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Explanation:

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T (°F) = 1700kJ/h.°C

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