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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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True or False? A constricting nozzle is used
marissa [1.9K]

Answer:

True

Explanation:

Principles of plasma arc cutting, Uses a constricting nozzle to create, concentrate,  and direct the high-velocity plasma. Plasma gas is always used

in plasma arc cutting When shielding gas is also used, the process is called dual flow plasma arc cutting. I hope this helps.

 

6 0
3 years ago
True or False - Blueprints can be drawn both by hand and on a computer
Orlov [11]

Answer:

The answer is true.

Explanation:

Most of it is done in computer, but it can also be drawn out by hand.

6 0
3 years ago
Read 2 more answers
Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒  P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0
4 years ago
If the atomic radius of copper is 0.128 nm, calculate the volume of its unit cell in cubic meters.
Alex_Xolod [135]

Answer:

Volume of face centered cubic cell=4.74531*10^{-29} m^3

Explanation:

Consider the face centered cubic cell:

1 atom at each corner of cube.

1 atom at center of each face.

Consider the one face (ABCD) as shown in attachment for calculation:

Length of the all sides of face centered cubic cell is L.

Volume of face centered cubic cell= L^3

Now Consider the figure shown in attachment:

According to Pythagoras theorem on ΔADC.

L^{2}+L^2=(4a)^2     (a is the atomic radius)

L=\frac{4a}{\sqrt{2}} (Put in the formula of Volume)

Volume of face centered cubic cell= L^3

Volume of face centered cubic cell= (\frac{4a}{\sqrt{2}})^3

Volume of face centered cubic cell= (\frac{4(0.128*10^{-9}}{\sqrt{2}})^3

Volume of face centered cubic cell=4.74531*10^{-29} m^3

3 0
4 years ago
(a) Sabbir usually (sit)______ in the front bench.
klasskru [66]

Answer:

a)sits

b)is shining

d)have been living

Explanation:

5 0
3 years ago
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