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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 smooth concrete pipe (1.5-ft diameter) carries water from a reservoir to an industrial treatment plant 1 mile away and dischar
Kamila [148]

ANSWER:

Q = 0.17ft3/s

EXPLANATION: since the water runs downhill on a 1:100 slope, that means the flow is laminar.

Using poiseuille equation:

Q = (π × D^4 × ∆P) ÷ (128 × U × ∆X)

Q is the volume flow rate.

π is pie constant value at 3.142

D is the diameter of the pipe

∆P is the pressure drop

U is the viscosity

∆X is the length of the pipe or distance of flow.

Form the question, we are to determine U then Find Q

Therefore;

D = 1.5ft

∆P = 1pa since the minor losses are negligible.

∆X = 1mile = 5280ft.

STEP1: FIND U

Viscosity is a function of the temperature of the liquid. An increase in temperature increases the viscosity of the liquid.

We know that at room temperature, which is 25°C the viscosity of water is 8.9×10^-4pa.s . We can find the viscosity of water at 4°C by cross multiplying.

Therefore;

25°C = 8.9×10^-4pa.s

4°C = U

Cross multiply

U25°C = 4°C × 8.9×10^-4pa.s

U25°C = 0.00356°C.pa.s

Therefore;

U = 0.00356°C.pa.s ÷ 25°C

U = 1.424×10^-4pa.s

Therefore at 4°C the viscosity of water in the pipe is 1.424×10^-4pa.s

STEP2: FIND Q

Imputing the values into poiseuille equation above.

Q = (3.142 × (1.5ft)^4 × 1pa) ÷ (128 × 1.424×10^-4pa.s × 5280ft)

Q = 15.906375pa.ft4 ÷ 96.239616pa.s.ft

Therefore;

Q = 0.16547887ft3/s

Approximately;

Q = 0.17ft3/s

6 0
3 years ago
Air enters the 1 m² inlet of an aircraft engine at 100 kPa and 20° C with a velocity of 180 m/s. Determine: a) The volumetric fl
Shkiper50 [21]

Answer:

a) 180 m³/s

b) 213.4 kg/s

Explanation:

A_1 = 1 m²

P_1 = 100 kPa

V_1 = 180 m/s

Flow rate

Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s

Volumetric flow rate = 180 m³/s

Mass flow rate

\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s

Mass flow rate = 213.4 kg/s

3 0
3 years ago
PLS HELP!!!
Advocard [28]

Answer:

I would say C

Explanation:

let me know if im right

3 0
2 years ago
Read 2 more answers
Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I
Snowcat [4.5K]

Answer:

Given, FDs are:

S -> D

I -> B

IS -> Q

B -> O

a)

"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.

The only candidate key is: IS

IS -> ISBDQO

b)

Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)

c)

Decomposition of R into BCNF:

Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).

R1 is in BCNF

Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).

R21is in BCNF

Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).

R221 is in BCNF.

R222 is in BCNF.

The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)

We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)

Explanation:

The answer above is rendered in a very explanatory way.

8 0
3 years ago
What two things must be included in your function definition?
mars1129 [50]

Explanation:

commands to be and function arguments

8 0
3 years ago
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