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Daniel [21]
3 years ago
6

A masonry facade consisting of 3,800 square feet is to be constructed for a building. The total cost per worker hour is estimate

d to be $31.50, and the total estimated cost of the labor for this task is $10,500. Assuming 8-hour work days and a crew of six workers, how many days should be allowed to complete this task? What is the production rate that the crew must attain to keep the project on schedule and within the budget?
Engineering
1 answer:
lana66690 [7]3 years ago
6 0

Answer:

Days: 6.9444 days

Production rate: 547.2035 ft²/s

Explanation:

the solution is attached in the Word file

Download docx
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According to fire regulations in a town, the pressure drop in a commercial steel, horizontal pipe must not exceed 2.0 psi per 25
bonufazy [111]

Answer:

6.37 inch

Explanation:

Thinking process:

We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.

To determine the pressure drop in the pipe:

Using the Bernoulli equation for mass conservation:

\frac{P1}{\rho } + \frac{v_{2} }{2g} +z_{1}  = \frac{P2}{\rho } + \frac{v2^{2} }{2g} + z_{2} + f\frac{l}{D} \frac{v^{2} }{2g}

thus

\frac{P1-P2}{\rho }  = f\frac{l}{D} \frac{v^{2} }{2g}

The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.

Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F

from the tables

Re= 2.01 × 10⁵

Hence, f = 0.018

Therefore, pressure drop, (P1-P2)/p = 2.70 ft

This occurs at ae presure change of 1.17 psi

Correlating with the chart, we find that the diameter will be D= 0.513

                                                                                                      = <u>6.37 in Ans</u>

7 0
4 years ago
2.4: Add a method called setValue(), and the description of setValue is: public int setValue(long searchKey) In this method, the
Yanka [14]

Answer:

Below is java code that must be used for the given question:

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

  {

  private long[] a;                 // ref to array a

  private int nElems;               // number of data items

  //-----------------------------------------------------------

  public HighArray(int max)         // constructor

     {

     a = new long[max];                 // create the array

     nElems = 0;                        // no items yet

     }

  //-----------------------------------------------------------

  public setValue find(long searchKey)

     {                              // find specified value

     int j;

     for(j=0; j<nElems; j++)            // for each element,

        if(a[j] == searchKey)           // found item?

           break;                       // exit loop before end

     if(j == nElems)                    // gone to end?

        return false;                   // yes, can't find it

     else

        return true;                    // no, found it

     }  // end find()

  //-----------------------------------------------------------

  public void insert(long value)    // put element into array

     {

     a[nElems] = value;             // insert it

     nElems++;                      // increment size

     }

  //-----------------------------------------------------------

  public void display()             // displays array contents

     {

     for(int j=0; j<nElems; j++)       // for each element,

        System.out.print(a[j] + " ");  // display it

     System.out.println("");

     }

  //-----------------------------------------------------------

  }  // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

  {

  public static void main(String[] args)

     {

     int maxSize = 100;            // array size

     HighArray arr;                // reference to array

     arr = new HighArray(maxSize); // create the array

     arr.insert(77);               // insert 10 items

     arr.insert(99);

     arr.insert(44);

     arr.insert(55);

     arr.insert(22);

     arr.insert(88);

     arr.insert(11);

     arr.insert(00);

     arr.insert(66);

     arr.insert(33);

     arr.display();                // display items

     int searchKey = 35;           // search for item

     if( arr.find(searchKey) )

        System.out.println("Found " + searchKey);

     else

        System.out.println("Can't find " + searchKey);

     }  // end main()

  }  // end class HighArrayApp

Explanation:

6 0
3 years ago
Which type of engineers were the designers of the Great Pyramids of Egypt and the Great Wall of China?
Gemiola [76]
I think it’s structural engineers but still check with the others
8 0
3 years ago
Read 2 more answers
Answer true or false 3.Individual people decide what will be produced in a command<br> oconomy
Pie

Answer:

False

Explanation:

The government decides the productions.

7 0
4 years ago
Read 2 more answers
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
4 years ago
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