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Lyrx [107]
2 years ago
14

Please help me on this it’s due now

Engineering
1 answer:
lbvjy [14]2 years ago
5 0

Answer:

road closures on deliver routes.. i think..

You might be interested in
Which statement about lean manufacturing is true when you compare it to mass production?
Len [333]
Where are the statements then bbs lol
6 0
3 years ago
A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft an
Zigmanuir [339]

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y  ----- ( 1 )

M ( bending moment ) = R * length of span / 2

                                     = (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I ( moment of inertia ) = bd^3 / 12

                                    = ( 2 )*( d )^3  / 12 =  2d^3 / 12

b = 2 in ,  d = ?

length of span = 4 * 12 = 48 inches

R = P  / 2 =  240 * 10^3 / 2 =   120 * 10^3 Ib

y ( centroid distance ) = d / 2  inches

back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

                = ( 288 * 10^4 * 12 ) / 2d^3 )  * d / 2

940300  = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d  ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188   ∴ Value of d ≈ 3.03 in

8 0
3 years ago
A loss in value caused by an undesirable or hazardous influence offsite is which type of depreciation?
Lubov Fominskaja [6]

External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.

<h3>What is depreciation?</h3>

Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.

External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.

Therefore, it is well described above.

To learn more about Depreciation, refer to the link:

brainly.com/question/1203926

#SPJ1

5 0
2 years ago
For a project in C++ we are supposed toDesign a class named Month. The class should have the following private members:-name: a
mart [117]

Answer:

include <iostream>

using namespace std;

 

class Month

{

public:

 Month (char firstLetter, char secondLetter, char thirdLetter);

 

 Month (int monthNum);

.

 

 Month();

 void outputMonth_num();

 

 

 void outputMonthLetters();

private:

 int month;

};

 

 

int main ()

{

 //

 // Variable declarations

 //

 int monthNum;

 char firstLetter, secondLetter, thirdLetter;    

 char testAgain;              

 

 do {

 

   cout << endl;

   cout << "Testing the default constructor ..." << endl;

   Month defaultMonth;

   defaultMonth.outputMonth_num();

   defaultMonth.outputMonthLetters();

 

   //

   // Construct a month using the constructor with one integer argument

   //

   cout << endl;

   cout << "Testing the constructor with one integer argument..." << endl;

   cout << "Enter a month number: ";

   cin >> monthNum;

 

   Month testMonth1(monthNum);

   testMonth1.outputMonth_num();

   testMonth1.outputMonthLetters();

 

   //

   // Construct a month using the constructor with three letters as arguments

   //

   cout << endl;

   cout << "Testing the constructor with 3 letters as arguments ..." << endl;

   cout << "Enter the first three letters of a month (lowercase): ";

   cin >> firstLetter >> secondLetter >> thirdLetter;

   cout << endl;

 

   Month testMonth2(firstLetter, secondLetter, thirdLetter);

   testMonth2.outputMonth_num();

   testMonth2.outputMonthLetters();

 

   //

   // See if user wants to try another month

   //

   cout << endl;

   cout << "Do you want to test again? (y or n) ";

   cin >> testAgain;

 }

 while (testAgain == 'y' || testAgain == 'Y');

 

 return 0;

}

 

 

Month::Month(char firstLetter, char secondLetter, char thirdLetter)

{

if ((firstLetter == 'j')&&(secondLetter == 'a')&&(thirdLetter == 'n'))

  outputMonth_num = 1;

if ((firstLetter == 'f')&&(secondLetter == 'e')&&(thirdLetter == 'b'))

  outputMonth_num = 2;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'r'))

  outputMonth_num = 3;

if ((firstLetter = 'a')&&(secondLetter == 'p')&&(thirdLetter == 'r'))

  outputMonth_num = 4;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'y'))

  outputMonth_num = 5;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(thirdLetter == 'n'))

  outputMonth_num = 6;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(.thirdLetter == 'l'))

  outputMonth_num = 7;

if ((firstLetter == 'a')&&(secondLetter == 'u')&&(thirdLetter == 'g'))

  outputMonth_num = 8;

if ((firstLetter == 's')&&(secondLetter == 'e')&&(thirdLetter == 'p'))

  outputMonth_num = 9;

if ((firstLetter == 'o')&&(secondLetter == 'c')&&(thirdLetter == 't'))

  outputMonth_num = 10;

if ((firstLetter == 'n')&&(secondLetter == 'o')&&(thirdLetter == 'v'))

 outputMonth_num = 11;

if ((firstLetter == 'd')&&(secondLetter == 'e')&&(thirdLetter == 'c'))

 outputMonth_num = 12;

}

 

Month::inputMonthByNumber

{

if (Month_num > 12 && Month_num < 1)

cout << "Invalid number for Month, please choose 1-12)\n";

}

 

void Month::outputMonth_num()

{

 if (month >= 1 && month <= 12)

   cout ><< "Month: " << month << endl;

 else

   cout << "Error - The month is not a valid!" << endl;

}

 

void Month::outputMonthLetters()

{

 switch (month)

   {

   case 1:

     cout << "Jan" << endl;

     break;

   case 2:

     cout << "Feb" << endl;

     break;

   case 3:

     cout << "Mar" << endl;

     break;

   case 4:

     cout << "Apr" << endl;

     break;

   case 5:

     cout << "May" << endl;

     break;

   case 6:

     cout << "Jun" << endl;

     break;

   case 7:

     cout << "Jul" << endl;

     break;

   case 8:

     cout << "Aug" << endl;

     break;

   case 9:  

     cout << "Sep" << endl;

     break;

   case 10:

     cout << "Oct" << endl;

     break;

   case 11:

     cout << "Nov" << endl;

     break;

   case 12:

     cout << "Dec" << endl;

     break;

   default:

     cout << "Error - the month is not a valid!" << endl;

   }

}

7 0
3 years ago
Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a
Arada [10]

Answer:

a) h_c = 0.1599 W/m^2-K

b) H_{loss} = 5.02 W

c) T_s = 302 K

d) \dot{Q} = 25.125 W

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, T_a = 25^0 C = 298 K

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600}  = 1.25 kg/sec

Rate of heat transfer,

\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W

a) To calculate the convection coefficient relationship for heat transfer by convection:

\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, k_c = 0.8

\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K

b) Heat loss from the pipe to the environment:

H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W

d) The required fan control power is 25.125 W as calculated earlier above

5 0
3 years ago
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