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Svetach [21]
3 years ago
11

Why might there be multiple foremen on one work site?

Engineering
1 answer:
IrinaK [193]3 years ago
8 0

Answer:

4

Explanation:

You might be interested in
C++ - Green Crud Fibonacci programThe following program is to be written with a loop. You are to write this program three times
Fynjy0 [20]

Answer:

Below is the required code:

Explanation:

Using for loop

#include <iostream>

using namespace std;

int main()

{

    //Initial crud size

    int init = 0;

    int newCrud;

    int next=0;

    //Number of days to simulate

    int no_days;

    int day;

    cout << "Enter initial amount of green crud: ";

    cin >> newCrud;

    cout << "Enter number of days to simulate: ";

    cin >> no_days;

    for (day = 10; day<=no_days; day++)

    {

         if (day % 10 == 0)

         {

             next = newCrud + init;

         }

             newCrud = init;

             init = next;

    }

    if (no_days < 5)

    cout << "\nCrud reproduce only after 5 days minimum.Hence the current amount is "

    << newCrud << " pounds.";

    else

    cout << "On day " << no_days << " you have " << init

    << " pounds of green crud." << endl;

    cout << "\nWould you like to continue? (y or n): ";

    cin >> ans;

         return 0;

}

Output:

         Enter initial amount of green crud: 5

         Enter number of days to simulate: 220

    On day 220 you have 10485760 pounds of green crud.

Using while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans='y';

    while (ans == 'Y' || ans == 'y')

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud:

         ";

         cin >> newCrud;

         cout << "Enter number of days to simulate:

         ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         cout << "On day " << no_days << " you have "

         << init

         << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or

         n): ";

         cin >> ans;

    }

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

Using do while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans;

    do

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud: ";

         cin >> newCrud;

         cout << "Enter number of days to simulate: ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         

         cout << "On day " << no_days << " you have " <<

         init << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or n):

         ";

         cin >> ans;

    } while (ans == 'Y' || ans == 'y');

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

7 0
3 years ago
Which of these is the BEST description of
strojnjashka [21]

Answer:

i would say C but i may be wrong have a great day

Explanation:

3 0
3 years ago
H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate
Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0 (1)

Where:

\dot m - Mass flow, in kilograms per second.

h_{in}, h_{out} - Specific enthalpies at inlet and outlet, in kilojoules per second.

v_{in}, v_{out} - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

p = 2000\,kPa

T = 300\,^{\circ}C

h_{in} = 3024.2\,\frac{kJ}{kg}

\nu_{in} = 0.12551\,\frac{m^{3}}{kg}

Where \nu_{in} is the specific volume of water at inlet, in cubic meters per kilogram.  

Outlet (Superheated steam)

p = 600\,kPa

T = 160\,^{\circ}C

h_{out} = 2758.9\,\frac{kJ}{kg}

If we know that \dot m = 50\,\frac{kJ}{kg}, h_{in} = 3024.2\,\frac{kJ}{kg}, h_{out} = 2758.9\,\frac{kJ}{kg} and v_{in} = 30\,\frac{m}{s}, then the flow speed at outlet is:

35765-25\cdot v_{out}^{2} = 0 (2)

v_{out} \approx 37.823\,\frac{m}{s}

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius (r_{in}), in meters, by this expression:

\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}} (3)

If we know that \dot m = 50\,\frac{kJ}{kg}, v_{in} = 30\,\frac{m}{s} and \nu_{in} = 0.12551\,\frac{m^{3}}{kg}, then the inlet radius is:

r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}

r_{in}\approx 0.258\,m

The inlet radius of the nozzle is approximately 0.258 meters.  

7 0
2 years ago
The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.91 and 0.09, respectively. Is this a hypoe
WINSTONCH [101]

Answer:

hypoeutectoid

Explanation:

ferrite:  pure form of iron

cementite: It is iron carbide with 93.3% iron and 6.67% carbon

hypoeutectoid: Eutectoid steel with carbon fraction less than 0.8%

hypereutectoid: Eutectoid Steel with carbon content more than 0.8%

For the mentioned iron-carbide alloy,

% of carbon in iron-carbide alloy= percentage of cementite × percentage of carbon in cementite

% of carbon in iron-carbide alloy= 0.09× 0.0667

                                                     = 0.6%

so the alloy is hypoeutectoid

4 0
3 years ago
Water that has evaporated returns to earth as
denis23 [38]

Answer:

rain

Explanation:

evaoration causes clouds

clouds condense and rain

4 0
3 years ago
Read 2 more answers
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