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Andrei [34K]
2 years ago
6

PLEASE HURRY!!!

Engineering
2 answers:
Naily [24]2 years ago
3 0

Answer:

A

Explanation:

He should get a job in engineering to see what it's like to work in the field.

qaws [65]2 years ago
3 0

I can confirm that the answer is "Seth should work for two years in the engineering field to gain experience." :)

<em>( On my own, I had put </em>"Seth should obtain a graduate degree in civil engineering."<em> as my answer and got the question wrong. It showed me that this was the correct answer. ) </em>

You might be interested in
1. True/False The Pressure Relief valve maintains the minimum pressure in the hydraulic circuit​
elena55 [62]
Yeah it’s true. Good luck!!
3 0
3 years ago
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
Assume that the water temperature is 10°C and the depth of the settling tank is 3.0 m (9.80 ft). Calculate the theoretical settl
Lelu [443]

Explanation:

See attached file

6 0
3 years ago
1. A thin plate of a ceramic material with E = 225 GPa is loaded in tension, developing a stress of 450 MPa. Is the specimen lik
mina [271]

Answer:

fracture will occur as the value is less than E/10 (= 22.5)

Explanation:

If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}

=  2\times 750 (\frac{\frac{0.2mm}{2}}{0.001 mm}})^{1/2}

                 = 15 GPa

fracture will occur as the value is less than E/10 = 22.5

7 0
3 years ago
At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r
ludmilkaskok [199]

Answer:

a) COP_{R} = 25.014, b) T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}

COP_{R} = 25.014

b) The respective coefficient of performance is determined:

COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}

COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}

COP_{R} = 5

But:

COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}

The temperature at hot reservoir is found with some algebraic help:

COP_{R} \cdot (T_{H}-T_{L})=T_{L}

T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}

T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}}  \right)

T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5}  \right)

T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

8 0
3 years ago
Read 2 more answers
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