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Gekata [30.6K]
1 year ago
13

Q5

Engineering
1 answer:
Klio2033 [76]1 year ago
5 0

The C++ code that would draw all the iterations in the selection sort process on the array is given below:

<h3>C++ Code</h3>

#include <stdio.h>

#include <stdlib.h>

int main() {

   int i, temp1, temp2;

   int string2[16] = { 0, 4, 2, 5, 1, 5, 6, 2, 6, 89, 21, 32, 31, 5, 32, 12 };

   _Bool check = 1;

   while (check) {

       temp1 = string2[i];

       temp2 = string2[i + 1];

       if (temp1 < temp2) {

           string2[i + 1] = temp1;

           string2[i] = temp2;

           i = 0;

       } else {

           i++;

           if (i = 15) {

               check = !check;

           }

       }

   }

   

   return 0;

}

Read more about C++ programming here:

brainly.com/question/20339175

#SPJ1

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Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el
Allisa [31]

Answer:

theoretical fracture strength  = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength  =   s_{0} \times \sqrt{\frac{L}{r} }   .............................1

put here value and we get

theoretical fracture strength  =    1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }  

theoretical fracture strength  =  16919.98 \times 10^6  

theoretical fracture strength  = 16919.98 MPa

3 0
3 years ago
What does a graphic designer need to do to monitor the progress of the project<br>​
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Which option identifies the next step Kristin should take to test the design in the following scenario?
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134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.
vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = - 10^{\circ}C = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = 60^{\circ}C = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T =60^{\circ}C:

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at  P = 140 kPa, T =- 10^{\circ}C:

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = s'_{s} and h'_{s} = 278.06 kJ/kg.K at 700kPa

(a) Isentropic efficiency of the compressor, \eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}

\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%

(b) The temperature of the environment, T_{e} = 27^{\circ}C = 273 + 27 = 300 K

Availability at state 1, \Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg

Similarly for state 2, \Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg

Now, the efficiency of the compressor as per the second law;

\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757 = 67.57%

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(HAVE A GOOD DAY!!!)

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