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;
}
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Answer:The Wright brothers invented and flew the first airplane in 1903, recognized as "the first sustained and controlled heavier-than-air powered flight". ... Airplanes had a presence in all the major battles of World War II. The first jet aircraft was the German Heinkel He 178 in 1939.
Explanation:
Answer:
A famous example of concurrent engineering is the development of the Boeing 777 commercial aircraft. The aircraft was designed and built by geographically distributed companies that worked entirely on a common product database of C A TIA without building physical mock-ups but with digital product definitions.
Answer:
B A and C
Explanation:
Given:
Specimen σ
σ
A +450 -150
B +300 -300
C +500 -200
Solution:
Compute the mean stress
σ
= (σ + σ)/2
σ
= (450 + (-150)) / 2
= (450 - 150) / 2
= 300/2
σ = 150 MPa
σ
= (300 + (-300))/2
= (300 - 300) / 2
= 0/2
σ = 0 MPa
σ
= (500 + (-200))/2
= (500 - 200) / 2
= 300/2
σ = 150 MPa
Compute stress amplitude:
σ
= (σ - σ)/2
σ
= (450 - (-150)) / 2
= (450 + 150) / 2
= 600/2
σ = 300 MPa
σ
= (300- (-300)) / 2
= (300 + 300) / 2
= 600/2
σ = 300 MPa
σ
= (500 - (-200))/2
= (500 + 200) / 2
= 700 / 2
σ = 350 MPa
From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.
Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.
In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.
