Consider the following program:
1 answer:
Answer:
The solution of this question is given below in explanation section.
Explanation:
The correct answer to this question is A i.e. 112002 .
the correct code of this question is given below
#include <stdio.h>
#include <string.h>
#include <sys/types.h>
//#include "csapp.h"
void end(void)
{
printf("2");
}
int main()
{
if (fork() == 0)
atexit(end);
if (fork() == 0)
printf("0");
else
printf("1");
exit(0);
}
/* $end forkprob2 */
when you run this program, fork function print different values.
However, it is noted that when you run the program repeatedly, it will show you different values. But the most possible value that this program will show is A.
The picture is attached of the running program to get the idea of code.
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Answer:
For [1 1 0] and [1 0 1] plane, σₓ = 6.05 MPa
For [0 1 1] plane, σ = 0; slip will not occur
Explanation:
compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.
Given;
Stress direction: [1 0 0] ⇒ A
Slip direction: [1 1 1]
Normal to slip direction: [1 1 1] ⇒ B
∅ is the angle between A & B
Step 1: cos∅ = A·B/|A| |B| = ⇒ cos∅ = 1/
σₓ = τ/cos ∅·cosλ
where τ is the critical resolved shear stress given as 2.47MPa
Step 2: Solve for the slip along each plane
(a) [1 1 0]
cosλ = [1 1 0]·[1 0 0]/(·
)
note: cosλ = slip D·stress D/|slip D||stress D|
cosλ = 1/
∵ σₓ = τ/ ·
=
* 2.47MPa = 6.05MPa
Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa
(b) [0 1 1]
cosλ = [0 1 1]·[1 0 0]/(·
) = 0
∵ σₓ = 2.47MPa/0, which is not defined
Hence, for stress along [1 0 0], slip will not occur along [0 1 1]
(c) [1 0 1]
cosλ = [0 1 1]·[1 0 0]/(·
)
cosλ = 1/
∵ σₓ = τ/ ·
=
* 2.47MPa = 6.05MPa
See attachment for the space diagram
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
