Describe three advantages and three disadvantages of JIT?
1 answer:
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Answer:
See explanation
Explanation:
//Include the
//required header files.
#include <stdio.h>
//Define the
//main() function.
int main(void) {
//Declare the
//required variables.
char input_char;
int input_int;
double input_double;
char input_string[100];
//Prompt the user
//to enter an integer.
printf("Enter integer: ");
//Read and store
//the integer.
scanf("%d", &input_int);
//Prompt the user
//to enter a double value.
printf("Enter double: ");
//Read and store
//the double value.
scanf("%lf", &input_double);
//Prompt the user
//to enter a character.
printf("Enter character: ");
//Read and store
//the character.
scanf(" %c", &input_char);
//Prompt user to
//enter the string
printf("Enter string: ");
//Read and
//store the string.
scanf("%s", input_string);
//(1)
//Display the values.
printf("%d %lf %c %s\n",
input_int, input_double,
input_char, input_string);
//(2)
//Display the values
//in reverse order.
printf("%s %c %lf %d\n",
input_string, input_char,
input_double, input_int);
//(3)
//Cast the double to
//an integer and display it.
printf("%lf cast to an integer is %d",
input_double, (int)(input_double));
//Return from the
//main() function.
return 0;
}
Answer:
i) 796.18 N/mm^2
ii) 1111.11 N/mm^2
Explanation:
Initial diameter ( D ) = 12 mm
Gage Length = 50 mm
maximum load ( P ) = 90 KN
Fractures at = 70 KN
minimum diameter at fracture = 10mm
<u>Calculate the engineering stress at Maximum load and the True fracture stress</u>
<em>i) Engineering stress at maximum load = P/ A </em>
= P / = 90 * 10^3 / ( 3.14 * 12^2 ) / 4
= 90,000 / 113.04 = 796.18 N/mm^2
<em>ii) True Fracture stress = P/A </em>
= 90 * 10^3 / ( 3.24 * 10^2) / 4
= 90000 / 81 = 1111.11 N/mm^2