An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
1 answer:
Answer:
Activation energy for creep in this temperature range is Q = 252.2 kJ/mol
Explanation:
To calculate the creep rate at a particular temperature
creep rate,
Creep rate at 800⁰C,
.........................(1)
Creep rate at 700⁰C
.................(2)
Divide equation (1) by equation (2)
Take the natural log of both sides
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Answer:
Velocity component in x-direction .
Explanation:
v=3xy+y
We know that for incompressible flow
So
By integrate with respect to x,we will find
+C
So the velocity component in x-direction .
Answer:
the value of horizontal force P is 170.625 N
the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.
Explanation:
The first diagram attached below shows the free body diagram of the tool chest when it is sliding.
Let start out by calculating the friction force
where :
friction force
= coefficient of friction
= normal friction
Given that:
= 0.3
0.3
Using the equation of equilibrium along horizontal direction.
P - 0
P = 0.3 ----- Equation (1)
To determine the moment about point B ; we have the expression
0 =
where;
P = horizontal force
= normal force at support A
W = self- weight of tool chest
Replacing W = 650 N
0 =
Replacing for P in equation (1)
Plugging the value of in equation (2)
P =170.625 N
Thus; the value of horizontal force P is 170.625 N
b) From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:
Taking the moments about point A:
-(P × 100)+ (W×35) = 0
P =
Replacing 650 N for W
P = 227.5 N
Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N
We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N
However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.
Answer:
// Scanner class is imported to allow program
// receive input
import java.util.Scanner;
// RomanNumerals class is defined
public class RomanNumerals {
// main method that signify beginning of program execution
public static void main(String args[]) {
// Scanner object scan is created
// it receive input via keyboard
Scanner scan = new Scanner(System.in);
// Prompt is display asking the user to enter number
System.out.println("Enter your number: ");
// the user input is stored at numberOfOrder
int number = scan.nextInt();
// switch statement which takes number as argument
// the switch statement output the correct roman numeral
// depending on user input
switch(number){
case 1:
System.out.println("I");
break;
case 2:
System.out.println("II");
break;
case 3:
System.out.println("III");
break;
case 4:
System.out.println("IV");
break;
case 5:
System.out.println("V");
break;
case 6:
System.out.println("VI");
break;
case 7:
System.out.println("VII");
break;
case 8:
System.out.println("VIII");
break;
case 9:
System.out.println("IX");
break;
case 10:
System.out.println("X");
break;
// this part is executed if user input is not between 1 to 10
default:
System.out.println("Error. Number must be between 1 - 10.");
}
}
}
Explanation:
The program is well commented. A sample image of program output is attached.
The switch statement takes the user input (number) as argument as it goes through each case block in the switch statement and match with the corresponding case to output the roman version of that number. If the number is greater 10 or less than 1; the default block is executed and it display an error message telling the user that number must be between 1 - 10.
