Answer:
option B is correct. Fracture will definitely not occur
Explanation:
The formula for fracture toughness is given by;
K_ic = σY√πa
Where,
σ is the applied stress
Y is the dimensionless parameter
a is the crack length.
Let's make σ the subject
So,
σ = [K_ic/Y√πa]
Plugging in the relevant values;
σ = [50/(1.1√π*(0.5 x 10^(-3))]
σ = 1147 MPa
Thus, the material can withstand a stress of 1147 MPa
So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.
Answer:
- using System;
- public class Program
- {
- public static void Main()
- {
- Console.WriteLine("Enter number of students: ");
- int num = Convert.ToInt32(Console.ReadLine());
- string [] firstName = new string[num];
- string [] lastName = new string[num];
-
- for(int i=0 ; i < num; i++){
- Console.WriteLine("Enter first name: ");
- firstName[i] = Console.ReadLine();
-
- Console.WriteLine("Enter last name: ");
- lastName[i] = Console.ReadLine();
- }
-
- for(int j=0; j < num; j++){
- Console.WriteLine(lastName[j] + "," + firstName[j]);
- }
- }
- }
Explanation:
Firstly, prompt user to enter number of student to be stored (Line 6- 7). Next, create two array, firstName and lastName with num size (Line 8-9).
Create a for-loop to repeat for num times and prompt user to enter first name and last name and then store them in the firstName and lastName array, respectively (Line 11 - 17).
Create another for loop to traverse through the lastName and firstName array and display the last name and first name by following the format given in the question (Line 19 - 21).
Answer:
5.1 Personnel Security. ...
5.2 Physical and Environmental Protection. ...
5.3 Production, Input and Output Controls. ...
5.4 Contingency Planning and Disaster Recovery. ...
5.5 System Configuration Management Controls. ...
5.6 Data Integrity / Validation Controls. ...
5.7 Documentation. ...
5.8 Security Awareness and Training.
Answer:
c. Alto módulo de elasticidad
Explanation:
The correct answer to the given question is c. Alto modulo de elasticidad
A Youngs modulus measures the resistance of any material to elastic deformation. It is basically the ratio of the stress applied to a body to the results of the stress which is the response of the body over the pressure applied. This is to test the stiffness of any material and most of time the material stays constant over stressing.
