What are the five basic elements of a system?
2 answers:
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Answer:
Here is the JAVA program:
import java.util.Scanner; // to take input from user
public class VectorElementOperations {
public static void main(String[] args) {
final int NUM_VALS = 4; // size is fixed that is 4 and assigned to NUM_VALS
int[] origList = new int[NUM_VALS];
int[] offsetAmount = new int[NUM_VALS];
int i;
//two lists origList[] and offsetAmount[] are assigned values
origList[0] = 20;
origList[1] = 30;
origList[2] = 40;
origList[3] = 50;
offsetAmount[0] = 4;
offsetAmount[1] = 6;
offsetAmount[2] = 2;
offsetAmount[3] = 8;
String product=""; // product variable to store the product of 2 lists
for(i = 0; i <= origList.length - 1; i++){
/* loop starts with i at 0th position or index and ends when the end of the origList is reached */
/* multiples each element of origList to corresponding element of offsetAmount and stores result in the form of character string in product*/
product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }
System.out.println(product); }} //displays the product of both lists
Output:
80 180 80 400
Explanation:
If you want to print the product of origList with corresponding value in offsetAmount in vertical form you can do this in the following way:
import java.util.Scanner;
public class VectorElementOperations {
public static void main(String[] args) {
final int NUM_VALS = 4;
int[] origList = new int[NUM_VALS];
int[] offsetAmount = new int[NUM_VALS];
int i;
origList[0] = 20;
origList[1] = 30;
origList[2] = 40;
origList[3] = 50;
offsetAmount[0] = 4;
offsetAmount[1] = 6;
offsetAmount[2] = 2;
offsetAmount[3] = 8;
for(i = 0; i <= origList.length - 1; i++){
origList[i] *= offsetAmount[i];
System.out.println(origList[i]); } }}
Output:
80
180
80
400
The program along with the output is attached as screenshot with the input given in the example.
To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as
Where
T = Static temperature
V = Velocity of Fluid
Specific Heat
Re-arrange to find the static temperature we have that
Now the pressure of helium by using the Adiabatic pressure temperature is
Where,
= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that
Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>