Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:
where;
m = mass flow rate = 120 kg/min
the pressure at the inlet 
= 96 kPa
the pressure at the exit 
= ???
the pressure 
= 1000 kg/m³
∴
400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
Answer:
Check the explanation
Explanation:
The loop invariant has to satisfy some amount of requirements to be of good use. Another complex factor as to why a loop is the question of loop termination. A loop that doesn’t terminate can’t invariably be correct, and in fact the computation in whatever form amounts to nothing. The total axiomatic description of a while construct will have to involve all of the following to be true, in which I is the loop invariant:
P => I
{I and B} S {I}
(I and (not B)) => Q
Then the loop terminates
Answer:
as all the people should go near stratosphere
Answer:
- public class Main {
- public static void main(String[] args) {
- String testString = "abscacd";
-
- String evenStr = "";
- String oddStr = "";
-
- for(int i=testString.length() - 1; i >= 0; i--){
-
- if(i % 2 == 0){
- evenStr += testString.charAt(i);
- }
- else{
- oddStr += testString.charAt(i);
- }
- }
-
- System.out.println(evenStr + oddStr);
- }
- }
Explanation:
Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).
Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.
Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to testString.length() - 1 (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).
At last, we print the concatenated evenStr and oddStr (Line 18).
