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2 answers:
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Answer:
The answer is "".
Explanation:
The amount of kilograms, which travel in a thick sheet of hydrogen:
calculating the value of
calculating the value of
Answer:
Hello your question is incomplete attached below is the missing part
a) p1 = 454.83 kPa, p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa
b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ
c) 5
Explanation:
Given data:
mass of air ( m ) = 1/10 kg
adiabatic index ( k ) = 1.4
temperature for isothermal expansion = 250K
rate of heat transfer ( Q12 ) = 3.4 KJ
temperature for Isothermal compression ( T4 ) = 300k
final volume ( V4 ) = 0.01m ^3
a) Calculate the pressure, in Kpa, at each of the four principal states
from an ideal gas equation
P4V4 = mRT4 ( input values above )
hence P4 = 860.959kPa
attached below is the detailed solution
b) Calculate work done for each processes
attached below is the detailed solution
C) Calculate the coefficient of performance
attached below is detailed solution
Answer and Explanation:
// code
class Main {
public static void main(String[] args) {
/*
*
*
* your code
*
*/
System.out.println(inDogish("aplderogad"));
System.out.println(inXish("aplderogad", "dog"));
}
// returns true if the word is in dog-ish
// returns false if word is not in dog-ish
public static boolean inDogish(String word) {
// first find d
if (dogishHelper(word, 'd')) {
// first find string after d
String temp = word.substring(word.indexOf("d"));
// find o
if (dogishHelper(temp, 'o')) {
// find string after o
temp = temp.substring(temp.indexOf("o"));
// find g
if (dogishHelper(temp, 'g'))
return true;
}
The output is attached below
}
return false;
}
// necessary to implement inDogish recursively
public static boolean dogishHelper(String word, char letter) {
// end of string
if (word.length() == 0)
return false;
// letter found
if (word.charAt(0) == letter)
return true;
// search in next index
return dogishHelper(word.substring(1), letter);
}
// a generalized version of the inDogish method
public static boolean inXish(String word, String x) {
if (x.length() == 0)
return true;
if (word.length() == 0)
return false;
if (word.charAt(0) == x.charAt(0))
return inXish(word.substring(1), x.substring(1));
return inXish(word.substring(1), x.substring(0));
}
}
PS E:\fixer> java Main true true ne on
PS E:\fixer> java Main true true ne on
