Answer:
import java.util.Scanner;
public class FindMatchValue {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_VALS = 4;
int[] userValues = new int[NUM_VALS];
int i;
int matchValue;
int numMatches = -99; // Assign numMatches with 0 before your for loop
matchValue = scnr.nextInt();
for (i = 0; i < userValues.length; ++i) {
userValues[i] = scnr.nextInt();
}
/* Your solution goes here */
numMatches = 0;
for (i = 0; i < userValues.length; ++i) {
if(userValues[i] == matchValue) {
numMatches++;
}
}
System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);
}
}
Answer:
hello your question is incomplete attached below is the complete question
A) optimum compressor ratio = 9.144
B) specific thrust = 2.155 N.s /kg
C) Thrust specific fuel consumption = 1670.4 kg/N.h
Explanation:
Given data :
Mo = 2.1 , To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k
γ = 1.4
attached below is the detailed solution
Answer:
a)We know that acceleration a=dv/dt
So dv/dt=kt^2
dv=kt^2dt
Integrating we get
v(t)=kt^3/3+C
Puttin t=0
-8=C
Putting t=2
8=8k/3-8
k=48/8
k=6
Answer: Let us use the pickled file - DeckOfCardsList.dat.
Explanation: So that our possible outcome becomes
7♥, A♦, Q♠, 4♣, 8♠, 8♥, K♠, 2♦, 10♦, 9♦, K♥, Q♦, Q♣
HPC (High Point Count) = 16
Answer:
Explanation:
Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:
![Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]](https://tex.z-dn.net/?f=Q_%7Bwater%7D%20%3D%20%281.4%5C%2CL%29%5Ccdot%28%5Cfrac%7B1%5C%2Cm%5E%7B3%7D%7D%7B1000%5C%2CL%7D%20%29%5Ccdot%20%281000%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%29%5Ccdot%20%5B%284.187%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20%5E%7B%5Ctextdegree%7DC%7D%20%29%5Ccdot%20%28100%5E%7B%5Ctextdegree%7DC-25%5E%7B%5Ctextdegree%7DC%29%2B2257%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%5D)
The heat liberated by the LP gas is:
A kilogram of LP gas has a minimum combustion power of 
. Then, the required mass is:
