Answer:
I hope this helps the tops of the waves are close together
Answer:
a) specific impulse = 203 s
(b) mass flow rate = 2511 kg/s
(c) throat diameter = 0.774 m
(d) exit diameter = 1.95 m
(e) thrust = 5.3*10^6 N/m^2
(f) thrust at sea level = 15.6*10^6 N
(g) thrust with hydrogen = 5*10^6 N
(h) thrust with stagnation = 5*10^6 N
Explanation:
See the attached file for explanation.
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:
The Space Needle is a cut away with minimal residual deflection due to load transfer.
Answer:
μb = 0.096
μc = 0.073
Explanation:
member AB:
-800( 4/3 ) + Nb (2) = 0
Nb (2) = 3200/3
Nb = 533.3N
Post BC:
summation of force along the y axis=0
Nc + Nb + 150(3/5 ) -50(9.81)=0
Nc + 533.3 + 150(3/5 ) -50(9.81)=0
Nc = 933.83 N
Also (-4/5)(150)(3) + Fb(0.7)= 0
Fb = (4/5)(150)(3)/0.7 = 51.429 N
Likewise alog the x axis,
4/5(150) - Fc -Fb = 0
4/5(150) - Fc -51.429 = 0
Fc = 4/5(150) -51.429 =68.571 N
μb = Fb/Nb = 51.429/533.3 = 0.096
μc = Fc/Nc = 68.571 / 933.83 = 0.073