Explanation:
<u>Filler:</u>
Filler is the material rod is used when we are joining two material by using welding process.If thickness of work piece is more so it will become compulsory to provide some filler material for making the welding join to withstand high stresses.
<u>Electrode:</u>
Electrode is the element which is used to complete the electric circuit in welding .Some time electrode is connected with positive terminal and some time with negative terminal ,it depends on the requirement of welding process.In Tungsten inert gas welding electrode is connected negative terminal but on the other hand Metal inert gas welding electrode is connected with positive terminal.Electrode can be consumable non-consumable depends on the condition.
Yes electrode can be work as filler material ,in Metal inert gas welding wire is used as electrode as well as filler material.In Metal inert gas welding consumable electrode is used on the other hand Tungsten inert gas welding non-consumable electrode is used.In Tungsten inert gas welding if thickness of work pieces is less than 5 mm then no need to used any filler material but if thickness is more than 5 mm then we have to use filler material.
Answer:
to make the bace of a building more sturdy
Explanation:
example: the bace of the empire state building is stone very sturdy
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:
mechanical engineer is the best answer
Answer:
Given that
Mass flow rate ,m=2.3 kg/s
T₁=450 K
P₁=350 KPa
C₁=3 m/s
T₂=300 K
C₂=460 m/s
Cp=1.011 KJ/kg.k
For ideal gas
P V = m R T
P = ρ RT
ρ₁=2.71 kg/m³
mass flow rate
m= ρ₁A₁C₁
2.3 = 2.71 x A₁ x 3
A₁=0.28 m²
Now from first law for open system
For ideal gas
Δh = CpΔT
by putting the values
Q= - 45.49 KJ/kg
Q =- m x 45.49 KW
Q= - 104.67 KW
Negative sign indicates that heat transfer from air to surrounding
