Answer:
1.444 psi
Explanation:
Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routin
1 answer:
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Answer:
a) The phase before eutectoid is commonly referred to as cementite, with the chemical formula Fe₃C.
b) The total mass of ferrite obtained is 0.8311 kg.
The total cementite mass equals 0.1689 kg.
c) The total cementite mass accounts for 0.9343 kg.
Explanation:
Provided:
1 kg of austenite
a carbon content of 1.15 wt%
Cooled below 727°C
Questions:
a) Identify the proeutectoid phase.
b) Calculate the mass of total ferrite and cementite, Wf =?, Wc =?
c) Determine the mass of both pearlite and the proeutectoid phase, Wp =?
d) Create a schematic to illustrate the resulting microstructure.
a) The proeutectoid phase is referred to as cementite with the formula Fe₃C.
b) To find the total mass of formed ferrite:
With:
Ccementite = composition of cementite = 6.7 wt%
C₁ = composition of phase 1 = 0.022 wt%
C₂ = overall composition = 1.15 wt%
Inserting the values yields:
For the total mass of cementite:
c) The mass of pearlite:
d) The diagram illustrates the different compositions: (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)
Response:
Refer to the explanation
Clarification:
Code:
import java.util.Scanner;
public class NumberPattern {
public static int x, count;
public static void displayNumPattern(int num1, int num2) {
if (num1 > 0 && x == 0) {
System.out.print(num1 + " ");
count++;
displayNumPattern(num1 - num2, num2);
} else {
x = 1;
if (count >= 0) {
System.out.print(num1 + " ");
count--;
if (count < 0) {
System.exit(0);
}
displayNumPattern(num1 + num2, num2);
}
}
}
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int num1;
int num2;
num1 = scnr.nextInt();
num2 = scnr.nextInt();
displayNumPattern(num1, num2);
}
}
See attached example output
Answer:
Change in length = 0.0913 in
Explanation:
Given data:
Length = 6 ft
Diameter = 0.2 in
Load w = 200 lb/ft
Solution:
We start by applying the equilibrium moment about point C, expressed as
∑M(c) = 0.............1
This can be used to find the force in AB.
10× 200 × ( 5) - (T cos(30)) × 10 = 0
Solving gives us
Tension in wire T(AB) = 1154.7 lb
We also know the modulus of elasticity for A992 is
E = 29000 ksi
And the area will be
Area =
The change in length is expressed as
Change in length = .........2
Substituting values results in
Change in length =
Change in length = 0.0913 in