Identify the four engineering economy symbols and their values from the following problem statement. Use a question mark with th

Answer:

a) q = 7671 W

T0 = 47.6°C

b) ΔP = 202.3 N/m²

P = 58.2 W

c) hDarray = 2 times hD of an isolated element.

Explanation:

see the image for the solution.

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)

Answer:

Here is the JAVA program:

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed to 4 assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two arrays origList[] and offsetAmount[] get assigned their values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // variable for storing the product results

for(i = 0; i <= origList.length - 1; i++){

/* iterates from 0 to the end of origList */

/* multiplies each origList entry with the corresponding offsetAmount entry, stores results in product */

product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }

System.out.println(product); }}

Output:

80 180 80 400

Explanation:

If you wish to print the product of origList alongside offsetAmount values vertically, this can be done in this manner:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

origList[i] *= offsetAmount[i];

System.out.println(origList[i]);}

}}

Output:

80

180

80

400

The program is shown with the output as a screenshot along with the example's input.

Answer:

Refer to the attached document

1512 ft

Explanation:

Because the acceleration is constant or zero, the acceleration-time graph consists of horizontal segments. The values for t2 and a4 are derived as follows:

Acceleration - Time

0 < t < 6: Velocity change = area beneath the a–t graph

V_6 - 0 =  (6 s)(4 ft/s²) = 24 ft/s

6 < t < t2: The velocity rises from 24 to 48 ft/s,

Velocity change = area beneath the a–t graph

48 - 24 = (t2 - 6) * 6

t2 = 10 s

t2 < t < 34: The velocity remains constant, meaning acceleration is zero.

34 < t < 40: Velocity change = area beneath the a–t graph

0 - 42 = 6*a4

a4 = - 8 ft / s²

A negative acceleration shows the area lies below the t-axis, indicating a decrease in speed.

Velocity - Time

Since acceleration remains constant or zero, the v−t graph is made up of linear segments connecting the calculated points.

Position change = area beneath the v−t graph

0 < t < 6:  x6 - 0 = 0.5*6*24 = 72 ft

6 < t < 10:    x10 - x6 = 0.5*4*(24 + 48) = 144 ft

10 < t < 34: x34 - x10 = 48*24 = 1152 ft

34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft

Summing these position changes yields the distance from A to B:

d = x40 - 0 = 1512 ft