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1 month ago

Who want to play "1v1 lol unblocked games 76"

Engineering

2 answers:

choli [298]1 month ago

7 0

Answer:

Corey, go assist Jade; her ex is causing trouble.

Explanation:

Daniel [329]1 month ago

3 0

Answer:I want to know what game to play?

Explanation:

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An air duct heater consists of an aligned array of electrical heating elements in which the longitudinal and transverse pitches

Kisachek [356]

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.

4 0

1 month ago

Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727C (1341F). (a) What is the proeutectoid phase? (b) How

pantera1 [306]

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:

$W_{f} =\frac{C_{cementite}-C_{2} }{C_{cementite}-C_{1} }$

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:

$W_{f} =\frac{6.7-1.15}{6.7-0.022} =0.8311kg$

For the total mass of cementite:

$W_{c} =\frac{C_{2}-C_{1}}{C_{cementite}-C_{1} } =\frac{1.15-0.022}{6.7-0.022} =0.1689kg$

c) The mass of pearlite:

$W_{p} =\frac{6.7-1.15}{5.94} =0.9343kg$

d) The diagram illustrates the different compositions: (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)

6 0

1 month ago

Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or

Kisachek [356]

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:

180

400

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

8 0

1 month ago

The first step to merging is entering the ramp and _____.

alex41 [359]

Answer:

D.informing your passengers of your destination

7 0

1 month ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

Mrrafil [318]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Inertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy =?

(b) rotor acceleration =?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

Explanation:

(a) Calculate the rotor's stored energy at synchronous speed.

The stored energy is represented as

$E = G \times H$

Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If we suddenly increase the mechanical input to 80 MW against an electrical load of 50 MW, we shall find the rotor's acceleration while ignoring mechanical and electrical losses.

The formula for rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is defined as

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration derived in part (b) persists over 10 cycles, we will calculate both the change in torque angle and the rotor speed in revolutions per minute at the end of this duration.

The change in torque angle is expressed as

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$