An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity kA while the other is fabricated from the material whose thermal conductivity kB is desired. Both rods are attached at one end to a heat source of fixed temperature Tb, are exposed to a fluid of temperature [infinity] T[infinity], and are instrumented with thermocouples to measure the temperature at a fixed distance x1 from the heat source. If the standard material is aluminum, with kA= 200 W/m·K, and measurements reveal values of TA= 75°C and TB= 70°C at x1 for Tb= 100°C and [infinity] T[infinity]= 25°C, what is the thermal conductivity kB of the test material?

The annual quantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

To determine the break-even point between the manual arc welding cell and the robotic cell, we need to calculate the total costs for each method and then equate them.

For the manual arc welding cell:

Labor cost per hour = (welder's hourly rate x arc-on time) + (fitter's hourly rate x fitter's participation in the cycle) = ($30 x 0.25) + ($25 x 0.3) = $11.25

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $11.25 x 15.4 / 60 = $2.89

Overhead cost per welded assembly = (labor cost per hour x (1 - arc-on time - fitter's participation in the cycle)) x cycle time per assembly / 60 = ($30 x 0.45) x 15.4 / 60 = $4.68

Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly = $2.89 + $4.68 = $7.57

Total cost per hour = total cost per welded assembly x production rate = $7.57 x 8 = $60.56

Total cost per year = total cost per hour x hours of operation per year = $60.56 x 2,000 = $121,120

For the robotic arc welding cell:

Labor cost per hour = fitter's hourly rate x fitter's participation in the cycle = $25 x 0.62 = $15.50

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $15.50 x 15.4 / 60 = $3.97

Overhead cost per welded assembly = power and utility cost per hour + annual maintenance cost / production rate = $3.80 + $3,500 / (8 x 2,000) = $3.80 + $0.22 = $4.02

Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly + (installed cost / (production rate x service life)) = $3.97 + $4.02 + ($158,000 / (8 x 3)) = $3.97 + $4.02 + $6,208.33 = $14.19

Total cost per hour = total cost per welded assembly x production rate = $14.19 x 8 = $113.52

Total cost per year = total cost per hour x hours of operation per year = $113.52 x 2,000 = $227,040

To find the break-even point, we set the total cost of the manual arc welding cell equal to the total cost of the robotic arc welding cell and solve for the annualquantity of welded assemblies:

$121,120 + x($7.57) = $227,040 + x($14.19)

$7.57x - $14.19x = $227,040 - $121,120

$-6.62x = $105,920

x = $105,920 / $6.62

x = 15,982.7

Therefore, the annualquantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

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Answer:

B) gate-source junction is reverse-biased

Explanation:

FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow"   in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".

In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".

The four relevant pressures in a Rankine cycle with one stage of reheat are P1, P2, P3, and P4.

For a Rankine cycle with one stage of reheat between turbines, there are typically four relevant pressures:

1. Boiler pressure (P1): This is the pressure at which the water is heated in the boiler before entering the first turbine.
2. High-pressure turbine outlet pressure (P2): This is the pressure at the outlet of the first turbine before the steam is sent to the reheater.
3. Reheat pressure (P3): This is the pressure at which the steam is reheated before entering the second turbine.
4. Low-pressure turbine outlet pressure (P4): This is the pressure at the outlet of the second turbine, which is also the condenser pressure.

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Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

Answer:

Output:-

Enter the five digit lottery number

Enter the digit 1 : 23

Enter the digit 2 : 44

Enter the digit 3 : 43

Enter the digit 4 : 66

Enter the digit 5 : 33

YOU LOSS!!

Computer Generated Lottery Number :

|12|38|47|48|49|

Lottery Number Of user:

|23|33|43|44|66|

Number Of digit matched: 0

Code:-

import java.util.Arrays;

import java.util.Random;

import java.util.Scanner;

public class Lottery {

int[] lotteryNumbers = new int[5];

public int[] getLotteryNumbers() {

return lotteryNumbers;

}

Lottery() {

Random randomVal = new Random();

for (int i = 0; i < lotteryNumbers.length; i++) {

lotteryNumbers[i] = randomVal.nextInt((50 - 1) + 1);

}

}

int compare(int[] personLottery) {

int count = 0;

Arrays.sort(lotteryNumbers);

Arrays.sort(personLottery);

for (int i = 0; i < lotteryNumbers.length; i++) {

if (lotteryNumbers[i] == personLottery[i]) {

count++;

}

}

return count;

}

public static void main(String[] args) {

int[] personLotteryNum = new int[5];

int matchNum;

Lottery lnum = new Lottery();

Scanner input = new Scanner(System.in);

System.out.println("Enten the five digit lottery number");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.println("Enter the digit " + (i + 1) + " :");

personLotteryNum[i] = input.nextInt();

}

matchNum = lnum.compare(personLotteryNum);

if (matchNum == 5)

System.out.println("YOU WIN!!");

else

System.out.println("YOU LOSS!!");

System.out.println("Computer Generated Lottery Number :");

System.out.print("|");

for (int i = 0; i < lnum.getLotteryNumbers().length; i++) {

System.out.print(lnum.getLotteryNumbers()[i] + "|");

}

System.out.println("\n\nLottery Number Of user:");

System.out.print("|");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.print(personLotteryNum[i] + "|");

}

System.out.println();

System.out.println("Number Of digit matched: " + matchNum);

}

}

Explanation: