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3 years ago

Please help me: Use the Node analysis to find the power of all resistors

Engineering

1 answer:

Viktor [21]3 years ago

3 0

‘Politics and planning are increasingly gaining prominence in contemporary urban and regional planning debates’. Using relevant examples, discuss this assertion reflecting on the critical success factors for the successful implementation of the land reform program in South Africa.

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Who is responsible for keeping your facility in compliance <br>

inessss [21]

<h2>Answer:</h2><h2>The safety manager is usually the person responsible for ensuring whether the company is in compliance with the OSHA employer requirements . These type of requirements include; Fatal accidents that result in the hospitalisation of three or more employees,must be reported to the OSHA nearest office within 8 hours.</h2>

4 0

3 years ago

Calculate the unit cell edge length for an 80 wt% Ag−20 wt% Pd alloy. All of the palladium is in solid solution, the crystal str

alisha [4.7K]

Answer:

λ^3 = 4.37

Explanation:

first let us to calculate the average density of the alloy

for simplicity of calculation assume a 100g alloy

80g --> Ag

20g --> Pd

ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)

= 100*10^-3/(20/11.9*10^6+80/10.44*10^6)

= 10744.62 kg/m^3

now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate

total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27

= 1.12*10^23 unit cells

mass of Pd in 1 unit cell = 20/1.12*10^23

Now,

ρ_avg= mass of unit cell/volume of unit cell

ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3

λ^3 = 4.37

6 0

3 years ago

Consider steady one-dimensional heat transfer through a plane wall exposed to convection from both sides to environments at know

Leona [35]

Answer:

Now find the temperature of each surface, we have that the the temperature on the left side of the wall is T∞₁ - Q/h₁A and the temperature on the right side of the wall is T∞₂ + Q/h₂A.

Note: kindly find an attached diagram to the complete question given below.

Sources: The diagram/image was researched and taken from Slader website.

Explanation:

Solution

Let us consider the rate of heat transfer through the plane wall which can be obtained from the relations given below:

Q = T∞₁ -T₁/1/h₁A = T₁ -T₂/L/kA =T₂ -T∞₂/1/h₂A

= T∞₁ - T∞₂/1/h₁A + L/kA + 1/h₂A

Here

The convective heat transfer coefficient on the left side of the wall is h₁, while the convective heat transfer coefficient on the right side of the wall is h₂. the thickness of the wall is L, the thermal conductivity of the wall material is k, and the heat transfer area on one side of the wall is A. Q is refereed to as heat transfer.

Thus

Let us consider the convection heat transfer on the left side of the wall which is given below:

Q = T∞₁ -T₁/1/h₁A

T₁ = T∞₁ - Q/h₁A

Therefore the temperature on the left side of the wall is T∞₁ - Q/h₁A

Now

Let us consider the convection heat transfer on the left side of the wall which is given below:

Q= T₂ -T∞₂/1/h₂A

T₂ = T∞₂ + Q/h₂A

Therefore the temperature on the right side of the wall is T∞₂ + Q/h₂A

4 0

3 years ago

Consider a junction that connects three pipes A, B and C. What can we say about the mass flow rates in each pipe for steady flow

Elis [28]

Answer:

The statement regarding the mass rate of flow is mathematically represented as follows $\Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}$

Explanation:

A junction of 3 pipes with indicated mass rates of flow is indicated in the attached figure

As a basic sense of intuition we know that the mass of the water that is in the pipe junction at any instant of time is conserved as the junction does not accumulate any mass.

The above statement can be mathematically written as

$Mass_{Junction}=Constant\\\\\Rightarrow Mass_{in}=Mass_{out}$

this is known as equation of conservation of mass / Equation of continuity.

Now we know that in a time 't' the volume that enter's the Junction 'O' is

1) From pipe 1 = $V_{1}=Q_{1}\times t$

1) From pipe 2 = $V_{2}=Q_{2}\times t$

Mass leaving the junction 'O' in the same time equals

From pipe 3 = $V_{3}=Q_{3}\times t$

From the basic relation of density, volume and mass we have

$\rho =\frac{mass}{Volume}$

Using the above relations in our basic equation of continuity we obtain

$\rho \times V_{3}=\rho \times V_{1}+\rho \times V_{2}\\\\Q_{3}\times t=Q_{1}\times t+Q_{2}\times t\\\\\Rightarrow Q_{3}=Q_{1}+Q_{2}$

Thus the mass flow rate equation becomes $\Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}$

6 0

4 years ago

In Java; Given numRows and numColumns, print a list of all seats in a theater. Rows are numbered, columns lettered, as in 1A or

baherus [9]

Answer:

Explanation:

import java.util.Scanner;

public class NestedLoops {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

int numRows;

int numColumns;

int currentRow;

int currentColumn;

char currentColumnLetter;

numRows = scnr.nextInt();

numColumns = scnr.nextInt();

for (currentRow = 0; currentRow < numRows; currentRow++) {

currentColumnLetter = 'A';

for (currentColumn = 0; currentColumn < numColumns; currentColumn++) {

System.out.print(currentRow + 1);

System.out.print(currentColumnLetter + " ");

currentColumnLetter++;

}

System.out.println("");

}

1A 1B 1C 2A 2B 2C

thanks

5 0

3 years ago