In this lab, you add a loop and the statements that make up the loop body to a Java program that is provided. When completed, the program should calculate two totals: the number of left-handed people and the number of right-handed people in your class. Your loop should execute until the user enters the character X instead of L for left-handed or R for right-handed.The inputs for this program are as follows: R, R, R, L, L, L, R, L, R, R, L, XVariables have been declared for you, and the input and output statements have been written.

Answer:

Attached is a solution hope it helps:

Answer:

The missing Table is;

Month  1  2  3  4  5  6  7

Value  24  13  21  14  20  23  15

MSE = 20.56

Forecast = 19.33

Explanation:

Table Formation

Month Value Forecast Error  Absolute Error Error Square

1          24    

2          13    

3          21    

4          14         19.33333 -5.33333 5.333333 28.44444

5          20           16                       4              4             16

6          23          18.33333 4.666667 4.666667 21.77778

7          15           19                   -4                       4             16

    20.55

    19.33

Table Explanation

In the above table, we first calculate the forecast column. A forecast is taken as the sum of the three entries in order and then taking out the average.

Error is calculated by subtracting the Actual value from the forecast value.

Then we take the absolute of the forecast value.

Then we take the square of the absolute of the forecasted values.

Now we calculate MSE as;

MSE = ( 28.44 + 16 + 21.78 + 16 ) / 4

MSE = 20.56

Forecast for 8th month  = ( 20 + 23 + 15 ) / 3

Forecast for 8th month  = 19.33

Answer:

The answer is in the attachment. Thanks

Explanation:

Answer:

There is very little possibility of ionic bonding.

Explanation:

First we need to understand the meaning of ionic bonding. ionic bonding means the complete transfer of valance electrons between atoms.

The electronegativity of Al is 1.5 whereas electronegativity of nickel "Ni" is 1.8-1.9. As we can see from these values of electronegativity of Al & Ni that they are closer to each other & the difference of electonegativity of these metals is very little which decreases the possibility of ionic bonding between them so it is not expected that there will be much ionic bonding. Moreover it is noticeable here that these both are metals and they prefer to give up their electrons rather to gain them.

The two elements can not form ionic bonds because ionic bonds are formed when the electronegativity difference is greater than 1.5.

The term intermetallic compound refers to a compound that is composed of two metallic elements. Let us recall that metals often have a low value of electronegativity.

Owing to the fact that the electronegativity difference between nickel and aluminium is small (about 0.1). An ionic bond can not be formed between the two elements because ionic bonds are formed when the electronegativity difference is greater than 1.5.

Learn more about ionic bonds: https://brainly.com/question/867804

Answer:

The gm required is 0.1 [tex]\frac{mA}{V}[/tex]  

Explanation:

Knowing

Vpeak = 0.50 V

Rd = 50 kΩ

Av = 5 V/V

To find the transconductance we should to use the equation

Av = gm * Rd

so,

gm = Av/Rd

gm = 5/50 = 0.1 [tex]\frac{mA}{V}[/tex]  

Answer:

(a)

dQ = mdq

dq = [tex]C_p[/tex]dT

[tex]q = \int\limits^{T_2}_{T_1} {C_p} \, dT[/tex]   = [tex]C_p[/tex] (T₂ - T₁)

From the above equations, the underlying assumption is that  [tex]C_p[/tex] remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ    ,   Q₂ = 6.14 KJ

ΔT = 3.10 K       ,   ΔT₂ = 3.10 K  for calorimeter

Let [tex]C_{cal}[/tex] be heat constant of calorimeter

Q₂ = [tex]C_{cal}[/tex] ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m [tex]C_p[/tex] ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³  at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m [tex]C_p[/tex]' ΔT

[tex]C_p[/tex] = (16.73 - 6.14) / (15.085 x 3.10)

[tex]C_p[/tex] = 0.22646 KJ mol⁻¹ k⁻¹

The maximum transmitted power throughout the solid circular rod is approximately 86.066 horse power.

Let consider that the solid circular rod experiments an uniform torsion and that the material represents a ductile material.

The solid circular rod rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, and all losses throughout the system are negligible. Net power ([tex]\dot W[/tex]), in horse power, is calculated as the product of the torque ([tex]T[/tex]), in kilopound-force-inches, and the angular speed. That is to say:

[tex]\dot W = T \cdot \omega[/tex]   (1)

In addition, we add a formula that relates the torque with the shear stress experimented by the solid circular rod:

[tex]\tau = \frac{T \cdot D}{2 \cdot J}[/tex]   (2)

Where [tex]J[/tex] is the torsion module, in quartic inches, whose formula for a solid circular cross section is described below:

[tex]J = \frac{\pi \cdot D^{4}}{32}[/tex]   (3)

The torsion module is calculated herein: ([tex]D = 2.5\,in[/tex])

[tex]J = 3.835\, in^{4}[/tex]

The maximum allowable torsion is found by isolating the corresponding variable in (2): ([tex]J = 3.835 in^{4}[/tex], [tex]D = 2.5\,in[/tex], )

[tex]T_{max} = \frac{2 \cdot J \cdot \tau_{max}}{D}[/tex]

[tex]T_{max} = 31.907\, kip \cdot in\,(2.659 \,kip \cdot ft)[/tex]

Then, the maximum transmitted power is determined by (1):

[tex]\dot W = (2.659 kip \cdot ft) \cdot (2)\cdot (\pi) \cdot (170 rpm)\\\dot W \approx 2840.188 \,\frac{kip \cdot ft}{min} \\\dot W \approx 86.066 \,h.p.[/tex]

The maximum transmitted power throughout the solid circular rod is approximately 86.066 horse power. [tex]\blacksquare[/tex]

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Answer:

the diffusion coefficient at 900°C is D₂=8.9*10⁻¹⁸ m²/s

Explanation:

The dependence of the diffusion coefficient is similar to the dependence of chemical reaction rate with respect to temperature , where

D=D₀*e^(-Q/RT)

where

D₀= diffusion energy at T=∞

Q= activation energy

T= absolute temperature

D= diffusion coefficient at temperature T

R=ideal gas constant

for temperatures T₁ and T₂

D₁=D₀*e^(-Q/RT₁)

D₂=D₀*e^(-Q/RT₂)

dividing both equations and rearranging terms we get

D₂=D₁*e^[-Q/R(1/T₂-1/T₁)]

replacing values

D₂=D₁*e^[-Q/R(1/T₂-1/T₁)] = 1.3*10⁻²² m²/s*e^(-256,000 J/mol/8.314 J/(mol K)*(1/1073K-1/773K) =  8.9*10⁻¹⁸ m²/s

Answer:

Explanation:

Answer:

Explanation:

Answer:  

Explanation:  

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.  

(a). here we are asked to determine the Temperature and Pressure.  

Given that the properties of Air;  

ha = 230.02 KJ/Kg  

Ta = 230 K  

Pra = 0.5477  

From the energy balance equation for a diffuser;  

ha + Va²/2 = h₁ + V₁²/2  

h₁ = ha + Va²/2 (where V₁²/2 = 0)  

h₁ = 230.02 + 220²/2 ˣ 1/10³  

h₁ = 254.22 KJ/Kg  

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg  

from this we have;  

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]  

Pr₁ = 0.77759  

therefore T₁ = 254.15K  

P₁ = (Pr₁/Pra)Pa  

= 0.77759/0.5477 ˣ 26  

P₁ = 36.91 kPa  

now we calculate Pr₂  

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349  

⇒ now we obtain properties of air at  

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg  

calculating the enthalpy of air at state 2  

ηc = h₁ - h₂ / h₁ - h₂  

0.85 = 254.22 - 505.387 / 254.22 - h₂  

h₂ = 549.71 KJ/Kg  

to obtain the properties of air at h₂ = 549.71 KJ/Kg  

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913  

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at  

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁  

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄  

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃       i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄  + 0  = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

Answer:

Follow step by step explabation to get answer to the question and also see attachments for output.

Explanation:

code:

def normalize(input_string):

l=list();

for i in input_string:

if (i>='a' and i<='z') or (i>='A' and i<='Z'):

l.append(i.lower())

return l

def find_missing_letters_algorithm(sentence,prevsentece):

l=list();

for i in prevsentece:

i=i.lower();

if i not in sentence:

l.append(i)

return l

print(find_missing_letters_algorithm(normalize("O.K. #1 Python!"),"O.K. #1 Python!"))

attached below is the answer

I would recommend installing both a 100-Base T network and a Wi-Fi network in the new office building for Eureka.

Based on the information provided, I would recommend installing both a 100-Base T network as well as a Wi-Fi network in the new office building for Eureka.

A 100-Base T network, also known as Ethernet, provides reliable and high-speed wired connections, perfect for the staff members who spend a significant amount of time on their computers searching the Internet. It ensures a stable and fast connection for their work.

On the other hand, a Wi-Fi network allows for flexible and convenient wireless connectivity, which is beneficial for staff members who frequently use their phones or move around the office space. It provides them with the freedom to connect wirelessly wherever they are in the building.

https://brainly.com/question/33827192

#SPJ3

Answer:

The correct option is A

Explanation:

Top down budgeting is a type of budget in which the senior management prepares a high level budget for the company. The senior management then allocates the amounts for the individual departments, which then use this number in preparing their own budget.

Answer:

C++ code explained below

Explanation:

#include "hw6.h"

//---------------------------------------------------

// Constructor function

//---------------------------------------------------

Connect4::Connect4()

{

ClearBoard();

}

//---------------------------------------------------

// Destructor function

//---------------------------------------------------

Connect4::~Connect4()

{

// Intentionally empty

}

//---------------------------------------------------

// Clear the Connect4 board

//---------------------------------------------------

void Connect4::ClearBoard()

{

// Initialize Connect4 board

for (int c = 0; c < COLS; c++)

for (int r = 0; r < ROWS; r++)

board[r][c] = ' ';

// Initialize column counters

for (int c = 0; c < COLS; c++)

count[c] = 0;

}

//---------------------------------------------------

// Add player's piece to specified column in board

//---------------------------------------------------

bool Connect4::MakeMove(int col, char player)

{

// Error checking

if ((col < 0) || (col >= COLS) || (count[col] >= ROWS))

return false;

// Make move

int row = count[col];

board[row][col] = player;

count[col]++;

return true;

}

//---------------------------------------------------

// Check to see if player has won the game

//---------------------------------------------------

bool Connect4::CheckWin(char player)

{

// Loop over all starting positions

for (int c = 0; c < COLS; c++)

for (int r = 0; r < ROWS; r++)

if (board[r][c] == player)

{

// Check row

int count = 0;

for (int d = 0; d < WIN; d++)

if ((r+d < ROWS) &&

(board[r+d][c] == player)) count++;

if (count == WIN) return true;

// Check column

count = 0;

for (int d = 0; d < WIN; d++)

if ((c+d < COLS) &&

(board[r][c+d] == player)) count++;

if (count == WIN) return true;

// Check first diagonal

count = 0;

for (int d = 0; d < WIN; d++)

if ((r+d < ROWS) && (c+d < COLS) &&

(board[r+d][c+d] == player)) count++;

if (count == WIN) return true;

// Check second diagonal

count = 0;

for (int d = 0; d < WIN; d++)

if ((r-d >= 0) && (c+d < COLS) &&

(board[r-d][c+d] == player)) count++;

if (count == WIN) return true;

}

return false;

}

//---------------------------------------------------

// Print the Connect4 board

//---------------------------------------------------

void Connect4::PrintBoard()

{

// Print the Connect4 board

for (int r = ROWS-1; r >= 0; r--)

{

// Draw dashed line

cout << "+";

for (int c = 0; c < COLS; c++)

cout << "---+";

cout << "\n";

// Draw board contents

cout << "| ";

for (int c = 0; c < COLS; c++)

cout << board[r][c] << " | ";

cout << "\n";

}

// Draw dashed line

cout << "+";

for (int c = 0; c < COLS; c++)

cout << "---+";

cout << "\n";

// Draw column numbers

cout << " ";

for (int c = 0; c < COLS; c++)

cout << c << " ";

cout << "\n\n";

}

//---------------------------------------------------

// Main program to play Connect4 game

//---------------------------------------------------

int main()

{

  int choice;

  int counter = 0;

  srand (time(NULL));

  Connect4 board;

  cout << "Welcome to Connect 4!" << endl << "Your Pieces will be labeled 'H' for human. While the computer's will be labeled 'C'" << endl;

  board.PrintBoard();

  cout << "Where would you like to make your first move? (0-6)";

  cin >> choice;

  while (board.MakeMove(choice,'H') == false){

  cin >> choice;

  }

  counter++;

  while (board.CheckWin('C') == false && board.CheckWin('H') == false && counter != 21){

  while (board.MakeMove(rand() % 7, 'C') == false){}

  board.PrintBoard();

  cout << "Where would you like to make your next move?" << endl;

  cin >> choice;

  board.MakeMove(choice,'H');

  while (board.MakeMove(choice,'H') == false){

  cin >> choice;

  }

  counter++;

  }

  if (board.CheckWin('C')){

  cout << "Computer Wins!" << endl;}

  else if (counter == 21){cout << "Tie Game!" << endl;}

  else {cout << "Human Wins!" << endl;}

  board.PrintBoard();

}

Answer:

The triangle() function is an inbuilt function in p5.js which is used to draw a triangle in a plane. This function accepts three vertices of triangle.

Syntax:

triangle(x1, y1, x2, y2, x3, y3)

Below programs illustrates the triangle() function in p5.js:

function setup() {  

   createCanvas(400, 400);  

}  

function draw() {  

   background(220);  

   fill('lightgreen');  

   // A triangle at (100, 250), (250, 170) and (330, 300)    

   triangle(100, 250, 250, 170, 330, 300);  

}  

Answer:

X1 = 2081.64

X2 = 523.91

X3 = 1394.45

Explanation:

See the attached pictures for detailed explanation.

Answer:

Maximum

Greater than

Explanation:

The maximum compressive load experienced by the same truss member would occur when the live load is maximum. The magnitude of the maximum compressive load would be greater than the magnitude of the maximum tensile load.

Explanation:

Find the difference between the simple interest and the compound interest on Rs 5000 for 2 years at 9% per annum

Answer:

What do i have to do

Explanation:

what do i do

Answer: The theoretical stopping distance is 1.245 miles (6572.193 ft).

Explanation:

First, the physical model is created by using Principle of Energy Conservation and Work-Energy Theorem. It is assumed that surface is horizontal, so there are no changes associated with potential energy. The car has an initial kinetic energy, which is completely dissipated by braking.

[tex]K_{1} = \Delta W_{loss}[/tex]

[tex]\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (\mu_{2} \cdot \Delta s_{2}+\mu_{2'} \cdot \Delta s_{2'})[/tex]

The distance is isolated from previous equation:

[tex]\Delta s_{2'} = \frac{1}{\mu_{2'}}\cdot (\frac{1}{2}\cdot \frac{v^{2}}{g} - \mu_{2} \cdot \Delta s_{2})[/tex]

By replacing variables, the distance is calculated herein:

[tex]\Delta s_{2'} = \frac{1}{\frac{0.02}{0.80}}\cdot (\frac{1}{2}\cdot \frac{(102.667\frac{ft}{s} )^2}{32.174 \frac{ft}{s^{2}}} - (0.02) \cdot (100 ft))[/tex]

[tex]\Delta s_{2'} = 6472.193 ft\\\Delta s_{2'} = 1.226 mi[/tex]

The theoretical stopping distance is:

[tex]\Delta s = \Delta s_{2} + \Delta s_{2'}\\\Delta s = 6572.193 ft\\\Delta s = 1.245 mi[/tex]

Answer:

maximum length to which it may be stretched without causing plastic deformation is 200 + 0.4782 = 2.4782 mm

Explanation:

given data

stress = 275 MPa

modulus of elasticity = 115 GPa

solution

we will apply here modulus of elasticity  formula that is

modulus of elasticity = stress ÷ strain   ....................1

here we consider original length is 200 mm

so strain is = [tex]\frac{\triangle L}{original\ length }[/tex]  ..........2

put here value in equation 1 we get

115 ×  [tex]10^{9}[/tex] = [tex]\frac{275 \times 10^{6}}{\frac{x}{200} }[/tex]

solve it we get

x = [tex]\frac{11}{23}[/tex]  

x = 0.4782

so maximum length to which it may be stretched without causing plastic deformation is 200 + 0.4782 = 2.4782 mm

Answer:

Explanation:

The principle and application of differential equation is applied and the steps with detailed and appropriate substitution is as carefully shown in the attached files.

m = mass attached

k = spring constant

c = damping coefficient

Answer:

Design rate of superelevation= 3.4%

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Answer:

a)  [tex]\Delta S = 0.386 \ \frac{kJ}{K}[/tex]

b)  [tex]\Delta S =0.395 \ \frac{kJ}{K}[/tex]

Explanation:

a)

Specific heat formula, we have:

[tex]mc(\theta_1-\theta_2)=Wt[/tex]

Where

m is mass

c is specific heat capacity of air at specific temp

[tex]\theta[/tex] is the temperature change

W is the power

t is the time

The ideal gas law is:

[tex]PV=nRT[/tex]

Where

P is the pressure

V is the volume

n is number of moles

T is temperature

R is the ideal gas constant

First, lets solve for [tex]\theta_2[/tex] in the 1st equation, remembering to use ideal gas law in it to have the variables that are given in the problem. Shown below:

[tex]mc(\theta_2-\theta_1)=Wt\\mc\theta_2-mc\theta_1=Wt\\mc\theta_2=mc\theta_1 + Wt\\\theta_2=\frac{mc\theta_1 + Wt}{mc}\\\theta_2=\theta_1+\frac{Wt}{mc}\\\theta_2=\theta_1+\frac{WtR\theta_1}{PVc}\\\theta_2=\theta_1(1+\frac{WtR}{PVc})[/tex]

Now, we know

Theta_1 is 17 celsius, which in Kelvin is 17 + 270 = 290K

Power is 200

Time is 15 mins = 15 * 60 = 900 seconds

R is 287 J/kg K

P is 120

V is 300 L

Specific heat of air at 290K is about 1005

Substituting we get:

[tex]\theta_2=\theta_1(1+\frac{WtR}{PVc})\\\theta_2=290(1+\frac{200*900*287}{120*300*1005})\\\theta_2=704 \ K[/tex]

Now, Entropy Change is:

[tex]\Delta S =mc Ln(\frac{\theta_2}{\theta_1})[/tex]

Again using the substitution equations, we have:

[tex]\Delta S = \frac{PV}{R\theta_1}cLn(1+\frac{WRt}{PVc})[/tex]

We know what the variables mean, we substitute the respective values, to get:

[tex]\Delta S = 0.386 \ \frac{kJ}{K}[/tex]

b)

For variable specific heats, we need entropy value from entropy table:

s_2 = 2.58044

s_1 = 1.66802  [1.05 * 290 K]

The formula is:

[tex]\Delta S = m(s_2-s_1)\\\Delta S = \frac{PV}{R\theta_1}(s_2-s_1)[/tex]

Substituting the values, we find the answer:

[tex]\Delta S = \frac{PV}{R\theta_1}(s_2-s_1)\\\Delta S = \frac{120*300}{287*290}(2.58044-1.66802)\\\Delta S =0.395 \ \frac{kJ}{K}[/tex]

Answer:

See the attached picture.

Explanation:

See the attached picture for explanation.

in this exercise we have to use the knowledge of software to explain its efficiency, as:

A) Relationship between the software performance level and the amount of resources used, under pre-defined usage conditions.

B) An algorithm is efficient if it doesn't waste time for nothing. In other words, an algorithm is efficient if it is faster than other algorithms for the same problem.

Software can be defined as a set of instructions that allow the user to control an electronic device. In a computer, for example, the physical parts and peripherals make up the hardware, but you need software so that the components know how they are supposed to work.

A) Greedy algorithms are a class of algorithm that will choose the local optimum at any time to find the optimum solution for the problem. Since only one super computer is there, it can process the jobs sequentially and its processing time is constant.

B) The selfish choice secondhand above is selecting the jobs accompanying greater value of PC occasion at each time. Thus the accomplishment period can be weakened because the jobs accompanying bigger PC time will be done first. Thus the overhead period after perfecting the excellent calculating processing maybe removed.

See more about software at brainly.com/question/1022352

Answer:

a) [tex]T_{out} = 2190.455 ^{\textdegree}R[/tex] b) [tex]\dot V_{air,out} = 8352.941 acfm[/tex]  c) [tex]\dot V_{air,out,corr} = 7350.588 afcm[/tex]

Explanation:

a) The heat lost by the power plant is:

[tex]\dot Q_{loss} = (0.67) \cdot (110 \frac{lbm}{s} ) \cdot (12000 \frac{BTU}{lbm} )\\\dot Q_{loss} = 884400 BTU[/tex]

The waste heat used by heat exchanger is:

[tex]\dot Q_{used} = (0.0012) \cdot \dot Q_{loss}\\\dot Q_{used} = 1061.28 BTU[/tex]

Assuming that air behaves as an ideal gas, density is given by following expression:

[tex]\rho_{air} = \frac{P \cdot M_{air}}{R_{u} \cdot T}[/tex]

[tex]\rho_{air} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(298 K)}\\\rho_{air} = 1.145 \frac{kg}{m^{3}}[/tex]

The density unit is converted to pounds per cubic feet:

[tex]\rho_{air} = 1.145 \frac{kg}{m^{3}} \cdot (\frac{1 lb}{0.453 kg}) \cdot (\frac{0.304 m}{1 ft} )^{3} \\\rho_{air} = 0.071 \frac{lb}{ft^{3}}[/tex]

The heat received by air flow through heat exchanger is:

[tex]\dot Q_{air} = (0.90) \cdot \dot Q_{used}\\\dot Q_{air} = 955.152 BTU[/tex]

Outlet temperature can isolated from the following formula:

[tex]\dot Q_{air} = \rho_{air} \cdot \dot V_{air} \cdot c_{p,air} \cdot (T_{out} - T_{in})[/tex]

[tex]T_{out} =T_{in} + \frac{\dot Q_{air}}{\rho_{air} \cdot \dot V_{air} \cdot c_{p,air}}[/tex]

[tex]T_{out} = 536.4 ^{\textdegree}R + \frac{955.152 BTU}{(0.071\frac{lb}{ft^{3}})\cdot (33.333 \frac{ft}{s} )\cdot(0.244 \frac{BTU}{lb ^{\textdegree}R})}\\T_{out} = 2190.455 ^{\textdegree}R[/tex]

b) Due to the compressibility of air, density has to be calculated by using the approach from section a).

[tex]\rho_{air,out} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(1217 K)}\\\rho_{air,out} = 0.280 \frac{kg}{m^{3}}[/tex]

[tex]\rho_{air,out} = 0.017 \frac{lb}{ft^{3}}[/tex]

Volumetric flow can be found by Principle of Mass Conservation:

[tex]\dot V_{air, out} = \frac{\rho_{air}}{\rho_{air,out}}\cdot \dot V_{air}[/tex]

[tex]\dot V_{air,out} = 8352.941 acfm[/tex]

c) The moisture component indicates that 88 percent of volume is occupied by dry-air. The moisture-corrected dry flow rate is:

[tex]\dot V_{air,out,corr} = 0.88 \cdot (8352.941 afcm)\\\dot V_{air,out,corr} = 7350.588 afcm[/tex]

Answer:

Please find attached

Explanation:

Answer:

a. 3.0 X 10^-10 C; 2.4 X 10 ^-8 C b. 1.0 X 10^-10 C; V= 240+3 =243 V c. 36 nJ  d. 12.2 nJ

Explanation:

a. C=Q/V = ε0εrA/d => Q=εoεrAV/d; since dielectric constant for free space is 8.85 x 10^-12 and that of distilled water is 80, this shows that other dielectric material is free space with εr=1, Q=5.21 x 10^-15 C; εr for water is 80;

c. Energy, E =1/2(QV)

To keep the center plate stationary, the bottom plate must move at 2 m/s to the left, counteracting the top plate's movement to the right at the same speed.

Three large plates are separated by thin layers of ethylene glycol and water. The top plate moves to the right at 2m/s. To hold the center plate stationary, the velocity profile of the fluid layers between the plates indicates that for every action, there is an equal and opposite reaction according to Newton's third law. Given that the top plate moves at 2 m/s, to keep the center plate stationary, the bottom plate must also move at 2 m/s but in the opposite direction, which is to the left.

This setup demonstrates the principles of fluid dynamics, where the velocity gradient across the thickness of the fluid layers causes a shear stress that is directly proportional to the velocity gradient according to Newton's law of viscosity. The equilibrium of forces ensures the central plate remains stationary if the bottom plate's movement perfectly counterbalances the top plate's movement.

The bottom plate should move left at 2 m/s to keep the center plate stationary relative to the top plate.

To hold the center plate stationary, the relative velocity between the top and bottom plates should be zero. Since the top plate moves to the right at 2 m/s, the bottom plate must move to the left at the same speed to cancel out this motion.

The velocity of the bottom plate  [tex](\(v_b\))[/tex] relative to the center plate is the sum of the velocities of the bottom plate relative to the top plate ( [tex]\(v_{b/t}\))[/tex] and the velocity of the top plate relative to the center plate

[tex](\(v_{t/c}\))[/tex] . Since [tex]\(v_{t/c}\)[/tex] is given as 2 m/s to the right, to make [tex]\(v_b\)[/tex] zero,  [tex]\(v_{b/t}\)[/tex] should be 2 m/s to the left.

So, the bottom plate must be moved to the left at 2 m/s to hold the center plate stationary relative to the top plate. This ensures that the relative velocity between the top and bottom plates is zero, maintaining the center plate stationary.

Answer:

The question is mentioned in the attachment.

Explanation: