A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps.

Calculate:

a) lenghtb) exit velocity of the final slab

Answer:

Technician B says that a ratchet is used to loosen fasteners that are very tight.

Explanation:

A ratchet is a common wrench device with a fastener component. A ratchet wrench is an essential tool that is used to fasten or loosen nuts and bolts.

Answer:

A ratchet wrench is usually used to loosen and tighten parts like steering linkages, tie rod end clamps and muffler clamps. Basically when nut is used on long thread a ratchet wrench is being used.

hence the technician B (option b) is correct.

Explanation:

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

Answer:

COP = 13.31

Explanation:

We have an allowed temperature difference of 2°C, thus, let's make use of temperature of 20°C in the evaporator.

Now, looking at table A-11 i have attached and looking at temperature of 20°C, we will see that the enthalpy(h1) = 261.59 Kj/Kg

While the enthropy(s1) = 0.92234 Kj/KgK

Now, the enthalpy at the second state will be gotten from the given condenser pressure under the condition s2 = s1.

Thus, looking at table A-13 which i have attached, direct 20°C is not there, so when we interpolate between the enthalpy values at 15.71°C and 21.55°C, we get an enthalpy of  273.18 Kj/Kg.

Now, the enthalpy at the third and fourth states is again obtained from interpolation between values at temperatures of 18.73 and 21.55 of the saturated liquid value in table A-12 i have attached.

Thus, h3=h4 = 107.34 Kj/kg

Formula for COP = QL/w = (h1- h4) / (h2 - h1)

COP = (261.59 - 107.34)/( 273.18 - 261.59) = 13.31

Answer:

Conventional single-level page table  [tex]2^{20}[/tex]  pages

Inverted page table are [tex]2^{17}[/tex] frame

Explanation:

given data

logical address = 32-bit = [tex]2^{32}[/tex]  Bytes

page size = 4-KB =   [tex]2^{12}[/tex] Bytes

physical memory = 512 MB  = [tex]2^{29}[/tex]  bytes

solution

we get here number of pages that will be

number of pages = [tex]\frac{logical\ address}{page\ size}[/tex]   ..............1

put here value

number of pages =   [tex]\frac{2^{32}}{2^{12}}[/tex]

number of pages = [tex]2^{20}[/tex]  pages

and

now we get number of frames  that is

number of frames  = [tex]\frac{physical\ memory}{page\ size}}[/tex]   ............2

number of frames  = [tex]\frac{2^{29}}{2^{12}}[/tex]

number of frames  = [tex]2^{17}[/tex] frame

so

Conventional single-level page table  [tex]2^{20}[/tex]  pages

and

Inverted page table are [tex]2^{17}[/tex] frame

To achieve an effluent concentration of 20 mg/L in a sewage lagoon receiving sewage with 100 mg/L of a contaminant, a precise biodegradation reaction rate coefficient, determined by the mass balance equation under steady-state conditions, must be achieved.

Calculating the Biodegradation Reaction Rate Coefficient

The question involves determining the biodegradation reaction rate coefficient necessary to reduce the concentration of a contaminant in a sewage lagoon, illustrating principles of environmental engineering. Given a lagoon with a surface area of 100,000 m2 and a depth of 1 m, receiving 8,640 m3/d of sewage that contains 100 mg/L of biodegradable contaminant, the goal is to lower the effluent concentration to no more than 20 mg/L.

To find the required biodegradation reaction rate coefficient (d-1), we must apply the mass balance concept in a steady-state condition, assuming the lagoon is well mixed. The mass balance equation for a contaminant undergoing a first-order degradation reaction can be expressed as: Input = Output + Decay. By substituting the given values and solving for the decay rate, we can find the coefficient that ensures the specified effluent concentration.

The calculation involves deriving relationships between the influent and effluent concentrations, the volume of the lagoon, and the decay process characterized by the reaction rate coefficient. For the lagoon described, achieving an effluent concentration of 20 mg/L from an influent concentration of 100 mg/L through biodegradation requires precise control of the treatment process and understanding of the kinetics of contaminant degradation.

To find an expression for the jet boat's steady speed, equate the water jet force (ρQVj) with the drag force (kV^2). Solve for V to get the equation for speed in terms of ρ, Q, k, and Vj. Then use known values for Vj and V to find Q and k and repeat the process for a different Vj to get new V and Q values.

To find an expression for the steady speed V of a freshwater jet boat in terms of the water density ρ, flow rate Q, constant k, and jet speed Vj, we can apply Newton's second law, assuming that the force provided by the water jet equals the drag force on the boat when at steady speed. The water jet force can be expressed as the rate of change of momentum of the water, which is ρQVj (since Q is the mass flow rate ρQ is the momentum flow rate), and the drag force is given as kV^2. Equating these two forces gives us:

ρQVj = kV^2

Solving for V will give us the steady speed expression we are looking for:

V = √(ρQVj / k)

To calculate the new flow rate Q given Vj = 15 m/s and V = 10 m/s, we plug these values into the expression obtained from above:

Q = kV^2 / (ρVj)

With given values, we would have:

Q = k * 10^2 / (ρ * 15)

The constant k can be determined using the known conditions and solving for k.

k = ρQVj / V^2

For (c) and (d), the same equations can be applied with the jet speed Vj changed to 25 m/s to find the new boat speed and flow rate.

Answer:

Note that writeln() add a new line after each statement

var txt = "<p>User-agent header: " + navigator.userAgent + "</p>";

$("#agent").writeln(txt);

Then Anywhere in the body tag of the html file

create a div tag and include an id="agent"

Answer:

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.

Explanation:

Answer:

[tex]\sigma_{max} = 275\,MPa[/tex], [tex]\sigma_{min} = - 175\,MPa[/tex]

Explanation:

Maximum stress:

[tex]\sigma_{max}=\overline \sigma + \sigma_{a}\\\sigma_{max}= 50\,MPa + 225\,MPa\\\sigma_{max} = 275\,MPa[/tex]

Minimum stress:

[tex]\sigma_{min}=\overline \sigma - \sigma_{a}\\\sigma_{min}= 50\,MPa - 225\,MPa\\\sigma_{min} = - 175\,MPa[/tex]

A fatigue test was conducted in which the mean stress was 50 MPa (7,250 psi) and the stress amplitude was 225 MPa (32,625 psi).

(a) Compute the maximum and minimum stress levels.

(b) Compute the stress ratio.

(c) Compute the magnitude of the stress range.

(a) The maximum and minimum stress levels are 275MPa and -175MPa respectively.

(b) The stress ratio is 0.6

(c) The magnitude of the stress range is 450MPa

(a )In fatigue, the mean stress ([tex]S_{m}[/tex]) is found by finding half of the sum of the maximum stress ([tex]S_{max}[/tex]) and minimum stress ([tex]S_{min}[/tex]) levels. i.e

[tex]S_{m}[/tex] = [tex]\frac{S_{max} + S_{min}}{2}[/tex]    ------------------------(i)

Also, the stress amplitude (also called the alternating stress), [tex]S_{a}[/tex], is found by finding half of the difference between the maximum stress ([tex]S_{max}[/tex]) and minimum stress ([tex]S_{min}[/tex]) levels. i.e

[tex]S_{a}[/tex] = [tex]\frac{S_{max} - S_{min}}{2}[/tex]    ------------------------(ii)

From the question,

[tex]S_{m}[/tex] = 50 MPa (7250 psi)

[tex]S_{a}[/tex] = 225 MPa (32,625 psi)

Substitute these values into equations(i) and (ii) as follows;

50 = [tex]\frac{S_{max} + S_{min}}{2}[/tex]

=> 100 = [tex]S_{max}[/tex] + [tex]S_{min}[/tex]          -------------------(iii)

225 = [tex]\frac{S_{max} - S_{min}}{2}[/tex]

=> 450 = [tex]S_{max}[/tex] - [tex]S_{min}[/tex]          -------------------(iv)

Now, solve equations (iii) and (iv) simultaneously as follows;

(1) add the two equations;

     100 = [tex]S_{max}[/tex] + [tex]S_{min}[/tex]

     450 = [tex]S_{max}[/tex] - [tex]S_{min}[/tex]

    ________________

     550 = 2[tex]S_{max}[/tex]               --------------------------------(v)

    _________________

(2) Divide both sides of equation (v) by 2 as follows;

     [tex]\frac{550}{2}[/tex] = [tex]\frac{2S_{max} }{2}[/tex]

     275 = [tex]S_{max}[/tex]

Therefore, the maximum stress level is 275MPa

(3) Substitute [tex]S_{max}[/tex] = 275 into equation (iv) as follows;

   450 = 275 - [tex]S_{min}[/tex]

   [tex]S_{min}[/tex] = 275 - 450

  [tex]S_{min}[/tex] = -175

Therefore, the minimum stress level is -175MPa

In conclusion, the maximum and minimum stress levels are 275MPa and -175MPa respectively.

===============================================================

(b) The stress ratio ([tex]S_{r}[/tex]) is given by;

[tex]S_{r}[/tex] = [tex]\frac{S_{min} }{S_{max} }[/tex]        ----------------------------(vi)

Insert the values of [tex]S_{max}[/tex] and [tex]S_{min}[/tex] into equation (vi)

[tex]S_{r}[/tex] = [tex]\frac{-175}{275}[/tex]

[tex]S_{r}[/tex] = 0.6

Therefore, the stress ratio is 0.6

===============================================================

(c) The magnitude of the stress range ([tex]S_{R}[/tex]) is given by

[tex]S_{R}[/tex] = | [tex]S_{max}[/tex] - [tex]S_{min}[/tex] |          ------------------------------(vii)

Insert the values of [tex]S_{max}[/tex] and [tex]S_{min}[/tex] into equation (vii)

[tex]S_{R}[/tex] = | 275 - (-175) |

[tex]S_{R}[/tex] =  450MPa

Therefore, the magnitude of the stress range is 450MPa

===============================================================

1 MPa = 145.038psi

Therefore, the values of the maximum and minimum stress levels, the stress range can all be converted from MPa to psi (pounds per inch square) by multiplying the values by 145.038 as follows;

[tex]S_{max}[/tex] = 275MPa = 275 x 145.038psi = 39885.45psi

[tex]S_{min}[/tex] = -175MPa = -175 x 145.038psi = 25381.65psi

[tex]S_{R}[/tex] =  450MPa = 450 x 145.038psi = 65267.1psi

Answer:

Explanation:

1) true ( reason is due to polarizability given in statement)

2)false ( nucleophile does not involve in rate determining step)

3)false ( it is an inverse attack , attacks from backside)

4) true

5) false ( it inverts stereo centres which are being attacked)

6) false (not H-)

7) true ( strong bases are good nucleophiles)

8) true ( because of steric hindrance)

9)false ( ther are stabilized by hyper conjugation)

10) false

Answer:

1) True

2) False

3) False

4)

5)

6) True

7)True

8) True

9) False

10) True

Explanation:

1) Generally, polarization increases down the column of the periodic table.

The maximum power delivered to the circuit is found to be 3.75W, occurring at the start (t = 0). Calculating the total energy requires integrating the power over time, involving exponentials reducing over time to approach a finite total energy delivered.

The question involves calculating the maximum power and the total energy delivered to a circuit element, where the voltage V and current I are given as time-dependent expressions.

Maximum Power Delivered

Power is calculated using P = IV. Inserting the given expressions for V and I,

P(t) = V(t) × I(t) = (75 - 75e^{-1000t}) × 50e^{-1000t} mA.

To find the maximum power, we would differentiate P(t) with respect to t and set the derivative equal to zero. However, because the setup includes exponential decay functions, their maximum product occurs at t = 0 for these specific functions.

Pmax = 75V × 50mA = 3.75W.

Total Energy Delivered

The total energy delivered can be found by integrating the power over time:

Energy = ∫ P(t) dt.

Considering the specific forms of V and I provided, this becomes an integral of the product of two exponentials, leading to an expression that evaluates the total energy consumed by the circuit over time.

Given the nature of exponential decay in V and I, the energy delivered approaches a finite value as t approaches infinity.

Answer:

Insulation Resistance Tests

Explanation:

An insulation resistance test is carried out when there are unstable readings and random changes in the zero balance point of the load cell. It is done by measuring the resistance between the load cell body and all its connected wires, as follows:

First, disconnect the load cell from the summing box and indicator panel.

Connect all the input, output and sense (if equipped) wires together.

Measure the insulator resistance between the connected wires and the load cell body with a mega-ohmmeter.

Measure the insulation resistance between the connected wires and the cable shield.

Measure the insulation resistance between the load cell body and the cable shield.

The insulation resistance should match the value in the product’s load cell datasheet. A lower value shows an electrical leakage caused by moisture; this causes short circuits, giving unstable load cell outputs.

Answer:

LI = Lifting Index = 0.71

No lifting risk is involved

Explanation:

NIOSH Lifting Index

LI = Load Weight / Recommended Weight Limit

LI = L / RWL                                                     ............. Eq (A)    

NIOSH Recommended Weight Limit equation is following,

RWL = LC * HM * VM * DM * AM * FM * CM   ........... Eq (B)

Where,

 LC = Load constant

 HM = Horizontal multiplier

 VM = Vertical multiplier

 DM = Distance multiplier

 AM = Asymmetric multiplier

 FM = Frequency multiplier

 CM = Coupling multiplier

Given data

V = 30 + (8/2) = 34 in

H = 7 in

D = 55 - 30 = 25 in

A = 0

F = 10 boxes/min

C = 1 = Good coupling

According to NIOSH lifting guide

 LC = 51 lb

 HM = 10/H

 VM = 1 - {0.0075*(v-30)}

 DM = 0.82 + (1.8/D)

 AM = 1 - (0.0032*A)

 FM = 0.45         (Table 5 from NIOSH lifting guide)

 CM = 1               (Table 7 from NIOSH lifting guide)

Solution:

RWL = 51 * (10/7) * [1-{0.0075(34-30)}] * (0.82+ 1.8/25) * (1-0.0032*0) * 0.45 * 1

RWL = 51 * (10/7) * 0.97 * 0.892 * 1 * 0.45 * 1

RWL = 28.37 lb

Using equation A

LI = L / RWL

LI = 20 / 28.37

LI = 0.71

According to NIOSH lifting guide LI <= 1

So No lifting risk is involved

Answer:

δ_AB = 0.0333 in

Explanation:

Given:

- The complete question is as follows:

" The rigid beam is supported by a pin at C and an A−36

steel guy wire AB. If the wire has a diameter of 0.5 in.

determine how much it stretches when a distributed load of

w=140 lb / ft acts on the beam. The material remains elastic."

- Properties for A-36 steel guy wire:

       Young's Modulus E = 29,000 ksi

       Yield strength σ_y = 250 MPa

- The diameter of the wire d = 0.5 in

- The distributed load w = 140 lb/ft

Find:

Determine how much it stretches under distributed load

Solution:

- Compute the surface cross section area A of wire:

                            A = π*d^2 / 4

                            A = π*0.5^2 / 4

                            A = π / 16 in^2

- Apply equilibrium conditions on the rigid beam ( See Attachment ). Calculate the axial force in the steel guy wire F_AB

                            Sum of moments about point C = 0

                            -w*L*L/2 + F_AB*10*sin ( 30 ) = 0

                             F_AB = w*L*L/10*2*sin(30)

                             F_AB = 140*10*10/10*2*sin(30)

                             F_AB = 1400 lb

- The normal stress in wire σ_AB is given by:

                            σ_AB = F_AB / A

                            σ_AB = 1400*16 / 1000*π

                            σ_AB = 7.13014 ksi

- Assuming only elastic deformations the strain in wire ε_AB would be:

                            ε_AB = σ_AB / E

                            ε_AB = 7.13014 / (29*10^3)

                            ε_AB = 0.00024

- The change in length of the wire δ_AB can be determined from extension formula:

                            δ_AB = ε_AB*L_AB

                            δ_AB = 0.00024*120 / cos(30)

                            δ_AB = 0.0333 in

Using the maximum stress limit and elongation criteria, the required diameter of the aluminum rod to withstand a 3-kN axial load without exceeding a 1 mm stretch and keeping the normal stress below 40 MPa is calculated to be approximately 9.8 mm.

Determining the Required Diameter of an Aluminum Rod

To ensure a 1.5-m-long aluminum rod does not stretch more than 1 mm (0.001 m) under a 3-kN (3000 N) axial load while keeping the normal stress below 40 MPa (40×106 N/m2), we first calculate the cross-sectional area required using the formula for stress (σ) which is σ = F/A, where F is the force applied and A is the cross-sectional area. Given that the maximum allowable stress σ is 40 MPa, we can reorganize the formula to solve for A, the required cross-sectional area of the rod. This gives us A = F/σ.

Substituting the given values, A = 3000 N / (40×106 N/m2) = 7.5×10-5 m2. To ensure the rod does not exceed the maximum stretch limit when this force is applied, we must also consider the modulus of elasticity (E) for aluminum, which is given as 70 GPa (70×109 N/m2). The formula for elongation (ΔL) under a force is ΔL = (FL)/(AE), where L is the original length of the rod. Given the requirements, the diameter can be calculated from the cross-sectional area (A = πd2/4), where d is the diameter of the rod.

From the area calculated earlier, we can determine the diameter is required to be sufficiently large to maintain stress and elongation within specified limits. Rearranging A = πd2/4 to solve for d, we find d to be approximately 9.8 mm, considering the area necessary to keep stress below 40 MPa while allowing for the specified elongation limit.

Answer:

Explanation:

The bolt is under double shear because it connects 3 plates.

Now the shear force resisted by each shear joint(V):

V=P/(No. of shear area)=185/2=92.5N.

Answer:

Explanation: see attachment

Complete Question

Sites like Zillow get input about house prices from a database and provide nice summaries for readers. Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. Ex: If the input is 200000 210000, the output is: This house is $200000. The change is $-10000 since last month. The estimated monthly mortgage is $750.

Use Coral Programming Language

Answer:

// Program is written in Coral Programming Language

// Comments are used for explanatory purpose

// Program starts here

// Variable declaration

int currentprice

int prevprice

int change

float mortgage

Put "Enter current price to output" to output

currentprice = Get next input

Put "Enter last month price to output" to output

prevprice = Get next input

// Calculate Change since last month

change = currentprice - prevprice

// Calculate Monthly Mortgage

mortgage = currentprice * 0.045 / 12

// Print Results

Put "This house is $" to output

Put currentprice to output

Put "\n" to output

Put "This change is $" to output

Put change to output

Put "\n" to output

Put "This house is $" to output

Put currentprice to output

Put "since last month\n" to output

Put "This estimated monthly mortgage is $" to output

Put mortgage to output

// End of Program

Answer:

The limit on the series resistance is R ≤ 400μΩ

Explanation:

Considering the circuit has a series of inductance and resistance. The current current in the current in the circuit in time is

[tex]i(t) = Iie^{\frac{R}{L} t}[/tex] (li = initial current)

So, the initial energy stored in the inductor is

[tex]Wi = \frac{1}{2} Li^{2}_{i}[/tex]

After 1 hour

[tex]w(3600) = \frac{1}{2} Li_{i}e^{-\frac{R}{L} 3600 }[/tex]

Knowing it is equal to 75

[tex]w(3600) = 0.75Wi = 0.75 \frac{1}{2} Li^{2}_{i} = \frac{1}{2} Li_{i}e^{-\frac{R}{L} 3600 }\\[/tex]

This way we have,

R = [tex]-10 \frac{ln 0.75}{2 * 3600} = 400[/tex] μΩ

Than, the resistance is R ≤ 400μΩ

Answer:

B. Two-factor authentication

Explanation:

As far as identity as been established, authentication must be carried out. There are various technologies and ways of implementing authentication, although most method falls under the same category.

The three major types of authentication.

a. Authentication through knowledge, what someone knows.

b. Authentication through possessions, what someone has.

c. Authentication by characteristic features, who a person is.

Logical controls that are in relations to these types of authentication are known as factors.

Two factor authentication has to do with the combination of 2 out of the 3 factors of authentication.

The general term used when more than one factor is adopted is Multi-party authentication

Answer:

The dependent variable is MEDV - Median value of owner-occupied homes in $1000's

Explanation:

The median value of the house has to be predicted, based on its properties and neighborhood properties, this can be done by using a linear regression model.

The dependent variable in Machine Learning is the output variable that we want to predict.

Therefore, according to the question given "MEDV" is the dependent variable.

Answer:

C++ code is explained below

Explanation:

#include<iostream>

using namespace std;

//Function Declarations

void add();

void sub();

void mul();

//Main Code Displays Menu And Take User Input

int main()

{

  int choice;

  cout << "\nMenu";

  cout << "\nChoice 1:addition";

  cout << "\nChoice 2:subtraction";

  cout << "\nChoice 3:multiplication";

  cout << "\nChoice 0:exit";

  cout << "\n\nEnter your choice: ";

  cin >> choice;

  cout << "\n";

  switch(choice)

  {

      case 1: add();

              break;

      case 2: sub();

              break;

      case 3: mul();

              break;

      case 0: cout << "Exited";

              exit(1);

      default: cout << "Invalid";      

  }

  main();  

}

//Addition Of Matrix

void add()

{

  int rows1,cols1,i,j,rows2,cols2;

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

   int m1[rows1][cols1];

  //Taking First Matrix

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  //Printing 1st Matrix

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

  int m2[rows2][cols2];

  //Taking Second Matrix

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  //Displaying second Matrix

  cout << "\n";

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m2[i][j] << " ";

      cout << "\n";

  }

  //Displaying Sum of m1 & m2

  if(rows1 == rows2 && cols1 == cols2)

  {

      cout << "\n";

      for(i=0;i<rows1;i++)

      {

          for(j=0;j<cols1;j++)

              cout << m1[i][j]+m2[i][j] << " ";

          cout << "\n";  

      }

  }

  else

      cout << "operation is not supported";

  main();

}

void sub()

{

  int rows1,cols1,i,j,k,rows2,cols2;

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

   int m1[rows1][cols1];

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

  int m2[rows2][cols2];

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

  cout << "\n";

  //Displaying Subtraction of m1 & m2

  if(rows1 == rows2 && cols1 == cols2)

  {

      for(i=0;i<rows1;i++)

      {

          for(j=0;j<cols1;j++)

              cout << m1[i][j]-m2[i][j] << " ";

          cout << "\n";  

      }

  }

  else

      cout << "operation is not supported";

  main();

}

void mul()

{

  int rows1,cols1,i,j,k,rows2,cols2,mul[10][10];

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

   int m1[rows1][cols1];

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  cout << "\n";

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

  int m2[rows2][cols2];

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  cout << "\n";

  //Displaying Matrix 2

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m2[i][j] << " ";

      cout << "\n";

  }

  if(cols1!=rows2)

      cout << "operation is not supported";

  else

  {

      //Initializing results as 0

      for(i = 0; i < rows1; ++i)

  for(j = 0; j < cols2; ++j)

  mul[i][j]=0;

// Multiplying matrix m1 and m2 and storing in array mul.

  for(i = 0; i < rows1; i++)

  for(j = 0; j < cols2; j++)

  for(k = 0; k < cols1; k++)

  mul[i][j] += m1[i][k] * m2[k][j];

// Displaying the result.

  cout << "\n";

  for(i = 0; i < rows1; ++i)

      for(j = 0; j < cols2; ++j)

      {

      cout << " " << mul[i][j];

      if(j == cols2-1)

      cout << endl;

      }

      }  

  main();

 }

Answer: The average distance the electron can travel in microns is 1.387um/s

Explanation: The average distance the electron can travel is the distance an exited electron can travel before it joins together. It is also called the diffusion length of that electron.

It is gotten, using the formula below

Ld = √DLt

Ld = diffusion length

D = Diffusion coefficient

Lt = life time

Where

D = 25cm2/s

Lt = 7.7

CONVERT cm2/s to um2/s

1cm2/s = 100000000um2/s

Therefore D is

25cm2/s = 2500000000um2/s = 2.5e9um2/s

Ld = √(2.5e9 × 7.7) = 138744.37um/s

Ld = 1.387e5um/s

This is the average distance the excited electron can travel before it recombine

Answer: 9.9KJ

Explanation: Q = U + W + losses

Q is heat transfered to the water

U is the change in energy of the system

W is work done by the system = 2J

Losses = 80J

Heat into system is 10kJ = 10000KJ

Therefore

U = Q - W - losses

U = 10000 - 2 - 80 = 9990J

= 9.9kJ

Answer:

The net work output from the cycle = 520.67 [tex]\frac{KJ}{kg}[/tex]

The efficiency of Brayton cycle = 0.448

Explanation:

Compressor inlet temperature [tex]T_{1}[/tex] = 300 K

Turbine inlet temperature [tex]T_{3}[/tex] = 1700 K

Pressure ratio [tex]r_{p}[/tex] = 8

For the compressor the temperature - pressure relation is given by the formula,

⇒ [tex]\frac{T_{2} }{T_{1} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]

⇒ [tex]\frac{T_{2} }{300} = 8^{\frac{1.4 - 1}{1.4} }[/tex]

⇒ [tex]T_{2}[/tex] = 543.42 K

This is the temperature at compressor outlet.

Same relation for turbine we can write this as,

⇒ [tex]\frac{T_{3} }{T_{4} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]

⇒[tex]\frac{1700 }{T_{4} }[/tex] = [tex]8^{0.2857}[/tex]

⇒ [tex]T_{4}[/tex] = 938.5 K

This is the temperature at turbine outlet.

Now the work output from the turbine [tex]W_{T}[/tex] = [tex]m C_{p} (T_{3} - T_{4} )[/tex]

Put all the values in above formula we get,

⇒ [tex]W_{T}[/tex] = 1 × 1.005 × ( 1700 - 938.5 )

    [tex]W_{T} = 765.3 \frac{KJ}{kg}[/tex]

This is the work output from the turbine.

Now the work input to the compressor is [tex]W_{C}[/tex] = [tex]m C_{p} (T_{2} - T_{1} )[/tex]

Put all the values in above formula we get,

⇒ [tex]W_{C}[/tex] = 1 × 1.005 × ( 543.42 - 300 )

⇒ [tex]W_{C}[/tex] = 244.63 [tex]\frac{KJ}{kg}[/tex]

This is the work input to the compressor.

Net work output from the cycle [tex]W_{net} = W_{T} - W_{C}[/tex]

⇒ [tex]W_{net}[/tex] = 765.3 - 244.63

   [tex]W_{net} = 520.67\frac{KJ}{kg}[/tex]

This is the net work output from the cycle.

The thermal efficiency is given by

[tex]E_{cycle} =1 - \frac{1}{r_{p}^{\frac{\gamma - 1}{\gamma} } }[/tex]

[tex]E_{cycle} =1 - \frac{1}{8^{\frac{1.4 - 1}{1.4} } }[/tex]

[tex]E_{cycle} = 0.448[/tex]

This is the efficiency of Brayton cycle.

(b). the graph between plot the net work developed per unit mass flowing and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50 is shown in the image below.

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

There is a part of the question missing and it says;

The velocity field of a flow is given by V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s, where x and y are in feet.

Answer:

A) At (1,5),angle is -11.31°

B) At (5,2),angle is -68.2°

C) At (1,0), angle is -90°

Explanation:

From the question ;

V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s

Let us assume that u and v are the flow velocities in x and y directions respectively.

Thus we have the expression;

u = [20y/(x² + y²)^(1/2)]

and v = - [20x/(x² + y²)^(1/2)]

Thus, V = √(u² + v²)

V = √[20y/(x² + y²)^(1/2)]² + [-20x/(x² + y²)^(1/2)]²

V = √[400y²/(x²+y²)] +[400x²/(x²+y²)

V = √(400x² + 400y²)/(x²+y²)

Now for the angle;

tan θ = opposite/adjacent

And thus, in this question ;

tan θ = v/u

tan θ = [-20x/(x² + y²)^(1/2)]/ [20y/(x² + y²)^(1/2)]

Simplifying this, we have;

tan θ = - 20x/20y = - x/y

so the angle is ;

θ = tan^(-1)(-x/y)

So let's now find the angle at the various coordinates.

At, 1,5

θ = tan^(-1)(-1/5) = tan^(-1)(-0.2)

θ = -11.31°

At, 5,2;

θ = tan^(-1)(-5/2) = tan^(-1)(-2.5)

θ = -68.2°

At, 1,0;

θ = tan^(-1)(-1/0) = tan^(-1)(-∞)

θ = -90°

At,

Answer:

Explanation:

AVERAGE: the average value is given as

[tex]\frac{1}{0.1} \int\limits^\frac{1}{10} _0 {25+10sint} \, dt = \frac{1}{0.1} [ 25t- 10cos\ t]_0^{0.1}[/tex]

=[tex]\frac{1}{0.1} ([2.5-10]-10)=-175[/tex]

RMS= [tex]\sqrt{\int\limits^\frac{1}{3} _0 {y(t)^2} \, dt }[/tex]

[tex]y(t)^2 = (25 + 10sin \ t)^2 = 625 +500sin \ t + 10000sin^2 \ t[/tex]

[tex]\frac{1}{\frac{1}{3} } \int\limits^\frac{1}{3} _0 {y(t)^2} \, dt =3[ 625t -500cos \ t + 10000(\frac{t}{2} - \frac{sin2t}{4} )]_0^{\frac{1}{3} }[/tex]

=[tex]3[[\frac{625}{3} - 500 + 10000(\frac{1}{6} - 0.002908)] + 500] = 2845.92\\[/tex]

therefore, RMS = [tex]\sqrt{2845.92} = 53.3[/tex]

The question involves applying thermodynamics to calculate work and heat transfer for propane in a piston-cylinder assembly undergoing a process with a specific pressure-volume relationship. It requires integrating over the process path for work and applying the first law of thermodynamics for heat transfer, considering the neglect of kinetic and potential energy effects.

A piston–cylinder assembly containing propane undergoes a process where the pressure-volume relationship is given as pV1.1 = constant. To evaluate the work and heat transfer for the propane, we apply the principles of thermodynamics, specifically the first law of thermodynamics, and the properties of processes adhering to specific equations of state. The work done in such processes can be calculated using the integral of p dV, considering the pressure-volume relationship provided. The heat transfer can then be inferred by applying the first law of thermodynamics, which equates the change in internal energy to the net heat added to the system minus the work done by the system.

The initial and final states of the propane provide the necessary boundary conditions to evaluate these quantities. However, without specific values for the molar mass or specific heat capacities of propane at constant pressure and volume, exact numerical answers cannot be provided. Generally, for processes described by a polytropic equation (pVn = constant), the work done is W = (p2V2 - p1V1)/(1-n) for an ideal gas, where p1, V1 are the initial pressure and volume, and p2, V2 are the final conditions. Heat transfer, Q, requires specific thermal properties of propane and can be approached via Q = ΔU + W, with ΔU denoting the change in internal energy of the gas.

For a precise evaluation, one would typically reference thermodynamic tables for propane or apply real gas equations of state considering the polytropic process specifics. It is essential to note that kinetic and potential energy changes are negligible, focusing the analysis solely on the thermodynamic work and heat transfer.

Answer:

a = 30.1 kj

b = 115 kj

Explanation:

To determine the mass we use the formula m = V/v1

v1 =1.08m3/kg, and V = 20L

m = 20/1000 × 1.08 = 0.0185kg

Next we determine the initial specific internal energy, u1.

Using softwares and appropriate values of T1 and p1, we get

u1 = 2650kj/kg.

After this we determine the final specific internal energy, u2 using the formula u2 = uf + x2 × ufg

Therefore we need to find x2 first.

x2 = u2 - uf/ug - uf

x2 = 1.08 - 0.001029/3.4053 -0.001029

x2 = 0.3180

But u2 = uf + x2× uf=334.97 + 0.3180×2146.6 = 1017.59 kj/kg

Now heat transfer Q= DU

Q = m x (u1 - u2)

Q = 0.0185(2650-1017.59

Q = 30.1 kj

Calculating the b part of the question we use the formula

W = m( u1-u2) - m. To. (s1 - s2)

Where s1 = 7.510kj/kgk

And s2 = 3.150 kj/kgk

We need to convert To and Ta to k values by adding 273 to 0 and 21 respectively.

Putting the values into the formula, we get W = 30.1 - 0.0185 × 273 (7.510-3.150)

W = 8.179kj

Finally maximum heat transfer

Qm = W/1 - to/ta

Qm = 8.179/1 - 273/294

Qm = 115kj

Answer:

The code is given which can be pasted in the Javascript file

Explanation:

The code is given as below

package Sorters;

public class javasort

{

private int[] arr=new int[13];

public void bubbleSort(int[] a){

int c,d,temp;

for (c = 0; c < ( 13 - 1 ); c++) {

for (d = 0; d < 13- c - 1; d++) {

if (a[d] > a[d+1]) /* For descending order use < */

{

temp = a[d];

a[d] = a[d+1];

a[d+1] = temp;

}

}

System.out.print("\n[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void insertionSort(int[] a){

int temp;

for (int i = 1; i < 13; i++) {

for(int j = i ; j > 0 ; j--){

if(a[j] < a[j-1]){

temp = a[j];

a[j] = a[j-1];

a[j-1] = temp;

}

}

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void shellSort(int[] a){

int increment = a.length / 2;

while (increment > 0)

{

for (int i = increment; i < a.length; i++)

{

int j = i;

int temp = a[i];

while (j >= increment && a[j - increment] > temp)

{

a[j] = a[j - increment];

j = j - increment;

}

a[j] = temp;

}

if (increment == 2)

increment = 1;

else

increment *= (5.0 / 11);

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

}

System.out.println("\nSorted list of numbers");

System.out.print("[");

for (int i = 0; i < 13; i++){

  System.out.print(a[i]+" ");

}

System.out.print("]");

}

public void MergeSort(int[] a, int low, int high){

int N = high - low;  

if (N <= 1)

return;

int mid = low + N/2;

// recursively sort

MergeSort(a, low, mid);

MergeSort(a, mid, high);

// merge two sorted subarrays

int[] temp = new int[N];

int i = low, j = mid;

for (int k = 0; k < N; k++)

{

if (i == mid)

temp[k] = a[j++];

else if (j == high)

temp[k] = a[i++];

else if (a[j]<a[i])

temp[k] = a[j++];

else

temp[k] = a[i++];

}

for (int k = 0; k < N; k++)

a[low + k] = temp[k];

System.out.print("\n[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

printM(a);

}

public void quickSort(int[] a,int low,int high){

  System.out.print("\n[");

  for (int c = 0; c < 13; c++){

      System.out.print(a[c]+" ");

  }

  System.out.print("]");

int i =low, j = high;

int temp;

int pivot = a[(low + high) / 2];

/** partition **/

while (i <= j)

{

while (a[i] < pivot)

i++;

while (a[j] > pivot)

j--;

if (i <= j)

{

/** swap **/

temp = a[i];

a[i] = a[j];

a[j] = temp;

i++;

j--;

}

}

/** recursively sort lower half **/

if (low < j)

quickSort(a, low, j);

/** recursively sort upper half **/

if (i < high)

quickSort(a, i, high);

printM(a);

}

public void printM(int[] a){

arr=a;

}

public void fPrint(){

  System.out.println("\nSorted list:");

  System.out.print("\n[");

  for (int c = 0; c < 13; c++){

      System.out.print(arr[c]+" ");

  }

  System.out.print("]");

}

}

package mani;

import java.util.Random;

import java.util.Scanner;

public class javasorttest

{

public static void main(String[] args){

int[] a=new int[13];

Random r=new Random();

for(int i=0;i<13;i++){

a[i]=r.nextInt(99)+1;

}

System.out.print("[");

for (int c = 0; c < 13; c++){

  System.out.print(a[c]+" ");

}

System.out.print("]");

javasort j=new javasort();

System.out.println("\nSelect the sorting algo.\n1.bubbleSort\n2.insertionSort\n3.shellSort\n4.MergeSort\n5.QuickSort.");

Scanner s=new Scanner(System.in);

int opt=s.nextInt();

switch(opt){

case 1:

j.bubbleSort(a);

break;

case 2:

j.insertionSort(a);

break;

case 3:

j.shellSort(a);

break;

case 4:

j.MergeSort(a, 0, 13);

j.fPrint();

break;

case 5:

j.quickSort(a ,0, 12);

j.fPrint();

break;

}

}

}