c++ If your company needs 200 pencils per year, you cannot simply use this year’s price as the cost of pencils 2 years from now. Because of inflation the cost is likely to be higher than it is today.

Answer:

Explanation: see the pictures attached

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: see attachment

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:

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

The getDenom() and getNum() methods would be sufficient functionality for a user of the Fraction class to determine if two fractions are equivalent, provided that the class also includes methods to compare fractions based on their numerators and denominators.

The getDenom() and getNum() methods are essential for accessing the denominator and numerator of a fraction. With these methods, users can directly compare the numerators and denominators of two fractions to determine if they are equivalent.

By retrieving the numerator and denominator values separately, users can perform mathematical operations or comparisons to check for equivalence without needing additional methods specifically for equivalence testing.

The Complete Question

Explain the importance of the getDenom() and getNum() methods in the Fraction class for determining if two fractions are equivalent.

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 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:

Please find the attached file for the calculation of the 4 environment solutions:

Given:

[tex]f_c = 900\ MHz\\\\ h_t = 20\ m\\\\ h_r = 5\ m\\\\ d = 100\ m\\[/tex]

To find:

environments=?

Solution:

Please find the attached file.

Learn more about the Qualitative:

brainly.com/question/276942

brainly.com/question/18011951

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

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

[tex]F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}[/tex]

The bouyancy force is:

[tex]F_{B} =V*p_{w} *g[/tex]

The weight is:

[tex]W=V*p_{c} *g[/tex]

Laminar flow:

[tex]v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s[/tex]

Reynold number:

[tex]Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1[/tex]

Not flow region

For Newton flow region:

[tex]v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s[/tex]

[tex]Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4[/tex]

[tex]Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31[/tex]

[tex]Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99[/tex]

[tex]Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K[/tex]

[tex]\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s[/tex]

[tex]e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m[/tex]

1225

Explanation:

This segment helps initialize sum as 0. The for loop is used to increment with every execution and it is added to the sum. The loop runs 49 times and every time the count is added to the sum. In short it is the sum of first 49 natural numbers i.e 1+2+3+......+49.

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:

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:

Explanation:

The detailed steps and calculations is as shown in the attachment.

where s = flow rate of incoming stream

Cs = concentration of pollutant in stream

Qf = flow rate from factory

Cp = concentration of pollutant from factory

Qo = flow rate out of lake

Co = concentration of pollutant in the lake

V = volume of lake

k = reaction rate coefficient

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:

The answer is True.

Explanation:

SED is a command in UNIX for stream editors that parses and transforms text, using a simple, compact programming language. So the answer is TRUE that sed is a multipurpose tool that combines the work of several filters and performs noninteractive operations on a data stream. sed has a host of features that allow you ti select lines and run instructions on them.

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:

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

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();

 }

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:

Explanation: see attachment below

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;

}

}

}

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

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.

Explanation:

Please refer to the attached image

Python Code:

Please refer to the attached image

Output:

Please refer to the attached image

Following are the program to print the first 128 ASCII values:

#include <iostream>//header file

using namespace std;

int main()//main method

{

int i=1;//defining integer variable

for(i=1;i<=128;i++) //defining loop that prints the first 128 ASCII values

{

   char x=(char)i;//defining character variable that converts integer value into ASCII code value

   printf("%d=%c\n",i ,x);//print converted ASCII code value

}

   return 0;

}

Program Explanation:

Output:

Please find the attached file.

Find out more information about the ASCII values here:

brainly.com/question/3115410

Determining the radius of a brass cylinder based on its elongation under tensile load involves understanding stress, strain, and the material's properties. Without the elastic modulus of brass or additional details, an exact calculation can't be provided. Theoretically, one uses the relationship between stress, strain, and Young's modulus, and the formula for the area of a circle to find the radius.

To determine the radius of a cylindrical specimen of brass that must elongate a specific amount under a given tensile load, one must approach the problem by considering the relationship between stress, strain, and the elastic modulus of the material. Given that the specimen is cylindrical, the cross-sectional area is crucial in these calculations, which is directly related to the radius of the cylinder.

The formula for stress (σ) is defined as the force (F) divided by the area (A), σ = F/A. Strain (ε) is the deformation (change in length) divided by the original length (L), ε = ΔL/L. However, without the elastic modulus of brass or a direct way to calculate the cross-sectional area from the provided data, finding the exact radius requires assuming or knowing additional properties of the material.

Without these specifics, a more detailed calculation cannot be accurately provided. In practice, one would use the known properties of brass and the relationship between stress, strain, and Young's modulus (Y) of the material (Y = stress/strain) to find the required dimensions. Typically, this involves rearranging the formulas to solve for the radius, given that the area (A) can be expressed in terms of the radius (r) for a circle (A = πr²).

Answer:

use this to help www.code.org

Explanation:

this helped me alot

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: 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

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.