For the first option your program should then ask the user for which row and column in the array to replace, and what value will it be replaced with. For the second option your program should run through all values held in the 2D array and calculate their summation. For the third option your program should print out the contents of the 2D array one row at a time. When given the fourth option your program should simply exit.

Answer:

A) I attached the diagrams

B)W_net = 0.5434 KJ

C) η_th = 0.262

Explanation:

A) I've attached the P-v and T-s diagrams

B) The temperature at state 2 can be calculated from ideal gas equation at constant specific volume;

So; P2/P1 = T2/T1

Thus, T2 = P2•T1/P1

We are given that;

P2 = 380 KPa

P1 = 95 KPa

T1 = 17 °C = 17 + 273K = 290K

Thus,

T2 = (380 x 290)/95

T2 = 1160 K

While the temperature at state 3 will be gotten from;

T3 = T2 x (P3/P2)^((γ - 1)/γ)

Where γ = cp/cv = 1.005/0.718 = 1.4

Thus;

T3 = 1160 (95/380)^((1.4 - 1)/1.4)

T3 = 780.6 K

Now, net work done is given by the formula;

W_net = Q_in - Q_out

W_net = Q_1-2 - Q_3-1

W_net = m(u2 - u1) - m(h3 - h1)

W_net = m(u2 - u1 - h3 + h1)

From the first table i attached,

At T1 = 290K, u1 = 206.91 KJ/Kg and h1 = 290.16 KJ/Kg

At T2 = 1160K,u2 = 897.91 KJ/Kg

At T3 = 780K, h3 = 800.03 KJ/Kg

We are also given that m = 0.003 kg

Thus;

W_net = 0.003(897.91 - 206.91 - 800.03 + 290.16)

W_net = 0.5434 KJ

C) The thermal efficiency is given by the formula ;

η_th = W_net/Q_in

η_th = 0.5434/(m(u2 - u1))

η_th = 0.5434/(0.003(897.91 - 206.91))

η_th = 0.262

Answer:

The factor of safety is 1.186

Explanation:

Find the attachment

The calculation of normal and shearing stress on a welded pipe requires applying the principles of stress analysis involving internal pressure and axial forces. Specific formulas are used to determine hoop and longitudinal stresses in relation to the pipe's dimensions and applied loads.

The question pertains to the calculation of normal stress and shearing stress on a spirally welded pipe subject to internal pressure and axial forces. To find the normal stress perpendicular to the weld, one would consider the internal pressure exerted on the cross-section of the vessel and the area of the cross-section. The shearing stress parallel to the weld can be found by applying the principles of equilibrium and mechanics of materials, potentially involving the computation of forces along the spiral path of the weld.

To adequately address the question, we would need additional information, particularly regarding the location and orientation of the weld in relation to the applied forces. Without this, it is not possible to provide a detailed solution. However, such calculations would typically involve using the formulas for hoop stress ( extit{ extsigma_h = p  imes r / t}) and longitudinal stress ( extit{ extsigma_l = p  imes r / (2t)}), where p is the internal pressure, r is the radius, and t is the thickness of the pipe wall.

Answer:

Rate of heat transfer from plate

6796.16 W

Corresponding rate of change of plate temperature

-2634 degrees.Celcius/sec

Explanation:

In this question, we are asked to calculate the rate of heat transfer and the corresponding rate of change of the plate temperature.

Please check attachment for complete solution and step by step explanation

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

(a) Jn = 64.08 μA/m²

(b) Jp = 80.1 μA/m²

(c) Jp = 22.94 μA/m²

(d) Jp = 3.6357 пA/m²

(e) Jn = 22.63 μA/m²

Explanation:

See the attached file for the explanation.

Answer:

(a) The velocity of the exhaust gases. is 832.7 m/s

(b) The rate of fuel consumption is 0.6243 kg/s

Explanation:

For the given turbojet engine operating on an ideal cycle, the pressure ,temperature, velocity, and specific enthalpy of air at [tex]i^{th}[/tex] state are [tex]P_i[/tex] , [tex]T_i[/tex] , [tex]V_i[/tex] , and [tex]h_i[/tex] , respectively.

Use "ideal-gas specific heats of various common gases" to find the properties of air at room temperature.

Specific heat at constant pressure, [tex]c_p[/tex] = 1.005 kJ/kg.K

Specific heat ratio, k = 1.4

Answer:

Explanation:

See the attachment for a step by step solution to the question.

Answer:

determine the pipe diameter.  = 0.41ft

Explanation:

check the attached file for answer explanation

Answer: you need to do it on your own first

Explanation:

Answer:

See explaination

Explanation:

Code;

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void printNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

printNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

printNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

printNumPattern(num1, num2);

}

}

See attachment for sample output

A recursive method in Python to output the following number pattern:

def printPattern(n, m):

 # Base case: if n is 0 or negative, return

 if n <= 0:

   return

 # Print the current number

 print(n)

 # Recursively call the function with n subtracted by m

 printPattern(n - m, m)

 # Recursively call the function with n added by m

 printPattern(n + m, m)

# Example usage:

printPattern(12, 3)

Output:

12

9

12

15

12

...

The method works by recursively calling itself twice, once with n subtracted by m and once with n added by m. This process continues until n reaches 0 or a negative value. At that point, the base case is reached and the function returns.

The following is a breakdown of the recursive calls for the example input of 12 and 3:

printPattern(12, 3)

 printPattern(9, 3)

   printPattern(6, 3)

     printPattern(3, 3)

       printPattern(0, 3)

         # Base case reached, return

       printPattern(3, 3)

     printPattern(6, 3)

   printPattern(9, 3)

 printPattern(15, 3)

   printPattern(12, 3)

The output of the method is the sequence of numbers that are printed in the recursive calls.

For such more question on Python

https://brainly.com/question/29563545

#SPJ3

Answer:

The total compressor work is 234.8 kJ/kg for a isentropic compression

Explanation:

Please look at the solution in the attached Word file

The total compressor work per unit of mass flow in a two-stage compressor with intercooling can be calculated by summing the compressor work in each stage and the work done in the intercooler.

The total compressor work per unit of mass flow in a two-stage compressor with intercooling can be calculated by summing the compressor work in each stage and the work done in the intercooler.

In the first stage, the air is compressed from 100 kPa to 300 kPa. The work done in this stage can be calculated using the isentropic compression process. In the second stage, the air is further compressed from 300 kPa to 1000 kPa. The work done in this stage can also be calculated using the isentropic compression process.

To calculate the total compressor work per unit of mass flow, you need to sum the work done in each stage and the work done in the intercooler.

Answer:

Explanation:

Find attach the solution

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

[tex]\eta_{th} = 30\,\%[/tex], [tex]\eta_{th,max} = 40\,\%[/tex], [tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex], The cycle is irreversible.

Explanation:

The real cycle efficiency is:

[tex]\eta_{th} = \frac{1000\,kJ-700\,kJ}{1000\,kJ} \times 100\,\%[/tex]

[tex]\eta_{th} = 30\,\%[/tex]

The theoretical cycle efficiency is:

[tex]\eta_{th,max} = \frac{500\,K-300\,K}{500\,K} \times 100\,\%[/tex]

[tex]\eta_{th,max} = 40\,\%[/tex]

The reversible and real versions of the power cycle are described by the Clausius Inequalty:

Reversible Unit

[tex]\frac{1000\,kJ - 600kJ}{300\,K}= 0[/tex]

Real Unit

[tex]\Delta S = \frac{1000\,kJ-600\,kJ}{300\,K} -\frac{1000\,kJ-700\,kJ}{300\,K}[/tex]

[tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex]

The cycle is irreversible.

The cycle efficiency using clausius inequality is;

σ_cycle = 0.333 kJ/kg and is internally irreversible

η = 1 - Q_c/Q_h

Thus;

Q_c = (1 - η)Q_h

σ_cycle = -∫(dQ/dt)ₐ

Thus; σ_cycle = -[(Q_h/T_h) - (Q_c/T_c)]

We are given;

Q_h = 1000 kJ

T_h = 500 k

T_c = 300 k

Q_c = 700 kJ

σ_cycle = -[(1000/500) - (700/300)]

σ_cycle = -(2 - 2.333)

σ_cycle = 0.333 kJ/kg

Since σ_cycle > 0, then the cycle is internally irreversible

Read more about cycle efficiency at; https://brainly.com/question/16014998

Answer:

decreased

Explanation:

when impaired you react slower then you would sober.

Alcohol adversely affects the complex reaction time, considerably decreasing the individual’s ability to respond swiftly and adequately in emergencies or unexpected situations. This impairment is attributed to alcohol's impact on the brain causing slow information processing, poor motor control, and a decrease in focus.

When a person's complex reaction time is compromised by alcohol, their ability to respond to unexpected situations or emergencies is greatly diminished. Alcohol's impact on the brain leads to slower processing of information, reduced concentration, and poorer motor control. As a result, they may not react as quickly or efficiently as they would if they were sober to changes in their environment. For example, if a situation arises that requires quick decision-making, such as stopping abruptly while driving to avoid a pedestrian, an intoxicated individual may not respond in time, leading to catastrophic outcomes.

Learn more about Alcohol impact on reaction time here:

https://brainly.com/question/1002220

#SPJ2

Answer:

See Explaination

Explanation:

# copy the function you have this is just for my convenniece

def finalGrade(scoresList):

weights = [0.05, 0.05, 0.40, 0.50]

grade = 0

for i in range(len(scoresList)):

grade += float(scoresList[i]) * weights[i]

return grade

import csv

def main():

with open('something.csv', newline='') as csvfile:

spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')

file = open('something.txt')

for row in spamreader:

student = file.readline().strip()

scores = row

print(student, finalGrade(scores))

if __name__ == "__main__":

main()

Answer:

a) y (n + 1) = 1.005 y(n) + 5U n

y (n + 1) - 1.005 y(n) = 5U (n)

b) Z^-1(Z(y0)=y(n) = [1010(1.005)^n - 1000(1)^n] U(n)

c) h(n) = (1.005)^n U(n - 1) + 10(1.005)^n U(n)

Explanation:

Her bank account can be modeled as:

y (n + 1) = y (n) + 0.5% y(n) + $5

y (n + 1) = 1.005 y(n) + 5U n

Given that y (0) = $10

y (n + 1) - 1.005 y(n) = 5U (n)

Apply Z transform on both sides

= ZY ((Z) - Z(y0) - 1.005) Z = 5 U (Z)

U(Z) = Z {U(n)} = Z/ Z - 1

Y(Z) [Z- 1.005] = Z y(0) + 5Z/ Z - 1

= 10Z/ Z - 1.005 + 5Z/(Z - 1) (Z - 1.005)

Y(Z) = 10Z/ Z - 1.005 + 1000Z/ Z - 1.005 + 1000Z/ Z - 1

= 1010Z/Z- 1.005 - 1000Z/Z-1

Apply inverse Z transform

Z^-1(Z(y0)) = y(n) = [1010(1.005)^n - 1000(1)^n] U(n)

Impulse response in output when input f(n) = S(n)

That is,

y(n + 1)= 1. 005y (n) + 8n

y(n + 1) - 1.005y (n) = 8n

Apply Z transform

ZY (Z) - Z(y0) - 1.005y(Z) = 1

HZ (Z - 1.005) = 1 + 10Z [Therefore y(Z) = H(Z)]

H(Z) = 1/ Z - 1.005 + 10Z/Z - 1. 005

Apply inverse laplace transform

= h(n) = (1.005)^n U(n - 1) + 10(1.005)^n U(n)

To have a 99.9% chance that the component does not fail before 1,000,000 cycles, the maximum stress level (σmax) should be set to 17.15 ksi or lower.

Given:

- Material: 2024-T4 aluminum

- Ultimate tensile strength: 70 ksi

- Minimum stress level: 0 ksi

- Desired reliability: 99.9% for 1,000,000 cycles

Step 1: Determine the endurance limit (σe) for the material.

For aluminum alloys, the endurance limit is typically taken as the stress level at which the S-N curve (stress vs. number of cycles to failure) becomes horizontal.

A common approximation for the endurance limit of aluminum alloys is:

σe ≈ 0.35 × Ultimate tensile strength

σe ≈ 0.35 × 70 ksi = 24.5 ksi

Step 2: Adjust the endurance limit for the desired reliability.

The endurance limit is typically determined for a reliability of 50%. To achieve a higher reliability, we need to apply a correction factor.

For a reliability of 99.9%, the correction factor is approximately 0.7.

Adjusted endurance limit = σe × 0.7 = 24.5 ksi × 0.7 = 17.15 ksi

Step 3: Determine the maximum stress level (σmax) based on the adjusted endurance limit.

For a fully reversed stress cycle (minimum stress = 0), the maximum stress level should not exceed the adjusted endurance limit.

σmax ≤ 17.15 ksi

Therefore, to have a 99.9% chance that the component does not fail before 1,000,000 cycles, the maximum stress level (σmax) should be set to 17.15 ksi or lower.

Answer:

resistor.h

//circuit class template

#ifndef TEST_H

#define TEST_H

#include<iostream>

#include<string>

#include<vector>

#include<cstdlib>

#include<ctime>

#include<cmath>

using namespace std;

//Node for a resistor

struct node {

  string name;

  double resistance;

  double voltage_across;

  double power_across;

};

//Create a class Ohms

class Ohms {

//Attributes of class

private:

  vector<node> resistors;

  double voltage;

  double current;

//Member functions

public:

  //Default constructor

  Ohms();

  //Parameterized constructor

  Ohms(double);

  //Mutator for volatage

  void setVoltage(double);

  //Set a resistance

  bool setOneResistance(string, double);

  //Accessor for voltage

  double getVoltage();

  //Accessor for current

  double getCurrent();

  //Accessor for a resistor

  vector<node> getNode();

  //Sum of resistance

  double sumResist();

  //Calculate current

  bool calcCurrent();

  //Calculate voltage across

  bool calcVoltageAcross();

  //Calculate power across

  bool calcPowerAcross();

  //Calculate total power

  double calcTotalPower();

  //Display total

  void displayTotal();

  //Series connect check

  bool seriesConnect(Ohms);

  //Series connect check

  bool seriesConnect(vector<Ohms>&);

  //Overload operator

  bool operator<(Ohms);

};

#endif // !TEST_H

resistor.cpp

//Implementation of resistor.h

#include "resistor.h"

//Default constructor,set voltage 0

Ohms::Ohms() {

  voltage = 0;

}

//Parameterized constructor, set voltage as passed voltage

Ohms::Ohms(double volt) {

  voltage = volt;

}

//Mutator for volatage,set voltage as passed voltage

void Ohms::setVoltage(double volt) {

  voltage = volt;

}

//Set a resistance

bool Ohms::setOneResistance(string name, double resistance) {

  if (resistance <= 0){

      return false;

  }

  node n;

  n.name = name;

  n.resistance = resistance;

  resistors.push_back(n);

  return true;

}

//Accessor for voltage

double Ohms::getVoltage() {

  return voltage;

}

//Accessor for current

double Ohms::getCurrent() {

  return current;

}

//Accessor for a resistor

vector<node> Ohms::getNode() {

  return resistors;

}

//Sum of resistance

double Ohms::sumResist() {

  double total = 0;

  for (int i = 0; i < resistors.size(); i++) {

      total += resistors[i].resistance;

  }

  return total;

}

//Calculate current

bool Ohms::calcCurrent() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  current = voltage / sumResist();

  return true;

}

//Calculate voltage across

bool Ohms::calcVoltageAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double voltAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      voltAcross += resistors[i].voltage_across;

  }

  return true;

}

//Calculate power across

bool Ohms::calcPowerAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double powerAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      powerAcross += resistors[i].power_across;

  }

  return true;

}

//Calculate total power

double Ohms::calcTotalPower() {

  calcCurrent();

  return voltage * current;

}

//Display total

void Ohms::displayTotal() {

  for (int i = 0; i < resistors.size(); i++) {

      cout << "ResistorName: " << resistors[i].name << ", Resistance: " << resistors[i].resistance

          << ", Voltage_Across: " << resistors[i].voltage_across << ", Power_Across: " << resistors[i].power_across << endl;

  }

}

//Series connect check

bool Ohms::seriesConnect(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  vector<node> temp = ohms.getNode();

  for (int i = 0; i < temp.size(); i++) {

      this->resistors.push_back(temp[i]);

  }

  return true;

}

//Series connect check

bool Ohms::seriesConnect(vector<Ohms>&ohms) {

  if (ohms.size() == 0) {

      return false;

  }

  for (int i = 0; i < ohms.size(); i++) {

      this->seriesConnect(ohms[i]);

  }

  return true;

}

//Overload operator

bool Ohms::operator<(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  if (this->sumResist() < ohms.sumResist()) {

      return true;

  }

  return false;

}

main.cpp

#include "resistor.h"

int main()

{

   //Set circuit voltage

  Ohms ckt1(100);

  //Loop to set resistors in circuit

  int i = 0;

  string name;

  double resistance;

  while (i < 3) {

      cout << "Enter resistor name: ";

      cin >> name;

      cout << "Enter resistance of circuit: ";

      cin >> resistance;

      //Set one resistance

      ckt1.setOneResistance(name, resistance);

      cin.ignore();

      i++;

  }

  //calculate totalpower and power consumption

  cout << "Total power consumption = " << ckt1.calcTotalPower() << endl;

  return 0;

}

Output

Enter resistor name: R1

Enter resistance of circuit: 2.5

Enter resistor name: R2

Enter resistance of circuit: 1.6

Enter resistor name: R3

Enter resistance of circuit: 1.2

Total power consumption = 1886.79

Explanation:

Note

Please add all member function details.Its difficult to figure out what each function meant to be.

Answer:

(c) Generator, 95.5 %

Explanation:

given data

voltage va = 120V

current Ia = - 5.5 A

Tmech =  5.5Nm

ns = 1200 RPM

solution

first we get here electric input power that is express as

electric input power = va × Ia    ......1

put here value and we get

electric input power = 120 × -5-5

electric input power -660 W

here negative mean it generate power

and here

Pin ( mech) will be

Pin ( mech) = Tl × ω   .........2

Pin ( mech) = 5.5 × [tex]\frac{2\pi N}{60}[/tex]    

Pin ( mech) =  5.5 × [tex]\frac{2\pi 1200}{60}[/tex]  

Pin ( mech) = 691.150 W

and

efficiency will be here as

efficiency = [tex]\frac{660}{691.150}[/tex]    

efficiency = 95.5 %

so correct option is (c) Generator, 95.5 %

Answer:

h_f = 15 ft, so option A is correct

Explanation:

The formula for head loss is given by;

h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))

Where;

h_f is head loss due to friction in ft

L is length of pipe in ft

Q is flow rate of water in gpm

C is hazen Williams constant

D is diameter of pipe in inches

We are given;

L = 1,800 ft

Q = 600 gpm

C = 120

D = 8 inches

So, plugging in these values into the equation, we have;

h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))

h_f = 14.896 ft.

So, h_f is approximately 15 ft

Answer:

a. BandStop filter

Explanation:

A band-stop filter or band-rejection filter is a filter that eliminates at its output all the signals that have a frequency between a lower cutoff frequency and a higher cutoff frequency. They can be implemented in various ways.  One of them is to implement a notch filter, which is characterized by rejecting a certain frequency that is interfering with a circuit. The transfer function of this filter is given by:

[tex]H(s)=\frac{s^2+\omega_o^2}{s^2+\omega_cs+\omega_o^2} \\\\Where:\\\\\omega_o=Central\hspace{3} rejected\hspace{3} frequency\\\omega_c=Width\hspace{3} of\hspace{3} the\hspace{3} rejected\hspace{3} band[/tex]

I attached you a graph in which you can see how the filter works.

Answer:

1.12kw is the heat transfer

Explanation:

Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force.

Potential energy, stored energy that depends upon the relative position of various parts of a system.

See attachment for the step by step solution.

The largest load P that can be applied to the steel bar, ensuring a factor of safety of 2.0 with respect to the weld material's tensile yield strength, is 32.5 kN. This is calculated based on the allowable tensile strength of the weld after applying the factor of safety and the cross-sectional area of the steel bar.

The student's question relates to the structural engineering concept of material strength and loading. The maximum load P that can be safely applied to a welded steel bar can be determined by considering the tensile and shear yield strengths of the two materials involved, i.e., the weld material and the bar material. Using the given factor of safety of 2.0, the tensile strength of the weld and the bar, and the cross section of the bar, we can calculate the allowable tensile stress and then the maximum load P by dividing the allowable tensile stress by the area of the cross section.

To begin with, we find the allowable tensile strength by dividing the weld material's yield strength (which is the weaker material concerning tensile strength) by the factor of safety:

Next, we need to calculate the area of the cross-section of the bar:

Now, we convert the units of the allowable tensile strength to N/m² (because 1 MPa = 1 x 10⁶ N/m²) and calculate the maximum tensile load (Ptensile) that the weld can sustain:

Thus, the largest load P that can be applied to the steel bar to satisfy the safety criteria is 32.5 kN.

Answer:

The static thrust of the engine is 20,856.44N

Explanation:

In this question, we are asked to calculate the static thrust of the turbojet.

Please check attachment for complete solution and step by step explanation

Answer:

Final velocities are:

Wedge B: v = 2.334 m/s

Sphere A: v = 5.386 m/s

Explanation:

Given:-

- The mass of sphere A,  mA = 2-kg

- The mass of the wedge B,  mB = 6-kg

- The sphere collides with " normal " to the wedge face.

- The coefficient of restitution , e = 0.5

- The wedge inclination angle, θ=40 degrees with the horizontal.

- The initial speed of sphere A, vA = 4m/s

- The initial speed of wedge B, vB = 0 m/s ... ( rest )

Find:-

- First step would be to sketch a system of sphere A and wedge B as ( FBD ).

- We will add a sketch of "two" coordinate axes on the ( FBD ).

   First coordinate system ( normal ( n ) - tangent ( t ) )

    Second coordinate system ( horizontal ( x ) - vertical ( y ) )

Note:- All the above directions of coordinate axes denote the positive direction of vectors.

- Resolve the angle ( α ) between the normal - ( n ) or velocity vector vA and the horizontal - ( x ) axis.

                            α = 90° - θ = 90° - 40°

                            α = 50°  

- We will denote the final velocity components of sphere A as ( v'An , v'At ):

And, final velocity components of wedge B as ( v'B ).

- Transform the the final velocity of wedge ( vB' ) in the (n-t) coordinate axis using the resolved coordinate transformation angle ( α ). The normal  ( n ) and tangential components of the velocity vector ( vB' ) are: ( vB'n , vB't ).

                     vB'n = vB'*cos ( α )             vB't = vB'*sin ( α° )

                    vB'n = vB'*cos ( 50° )         vB't = vB'*sin ( 50° )

- Note: There is no y-component of velocity of wedge. This is because the motion of wedge in the vertical direction is restricted by the ground - equilibrium conditions. Hence, v'By = 0.

- Consider the system of sphere A and wedge B to be isolated with no external forces acting on the system. For such conditions the principle of conservation of momentum ( P ) is valid. Which states:

                        PA,i + PBi = PA,f + PB,f

Where,

           PA,i : The initial momentum of the sphere A

           PB,i : The initial momentum of wedge B

           PA,f : The final momentum of the sphere A

           PB,f : The final momentum of wedge B.

- Using the data given and relations computed in the ( n-t ) coordinate system. We will use the principle of conservation of linear momentum in both axis ( n and t ) combined in vector momentum of system.

Conservation of linear momentum:

                      [tex]m_A*v_A + m_B*v_B = m_A*v_A' + m_B*v_B'[/tex]

- Using vector notations we have, Taking ( i unit vector in tangential direction and j unit vector in the normal direction ):

                       [tex]m_A*v_A_n j = m_A*( v_A'_t i + v_A'_n j )+ m_B* ( v_B'_n j + v_B'_t i )\\\\2*4 j = 2*( v_A'_t i + v_A'_n j )+ 6* ( v_B'_n j + v_B'_t i )\\\\0 i + 8 j = ( 2*v_A'_t + 6*v_B'*sin ( 50 ) ) i + ( 2*v_A'_n + 6*v_B'* cos (50 ) ) j[/tex]

- Equate all the ( i - j ) vectors on left and right hand side of the equation respectively,

               [tex]0 = 2*v_A'_t + 6*v_B'*sin ( 50 )\\\\\\ 8 = 2*v_A'_n + 6*v_B'* cos(50 )[/tex]         .... Eq 1

- The coefficient of restitution ( e ) is a squared loss of kinetic energy of the system that can be expressed in terms of velocities of two objects as relative change in velocity of two objects after and before impact.:

                             [tex]e = \frac{v_B'_n - v_A'_n}{v_A_n - v_B_n}[/tex]

Note: We have only used the velocities normal to the surface of wedge. This is because the kinetic loss is a scalar dimension; hence, the normal direction to the surface of impact is assumed to cater all the loss in kinetic energy.

We have,

                            [tex]0.5 = \frac{v_B'*cos ( 50 ) - v_A'_n}{ 4 - 0}\\\\2 = v_B'*cos ( 50 ) - v_A'_n \\\\v_B'*cos ( 50 ) = v_A'_n + 2[/tex]       ..... Eq 2

- Now substitute equation 2 into equation 1:

                             [tex]8 = 2*v_A'_n + 6*( v_A'_n + 2 )\\\\-4 = 8*v_A'_n\\\\v_A'_n = -0.5 \frac{m}{s}[/tex]

- Using Equation 2 compute v'B:

                             [tex]v_B'*cos ( 50 ) = -0.5 + 2\\\\v_B'= \frac{1.5}{cos ( 50 ) } \\\\v_B'= 2.334 \frac{m}{s}[/tex]

- Using Equation 1 compute v'At:

                             [tex]0 = 2*v_A'_t + 6*2.334*sin ( 50 )\\\\v_A'_t = - \frac{6*2.334*sin ( 50 )}{2} \\\\v_A'_t = - 5.363 \frac{m}{s}[/tex]

- The final velocity of wedge B along the horizontal direction ( x ) is:

                          vB' = 2.334 m/s   .... x-direction

- The velocity of sphere A after impact is given by:

                          [tex]v_A' = \sqrt{v_A't^2 + v_A'n^2}\\\\v_A' = \sqrt{5.363^2 + 0.5^2}\\\\v_A' = \sqrt{29.011769}\\\\v_A' = 5.386 \frac{m}{s}[/tex]

Answer:

A) τ_max = 59.139 x 10^(6) Pa

B) θ = 0.0228 rad.

Explanation:

A) In the left half of the shaft we have 25 hp which corresponds to a torque T1 given by;

P = Tω

Where P is power and ω is angular speed.

Power = 25 HP = 25 x 746 W = 18650W

ω = 200 rev/min = 200 x 0.10472 rad/s = 20.944 rad/s

P = T1•ω

T1 = P/ω = 18650/20.944

T1 = 890.47 N.m

Similarly, in the right half we have 40 hp corresponding to a torque T2

given by;

P = T2•ω

T2 = P/ω

Where P = 40 x 760 = 30,400W

T2 = 30400/20.944 = 1451.49 N.m

The maximum shearing stress consequently occurs in the outer fibers in the right half and is given by;

τ_max = Tρ/J

Where J is polar moment of inertia and has the formula ;J = πd⁴/32

d = 5cm = 0.05m

J = π(0.05)⁴/32 = 6.136 x 10^(-7) m⁴

ρ = 0.05/2 = 0.025m

T will be T2 = 1451.49 N.m

Thus,

τ_max = Tρ/J

τ_max = 1451.49 x 0.025/6.136 x 10^(-7)

τ_max = 59139022.94 N/m² = 59.139 x 10^(6) Pa

B) The angles of twist of the left and right ends relative to the center are, respectively, using θ = TL/GJ

G = 80 Gpa = 80 x 10^(9) Pa

θ1 = (890.47 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0363 rad

Similarly;

θ2 = (1451.49 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0591 rad

Since θ1 and θ2 are in the same direction, the relative angle of twist between the two ends of the shaft is

θ = θ2 – θ1

θ = 0.0591 - 0.0363

θ = 0.0228 rad.

Answer:

The flux into the tissue at x=0 and t=5 s is 0.4476 uM/cm²s

Explanation:

Flux = [tex]\frac{Quantity}{Area * Time}[/tex]

given:

Area = 4.2 cm²

Time = 5 sec

Quantity ( concentration) = 9.4 uM

∴ Flux = Quantity / (Area × Time)

Flux = 9.4 uM / (4.2 cm² × 5 s)

Flux = 9.4 uM / 21 cm²s

Flux = 0.4476 uM/cm²s

Answer:

Explanation:

Given that,

Mass of boron fiber in unidirectional orientation

Mb = 5kg = 5000g

Mass of aluminum fiber in unidirectional orientation

Ma = 8kg = 8000g

A. Density of the composite

Applying rule of mixing

ρc = 1•ρ1 + 2•ρ2

Where

ρc = density of composite

1 = Volume fraction of Boron

ρ1 = density composite of Boron

2 = Volume fraction of Aluminum

ρ2 = density composite of Aluminum

ρ1 = 2.36 g/cm³ constant

ρ2 = 2.7 g/cm³ constant

To Calculate fractional volume of Boron

1 = Vb / ( Vb + Va)

Vb = Volume of boron

Va = Volume of aluminium

Also

To Calculate fraction volume of aluminum

2= Va / ( Vb + Va)

So, we need to get Va and Vb

From density formula

density = mass / Volume

ρ1 = Mb / Vb

Vb = Mb / ρ1

Vb = 5000 / 2.36

Vb = 2118.64 cm³

Also ρ2 = Ma / Va

Va = Ma / ρ2

Va = 8000 / 2.7

Va = 2962.96 cm³

So,

1 = Vb / ( Vb + Va)

1 = 2118.64 / ( 2118.64 + 2962.96)

1 = 0.417

Also,

2= Va / ( Vb + Va)

2 = 2962.96 / ( 2118.64 + 2962.96)

2 = 0.583

Then, we have all the data needed

ρc = 1•ρ1 + 2•ρ2

ρc = 0.417 × 2.36 + 0.583 × 2.7

ρc = 2.56 g/cm³

The density of the composite is 2.56g/cm³

B. Modulus of elasticity parallel to the fibers

Modulus of elasticity is defined at the ratio of shear stress to shear strain

The relation for modulus of elasticity is given as

Ec = = 1•Eb+ 2•Ea

Ea = Elasticity of aluminium

Eb = Elasticity of Boron

Ec = Modulus of elasticity parallel to the fiber

Where modulus of elastic of aluminum is

Ea = 69 × 10³ MPa

Modulus of elastic of boron is

Eb = 450 × 10³ Mpa

Then,

Ec = = 1•Eb+ 2•Ea

Ec = 0.417 × 450 × 10³ + 0.583 × 69 × 10³

Ec = 227.877 × 10³ MPa

Ec ≈ 228 × 10³ MPa

The Modulus of elasticity parallel to the fiber is 227.877 × 10³MPa

OR Ec = 227.877 GPa

Ec ≈ 228GPa

C. modulus of elasticity perpendicular to the fibers?

The relation of modulus of elasticity perpendicular to the fibers is

1 / Ec = 1 / Eb+ 2 / Ea

1 / Ec = 0.417 / 450 × 10³ + 0.583 / 69 × 10³

1 / Ec = 9.267 × 10^-7 + 8.449 ×10^-6

1 / Ec = 9.376 × 10^-6

Taking reciprocal

Ec = 106.66 × 10^3 Mpa

Ec ≈ 107 × 10^3 MPa

Note that the unit of Modulus has been in MPa,

Answer:

Number of Trays = Six (6)

Explanation:

Given that: y' = 25x' , in terms of molecular ratio, we can write it as

[tex]\frac{Y'}{1 + Y'} =25 \frac{X'}{1 + X'}[/tex]  ......... 1

after plotting this we get equilibrium curve as shown in the attached picture.

inlet concentration and outlet concentration of liquid phase is

x₂ = 4% = 0.04 (inlet)

so that can be converted into molar

[tex]X_2 = \frac{x_2}{1-x_2} = \frac{0.04}{1-0.04} = 0.04167[/tex]

and

x₁ = 0.2% = 0.002

[tex]X_1 = \frac{x_1}{1-x_1} = \frac{0.002}{1-0.002} = 2.004*10^{-3}[/tex]

Now we have to use the balance equation a

[tex]\frac{G_s}{L_s} = \frac{X_2-X_1}{Y_2-Y_1}[/tex] .............. a

here amount of solute is comparably lower than

Here we have

L = 300 kmol (total)

[tex]L_s[/tex] = 300(1 - 0.04) = 288 kmol pure oil

G = [tex]G_s[/tex] = 11.42 kmol

[tex]Y_1[/tex] = 0 , solvent free steam

substitute into the equation a

[tex]\frac{11.42}{288} = \frac{0.04167 - 2*10^{-3}}{Y_2 - 0}[/tex]

Y₂ = 1.0003

Now plot the point A(X₁ , Y₁) and B(X₂ , Y₂) and join them to construct operating line AB.

Starting from point B, stretch horizontal line up to equilibrium curve and from there again go down to operating line as shown in the picture attached. This procedure give one count of tray and continue the same procedure up to end of operating.

at last count, the number of stage, gives 6.

∴ Number of trays = 6

The number of theoretical trays needed is 20.

From the given data,

Equilibrium relation

y = 25x

[tex]L_s=L_2(1-x_2)\\L_s=300(1-0.04)= 288Kmol[/tex]

[tex]G_sy_1+L_sx_2=G_sy_2+L_sx_1\\G_s(y_1-y_2)=L_s(x_1-x_2)\\x_1=\frac{0.002}{1-0.002}\\x_1=0.002\\y_1=0, y_2=?\\x_2=\frac{0.04}{1-0.04}; x_2=0.0417[/tex]

substituting the values into the equation;

[tex]11.42(0-y_2)=288(0.002-0.0417)\\0-11.42y_2=-11.4048\\y_2=0.998[/tex]

[tex]N=\frac{In[(\frac{(x_2-y_1)/m}{(x_1-y_1)/m}(1-A)+A }{In(1/A)}\\[/tex]

But [tex]A=\frac{L_s}{mGs}[/tex]

[tex]N=\frac{x_2-x_1}{(x_1-y_1)/m}=\frac{0.0417-0.002}{0.002}\\N=19.85[/tex]

The numbers of tray is approximately 20.

learn more about steam stripping and numbers of tray here

https://brainly.com/question/9349349