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ExtremeBDS [4]
3 years ago
9

Derive the next state equations for each type (D, T, SR, and JK) of basic memory element. The next state equation is a symbolic

equation describing the next state (Q ) as a function of the inputs (D,T,SR, or JK) and state (Q). In order to determine the next state equations for a a JK memory element, build a 3-variable Kmap with Q, J, and K as the inputs. The entries in the Kmap should be Q . Solving this Kmap will yield the next state equation. Show all work for full credit.
Engineering
1 answer:
bagirrra123 [75]3 years ago
6 0

Answer:

Attached below is the derived next state equations

Explanation:

Attached below is the derived next state equations

used for the solution of the given problem.

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Which of the following explains the difference between rangeland management specialists and conservation biologists?
tekilochka [14]

Answer:

b

Explanation:

8 0
3 years ago
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You will create three classes, the first two being Student and LineAtOfficeHour. The instances of the first class defines a sing
White raven [17]

Answer:

Complete solution is given below:

Explanation:

//student class

class Student{

  private String firstname,lastname;

 

  //constructor

  Student(String first,String last){

      this.firstname=first;

      this.lastname=last;

  }

 

  //getters and setters

  public String getFirstname() {

      return firstname;

  }

  public void setFirstname(String firstname) {

      this.firstname = firstname;

  }

  public String getLastname() {

      return lastname;

  }

  public void setLastname(String lastname) {

      this.lastname = lastname;

  }

  //function to get the fullname of student

  public String fullName() {

      return this.firstname+" "+this.lastname;

  }

}

//class for line at office hour

class LineAtOfficeHour{

 

  private Student line[];

  private int N=0;

  private int back=0;

 

  //empty constructor

  LineAtOfficeHour() {

      line=new Student[5];

  }

  //parameterized constructor

  LineAtOfficeHour(Student st[]) {

      int i=0;

      line=new Student[5];

      while(i<st.length && i<5) {

          line[i]=st[i];

          i++;

      }

      this.N=i;

      this.back=i-1;

  }

  //function to check if line is empty or not

  public boolean isEmpty() {

      if(this.N==0)

          return true;

      else

          return false;

  }

  //function to check if line is full

  public boolean isFull() {

      if(this.N==5) {

          return true;

      }else

          return false;

  }

  ///function to get the size of line

  public int size() {

      return this.N;

  }

 

  //function to add a student to the line

  public void enterLine(Student s) {

      if(isFull())

          System.out.println("Line is full!!!!");

      else {

          line[++back]=s;

          this.N++;

      }

  }

  public Student seeTeacher() {

      Student result=null;

      if(this.N>=0) {

          result=line[0];

          int i=0;

          for(i=1;i<N;i++) {

              line[i-1]=line[i];

          }

          line[i-1]=null;

          this.N--;

          this.back--;

      }

     

     

      return result;

  }

  //function to print students in line

  public String whosInLine() {

      String result ="";

      for(int i=0;i<this.N;i++) {

          result+=line[i].fullName()+",";

      }

      return result;

  }

}

//driver method

public class TestLine {

  public static void main(String[] args) {

      LineAtOfficeHour list=new LineAtOfficeHour();

     

      if(list.isEmpty()) {

          System.out.println("Line is empty!!!!!!!!!");

      }

     

      Student s1[]=new Student[3];

      s1[0]=new Student("John","Smith");

      s1[1]=new Student("Sam","Zung");

      s1[2]=new Student("Peter","Louis");

      list=new LineAtOfficeHour(s1);

     

      if(list.isEmpty()) {

          System.out.println("Line is empty!!!!!!!!!");

      }else {

          System.out.println("Line is not empty.........");

      }

     

      System.out.println("Students in line: "+list.whosInLine());

     

      System.out.println("Student removed: "+list.seeTeacher().fullName());

     

      System.out.println("Students in line: "+list.whosInLine());

  }

}

6 0
4 years ago
Chemical engineering got is unofficial start around the time of the __________ __________ ________.
Gre4nikov [31]

Answer:

Option A,  World War II

Explanation:

During the period of industrial revolution around 1915-25, the chemical engineering has taken a new shape. During this period (i.e around the world war I), there was rise in demand for  liquid fuels, synthetic fertilizer, and other chemical products. This lead to development of chemistry centre in Germany . There was rise in use of synthetics fibres and polymers. World war II saw the growth of catalytic cracking, fluidized beds, synthetic rubber, pharmaceuticals production, oil & oil products, etc. and because of rising chemical demand, chemical engineering took a new shape during this period

Hence, option A is the right answer

4 0
3 years ago
What is a motor cycle motor made out of
Natali5045456 [20]

Explanation:

a motorcycle Motor is made out of iron

4 0
4 years ago
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The elementary liquid-phase series reaction
liraira [26]

Answer:

Concentration of A: \frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }

Concentration of B: \frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1}  } (e^{-k_{1}t } -e^{-k_{2}t } )

Concentration of C: \frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1}  } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1}  } e^{-k_{1} t}

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }

Where x=CAo and y=CA

The mass balance for B:

-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1}  } (e^{-k_{1}t }-ex^{-k_{2}t }  )

The mass balance for C:

\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1}  } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1}  }  e^{-k_{1}t }

The maximum concentration of C is:

C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1}  }}  =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}

and the maximum time is:

t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1}  } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h

6 0
3 years ago
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