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KengaRu [80]
4 years ago
15

For the following three (3) questions, assume that the link layer uses the flag bytes with byte stuffing approach for framing, w

ith the following details:
The beginning of a data frame is indicated by the special flag bytes: DLESTX
The end of a data frame is indicated by the special flag bytes: DLEETX
The special escape bytes DLE are used for escaping accidental occurrences of either the flag bytes or the escape bytes within the data.
For simplicity, assume that no other header/trailer information is added to the data.
The following byte streams represent data that needs to be framed by the link layer on the sender’s side. Derive the resultant byte streams by adding necessary flag bytes and performing byte stuffing.
1. PZDLEAFRGSTXGYK
2. PARSDLEETXZKPUM
3. ASTXDLELEFTDAOY
Engineering
1 answer:
pav-90 [236]4 years ago
3 0

I know it's either 1 2 or 3..

You might be interested in
When you are climbing a hill on a bike with different speeds, why does it seem like you are pedaling a lot to cover a short dist
ololo11 [35]

It seems like your pedaling a lot because it takes more energy and is way slower than a regular road. It you switch to the higher gear it will make you go slower because it is made for going down hill or going high speeds and a lower gear will help you more cause its easier and it will make it go faster.

5 0
3 years ago
The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along
JulsSmile [24]

Answer:

D=41.48 ft

a=54.43\ ft/s^2

Explanation:

Given that

y=0.5 x²                      

Vx= 2 t

We know that

V_x=\dfrac{dx}{dt}

At t= 0 ,x=0  

x=\int V_x.dt

At t= 3 s

x=\int_{0}^{3} 2t.dt

x=[t^2\left\right ]_0^3

x= 9 ft

When x= 9 ft then

y= 0.5 x 9²  ft

y= 40.5 ft

So distance from origin is

x= 9 ft ,y= 40.5 ft

D=\sqrt{9^2+40.5^2} \ ft

D=41.48 ft

a_x=\dfrac{dV_x}{dt}

Vx= 2 t

a_x= 2\ ft/s^2

At t= 3 s , x= 9 ft

y=0.5 x²    

a_y=\dfrac{d^2y}{dt^2}

y=0.5 x²    

\dfrac{dy}{dt}=x\dfrac{dx}{dt}

\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}

Given that

\dfrac{dx}{dt}=2t

\dfrac{dx}{dt}=2\times 3

\dfrac{dx}{dt}=6\ ft/s

a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2

a_y=54\ ft/s^2

a=\sqrt{a_x^2+a_y^2}\ ft/s^2

a=\sqrt{2^2+54^2}\ ft/s^2

a=54.43\ ft/s^2

7 0
3 years ago
For a project in C++ we are supposed toDesign a class named Month. The class should have the following private members:-name: a
mart [117]

Answer:

include <iostream>

using namespace std;

 

class Month

{

public:

 Month (char firstLetter, char secondLetter, char thirdLetter);

 

 Month (int monthNum);

.

 

 Month();

 void outputMonth_num();

 

 

 void outputMonthLetters();

private:

 int month;

};

 

 

int main ()

{

 //

 // Variable declarations

 //

 int monthNum;

 char firstLetter, secondLetter, thirdLetter;    

 char testAgain;              

 

 do {

 

   cout << endl;

   cout << "Testing the default constructor ..." << endl;

   Month defaultMonth;

   defaultMonth.outputMonth_num();

   defaultMonth.outputMonthLetters();

 

   //

   // Construct a month using the constructor with one integer argument

   //

   cout << endl;

   cout << "Testing the constructor with one integer argument..." << endl;

   cout << "Enter a month number: ";

   cin >> monthNum;

 

   Month testMonth1(monthNum);

   testMonth1.outputMonth_num();

   testMonth1.outputMonthLetters();

 

   //

   // Construct a month using the constructor with three letters as arguments

   //

   cout << endl;

   cout << "Testing the constructor with 3 letters as arguments ..." << endl;

   cout << "Enter the first three letters of a month (lowercase): ";

   cin >> firstLetter >> secondLetter >> thirdLetter;

   cout << endl;

 

   Month testMonth2(firstLetter, secondLetter, thirdLetter);

   testMonth2.outputMonth_num();

   testMonth2.outputMonthLetters();

 

   //

   // See if user wants to try another month

   //

   cout << endl;

   cout << "Do you want to test again? (y or n) ";

   cin >> testAgain;

 }

 while (testAgain == 'y' || testAgain == 'Y');

 

 return 0;

}

 

 

Month::Month(char firstLetter, char secondLetter, char thirdLetter)

{

if ((firstLetter == 'j')&&(secondLetter == 'a')&&(thirdLetter == 'n'))

  outputMonth_num = 1;

if ((firstLetter == 'f')&&(secondLetter == 'e')&&(thirdLetter == 'b'))

  outputMonth_num = 2;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'r'))

  outputMonth_num = 3;

if ((firstLetter = 'a')&&(secondLetter == 'p')&&(thirdLetter == 'r'))

  outputMonth_num = 4;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'y'))

  outputMonth_num = 5;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(thirdLetter == 'n'))

  outputMonth_num = 6;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(.thirdLetter == 'l'))

  outputMonth_num = 7;

if ((firstLetter == 'a')&&(secondLetter == 'u')&&(thirdLetter == 'g'))

  outputMonth_num = 8;

if ((firstLetter == 's')&&(secondLetter == 'e')&&(thirdLetter == 'p'))

  outputMonth_num = 9;

if ((firstLetter == 'o')&&(secondLetter == 'c')&&(thirdLetter == 't'))

  outputMonth_num = 10;

if ((firstLetter == 'n')&&(secondLetter == 'o')&&(thirdLetter == 'v'))

 outputMonth_num = 11;

if ((firstLetter == 'd')&&(secondLetter == 'e')&&(thirdLetter == 'c'))

 outputMonth_num = 12;

}

 

Month::inputMonthByNumber

{

if (Month_num > 12 && Month_num < 1)

cout << "Invalid number for Month, please choose 1-12)\n";

}

 

void Month::outputMonth_num()

{

 if (month >= 1 && month <= 12)

   cout ><< "Month: " << month << endl;

 else

   cout << "Error - The month is not a valid!" << endl;

}

 

void Month::outputMonthLetters()

{

 switch (month)

   {

   case 1:

     cout << "Jan" << endl;

     break;

   case 2:

     cout << "Feb" << endl;

     break;

   case 3:

     cout << "Mar" << endl;

     break;

   case 4:

     cout << "Apr" << endl;

     break;

   case 5:

     cout << "May" << endl;

     break;

   case 6:

     cout << "Jun" << endl;

     break;

   case 7:

     cout << "Jul" << endl;

     break;

   case 8:

     cout << "Aug" << endl;

     break;

   case 9:  

     cout << "Sep" << endl;

     break;

   case 10:

     cout << "Oct" << endl;

     break;

   case 11:

     cout << "Nov" << endl;

     break;

   case 12:

     cout << "Dec" << endl;

     break;

   default:

     cout << "Error - the month is not a valid!" << endl;

   }

}

7 0
3 years ago
You want a potof water to boil at 105 celcius. How heavy a
ankoles [38]

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = \frac{\pi }{4} d^2    ..................1

here d is diameter

put the value in equation 1

area = \frac{\pi }{4} 0.15^2

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = \frac{Fnet}{g}

mass  = \frac{350}{9.8}

mass = 35.7 kg

so 35.7 kg lid we put

3 0
3 years ago
One of the flaws in the engineers' reasoning for galloping gertie's design was that they attributed prior failures of suspension
Jlenok [28]
False i think it would be
3 0
3 years ago
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