Air is contained in a cylinder device fitted with a piston-cylinder. The piston initially rests on a set of stops, and a pressure of 300 kPa is required to move the piston. Initially, the air is at 100 kPa and 27°C and occupies a volume of 0.4 m^3. A) Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K. Assume air has constant specific heats evaluated at 300 K.

Answer:

The code is attached.

Explanation:

I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.

Then I used methods append(), insert() and extend() for adding elements to the list.

Finally I converted list into a string using join() and adding space in between the elements of the list.

Answer:

Output:-

Enter the five digit lottery number

Enter the digit 1 : 23

Enter the digit 2 : 44

Enter the digit 3 : 43

Enter the digit 4 : 66

Enter the digit 5 : 33

YOU LOSS!!

Computer Generated Lottery Number :

|12|38|47|48|49|

Lottery Number Of user:

|23|33|43|44|66|

Number Of digit matched: 0

Code:-

import java.util.Arrays;

import java.util.Random;

import java.util.Scanner;

public class Lottery {

int[] lotteryNumbers = new int[5];

public int[] getLotteryNumbers() {

return lotteryNumbers;

}

Lottery() {

Random randomVal = new Random();

for (int i = 0; i < lotteryNumbers.length; i++) {

lotteryNumbers[i] = randomVal.nextInt((50 - 1) + 1);

}

}

int compare(int[] personLottery) {

int count = 0;

Arrays.sort(lotteryNumbers);

Arrays.sort(personLottery);

for (int i = 0; i < lotteryNumbers.length; i++) {

if (lotteryNumbers[i] == personLottery[i]) {

count++;

}

}

return count;

}

public static void main(String[] args) {

int[] personLotteryNum = new int[5];

int matchNum;

Lottery lnum = new Lottery();

Scanner input = new Scanner(System.in);

System.out.println("Enten the five digit lottery number");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.println("Enter the digit " + (i + 1) + " :");

personLotteryNum[i] = input.nextInt();

}

matchNum = lnum.compare(personLotteryNum);

if (matchNum == 5)

System.out.println("YOU WIN!!");

else

System.out.println("YOU LOSS!!");

System.out.println("Computer Generated Lottery Number :");

System.out.print("|");

for (int i = 0; i < lnum.getLotteryNumbers().length; i++) {

System.out.print(lnum.getLotteryNumbers()[i] + "|");

}

System.out.println("\n\nLottery Number Of user:");

System.out.print("|");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.print(personLotteryNum[i] + "|");

}

System.out.println();

System.out.println("Number Of digit matched: " + matchNum);

}

}

Explanation:

Answer:

The answer is "0.147 nm and  99.63 mol %"

Explanation:

In point (a):

find:

d(062)=?

formula:

In point (b):

formula:

that's why the composition value equal to 99.63 %

Answer:

you too

Explanation:

thank you for the free points

This type of search would be a linear search. A linear search involves looking at each item in the array one by one until the desired item is found.

In this case, you would look at each square in the array until you find the one that matches the object pictured on the left.  This is a common search method for small arrays or when the array is unsorted.

The algorithm continues until the target element is found, or until it reaches the end of the array and fails to find the target element. Linear searches are useful in scenarios where the array is small and unsorted, as the algorithm does not need to compare every element in the array to find the target element.

Learn more about linear search:

brainly.com/question/30026271

#SPJ11

Answer:

It is important to add presentation layer after the physical layer, so that the data along with it's headers can be translated, when the receiver machine is applying a set of different characters.

Data compression is also required to reduce the space that is occupied by data during transmission, now once the presentation is added to the physical layer, data from the physical layer can be compressed at the presentation layer and sent by improving the throughput.

Explanation:

Solution

The presentation of data involves the following as shown below:

Presentation of data comprises of the task like translating between receiver and sender devices so that machines with different capabilities sets can communicate with one another.

It involves encoding  and decoding of data to provide data security that is been transmitted by different machines.

Data sometimes needs to compressed for efficiency improvement  for transmission.

The physical layer of the TCP/IP protocol suite is responsible or refers to the transmission of physical data over a physical medium

It is good or important to add presentation layer after the physical layer, so that the data along with it's headers can be translated, when the receiver machine is applying a set of different characters.

Data encryption at this stage is good for security instead of encrypting the data at upper/higher layers.

Hence, it is advisable to add presentation layer after the physical layer in the TCP/IP suite.

Answer:

The layer ought to be embedded between Layer 2 and 3.

Explanation:

Applications often communicate with each other. This cannot be successful if they don't see data the same way. The Presentation Layer in the Open Systems Interconnection defines how data is presented and is often processed in the TCP/IP applications.

While the Presentation Layer does not exist as a different layer in the TCP/IP protocol order of arrangement, it is important to note that the Network Layer is also known referred to as the TCP/IP’s Network Layer.

Therefore, if the presentation of the data layer will be separated, it should be between layer 2 and 3.  

Cheers!