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valentina_108 [34]
3 years ago
8

Why do electronic devices gather so much dust? Thanks

Computers and Technology
1 answer:
Fudgin [204]3 years ago
3 0
Any electrical device that produces any type of air kind of gets collected as the air flows in or out
You might be interested in
In this assignment you'll write a program that encrypts the alphabetic letters in a file using the Vigenère cipher. Your program
solniwko [45]

Answer:

C code is given below

Explanation:

// Vigenere cipher

#include <ctype.h>

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

/**

* Reading key file.

*/

char *readFile(char *fileName) {

   FILE *file = fopen(fileName, "r");

   char *code;

   size_t n = 0;

   int c;

   if (file == NULL) return NULL; //could not open file

   code = (char *)malloc(513);

   while ((c = fgetc(file)) != EOF) {

      if( !isalpha(c) )

          continue;

      if( isupper(c) )

          c = tolower(c);

      code[n++] = (char)c;

   }

   code[n] = '\0';

  fclose(file);

   return code;

}

int main(int argc, char ** argv){  

   // Check if correct # of arguments given

   if (argc != 3) {

       printf("Wrong number of arguments. Please try again.\n");

       return 1;

   }

 

  // try to read the key file

  char *key = readFile(argv[1]);

  if( !key ) {

      printf( "Invalid file %s\n", argv[1] );

      return 1;

  }

 

  char *data = readFile(argv[2]);

  if( !data ) {

      printf("Invalid file %s\n", argv[2] );

      return 1;

  }

 

   // Store key as string and get length

   int kLen = strlen(key);

  int dataLen = strlen( data );

 

  printf("%s\n", key );

  printf("%s\n", data );

 

  int paddingLength = dataLen % kLen;

  if( kLen > dataLen ) {

      paddingLength = kLen - dataLen;

  }

  for( int i = 0; i < paddingLength && dataLen + paddingLength <= 512; i++ ) {

      data[ dataLen + i ] = 'x';

  }

 

  dataLen += paddingLength;

 

   // Loop through text

   for (int i = 0, j = 0, n = dataLen; i < n; i++) {          

       // Get key for this letter

       int letterKey = tolower(key[j % kLen]) - 'a';

     

       // Keep case of letter

       if (isupper(data[i])) {

           // Get modulo number and add to appropriate case

           printf("%c", 'A' + (data[i] - 'A' + letterKey) % 26);

         

           // Only increment j when used

           j++;

       }

       else if (islower(data[i])) {

           printf("%c", 'a' + (data[i] - 'a' + letterKey) % 26);

           j++;

       }

       else {

           // return unchanged

           printf("%c", data[i]);

       }

      if( (i+1) % 80 == 0 ) {

          printf("\n");

      }

   }

 

   printf("\n");

 

   return 0;

}

4 0
3 years ago
Read 2 more answers
Write an application that accepts up to 20 Strings, or fewer if the user enters the terminating value ZZZ. Store each String in
notsponge [240]

Answer:

count = 20

i = 0

short_strings = []

long_strings = []

while(i<count):

   s = input("Enter a string: ")

   

   if s == "ZZZ":

       break

   

   if len(s) <= 10:

       short_strings.append(s)

   elif len(s) >= 11:

       long_strings.append(s)

   

   i += 1

choice = input("Enter the type of list to display [short/long] ")

if choice == "short":

   if len(short_strings) == 0:

       print("The list is empty.")

   else:

       print(short_strings)

else:

   if len(long_strings) == 0:

       print("The list is empty.")

   else:

       print(long_strings)

Explanation:

*The code is in Python.

Initialize the count, i, short_strings, and long_strings

Create a while loop that iterates 20 times.

Inside the loop:

Ask the user to enter a string. If the string is "ZZZ", stop the loop. If the length of the string is smaller than or equal to 10, add it to the short_strings. If the length of the string is greater than or equal to 11, add it to the long_strings. Increment the value of i by 1.

When the loop is done:

Ask the user to enter the list type to display.

If the user enters "short", print the short list. Otherwise, print the long_strings. Also, if the length of the  chosen string is equal to 0, print that the list is empty.

3 0
4 years ago
___ is the most important variable that is measured and controlled in a commercial hvac system.
Pavel [41]

The parameter being monitored and controlled is the controlled variable. The dry-bulb temperature of the air leaving the cooling coil is the controllable variable in this particular instance. The sensor assesses the state of the controlled variable and provides the controller with an input signal.

<h3>What is a HVAC ?</h3>

The employment of various technologies for heating, ventilation, and air conditioning is done to regulate the temperature, humidity, and cleanliness of the air in a closed environment. Its objective is to provide adequate indoor air quality and thermal comfort.

  • To cool or heat a building, the main unit of a ducted system forces air via a network of air ducts. On the other hand, ductless systems don't have air ducts and employ different techniques to spread cleaned air across a room.

Learn more about HVAC system here:

brainly.com/question/20264838

#SPJ4

6 0
2 years ago
What would be the consequences if the Memory Manager and the
Ann [662]

Answer:

Result could be a memory leak or a spaceleak.

Explanation:

The result if a memory supervisor and a processor administrator quit conveying would prompt a negative result as they have to impart all together for the procedure manager to process the data that is being conveyed in and out.

3 0
3 years ago
A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To ac
Gekata [30.6K]

Answer:

# include <iostream>

#include<stdio.h>

using namespace std;

bool IsLeapYear(int y)

int main()

{

int y;

cout<<"Enter the Year to check Leap or Not"<<endl;

cin>>y;

IsLeapYear(int y);

getch();

}

bool IsLeapYear(int y)

{

if (y%4==0)

{

   if (y%100==0)

    {

         if (y%400==0 )

         {

          cout<"The year is leap Year";

         }

         else

         {

         cout<<" The year is not Leap Year";

         }

     }

     else

     {

     cout<<"The year is Leap Year" ;

     }

}    

else

{

cout<<"The year is not Leap Year";

}    

}

Explanation:

In this program a function has been defined named as IfLeapYear, to check that whether the entered year is leap year or not. An year taken as integer data type named as y to enter the year to check. If the year is divisible by 4 but not divisible by 100 is the leap year. If the year is divisible by 4, divisible by 100 and also divisible by 400 is the century year and is also the leap year.

To check all the statements, Nested if-else conditions has been used to check multiple requirements of the leap year.

6 0
3 years ago
Read 2 more answers
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