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Triss [41]
3 years ago
9

You are NASA. Build a space station on Mars that could support humans to live in for an extended period of time.

Engineering
1 answer:
Ierofanga [76]3 years ago
3 0
Literally just do the project
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'Energy' has the potential to:
Kazeer [188]

answer:

Energy' has the potential to:<u>do work</u>

b.do work

6 0
2 years ago
If a pilot-operated check valve (POC) does not check flow, you will see
Elan Coil [88]

if a pilot-operated check valve (POC) does not check flow, you will see The symptoms and symptoms of a failing swing test valve are regularly audible.

<h3>What happens when the pilot valve reaches set strain?</h3>

When the pilot valve reaches set strain it opens and releases the strain from the dome. The piston is then unfastened to open and the principle valve exhausts the machine fluid. The manage pilot opens both to the principle valve exhaust pipe or to the atmosphere.

Pilot operated test valves paintings through permitting unfastened go with the drift from the inlet port thru the opening port. Supplying a pilot strain to the pilot port permits go with the drift withinside the contrary direction. Air strain on the pinnacle of the poppet meeting opens the seal permitting air to go with the drift freely.

Read more about the pilot valve :

brainly.com/question/15565472

#SPJ1

4 0
3 years ago
(Gas Mileage) Drivers are concerned with the mileage their automobiles get. One driver has kept track of several trips by record
Effectus [21]

Answer:

import java.util.*;

public class Main {

   

   public static void main(String[] args) {

     

       double milesPerGallon = 0;

       int totalMiles = 0;

       int totalGallons = 0;

       double totalMPG = 0;

       

       Scanner input = new Scanner(System.in);

 

       while(true){

           System.out.print("Enter the miles driven: ");

           int miles = input.nextInt();

           if(miles <= 0)

               break;

           else{

               System.out.print("Enter the gallons used: ");

               int gallons = input.nextInt();

               totalMiles += miles;

               totalGallons += gallons;

               milesPerGallon = (double) miles/gallons;

               totalMPG = (double) totalMiles / totalGallons;

               System.out.printf("Miles per gallon for this trip is: %.1f\n", milesPerGallon);

               System.out.printf("Total miles per gallon is: %.1f\n", totalMPG);

           }

       }

   }  

}

Explanation:

Initialize the variables

Create a while loop that iterates until the specified condition is met inside the loop

Inside the loop, ask the user to enter the miles. If the miles is less than or equal to 0, stop the loop. Otherwise, for each trip do the following: Ask the user to enter the gallons. Add the miles and gallons to totalMiles and totalGallons respectively. Calculate the milesPerGallon (divide miles by gallons). Calculate the totalMPG (divide totalMiles by totalGallons). Print the miles per gallon and total miles per gallon.

6 0
3 years ago
The current though a capacitor cannot change instantaneously.
malfutka [58]

Answer: The answer is (b) False

Explanation:

5 0
3 years ago
Read 2 more answers
Given an N x M matrix and a dictionary containing K distinct words of length between 1 and 30 symbols, the Word Search should re
Ket [755]

Answer:

import java.util.*;

public class Main {

  public static String[] wordSearch(char[][] matrix, String[] words) {

      int N = matrix.length;

      List<String> myList = new ArrayList<String>();

      int pos = 0;

      for(String word : words)

      {

      for (int i = 0; i < N; i++) {

          for (int j = 0; j < N; j++) {

              if (search(matrix, word, i, j, 0, N)) {

              if(!myList.contains(word))

              {

              myList.add(word);

              }

              }

          }

      }

      }

     

      String[] newDict = myList.toArray(new String[myList.size()]);

     

  return newDict;

  }

  public static boolean search(char[][] matrix, String word, int row, int col,

          int index, int N) {

      // check if current cell not already used or character in it is not not

      if (word.charAt(index) != matrix[row][col]) {

          return false;

      }

      if (index == word.length() - 1) {

          // word is found, return true

          return true;

      }

             

      // check if cell is already used

      if (row + 1 < N && search(matrix, word, row + 1, col, index + 1, N)) { // go

                                                                              // down

          return true;

      }

      if (row - 1 >= 0 && search(matrix, word, row - 1, col, index + 1, N)) { // go

                                                                              // up

          return true;

      }

      if (col + 1 < N && search(matrix, word, row, col + 1, index + 1, N)) { // go

                                                                              // right

          return true;

      }

      if (col - 1 >= 0 && search(matrix, word, row, col - 1, index + 1, N)) { // go

                                                                              // left

          return true;

      }

      if (row - 1 >= 0 && col + 1 < N

              && search(matrix, word, row - 1, col + 1, index + 1, N)) {

          // go diagonally up right

          return true;

      }

      if (row - 1 >= 0 && col - 1 >= 0

              && search(matrix, word, row - 1, col - 1, index + 1, N)) {

          // go diagonally up left

          return true;

      }

      if (row + 1 < N && col - 1 >= 0

              && search(matrix, word, row + 1, col - 1, index + 1, N)) {

          // go diagonally down left

          return true;

      }

      if (row + 1 < N && col + 1 < N

              && search(matrix, word, row + 1, col + 1, index + 1, N)) {

          // go diagonally down right

          return true;

      }

      // if none of the option works out, BACKTRACK and return false

         

      return false;

  }

  public static void main(String[] args) {

      char[][] matrix = { { 'j', 'a', 's' },

              { 'a', 'v', 'o'},

              { 'h', 'a', 'n'} };

 

  String[] arr_str = {"a", "java", "vaxn", "havos", "xsds", "man"};

 

     

      arr_str = wordSearch(matrix, arr_str);

     

      for(String str : arr_str)

      {

      System.out.println(str);

      }

  }

}

8 0
4 years ago
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