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hoa [83]
3 years ago
8

An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava

tion protection will address this situation?
Engineering
2 answers:
ivolga24 [154]3 years ago
7 0
Please mark me brainliest
N76 [4]3 years ago
4 0

Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.

Explanation:

You might be interested in
Kieran and Kurt spend most of their day performing physical labor, aligning materials, and inspecting projects for quality and s
xxMikexx [17]

Answer:

Kieran is a Carpenter, and Kurt is a Drafter

Explanation:

Among all the professions listed, Kurt can only possibly be a Drafter.  A Drafter is an engineering technician who makes detailed technical drawings or plans for machinery, buildings, electronics, infrastructure, sections, etc. Drafters make use of computer software and manual sketches to convert the designs, plans, and layouts of engineers and architects into a set of technical drawings. All the other professions will at some point need for the professional to stand and work at an height above the ground except for a drafter, whose feet will remain on the ground at almost all times.

5 0
3 years ago
Read 2 more answers
An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening
Viktor [21]

Given:

Pressure, P_{1} = 1300 kPa

Temperature,  T_{1} = 500^{\circ}

P_{2} = 100 kPa

T_{2} = 127^{\circ}  

velocity, v = 40 m/s

A = 1m^{2}

Solution:

For air propertiess at

P_{1} = 1300 kPa

T_{1} = 500^{\circ}

h_{1} = 793kJ/K

v_{1} = 0.172\frac{m^{3}}{kg}

and also at

P_{2} = 100 kPa

T_{2} = 127^{\circ}  

h_{2} = 401 KJ/K

v_{2} =  1.15\frac{m^{3}}{kg}

a) Mass flow rate is given by:

m' = \frac{Av}{v_{1}}

Now,

m = \frac{0.2\times 40}{0.172} = 46.51 kg/s

b) for the power produced by turbine, P = m'(h_{1} - h_{2})

P = 46.51\times(793 - 401) = 18.231 MW

5 0
4 years ago
EnQueue(X): Thêm phần tử X vào Queue
irina [24]
Yes that is right no matter what you are talking about I’m not sure tho
6 0
3 years ago
The following figure shows the tensile stress-strain curve for a brass alloy.
Aleksandr-060686 [28]
(A)-The Theoretical Strength Is Higher Than The Experimental Strength.

7 0
3 years ago
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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