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Alecsey [184]
3 years ago
5

If a motor driven by PWM using half-bridge operation has a speed of 1,426 rpm at a duty cycle of 100% (full forward), what will

its speed be at a duty cycle of 47%
Engineering
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

Speed of the motor proportional to duty ratio.

Since duty ratio is proportional to width of the pulse.

So with 61% duty ratio the speed is given by

N = D× Speed of motor with 100% duty ratio

N = 0.47× 1426

= 670.22 rpm

Explanation:

You might be interested in
The minimum fresh air requirement of a residential building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1
ra1l [238]

Answer:

3150 L/min

Explanation:

The volume of the room is the product of the height of the room and its area. It is given as:

Volume (V) = height (h) × Area (A)

V = h × A

Height = 2.7 m, Area = 200 m². Hence:

V = h × A = 2.7 × 200 = 540 m³

The flow capacity (\dot {V}) is given by

\dot {V}=ACH*Volume\ of\ room

ACH = air changes per hour = 0.35

\dot {V}=ACH*Volume\ of\ room\\\\\dot {V}=0.35*540\\\\\dot {V}=189\ m^3/h

But 1 m³ = 1000 L, 1 hr = 60 min

\dot{V}=189\ m^3/h=\frac{189\ m^3*\frac{1000\ L}{1\ m^3} }{1\ h*\frac{60\ min}{1\ h} } =3150\ L/min\\\\\dot{V}=3150 \ L/min

6 0
4 years ago
Study the graphs below what type of trend and pattern in observed .
hichkok12 [17]

Answer:

negative and linear

Explanation:

its a straight line going down

8 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
Should I buy this as my first truck ?
Arlecino [84]

Answer:

yes, I  think it looks good.

Explanation:

6 0
3 years ago
Read 2 more answers
Write multiple if statements. If car_year is 1969 or earlier, print "Few safety features." If 1970 or later, print "Probably has
Slav-nsk [51]

Answer:

Explanation along with code and output results is provided below.

C++ Code:

#include <iostream>

using namespace std;

int main()

{

   int year;

   cout<<"Enter the car model year."<<endl;

   cin>>year;    

  if (year<=1969)

  {

cout<<"Few safety features."<<endl;

  }

else if (year>=1970 && year<1989)

{

cout<<"Probably has seat belts."<<endl;

}

else if (year>=1990 && year<1999)

{

cout<<"Probably has antilock brakes."<<endl;

}

else if (year>=2000)

{

cout<<"Probably has airbags."<<endl;

   }

   return 0;

}

Explanation:

The problem was to print feature messages of a car given its model year.

If else conditions are being used incorporate the logic. The code has been tested with several inputs and got correct output results.

Output:

Enter the car model year.

1961

Few safety features.

Enter the car model year.

1975

Probably has seat belts.

Enter the car model year.

1994

Probably has antilock brakes.

Enter the car model year.

2005

Probably has airbags.

5 0
4 years ago
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