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wolverine [178]
4 years ago
5

3. Technician A says passive permanent

Engineering
1 answer:
Angelina_Jolie [31]4 years ago
6 0

Answer:

Both technician A and B

Explanation:

Passive permanent  magnet ABS wheel speed sensors produce an  A/C voltage signal. Wheel speed sensors are a necessary ABS component and sensor input. It is used to inform the ABS control module of rotational wheel speed. A passive sensor creates an AC signal that changes frequency as the wheel changes speed. Moreover, input  from wheel speed sensors are used for anti- lock brake, electronic traction control, and  electronic stability control systems. Therefore, both technicians are correct.

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More discussion about seriesConnect(Ohm) function In your main(), first, construct the first circuit object, called ckt1, using
STALIN [3.7K]

Answer:

resistor.h

//circuit class template

#ifndef TEST_H

#define TEST_H

#include<iostream>

#include<string>

#include<vector>

#include<cstdlib>

#include<ctime>

#include<cmath>

using namespace std;

//Node for a resistor

struct node {

  string name;

  double resistance;

  double voltage_across;

  double power_across;

};

//Create a class Ohms

class Ohms {

//Attributes of class

private:

  vector<node> resistors;

  double voltage;

  double current;

//Member functions

public:

  //Default constructor

  Ohms();

  //Parameterized constructor

  Ohms(double);

  //Mutator for volatage

  void setVoltage(double);

  //Set a resistance

  bool setOneResistance(string, double);

  //Accessor for voltage

  double getVoltage();

  //Accessor for current

  double getCurrent();

  //Accessor for a resistor

  vector<node> getNode();

  //Sum of resistance

  double sumResist();

  //Calculate current

  bool calcCurrent();

  //Calculate voltage across

  bool calcVoltageAcross();

  //Calculate power across

  bool calcPowerAcross();

  //Calculate total power

  double calcTotalPower();

  //Display total

  void displayTotal();

  //Series connect check

  bool seriesConnect(Ohms);

  //Series connect check

  bool seriesConnect(vector<Ohms>&);

  //Overload operator

  bool operator<(Ohms);

};

#endif // !TEST_H

resistor.cpp

//Implementation of resistor.h

#include "resistor.h"

//Default constructor,set voltage 0

Ohms::Ohms() {

  voltage = 0;

}

//Parameterized constructor, set voltage as passed voltage

Ohms::Ohms(double volt) {

  voltage = volt;

}

//Mutator for volatage,set voltage as passed voltage

void Ohms::setVoltage(double volt) {

  voltage = volt;

}

//Set a resistance

bool Ohms::setOneResistance(string name, double resistance) {

  if (resistance <= 0){

      return false;

  }

  node n;

  n.name = name;

  n.resistance = resistance;

  resistors.push_back(n);

  return true;

}

//Accessor for voltage

double Ohms::getVoltage() {

  return voltage;

}

//Accessor for current

double Ohms::getCurrent() {

  return current;

}

//Accessor for a resistor

vector<node> Ohms::getNode() {

  return resistors;

}

//Sum of resistance

double Ohms::sumResist() {

  double total = 0;

  for (int i = 0; i < resistors.size(); i++) {

      total += resistors[i].resistance;

  }

  return total;

}

//Calculate current

bool Ohms::calcCurrent() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  current = voltage / sumResist();

  return true;

}

//Calculate voltage across

bool Ohms::calcVoltageAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double voltAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      voltAcross += resistors[i].voltage_across;

  }

  return true;

}

//Calculate power across

bool Ohms::calcPowerAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double powerAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      powerAcross += resistors[i].power_across;

  }

  return true;

}

//Calculate total power

double Ohms::calcTotalPower() {

  calcCurrent();

  return voltage * current;

}

//Display total

void Ohms::displayTotal() {

  for (int i = 0; i < resistors.size(); i++) {

      cout << "ResistorName: " << resistors[i].name << ", Resistance: " << resistors[i].resistance

          << ", Voltage_Across: " << resistors[i].voltage_across << ", Power_Across: " << resistors[i].power_across << endl;

  }

}

//Series connect check

bool Ohms::seriesConnect(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  vector<node> temp = ohms.getNode();

  for (int i = 0; i < temp.size(); i++) {

      this->resistors.push_back(temp[i]);

  }

  return true;

}

//Series connect check

bool Ohms::seriesConnect(vector<Ohms>&ohms) {

  if (ohms.size() == 0) {

      return false;

  }

  for (int i = 0; i < ohms.size(); i++) {

      this->seriesConnect(ohms[i]);

  }

  return true;

}

//Overload operator

bool Ohms::operator<(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  if (this->sumResist() < ohms.sumResist()) {

      return true;

  }

  return false;

}

main.cpp

#include "resistor.h"

int main()

{

   //Set circuit voltage

  Ohms ckt1(100);

  //Loop to set resistors in circuit

  int i = 0;

  string name;

  double resistance;

  while (i < 3) {

      cout << "Enter resistor name: ";

      cin >> name;

      cout << "Enter resistance of circuit: ";

      cin >> resistance;

      //Set one resistance

      ckt1.setOneResistance(name, resistance);

      cin.ignore();

      i++;

  }

  //calculate totalpower and power consumption

  cout << "Total power consumption = " << ckt1.calcTotalPower() << endl;

  return 0;

}

Output

Enter resistor name: R1

Enter resistance of circuit: 2.5

Enter resistor name: R2

Enter resistance of circuit: 1.6

Enter resistor name: R3

Enter resistance of circuit: 1.2

Total power consumption = 1886.79

Explanation:

Note

Please add all member function details.Its difficult to figure out what each function meant to be.

8 0
3 years ago
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid
dmitriy555 [2]

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

T_{min} = 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, d_c = 0.5 mm

Number of passes necessary for this reduction, n = \frac{\triangle d}{d_c}

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec

minimum time required to reduce the depth of the plate by 20 mm:

T_{min} = number of passes * Time/pass

T_{min} = n * Time/pass

T_{min} = 40 * 40

T_{min} =  1600 = 26 mins 40 secs

3 0
4 years ago
Read 2 more answers
14. The load across a 50.0-V battery consists of a
Paladinen [302]
The answer you’re looking for is
A.
- I hope this helps, I got it right. Enjoy the rest of your day/night, please make me brainliest!
7 0
3 years ago
g . 3. For each of the following statements Write the statement as an English sentence that does not use the symbols for quan 2.
lubasha [3.4K]

Answer:

(a)

( ∃x ∈ Q) ( x > √2)

There exists a rational number x such that x > √2.

( ∀x ∈ Q) ( ( x ≤ √2)

For each rational number x,  x ≤ √2.

(b)

(∀x ∈ Q)(x² -  2 ≠ 0).

For all rational numbers x, x² -  2 ≠ 0

( ∃x ∈ Q ) ( x² - 2 = 0 )

There exists a rational number x such that  x² -  2 = 0

(c)

(∀x ∈ Z)(x is even or x is odd).

For each integer x, x is even or x is odd.

( ∃x ∈ Z ) (x is odd and x is even)

There exists an integer x such that x is odd and x is even.

(d)

( ∃x ∈ Q) ( √2 < x < √3 )

There exists a rational number x such that √2 < x < √3

(∀x ∈ Q) ( x ≤ √2 or x ≥ √3 )

For all rational numbers x,  x ≤ √2 or x ≥ √3.

3 0
3 years ago
A ½-hp motor requires 120 volts. What amperage is required?
Zolol [24]

Answer:

  3.11 A

Explanation:

Use the given relations between power, horsepower, voltage, and current to find the current requirement for the motor.

  P = hp(745.7 W) = (1/2)(745.7 W) = 372.85 W

  I = P/V = (372.85 W)/(120 V) = 3.1071 A

The amperage required is about 3.11 A.

5 0
3 years ago
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