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Gwar [14]
4 years ago
7

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

the air in and forces it through the resistors where it is heated. If the density of air is 1.18 kg/m3 at the inlet and 1.05 kg/m3 at the exit, determine the percent increase in the velocity of air as it flows through the hair dryer.
Engineering
1 answer:
nikitadnepr [17]4 years ago
4 0

Answer:

the percent increase in the velocity of air as it flows through the dryer is 12%

Explanation:

given data

density of air ρ  = 1.18 kg/m³

density of air ρ' = 1.05 kg/m³

solution

we know there is only one inlet and exit

so

ρ × A × v = ρ' × A × v'     ........................1

put here value and we get

\frac{v'}{v} = \frac{\rho }{\rho '}  

\frac{v'}{v} = \frac{1.18}{1.05}  

\frac{v'}{v}  = 1.12

so the percent increase in the velocity of air as it flows through the dryer is 12%

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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.

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3 years ago
50. You are not permitted to work on any equipment or machinery at any time if the
dexar [7]
I assume this is a wood shop question:

You may not work on any equipment if the TEACHER is not present in the room.
7 0
4 years ago
- Find the file that has the string "You Got Me" in your home directory. - When you have the correct command to do that, add the
Scilla [17]

Answer:

Explanation: see attachment below

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3 years ago
A house is on a hill 10 meters above the base of a 50 meter tall water tower. Calculate the gauge pressure at a tap on the groun
marin [14]

Answer:

Explanation:

Given: Pressure of a fluid column ΔP=ρgΔh

Density of water (ρ) = 1000 kg/m^3

Acceleration of gravity (g) = 9.8 m/sec^2

Δh = difference in height / elevation of water tower and tap

= 50 - 10

= 40

Substitute back into equation:

Pressure = 1000*9.8*40

= 392000 N/m^2

The answer is D.

7 0
3 years ago
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A part has been tested to have Sut = 530 MPa, f = 0.9, and a fully corrected Se = 210 MPa. The design requirements call for the
kobusy [5.1K]

Answer:

126984 cycles

Explanation:

Given data :

Sut = 530 MPa

f = 0.9

fully corrected Se = 210 MPa

using Miner's method attached below is the detailed solution of the given problem

when loaded with ± 225 MPa the number of cycles before it fails will be

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