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kumpel [21]
3 years ago
5

Wastewater flows into a _________ once it is released into a floor drain.

Engineering
1 answer:
vodomira [7]3 years ago
5 0

Answer:

a

Explanation:

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Vesna [10]
The Juglandaceae tree
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3 years ago
Read 2 more answers
Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil
Karo-lina-s [1.5K]

Answer:

E = \frac{3Q}{2A\epsilon_0}

Explanation:

By Gauss Law for electric field:

E = \frac{\sigma}{2\epsilon_0}

Where \sigma is the charge density Q/A. Since we have 2 parallel  plates with different charge, the electric field at point P in the gap would be the sum of 2 field

E = E_1 + E_2

E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}

E = \frac{3Q}{2A\epsilon_0}

5 0
3 years ago
5.11: Population Write a program that will predict the size of a population of organisms. The program should ask the user for th
fomenos

Answer:

// using c++ language

#include "stdafx.h";

#include <iostream>

#include<cmath>

using namespace std;

//start

int main()

{

  //Declaration of variables in the program

  double start_organisms;

  double daily_increase;

  int days;

  double updated_organisms;

  //The user enters the number of organisms as desired

  cout << "Enter the starting number of organisms: ";

  cin >> start_organisms;

  //Validating input data

  while (start_organisms < 2)

  {

      cout << "The starting number of organisms must be at least 2.\n";

      cout << "Enter the starting number of organisms: ";

      cin >> start_organisms;

  }

  //The user enters daily input, here's where we apply the 5.2% given in question

  cout << "Enter the daily population increase: ";

  cin>> daily_increase;

  //Validating the increase

  while (daily_increase < 0)

  {

      cout << "The average daily population increase must be a positive value.\n ";

      cout << "Enter the daily population increase: ";

      cin >> daily_increase;

  }

  //The user enters number of days

  cout << "Enter the number of days: ";

  cin >> days;

  //Validating the number of days

  while (days<1)

  {

      cout << "The number of days must be at least 1.\n";

      cout << "Enter the number of days: ";

      cin >> days;

  }

 

  //Final calculation and display of results based on formulas

  for (int i = 0; i < days; i++)

  {

      updated_organisms = start_organisms + (daily_increase*start_organisms);

      cout << "On day " << i + 1 << " the population size was " << round(updated_organisms)<<"."<<"\n";

     

      start_organisms = updated_organisms;

  }

  system("pause");

   return 0;

//end

}

4 0
3 years ago
If you install special sound-reflecting windows that reduce the sound intensity level by 33.0 dBdB , by what factor have you red
Sever21 [200]

Answer:

The sound intensity is reduced by a factor of 1995.26

Explanation:

When comparing two sound intensities, the intensity level is measured in the unit of decibel or dB. The intensity of the threshold of hearing for a human being is  10^−12 W/m^2 . When the intensity level is zero, it means that the sound intensity is the same as the threshold of hearing.

The reduced sound intensity level is given as 33 dB, so

10  log (I/Io)  = - 33

I  : intensity of the sound

Io ( =   10^−12 W/m^2): threshold of hearing

So, the intensity ratio is

I/Io = 10^-3.3

     = 5.01 x 10^-4

1/ 5.01 x 10^-4 = 1995.26

3 0
3 years ago
A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para
mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c)  54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

V_p=\frac{3810.5}{\sqrt{3} }=2200 V

The complex power S is given as:

S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA

where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA

The line current I is given as:

I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)}  =100\angle -36.87^o\ A

The phase voltage at the sending end is:

V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV

The magnitude of the line voltage at the source end of the line (V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V

b) The Total real and reactive power loss in the line is:

S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0
3 years ago
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