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11111nata11111 [884]
3 years ago
10

Do the coil resistances have any effect on the plots?

Engineering
1 answer:
PolarNik [594]3 years ago
8 0
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
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Question text
lisabon 2012 [21]

Answer:

That's a really nice question sadly I don't know the answer I'm replying to you cuz I'm tryna get points so... Sorry

3 0
4 years ago
How does a turbo charger work
Romashka-Z-Leto [24]

Answer:

The turbocharger on a car applies a very similar principle to a piston engine. It uses the exhaust gas to drive a turbine. This spins an air compressor that pushes extra air (and oxygen) into the cylinders, allowing them to burn more fuel each second

Explanation:

7 0
3 years ago
Read 2 more answers
**Please Help, ASAP**
Novay_Z [31]

Explanation:

We need to rearrange the following formula for the values given in parenthesis.

(1) x+xy = y, (x)

taking x common in LHS,

x(1+y)=y

x=\dfrac{y}{1+y}

(2) x+y = xy, (x)

Subrtacting both sides by xy.

x+y-xy = xy-xy

x+y-xy = 0

x-xy=-y

x(1-y)=-y

x=\dfrac{-y}{1-y}

(3) x = y+xy, (x)

Subrating both sides by xy

x-xy = y+xy-xy

x(1-y)=y

x=\dfrac{y}{1-y}

(4) E = (1/2)mv^2-(1/2)mu^2, (u)

Subtracting both sides by (1/2)mv^2

E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2

E-(1/2)mv^2 =-(1/2)mu^2

So,

2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}

(5) (x^2/a^2)-(y^2/b^2) = 1, (y)

\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}

(6) ay^2 = x^3, (y)

y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}

Hence, this is the required solution.

3 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 I have a scaffold that is 83' tall, and the base of my scaffold is 5' wide least base dimension, then I am allowed to space m
Minchanka [31]

In the scaffold, the intermediate bracing supports every 26 feet while the the lowest brace being no greater than 20 feet.

<h3>What is scaffold?</h3>

Scaffolding simply means a temporary structure that's used to support a work crew.

In this case, the height is 83 feet and the base width is 5 feet. The height to the width ratio will be:

= 83/5 = 16.6 > 4.

Therefore, the scaffold will be restrained. Lowest brace will be placed at 20 and intermediate support braces at 26 feet.

Learn more about scaffolding on:

brainly.com/question/20726451

8 0
2 years ago
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