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Marizza181 [45]
3 years ago
10

At what voltage would a () capacitor have the minimum energy to raise by ignoring all losses in the system? If needed, you may a

ssume a gravitational acceleration of 9.80 m/s2.

Engineering
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

V = 280.15 V

Explanation:

" The complete question is attached with figure"

Given:

- The capacitance of the capacitor C = 10 nF

- The amount of mass attached to motor m = 4 grams

- The amount of distance it is to be lifted h = 1 cm

- Ignore all other losses in the system

Find:

- The voltage required to lift the mass m through distance h?

Solution:

- The conservation of energy for the entire system is written as:

                            Work_gravity = U_c

Where,

           Work_gravity: Work done by gravity on mass m

           U_c: The amount of energy stored in a capacitor

                             m*g*h = 0.5*C*V^2

                             V^2 = 2*m*g*h / C

                             V = sqrt ( 2*m*g*h / C )

Plug in the values:

                             V = sqrt ( 2*0.004*9.81*0.01 / 10*10^-9 )

                             V = sqrt ( 78,480)

                             V = 28.15 V

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Write a program that prompts the user to enter time in 12-hour notation. The program then outputs the time in 24-hour notation.
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Answer:

THE CODE FOR THE PROGRAM IS GIVEN BELOW:

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int main()

{

convertTime convert;

int hr, mn, sc = 0;

 

cout << "Please input hours in 12 hr notation: ";

cin >> hr;

cout << "Please input minutes: ";

cin >> mn;

cout << "Please input seconds: ";

cin >> sc;

 

convert.invalidHr(hr);

convert.invalidMin(mn);

convert.invalidSec(sc);

convert.printMilTime();

 

system("Pause");

 

return 0;  

 

}

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int convertTime::invalidHr (int hour)

{

try{

 if (hour < 13 && hour > 0)

  {hour = hour + 12;

  return hour;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input hour again in correct 12 hour format: ";

  cin >> hour;

  invalidHr(hour);

  throw 10;

 }

   

}

catch (int c) { cout << "Invalid hour input!";}

}

int convertTime::invalidMin (int min)

{

try{

 if (min < 60 && min > 0)

  {return min;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input minutes again in correct 12 hour format: ";

  cin >> min;

  invalidMin(min);

  throw 20;

  return 0;

 }

   

}

catch (int e) { cout << "Invalid minute input!" << endl;}

}

int convertTime::invalidSec(int sec)

{

try{

 if (sec < 60 && sec > 0)

  {return sec;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input seconds again in correct 12 hour format: ";

  cin >> sec;

  invalidSec(sec);

  throw 30;

  return 0;

 }

   

}

catch (int t) { cout << "Invalid second input!" << endl;}

}

void convertTime::printMilTime()

{

cout << "Your time converted: " << hour << ":" << min << ":" << sec;

}

Explanation:

4 0
3 years ago
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Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)
ArbitrLikvidat [17]

Answer:

\mathbf{P_x =25 \ watts}

\mathbf{x_{rmx} = 5 \ unit}

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) +  sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal  Asin (ωt), Power can be expressed as:

P= \dfrac{A^2}{2}

For the number of sinosoidial signals;

Power can be expressed as:

P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...

As such,

For x(t), Power  P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}

P_x = \dfrac{25}{2}+ \dfrac{25}{2}

P_x = \dfrac{50}{2}

\mathbf{P_x =25 \ watts}

For the number of sinosoidial signals;

RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...

For x(t), the RMS value is as follows:

x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }

x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }

x_{rmx }=\sqrt{(\dfrac{50}{2} )}

x_{rmx} =\sqrt{25}

\mathbf{x_{rmx} = 5 \ unit}

8 0
3 years ago
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