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harina [27]
2 years ago
5

Project 8:The Harris-Benedict equation estimates the number of calories your body needs to maintain your weight if you do no exe

rcise. This is called your basal metabolic rate, or BMR.The calories needed for a woman to maintain her weight is:BMR = 655 + (4.3 Ã weight in pounds) + (4.7 Ã height in inches) â (4.7Ã age in years)The calories needed for a man to maintain his weight is:BMR = 66 + (6.3 Ã weight in pounds) + (12.9 Ã height in inches) â (6.8 Ã age in years)A typical chocolate bar will contain around 230 calories. Write a program that allows the user to input his or her weight in pounds, height in inches, and age in years. The program should then output the number of chocolate bars that should be consumed to maintain oneâs weight for both a woman and a man of the input weight, height, and age.
Repeat the calorie-counting program described in Programming Project 8 from Chapter 2. This time ask the user to input the string "M" if the user is a man and "W" if the user is a woman. Use only the male formula to calculate calories if "M" is entered and use only the female formula to calculate calories if "W" is entered. Output the number of chocolate bars to consume as before.

Engineering
1 answer:
lilavasa [31]2 years ago
4 0

Answer:

See explaination and attachment for the program code and output

Explanation:

#include <iostream>

using namespace std;

int main()

{

char gender; //details for gender and checking

char ans;

do

{

cout<<"Gender (M or F): ";

cin>>gender;

switch(gender)

{

case 'M':

//cout<<"Male."<<endl;

break;

case 'F':

//cout<<"Female."<<endl;

break;

default:

cout<<"Wrong Gender. Please enter again (M or F): ";

cin>>gender;

}

int Weight,Height,Age; //declaration of variables

double bmr;

cout<<"Weight: ";

cin>>Weight;

cout<<"Height (in inches): ";

cin>>Height;

cout<<"Age: ";

cin>>Age;

//bmr calculations for male and female

if (gender = 'M')

{

bmr = 66 + (6.3 * Weight) + (12.9 * Height) - (6.8 * Age);

cout<<"He needs "<<bmr<<" to maintain his weight."<<endl;

cout<<"He needs to eat "<<(bmr/230)<< " candy bars in one day."<<endl;

}

else if (gender = 'F')

{

bmr = 655 + (4.3 * Weight) + (4.7 * Height) - (4.7 *Age);

cout<<"She needs "<<bmr<<" to maintain her weight"<<endl;

cout<<"She needs to eat "<<(bmr/230)<< " candy bars in one day."<<endl;

}

cout<< "Do you want to do another one>continue (Y/N): ";

cin >> ans;

}while(ans=='y'||ans=='Y');

cout<<"\n Thanks for using my BMR calculator. Good Bye!.";

return 0;

}

Kindly check attachment for output.

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1 year ago
For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or
Alborosie

Answer:

Glass: Low-Loss dielectric

  α = 8.42*10^-11 Np/m

  β = 468.3 rad/m

  λ = 1.34 cm

  up = 1.34*10^8 m/s

  ηc = 168.5 Ω

Tissue: Quasi-Conductor

  α = 9.75 Np/m

  β = 12.16 rad/m

  λ = 51.69 cm

  up = 0.52*10^8 m/s

  ηc = 39.54 + j 31.72 Ω        

Wood: Good conductor

  α = 6.3*10^-4 Np/m

  β = 6.3*10^-4 Np/m

  λ = 10 km

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  ηc = 6.28*( 1 + j )

Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if  the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then  calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to determine category of the medium as follows:

                                σ / w*εr*εo

Where, w is the angular speed of wave

            εo is the permittivity of free space = 10^-9 / 36*pi

- Now we classify as follows:

    Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\  

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.

    Glass: Low-Loss dielectric

    Tissue: Quasi-Conductor

    Wood: Good conductor

- Now we will use categorized material base equations from Table 17-1 as follows:

     Glass: Low-Loss dielectric

          α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

          α = 8.42*10^-11 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

          β = 468.3 rad/m

          λ = 2*pi / β = 2*pi / 468.3

          λ = 1.34 cm

          up = λ*f = 0.0134*10^10

          up = 1.34*10^8 m/s

          ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

          ηc = 168.5 Ω

     Tissue: Quasi-Conductor

          α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

          α = 9.75 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

          β = 12.16 rad/m

          λ = 2*pi / β = 2*pi / 12.16

          λ = 51.69 cm

          up = λ*f = 0.5169*100*10^6

          up = 0.52*10^8 m/s

          ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

          ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

          ηc = 39.54 + j 31.72 Ω

     Wood: Good conductor

          α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

          β = α = 6.3*10^-4 Np/m

          λ = 2*pi / β = 2*pi / 6.3*10^-4

          λ = 10 km

          up = λ*f = 10,000*1*10^3

          up = 0.1*10^8 m/s

          ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

          ηc = 6.28*( 1 + j )

         

           

         

8 0
3 years ago
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