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fredd [130]
3 years ago
12

Solve using Matlab the problems:

Engineering
1 answer:
Firlakuza [10]3 years ago
7 0

Answer:

Explanation:

% Clears variables and screen

clear; clc

% Asks user for input

n = input('Total number of objects: ');

r = input('Size of subgroup: ');

% Computes and displays permutation according to basic formulas

p = 1;

for i = n - r + 1 : n

   p = p*i;

end

str1 = [num2str(p) ' permutations'];

disp(str1)

% Computes and displays combinations according to basic formulas

str2 = [num2str(p/factorial(r)) ' combinations'];

disp(str2)

=================================================================================

Example: check

How many permutations and combinations can be made of the 15 alphabets, taking four at a time?

The answer is:

32760 permutations

1365 combinations

==================================================================================

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Compared to arc welding, which of the following statements are true about<br> gas welding?
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Show than arc welding

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3 years ago
Water at 60°F passes through 0.75-in-internal diameter copper tubes at a rate of 1.2 lbm/s. Determine the pumping power per ft
Lelu [443]

Answer:

The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts

Explanation:

See calculation attached.

- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.

- Calculate the cross-sectional flow area.

- Calculate the average velocity of water in the copper tubes.

- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.

- Calculate the pressure drop in the copper tubes.

- Then finally calculate the power required for pumping.

8 0
3 years ago
The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
Implement this C program by defining a structure for each payment. The structure should have at least three members for the inte
Klio2033 [76]

Answer:

#include<stdio.h>

#include<math.h>

void output_amortized(float loan_amount,float intrest_rate,int term_years)

{

  int i,j;                       //Month

  int payments;                   //Number of payments  

  float loanAmount;               //Loan amount

  float anIntRate;               //Yealy interest Rate

  float monIntRate;               //Monthly interest rate

  float monthPayment;           //Monthly payment

  float balance;                   //Balance due

  float monthPrinciple;           //Monthly principle paid

  float monthPaidInt;           //Month interest paid

 

  balance=loan_amount;

  //Calculations

  //Monthly interest rate

  monIntRate = ((intrest_rate/(100*12)));

  //Monthly payment

  payments=term_years;  

  monthPayment = (loan_amount * monIntRate * (pow(1+monIntRate, payments)/(pow (1+monIntRate, payments)-1)));

  monthPaidInt = balance * monIntRate;

  //Amount paid to principle

  monthPrinciple = monthPayment-monthPaidInt;

  //New balance due

  balance = balance - monthPrinciple;

 

  printf("\n\nMonthly payment should be :%.2f\n\n",monthPayment);

  printf("============================AMORTIZATION SCHEDUAL==========================\n");

  printf("#\tPayment\t\tIntrest\t\tPrinciple\t\tBalance\n");

 

  for(i=0;i<payments;i++)

  {

      printf("%d%9c%.2f%9c%.2f%16c%.2f%14c%.2f\n",(i+1),'$',monthPayment,'$',monthPaidInt,'$',monthPrinciple,'$',balance);

      monthPaidInt = balance * monIntRate;

      //Amount paid to principle

      monthPrinciple = monthPayment-monthPaidInt;

      //New balance due

      balance = balance - monthPrinciple;

  }

}

int main()

{

  float principle,rate;

  int termYear;

  printf("Enter the loan amount: $");

  scanf("%f",&principle);

  printf("Enter the intrest rate :%");

  scanf("%f",&rate);

  printf("Enter the loan duration in years: ");

  scanf("%d",&termYear);

  output_amortized(principle,rate,termYear);

}

Explanation:

see output

6 0
3 years ago
What is local Technology​
Damm [24]

Answer:

Local Technology is a IT services that focused on building businesses up

Hope This Helped

5 0
2 years ago
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