1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alexus [3.1K]
3 years ago
6

Implement this C program by defining a structure for each payment. The structure should have at least three members for the inte

rest, principle and balance separately. And store all the payments in a structure array (the max size of which could be 100). Name this C program as loanCalcStruct.c. Note: . In your answer sheet, for each C program above please attach a screenshot of the output when amount of loan is $2000, interest rate per year is %7.5 and number of payment is 6. . Put the source code of each C program above into your answer sheet Upload the c file of each Cprogram above into the submission folder of iCollege.

Engineering
1 answer:
Klio2033 [76]3 years ago
6 0

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

You might be interested in
An important ethical concern for behavior analysts involves special cautions in interactions with a client in any capacity outsi
Phoenix [80]

Answer:A and B are correct. one must take care to avoid violation of ethical codes regarding conflicts of interest and dual or multiple relationships

Explanation:

A conflict of interest (COI) is a situation in which a person or organization is involved in multiple interests, financial or otherwise, and serving one interest could involve working against another. 

4 0
3 years ago
Where can you find free air pods that look real
Tema [17]
You can find air pods that look real on letgo. or you can go to wish.com but if you want a good pair jus get the real ones
7 0
3 years ago
Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba
Tpy6a [65]

Answer:

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Explanation:

The optimum cutting speed for the minimum cost

V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)

Where,

C,n = Taylor equation parameters

T_h =Tool changing time in minutes

C_e=Cost per grinding per edge

C_m= Machine and operator cost per minute

On comparing with the Taylor equation VT^n=C,

Tool life,

T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)

Given that,  

Cost of operator and machine time=\$40/hr=\$0.667/min

Batch setting time = 2 hr

Part handling time: T_h=2.5 min

Part diameter: D=73 mm =73\times 10^{-3} m

Part length: l=250 mm=250\times 10^{-3} m

Feed: f=0.30 mm/rev= 0.3\times 10^{-3} m/rev

Depth of cut: d=3.5 mm

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, C_e= \$20/15+2=\$3.33/edge

Tool changing time, T_t=3 min.

C= 80 m/min

n=0.130

(a) From equation (i), cutting speed for the minimum cost:

V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}

\Rightarrow 47.7 m/min

(b) From equation (ii), the tool life,

T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}

\Rightarrow T=53.4 min

(c) Cycle time: T_c=T_h+T_m+\frac{T_t}{n_p}

where,

T_m= Machining time for one part

n_p= Number of pieces cut in one tool life

T_m= \frac{l}{fN} min, where N=\frac{V_{opt}}{\pi D} is the rpm of the spindle.

\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc

So, the number of parts produced in one tool life

n_p=\frac {T}{T_m}

\Rightarrow n_p=\frac {53.4}{4.01}=13.3

Round it to the lower integer

\Rightarrow n_p=13

So, the cycle time

T_c=2.5+4.01+\frac{3}{13}=6.74 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc

(e) Total time to complete the batch= Sum of setup time and production time for one batch

=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times4.01}{457}=0.4387=43.87\%

Now, for the cemented carbide tool:

Cost per edge,

C_e= \$8/6=\$1.33/edge

Tool changing time, T_t=1min

C= 650 m/min

n=0.30

(a) Cutting speed for the minimum cost:

V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min [from(i)]

(b) Tool life,

T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min [from(ii)]

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc

n_p=\frac {7}{0.53}=13.2

\Rightarrow n_p=13 [ nearest lower integer]

So, the cycle time

T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc

(e) Total time to complete the batch=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.53}{275.5}=0.0962=9.62\%

Similarly, for the ceramic tool:

C_e= \$10/6=\$1.67/edge

T_t-1min

C= 3500 m/min

n=0.6

(a) Cutting speed:

V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}

\Rightarrow V_{opt}=2105 m/min

(b) Tool life,

T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc

n_p=\frac {2.33}{0.091}=25.6

\Rightarrow n_p=25 pc/tool\; life

So,

T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc

(e) Total time to complete the batch

=2\times60+ {50\times 2.63}=251.5 min=4.19 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.091}{251.5}=0.0181=1.81\%

3 0
4 years ago
A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion
kati45 [8]

Answer:

X_B = 1.8 \times 10^{-3} m = 1.8 mm

Explanation:

Given data:

Diffusion constant for nitrogen is = 1.85\times 10^{-10} m^2/s

Diffusion flux = 1.0\times 10^{-7} kg/m^2-s

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen  concentration is 0.5 kg/m^3   ......?

from fick's first law

J = D \frac{C_A C_B}{X_A X_B}

Take C_A as point  on which nitrogen concentration is 2 kg/m^3

x_B = X_A + D\frac{C_A -C_B}{J}

Assume X_A is zero at the surface

X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}

X_B = 1.8 \times 10^{-3} m = 1.8 mm

4 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

7 0
4 years ago
Other questions:
  • What is an Algorithm? *
    5·1 answer
  • Compare automation and autonomous
    12·1 answer
  • Describe the steps, tools, and technology needed in detail and
    12·1 answer
  • Part of the basic procedures is the vehicle check. What does that mean?
    7·1 answer
  • What kind of value should an employee possess when employees are expected to be responsible and fair?
    5·1 answer
  • Communication "works" to the degree that a wide variety of information is completely and thoroughly shared among the parties, an
    13·1 answer
  • How is the air delivery temperature controlled during A/C operation?
    8·1 answer
  • The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the av
    9·1 answer
  • 28. What is the value of a resistor in a series circuit if you measure 0.5 amps flowing through it and 15 volts
    10·1 answer
  • The distribution of ground shaking around the fault
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!