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FromTheMoon [43]
2 years ago
4

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

Engineering
1 answer:
Fantom [35]2 years ago
8 0

Answer:

Explanation:

check the attached files for the solution and output result.

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NEEDS TO BE IN PYTHON:ISBN-13 is a new standard for indentifying books. It uses 13 digits d1d2d3d4d5d6d7d8d910d11d12d13 . The la
kondor19780726 [428]

Answer:

Follows are the code to this question:

n=input("Enter the first 12 digits of an ISBN-13 as a string:")#defining a varaible isbn for input value

if len(n)!=12: #use if block to check input value is equal to 12 digits

   print("incorrect input") #print error message

elif n.isdigit()==False: #use else if that check input is equal to digit

   print("incorrect input") #print error message

else:# defining else block

   s=0 #defining integer vaiable s to 0

   for i in range(12):#defining for loop to calculate sum of digit

       if i%2==0: #defining if block to check even value

           s=s+int(n[i])#add even numbers in s vaiable  

       else: #use else block for odd numbers

           s=s+int(n[i])*3 #multiply the digit with 3 and add into s vaiable

   s=s%10#calculate the remainder value  

   s=10-s#subtract the remainder value with 10 and hold its value

   if s==10: #use if to check s variable value equal to 10  

       s=0#use s variable to assign the value 0

   n=n+s.__str__() #u

Output:

please find attached file.

Explanation:In the above Python code, the "n" variable is used for input the number into the string format uses multiple conditional statements for a check input value, which can be defined as follows:

  • In if block, it checks the length isn't equal to 12, if the condition true, it will print an error message.
  • In the else, if the block it checks input value does not digit, if the condition is true, it will print an error message.
  • In the else block, it uses the for loop, in which it calculates the even and odd number sum, and in the odd number, we multiply by 3 then add into s variable.
  • In this, the s variable is used to calculate its remainder and subtract from the value and use the if block to check, its value is not equal to 10 if it's true, it adds 0 into the last of n variable, otherwise, it adds its calculated value.    

4 0
2 years ago
The hot water needs of an office are met by heating tab water by a heat pump from 16 C to 50 C at an average rate of 0.2 kg/min.
Alex777 [14]

Answer:

option B

Explanation:

given,

heating tap water from 16° C to 50° C

at the average rate of 0.2 kg/min

the COP of this heat pump is 2.8

power output = ?

COP = \dfrac{Q_H}{W_{in}}\\W_{in} = \dfrac{Q_H}{COP}\\W_{in} = \dfrac{\dfrac{0.2}{60}\times 4.18\times (50-16)}{2.8}\\W_{in} = 0.169

the required power input is 0.169 kW or 0.17 kW

hence, the correct answer is option B

7 0
2 years ago
The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h
diamong [38]

Answer:

M=0.0411 kg/h or 4.1*10^{-2} kg/h

Explanation:

We have to combine the following formula to find the mass yield:

M=JAt

M=-DAt(ΔC/Δx)

The diffusion coefficient : D=6.0*10^{-8} m/s^{2}

The area : A=0.25 m^{2}

Time : t=3600 s/h

ΔC: (0.64-3.0)kg/m^{3}

Δx: 3.1*10^{-3}m

Now substitute the  values

M=-DAt(ΔC/Δx)

M=-(6.0*10^{-8} m/s^{2})(0.25 m^{2})(3600 s/h)[(0.64-3.0kg/m^{3})(3.1*10^{-3}m)]

M=0.0411 kg/h or 4.1*10^{-2} kg/h

8 0
2 years ago
Chapter 19: Diesel Engine Operation and Diagnosis -Chapter Quiz
Llana [10]

Answer: See explanation

Explanation:

1. How is diesel fuel ignited in a warm diesel engine?

B. Heat compression

2. Which type of diesel injection produces less noise?

A. Indirect injection (IDI)

3. Which diesel injection system requires the use of a glow plug?

A. Indirect injection (IDI)

4. The three phases of diesel ignition include:

C. Ignition delay, repaid combustion, controlled combustion.

5. What fuel system component is used in a vehicle equipped with a diesel engine that is seldom used on the same vehicle when it is equipped with a gasoline engine?

D. Water-fuel separator

6. The diesel injection pump is usually driven by a _________________.

A. Gear off the camshaft

7. Which diesel system supplies high-pressure diesel fuel to all the injectors all of the time?

C. High-pressure common rail

8. Glow plugs should have high resistance when _____________and lower resistance when __________________.

B. Warm/cold

9. Technician A says that glow plugs are used to help start a diesel engine and are shut off as soon as the engine starts. Technician B says that the glow plugs are turned off as soon as a flame is detected in the combustion chamber. Which Technician is correct?

D. Neither Technicians A NOR B

10. What part should be removed to test cylinder compression on a diesel engine?

D. A glow plug

6 0
2 years ago
For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active
Sonja [21]

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 =  0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 =  0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97  = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.

6 0
2 years ago
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