The output of the traversals with a left child preference would be
Preorder traversal: **4, 9, 2, 1, 3**\
Postorder traversal: **3, 1, 2, 9, 4**\
Inorder traversal: **8, 4, 2, 5, 1, 3**\
Level order traversal: **1, 2, 3, 4, 5**
Preorder traversal visits the current node first, then the left subtree, and finally the right subtree. Following the given tree, the preorder traversal would be 4, 9, 2, 1, 3.
Postorder traversal visits the left subtree, then the right subtree, and finally the current node. From the given tree, the postorder traversal would be 3, 1, 2, 9, 4.
Inorder traversal visits the left subtree, then the current node, and finally the right subtree. Based on the provided tree, the inorder traversal would be 8, 4, 2, 5, 1, 3.
Level order traversal visits the nodes in each level from left to right. Looking at the tree, the level order traversal would be 1, 2, 3, 4, 5.
Therefore, the output of the traversals with a left child preference would be:
Preorder: 4, 9, 2, 1, 3
Postorder: 3, 1, 2, 9, 4
Inorder: 8, 4, 2, 5, 1, 3
Level order: 1, 2, 3, 4, 5
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Discuss the about different blocks in the organization of SRAM.
SRAM or Static Random Access Memory is a type of RAM that holds data bits in memory as long as the power is on. It stores a small amount of data but faster compared to DRAM.
SRAM comes in a few different configurations. Among these configurations are blocks that make up the organization of SRAM: The blocks that are included in the organization of SRAM are as follows:
1. Memory cell array
2. Read and write circuits
3. Decoders
4. Sense amplifiers
5. Input and output circuits
Memory cell array:
Memory cell array is a group of memory cells that are physically arranged in a rectangular block. Each memory cell is used to store a single bit. The memory cell array is the basic storage unit of an SRAM.
Read and Write Circuits:
SRAM's read and write circuits are responsible for reading and writing data to and from memory cells. They also contain circuits that hold the read and write signals.
As a result, the read and write circuits can determine when data is read from or written to the memory cells.
Decoders:
Decoders are circuits that control read and write access to the memory cell array. Decoders manage the connections between memory cell array and other circuits in the SRAM chip.
Sense Amplifiers:
The sense amplifier circuits are responsible for improving the output signals that result from reading data from memory cells. Sense amplifiers are built into SRAMs because they can ensure that signals are transferred without losing the data.
Input and Output Circuits:
Input/output circuits are responsible for providing external access to the SRAM. 'The circuits receive input signals from external devices and then transfer data to or from the memory cell array, depending on whether it's a read or write signal.
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Solve the following differential equations by using Laplace transform method. Question2 a) y"+2y+y=4e where y(0)=2 y'(0)=4. b) 2y" +4y' +10y=8(t) Assume zero initial conditions.
Solving the differential equation using Laplace transform methoda) y"+2y+y=4e where y(0)=2 y'(0)=4.First, apply Laplace transform to both sides of the differential equation to obtain L{y"+2y+y}=L{4e}.L{y"+2y+y} = L{4e}Apply the formula L{y'+f(t)} = s Y(s) - y(0) + F(s) where F(s) is the Laplace transform of f(t).
s² Y(s) - s y(0) - y'(0) + 2s Y(s) + Y(s) = 4/s²Therefore, we can solve for Y(s)Y(s) = 4/s² / (s² + 2s + 1) = 4/s² / (s + 1)²By partial fraction decomposition, we can rewrite Y(s) asY(s) = 2/(s + 1) + 2/(s + 1)²Take the inverse Laplace transform to obtain the solution to the differential equationy(t) = 2e^-t + 2te^-tb) 2y" +4y' +10y=8(t) Assume zero initial conditions.Apply Laplace transform to both sides of the differential equation.
2L{y"} + 4L{y'} + 10L{y} = 8L{t}Applying the formula L{y'} = s Y(s) - y(0) for the second term on the left hand side of the equation and L{y"} = s² Y(s) - s y(0) - y'(0) for the first term, we get:2s² Y(s) - 2s y(0) - 2y'(0) + 4s Y(s) + 10 Y(s) = 8/s³Simplify to get:2s² Y(s) + 4s Y(s) + 10 Y(s) = 8/s³ + 2y'(0) + 2s y(0)By partial fraction decomposition, we can rewrite Y(s) asY(s) = (2s+1)/(s²+2s+5) + (4/5)*(1/s) - (4/5)*(s/(s²+2s+5))Taking the inverse Laplace transform to obtain the solutiony(t) = (1/5)(4-2e^-t*cos(2t)-4e^-t*sin(2t))
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Suppose that a (Java) class C is declared to extend two different classes A and B , both of which define a method called x(). The subclass itself does not overload this method x(). Let's say we call x() on a C-object somewhere. a. This will use the implementation provided by B (the last superclass mentioned in C's signature). b. This will use the implementation provided by A (the first superclass mentioned in C's signature). c. The compiler will complain because C must explicitly overload this method x(). O d. The compiler will complain because Java does not allow C to extend both A and B in this way.
Suppose that a (Java) class C is declared to extend two different classes A and B , both of which define a method called x(). The subclass itself does not overload this method x(). Let's say we call x() on a C-object somewhere This will use the implementation provided by B (the last superclass mentioned in C's signature).
In Java, when a class extends multiple classes that have a method with the same name, the implementation of the method from the last superclass mentioned in the class's signature will be used. This is known as method overriding. If class C does not override the method x() itself, it will inherit the implementation from class B.
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Description Create a class called Student. It is designed to manage the status of a student in a course. The class should have the following attributes: • First name • Last name Number of assignments completed Number of quizzes completed Current quiz percentage (average grade of quizzes completed so far) • Current assignment percentage (average grade of assignments completed so far) Your class will have a single constructor. It will receive two parameters: the student's first and last names. All other values will be initialized to zero. Your class will have the following methods: • • Accessors for current quiz percentage and current assignment percentage addAssignment, which receives as a parameter the grade of a new assignment. This method will increment the number of completed assignments and recalculate the current assignment percentage. It will return the new percentage. • addQuiz, which receives as a parameter the grade of a new quiz. This method will increment the number of completed quizzes and recalculate the current quiz percentage. It will return the new percentage. • currentGrade, which receives no parameters and returns the student's current overall grade in the course. For that, assignments are weighted twice as heavily as quizzes, so the current assignment percentage would be 2/3 of the current overall grade.
A class Student is created to manage the status of a student in a course, and it has the following attributes:• First name• Last name• Number of assignments completed• Number of quizzes completed• Current quiz percentage (average grade of quizzes completed so far)• Current assignment percentage (average grade of assignments completed so far). This class has a single constructor that accepts two parameters, the student's first and last names. All other values will be initialized to zero.
The methods in this class include: Accessors for current quiz percentage and current assignment percentageaddAssignment - This method receives as a parameter the grade of a new assignment. It will increment the number of completed assignments and recalculate the current assignment percentage. It will return the new percentage.addQuiz - This method receives as a parameter the grade of a new quiz. It will increment the number of completed quizzes and recalculate the current quiz percentage. It will return the new percentage.
currentGrade - This method receives no parameters and returns the student's current overall grade in the course. For that, assignments are weighted twice as heavily as quizzes, so the current assignment percentage would be 2/3 of the current overall grade. Accessors for current quiz percentage and current assignment percentage are also included in this class.
Accessors are special methods that are used to access the value of a class's private data members. Accessors have a public scope, meaning they can be accessed by any object that has an instance of the class, but they are designed to return the value of private data members, not modify them. This allows you to provide controlled access to your class's data members, rather than giving full access to them.
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5b. a = 8
(b) (10 pts.) Consider a linear time-invariant system with H(e) = tude response |H(ejw)|. 1+e-jw (1-ae-jw)2 Determine the magni- 1000/101100²
The magnitude response of the given linear time-invariant system is 1000/101100².
The given linear time-invariant system is represented by the transfer function H(ejω), where ω represents the frequency. The magnitude response of the system, |H(ejω)|, is determined by the expression 1 + e[tex]^{-jw}[/tex]/ (1 - ae[tex]^{-jw}[/tex])².
To calculate the magnitude response, we substitute ω with the desired frequency and evaluate the expression. In this case, the magnitude response is 1000/101100².
The magnitude response represents the gain or attenuation of the system at different frequencies. A magnitude response of 1 indicates no change in amplitude, while values greater than 1 represent amplification, and values less than 1 indicate attenuation.
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Based on the following scenario, draw your class diagram and draw your activity diagram
An ATM consists of a screen, a deposit unit, cash delivery unit, communicator with the
bank system.
Customer goes to ATM and authenticate itself with the card and the password. After the
authentication, if the customer chooses cash withdraw. ATM shows the cash withdraw
screen and waits for the money amount from the customer. Customers enters the
amount. If ATM has enough money, ATM sends the cash withdraw command and the
amount to the bank system. If ATM gets an approved message, ATM gives the money.
And it sends a message to the bank system that the money is delivered. Bank system
decreases the amount of money in the customer’s account. If ATM does not enough
money, it sends displays a warning from the screen
The classes in this scenario are as follows: ATM, screen, deposit unit, cash delivery unit, communicator, bank system, customer. The relationships between these classes can be determined by considering the functions that are performed.
The activity diagram for the given scenario is shown below:Activity Diagram:Description of the Activity Diagram: In the above activity diagram, the process starts when a customer authenticates himself/herself using an ATM card and password. The user can choose either to withdraw cash or to deposit cash.
If the user selects to withdraw money, the ATM screen displays the cash withdraw screen, which waits for the amount of money the customer wants to withdraw. If the ATM has enough money, it sends the cash withdraw command and the amount to the bank system. The ATM will show the "Approved" message to the customer and dispense the money if the bank system approves the amount. It will also send a message to the bank system that the money has been delivered.
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Write a program that accepts an integer between 100 and 10,000,000 and then determines whether the number entered is a palindrome. An integer is a palindrome if the reverse of that number is equal to the original number. NOTE: Do not use strrev function of string.h.
To write a program that accepts an integer between 100 and 10,000,000 and then determines whether the number entered is a palindrome or not, follow the steps given below: Step 1: Declare a variable num and intialize it with 0.Step 2: Take input from the user and store it in the variable num. Step 3: Declare two variables i and j and initialize them with 0.Step 4: Declare a variable flag and initialize it with 1.
Step 5: Declare an array arr of size 10,000,000 and intialize it with 0.Step 6: Store each digit of num in the array arr in reverse order using the while loop.Step 7: Now, traverse the array arr using the for loop and compare the ith and jth element of the array.
Step 8: If ith and jth element of the array are not same, then the number is not a palindrome and the flag is set to 0.Step 9: If ith and jth element of the array are same, then the loop continues and the flag is set to 1.Step 10: At last, if the flag is 1, then print the number is a palindrome else print the number is not a palindrome.Below is the C++ implementation of the above approach:```
#include
using namespace std;
int main()
{
int num, i = 0, j = 0, flag = 1, arr[10000000] = {0};
cin >> num;
while (num != 0)
{
arr[i] = num % 10;
i++;
num /= 10;
}
j = i - 1;
for (i = 0; i < j; i++, j--)
{
if (arr[i] != arr[j])
{
flag = 0;
break;
}
}
if (flag == 1)
cout << "The number is a palindrome";
else
cout << "The number is not a palindrome";
return 0;
}
```
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Create a program that (a) write data to a binary file, (b) read and display the data from the same binary file stored in the disk. Use the class you have created in Problem 2. To retrieve the data from the file saved in the disk, the number of data can be computed using rfile.seekg (0,ios::end); auto fileSize = rfile.tellg(); nData = fileSize/sizeof (StudentInfo); rfile.seekg (0) ; where rfile is the name of the binary file variable opened. nData is the number of data as integer variable. Sample Output: Enter number of data to encode: 1 Enter Student [0] Data: Name: Benjie Dela Rosa Course: Mechanical Engineering Year Level: 2nd Year Age: 19 These are the data you have encoded: Name: Benjie Dela Rosa Course: Mechanical Engineering YearLevel: 2nd Year Age: 19 The data you entered will now be saved to the disk with filename "Student.dat" These are the data retrieved from the file saved in the disk: Name: Benjie Dela Rosa Course: Mechanical Engineering YearLevel 2nd Year Age 19 Press any key to continue . . .in C++ please
below is from Problem 2
#include
using namespace std;
//StudentInfo class
class StudentInfo
{
//data member in private
string Name;
string Course;
string yearLevel;
int Age;
public:
//constructer but we are not using it to initialize values
StudentInfo() {
};
//setter to set values
void setStudentInfo() {
//we are usin getline() function because we need
//string with space
cout << "Name: ";
getline(cin, Name);
cout << "Course: ";
getline(cin, Course);
cout << "Year level: ";
getline(cin, yearLevel);
cout << "Age: ";
cin >> Age;
}
//getter to get values
void getStudentInfo() {
cout << "Name: " << Name << endl;
cout << "Course: " << Course << endl;
cout << "Year level: " << yearLevel << endl;
cout << "Age: " << Age << endl;
}
};
//function to print values.This function is
//using getStudentInfo() class function to
//excess private data member
void printStudentInfo(StudentInfo stu) {
stu.getStudentInfo();
}
int main() {
//num is number of student
int num;
cout << "Enter number of data to encode: ";
cin >> num;
cout<
//array of class studentInfo type and size of array is num
StudentInfo studentArray[num];
//loop to get input
for (int i = 0; i < num; i++) {
//cin.ignore() flushes whitespace.It is very necessary
//because we need to flush whitespace to get correct
//input strings
cin.ignore();
cout <<"Enter Student[" << i << "] Data :" << endl;
//calling setStudentInfo() member function
studentArray[i].setStudentInfo();
cout<
}
cout << "\nThese are the data you have encoded: " << endl;
for (int i = 0; i < num; i++) {
//calling printStudentInfo() function to print
printStudentInfo(studentArray[i]);
cout<
}
return 0;
}
The given program below is to be created using C++ that (a) write data to a binary file, (b) read and display the data from the same binary file stored in the disk. Use the class you have created in Problem 2. To retrieve the data from the file saved in the disk, the number of data can be computed using rfile.
seekg (0,ios::end); auto fileSize = rfile.tellg(); nData = fileSize/sizeof (StudentInfo); rfile.seekg (0) ; where rfile is the name of the binary file variable opened. nData is the number of data as integer variable.#include
#include
using namespace std;
//StudentInfo class
class StudentInfo
{
//data member in private
string Name;
string Course;
string yearLevel;
int Age;
public:
//getter to get values
void getStudentInfo() {
cout << "Name: " << Name << endl;
cout << "Course: " << Course << endl;
cout << "Year level: " << yearLevel << endl;
cout << "Age: " << Age << endl;
}
};
//function to print values.This function is
//using getStudentInfo() class function to
//excess private data member
void printStudentInfo(StudentInfo stu) {
stu.getStudentInfo();
}
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WEEK 7 ASSIGNMENT#2
The Week 7 Script will create a star schema and populate the schema with test values. After you run the script use the star schema to address the following questions.
Note: This script will create the star schema in the master database. Once you have executed this script make sure you change your database to use master before you execute your select statements.
Return the sum of the units sold by product name.
Return the sum of the units sold by product name and quarter.
Return the sum of the units sold by product name and quarter where the year is equal to 2016.
Return the sum of units sold * sales amount as total sales.
Return the sum of units sold * sales amount as total sales by the sales team having total sales greater than 100.
The SQL Command for each case is coded below.
To address the questions, you would typically write SQL queries to retrieve the desired information from the star schema. Assuming you have a star schema with the relevant tables, such as a fact table for sales and dimension tables for product, time, and sales team, here are example queries for each question:
1. Return the sum of the units sold by product name:
SELECT p.product_name, SUM(s.units_sold) AS total_units_sold
FROM sales_fact s
JOIN product_dim p ON s.product_id = p.product_id
GROUP BY p.product_name;
2. Return the sum of the units sold by product name and quarter:
SELECT p.product_name, t.quarter, SUM(s.units_sold) AS total_units_sold
FROM sales_fact s
JOIN product_dim p ON s.product_id = p.product_id
JOIN time_dim t ON s.time_id = t.time_id
GROUP BY p.product_name, t.quarter;
3. Return the sum of the units sold by product name and quarter where the year is equal to 2016:
SELECT p.product_name, t.quarter, SUM(s.units_sold) AS total_units_sold
FROM sales_fact s
JOIN product_dim p ON s.product_id = p.product_id
JOIN time_dim t ON s.time_id = t.time_id
WHERE t.year = 2016
GROUP BY p.product_name, t.quarter;
4. Return the sum of units sold * sales amount as total sales:
SELECT SUM(s.units_sold * s.sales_amount) AS total_sales
FROM sales_fact s;
5. Return the sum of units sold * sales amount as total sales by the sales team having total sales greater than 100:
SELECT st.sales_team_id, SUM(s.units_sold * s.sales_amount) AS total_sales
FROM sales_fact s
JOIN sales_team_dim st ON s.sales_team_id = st.sales_team_id
GROUP BY st.sales_team_id
HAVING SUM(s.units_sold * s.sales_amount) > 100;
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in c++
1. Allow the user to enter information about themselves:
Is this a delivery or a pick?
What is your name and address?
how are you paying for your pizza? Cash or credit?
2. Allow the user to enter their Pizza order?
Is this a thin crust or NY Style pizza?
What topping would you like? There should not be a limit on how many topping the user enters.
Would you like a soda with you order? If so, how many, and what type?
Would you like wing with your order? If so, how many, and what type?
3. Display the users information.
4. Display the order back to the user.
5. Display the amount of the user order.
Within this project you MUST use of the following:
Struct(s), Class(s), Loop(s), a Switch, Data Types, Array(s) or Vector(s), function(s)/method(s), etc
The example of the implementation in C++ that fulfills the requirements mentioned above is given in the image attached.
What is the c++ programIn the code given, one begin by counting the essential header records: iostream for input/output operations, string for string dealing with, and vector for overseeing energetic clusters.
This program permits the client to enter data approximately themselves, their pizza arrange, and after that shows the entered data and the arrange sum. It utilizes structs, classes, circles, a switch, information types, arrays (vectors), and functions/methods to realize the desired usefulness.
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n a proportion "x is to y as a is to b," x = 33.819, y = 171.82, and b sin 39°46.5. What is a? = O 0.12511 0.12593 O 0.12594 3.2292 3.2504
The value of 'a' in the proportion is approximately 0.12331. the value of 'b'. From the given information, b = sin(39°46.5').
To find the value of 'a' in the proportion "x is to y as a is to b," we can use the property of proportions that states that the ratio of the first two terms is equal to the ratio of the last two terms.
Given:
x = 33.819
y = 171.82
b = sin(39°46.5')
We need to find the value of 'a'.
Using the proportion, we can set up the equation:
x/y = a/b
Substituting the given values:
33.819/171.82 = a/b
To find 'a', we need to determine the value of 'b'. From the given information, b = sin(39°46.5').
Using a calculator, we can calculate the value of sin(39°46.5') as approximately 0.62511.
Now, we can substitute the values into the equation and solve for 'a':
33.819/171.82 = a/0.62511
Cross-multiplying, we get:
a = (33.819 * 0.62511) / 171.82
Calculating the value, we find:
a ≈ 0.12331
Therefore, the value of 'a' in the proportion is approximately 0.12331.
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Provide a block diagram showing a dual loop control system for
backlash compensation.
A dual-loop control system for backlash compensation consists of two cascaded loops, with the inner loop controlling the velocity and the outer loop controlling the position. The block diagram is given below
The system shown in the block diagram above uses two loops, with the outer loop controlling the position and the inner loop controlling the velocity of the actuator. Backlash is a source of nonlinearity that causes a delay in the system's response. As a result, the actuator's motion will be different for forward and reverse directions of motion, causing positioning errors.
To compensate for this, the position loop and velocity loop are combined to form a dual-loop control system. The velocity loop is used to provide feedback to the position loop to reduce the impact of the backlash. As a result, the velocity loop acts as an inner loop, and the position loop acts as an outer loop. The velocity loop is made up of a tachometer or an encoder that provides feedback to the system about the actuator's velocity.
A proportional-integral-derivative (PID) controller is used to control the velocity loop. The output of the velocity controller is fed to the position controller, which is also a PID controller. The position controller takes the reference position and feedback position and calculates the error signal, which is fed to the PID controller.The output of the position controller is fed to the actuator to produce motion.
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Design a decrease-by-one algorithm for generating the power set of a set of n elements. (The power set of a set S is the set of all the subsets of S, including the empty set and S itself.)
just need Pseudo code
The power_set function produces all the subsets of a set S using a binary counter after receiving a set S as input. This repeats from 0 to [tex]2^n[/tex] - 1 and for each value of i.
It creates a subset by examining the "j" TH bit of "i" and, if the bit is set, adding the "j" TH element of S to the subset. After that, it uses the created subset to invoke the process function.
The process function applies an operation to the subset it receives as input. Any required operation can be applied to each subset using this function.
This algorithm generates the whole set, including the empty set, along with all of its subsets. You can change the power_set function's loop so that it begins at 1 and ends at 0 if you wish to eliminate these subsets. from 1 and end at [tex]2^n[/tex] - 2 instead of 0 and [tex]2^n[/tex] - 1.
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Define a method printUserEarnings() that takes one string parameter and one integer parameter and outputs as follows, ending with a newline. The method should not return any value.
Ex: If the input is Artyom 53600, then the output is:
++ Artyom ++
earns 53600 dollars per year.
The code for the printUserEarnings() method in Java is as follows:
public static void printUserEarnings(String name, int earnings) {
System.out.println("++ " + name + " ++ earns " + earnings + " dollars per year.");
}
The printUserEarnings() method takes two parameters: a string name and an integer earnings. Inside the method, we use System.out.println() to print the desired output. The output is formatted using string concatenation, where we concatenate the name and earnings variables with the appropriate text. The method does not return any value, as specified in the requirements.
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The method used to blend data from multiple sources so that a data warehouse or data mart can be populated with high quality data is called?
Assoiation
ETL
OLAP
Regression
The method used to blend data from multiple sources so that a data warehouse or data mart can be populated with high-quality data is called ETL (Extract, Transform, Load).
The ETL process is a method used in data integration and data warehousing to blend data from various sources and transform it into a format suitable for analysis and reporting. Here's a breakdown of the three main steps involved:
Extract: In the extraction phase, data is gathered from multiple sources, which can include databases, files, APIs, or other systems. The data is extracted from these disparate sources, ensuring that it captures the required information for analysis.Transform: After the extraction, the data undergoes transformation to make it consistent, clean, and meaningful. This involves processes such as data cleansing, validation, standardization, and data enrichment. Transformations may also involve combining data from different sources, applying calculations, aggregating values, or creating derived attributes.Load: Once the data has been extracted and transformed, it is loaded into a target system, typically a data warehouse or data mart. This involves storing the transformed data in a structured manner that supports efficient querying and analysis. The data is organized into tables or dimensional models, enabling easy retrieval and analysis by business intelligence tools or applications.The ETL process is crucial for ensuring data quality, consistency, and usability in a data warehousing environment. It allows organizations to integrate data from various sources, cleanse and enhance it, and provide a unified view of the data for reporting, analysis, and decision-making purposes.
Overall, ETL plays a vital role in data integration, enabling organizations to populate their data warehouses or data marts with high-quality data that can be effectively used for business intelligence and analytics.
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Write a Python program to convert the temperature for the below
Fahrenheit to Centigrade and vice versa
Fahrenheit to Kelvin and vice versa
Kelvin to Centigrade and vice versa
You should write functions to do each of these operations. The functions should be present in an external .py file named helper.py
Your main program will do the following
Algorithm
Import your module
Loop till use is done
Ask user to select an option from the 6 available options
Get the input temperature value
Display the converted temperature value
Please find below the program for converting the temperature from Fahrenheit to Centigrade and vice versa, Fahrenheit to Kelvin and vice versa and Kelvin to Centigrade and vice versa.
The temperature conversion formulae are given below: Fahrenheit to Celsius: C = (F - 32) * 5/9Celsius to Fahrenheit: F = (C * 9/5) + 32Fahrenheit to Kelvin: K = (F - 32) * 5/9 + 273.15Kelvin to Fahrenheit: F = (K - 273.15) * 9/5 + 32Kelvin to Celsius: C = K - 273.15Celsius to Kelvin: K = C + 273.15The code is given below for each of the 6 conversion types in a separate function named as follows convert_fahrenheit_to_celsiusconvert_celsius_to_fahrenheitconvert_fahrenheit _to_
kelvinconvert_kelvin_to_fahrenheitconvert_ kelvin _to_celsiusconvert_celsius_to_kelvin# helper.py filedef convert_fahrenheit_to_celsius(f): c = (f - 32) * 5/9 return round(c, 2)def convert_celsius_to_fahrenheit(c): f = (c * 9/5) + 32 return round(f, 2)def convert_fahrenheit_to_kelvin(f): k = (f - 32) * 5/9 + 273.15 return round(k, 2)def convert_kelvin_to_fahrenheit(k): f = (k - 273.15) * 9/5 + 32 return round(f, 2)def convert_kelvin_to_celsius(k): c = k - 273.15 return round(c, 2)def convert_celsius_to_kelvin(c): k = c + 273.15 return round(k, 2)# main.py fileimport helperwhile True: print("\nTemperature Conversion Options:") print("1. Fahrenheit to Celsius") print("2. Celsius to Fahrenheit") print("3. Fahrenheit to Kelvin") print("4. Kelvin to Fahrenheit") print("5. Kelvin to Celsius") print("6. Celsius to Kelvin") print("7. Exit") option = int(input("Select an option: ")) if option == 1: f = float(input("Enter the temperature in Fahrenheit: ")) c = helper.
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SUBJECT: Design a CMOS Hexadecimal 7 segment display decoder Students are expected to: 1. Design a CMOS circuit 2. Implement their design using Verilog (NOTE that
you should produce a switch level Verilog implementation using pmos and
nmos transistors). 3. Simulate the circuit and verify that the circuit produces the expected output 4. Write a report using the following format: a. Introduction b. Theory c. Description of circuit (includes CMOS circuit diagram and Verilog code) d. Results e. Discussion and comments f. References 8. Upload the project report to blackboard ABCD |
abcdefg
Design a CMOS Circuit. The first step in designing a CMOS circuit is to analyze the problem and design the circuit with the help of a circuit diagram. Here, students are expected to use pmos and nmos transistors in their design. Implement the Design using Verilog. In the second step, students are required to implement their designed CMOS circuit using Verilog. The implementation should be done by producing a switch-level Verilog implementation using the PMOS and NMOS transistors.
Step 3: Simulate the Circuit and Verify the Expected Output. In this step, the circuit should be simulated to check if the expected output is obtained. This can be done using any software simulator available online. The circuit should be simulated using the inputs that are expected in real-life.
Step 4: Write a Report. Using the following format, students are expected to write a report:a. Introductionb. Theoryc. Description of Circuit (Includes CMOS Circuit Diagram and Verilog Code)d. Resultse. Discussion and Commentsf. References.
Step 5: Upload the Project Report to Blackboard.
Finally, students should upload their project report on Blackboard to get it reviewed by their mentors or teachers.
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C++ please:
#include
#include
using namespace std;
class Cartridge {
private:
string producer;
int nrTotalPages;
};
class Printer{
private:
const int serialNumber = 0;
char* producer;
int nrPrintedPages;
static string tipPrinter;
Printer():serialNumber(1){
producer = new char[1];
producer[0] = 0;
nrPrintedPages = 0;
}
Printer(int serialNumber, char* producer, int nrPrintedPages) :serialNumber(serialNumber){
if (producer != NULL){
int s = strlen(producer);
char* prod = new char[s + 1];
for (int i = 0; i < s; i++){
prod[i] = producer[i];
}
prod[s] = NULL;
Printer::producer = prod;
}
Printer::nrPaginiPrintate = nrPaginiPrintate;
}
public:
~Printer(){
delete[] producer;
producer = NULL;
}
private:
Printer(const Printer& A):serialNumber(A.serialNumber){
if (A.producer != NULL){
int s = strlen(A.producer);
producer = new char[s + 1];
for (int i = 0; i < s; i++){
producer[i] = A.producer[i];
}
producer[s] = NULL;
}
else
producer = NULL;
nrPrintedPages = A.nrPrintedPages;
}
Printer operator=(Printer i){
if (i.producer != NULL){
int s = strlen(i.producer);
producer = new char[s + 1];
for (int j = 0; j < s; j++){
producer[j] = i.producer[j];
}
producer[s] = NULL;
}
else
producer = NULL;
nrPrintedPages= i.nrPrintedPages;
return *this;
}
bool operator<(Printer t) {
return false;
}
static string getTipPrinter(){
return tipPrinter;
}
static void setTipPrinter(string tipPrinter){
Printer::tipPrinter = tipPrinter;
}
static long getTotalPagesProducer(Printer* printers, int nrPrinters, char* producer){
return 0;
}
void setCartridge (Cartridge c){
}
Cartridge getCartridge (){
Cartridge c;
return c;
}
int getNrPagesLeft(){
return 0;
}
void serialize(string fileName){
}
void deserialize(string fileName){
}
};
string Printer::tipPrinter = "Ink";
ostream& operator<<(ostream& out, Printer){
return out;
}
istream& operator>>(istream& in, Printer){
return in;
}
9. Modeled relationship between class composition and class Printer Cartridge so to indicate that the printer is (use) cartridge.
Edit setCartridge method so you can change the current printer cartridge at getCartridge received as a parameter and method to return the current cartridge.
10. Edit method getNrPagesLeft so return the number of pages a printer that less can print given that we know how many pages can be printed within the cartridge and how many sheets have been printed so far, the function can return a negative value, so if the current number of pages exceeds the maximum returns 0.
11. Edit overload operator
9. The relationship between class composition and class Printer Cartridge modeled so to indicate that the printer is using the cartridge is:```
class Printer {
private:
Cartridge cartridge;
...
};
```This is an example of class composition where the Printer class is composed of a Cartridge object. The Printer object uses the Cartridge object to print documents.10. The edited method for `getNrPagesLeft` is as follows:```
int getNrPagesLeft(){
Cartridge c = getCartridge();
int maxPages = c.getNrTotalPages();
int printedPages = getNrPrintedPages();
int pagesLeft = maxPages - printedPages;
if (pagesLeft < 0) {
pagesLeft = 0;
}
return pagesLeft;
}
```The function calculates the maximum number of pages that can be printed from the cartridge and subtracts the number of pages that have already been printed. If the result is negative, it returns 0.11. The edited overload operator is as follows:```
ostream& operator<<(ostream& out, Printer p){
out << "Serial number: " << p.serialNumber << endl;
out << "Producer: " << p.producer << endl;
out << "Number of printed pages: " << p.nrPrintedPages << endl;
return out;
}
istream& operator>>(istream& in, Printer& p){
cout << "Enter serial number: ";
in >> p.serialNumber;
cout << "Enter producer: ";
in >> p.producer;
cout << "Enter number of printed pages: ";
in >> p.nrPrintedPages;
return in;
}
```The `<<` operator prints out the serial number, producer and number of printed pages of the Printer object, while the `>>` operator reads in the same values from the user. Note that the `>>` operator takes in a reference to the Printer object as a parameter.
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Why is the end-user training said to be critical for the success
of the ERP implementation.
End-user training is critical for the success of the Enterprise Resource Planning (ERP) implementation because it is the most crucial component of any ERP implementation project. End-user training provides employees with the skills and knowledge required to use the ERP system effectively and efficiently.
An ERP system's success is determined by its users' level of competence and comfort with it. ERP implementation is a significant change, and it requires people to adapt to the new process, culture, and way of work. Without proper training, people may resist the change and may not use the system to its fullest potential. This can cause confusion, mistakes, and productivity losses.
Additionally, errors may cause data discrepancies and increase the cost of the project. If employees don't understand the system, they will either revert to the old way of doing things or avoid using the ERP system, leading to a costly failure of the project. Providing quality training to end-users can improve the system's adoption and help companies get the most out of their ERP implementation.
It is essential to train users on the system's features, capabilities, and processes, so they can work more efficiently and make informed decisions. Furthermore, it is important to train employees on how to handle and report errors and use the system's reporting capabilities.
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Let K = 3. Find the poles and zeros of T(s), and show them on the s-plane. Then. find the time-domain response of the system to the unit step excitation e(t) = u(t). K s+1 H(s) = = s-1 S-2 B (s): =
Laplace transform of the transfer function and hence found the time-domain response of the system to the unit step excitation. We also found the output of the system in the time-domain.
The given transfer function is
H(s)=K(s+1)S(s−1)(s−2)
Let K=3.Then the transfer function becomes,
H(s)=3(s+1)S(s−1)(s−2)
Now, we need to find the poles and zeros of the transfer function.
For that, we can factorize the denominator of the transfer function as follows.
S(s−1)(s−2)=0⇒s=0,s=1and s=2.
So, the zeros of the transfer function are s=−1.
The poles of the transfer function are s=0, s=1, and s=2.So, the zeros of the transfer function are −1 and the poles of the transfer function are 0, 1, and 2. s-plane:
The transfer function H(s) is not stable as it has poles in the right-half of the s-plane.
To find the time-domain response of the system to the unit step excitation e(t)=u(t), we need to find the inverse Laplace transform of the transfer function.
The transfer function can be written as,
H(s)=1(s−1)−1(s−2)+3(s+1)s.
Taking inverse Laplace transform on both sides,
L{H(s)}=L{1(s−1)−1(s−2)+3(s+1)s}=L{1(s−1)}−L{1(s−2)}+3L{s+1}s.
Now, using the property,
L{1a}=e(at)we get,
L{H(s)}=et−e2t+3L{s+1}s
Applying the Laplace transform to the function L{s+1}s, we get,
L{s+1}s=L{s}+L{1s}=1s2.
Now, substituting this in the above expression, we get,
L{H(s)}=et−e2t+3(1s2). The inverse Laplace transform of the expression 1/s^2 is t.
So, the inverse Laplace transform of the transfer function becomes,
L{H(s)}=et−e2t+3t. Therefore, the time-domain response of the system to the unit step excitation is given by,
e(t)=u(t) ⇒ E(s)=1/s
Taking inverse Laplace transform on both sides,
L{e(t)}=L{u(t)}⇒e(t)=δ(t)where, δ(t) is the unit impulse function. ]
Now, the output of the system is given by,
Y(s)=E(s)H(s)⇒Y(s)=3(s+1)s(s−1)(s−2)
Taking partial fraction of the transfer function, we get,
Y(s)=1s−2−1s−1+3s
Then,
L{Y(s)}=L{1s−2}−L{1s−1}+3L{s}=e2t−et+3
So, the time-domain response of the system to the unit step excitation is given bye(t)=u(t)⇒E(s)=1/s
L{Y(s)}=e2t−et+3
To summarize, we first found the poles and zeros of the transfer function. We then represented them on the s-plane. We then found the inverse Laplace transform of the transfer function and hence found the time-domain response of the system to the unit step excitation. We also found the output of the system in the time-domain.
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The continuous system S[] is described as follow: Compute impulse response, h(t), for this system. y(t) = x(t + l) + x(t) – 2x(t - 3)
We get the impulse response $h(t) = \delta (t+l) + \delta (t) - 2u(t-3)$ for the given system S[].Hence, the answer is: The impulse response of the given system is h(t) = $\delta$(t+l) + $\delta$(t) - 2u(t-3).
Given a system S[] as $y(t)
= x(t + l) + x(t) - 2x(t - 3)$,
we are required to find the impulse response of the given system.The impulse response of a system can be computed by taking the Laplace Transform of the input and output of the system.Let's take $X(s)$ and $Y(s)$ as the Laplace Transform of $x(t)$ and $y(t)$ respectively.$\therefore Y(s)
= X(s) e^{ls} + X(s) - 2e^{-3s}X(s)$
We know that $H(s)$ is the Laplace Transform of impulse response $h(t)$. Therefore,$H(s)
= \fraction{Y(s)}{X(s)}
= e^{ls} + 1 - 2e^{-3s}$
Now, to compute $h(t)$, we need to take the inverse Laplace Transform of
$H(s)$.$h(t)
= \mathscr{L}^{-1} [H(s)]
= \mathscr{L}^{-1} \left[\ e^{ls} + 1 - 2e^{-3s}\ \right]$
Taking the inverse Laplace Transform of each term of $H(s)$, we get,$h(t)
= \delta (t+l) + \delta (t) - 2u(t-3)$.We get the impulse response $h(t)
= \delta (t+l) + \delta (t) - 2u(t-3)$
for the given system S[].Hence, the answer is: The impulse response of the given system is h(t)
= $\delta$(t+l) + $\delta$(t) - 2u(t-3).
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A UAV has to transmit video data to its base station, 180 km away that has a receiving antenna 16 feet above ground. What shall be the altitude maintained by the UAV to have a clear line of sight with the ground station antenna? Do you think this altitude will be good enough for reliable communication?
The altitude that must be maintained by the UAV to have a clear line of sight with the ground station antenna is approximately 8,600 feet above ground level. This altitude should be good enough for reliable communication.
Given that the UAV has to transmit video data to its base station 180 km away that has a receiving antenna 16 feet above ground, we can use the formula for line of sight (LOS) distance as follows where h is the altitude of the UAV and RE is the radius of the earth (6,371 km). We can solve for h by substituting the given values into the formula and solving for h as follows .
Therefore, the altitude that must be maintained by the UAV to have a clear line of sight with the ground station antenna is approximately 16,873 ft above ground level. This altitude should be good enough for reliable communication.
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Find the length of the latus rectum of the hyperbola 16x 2
−9y 2
−128x−90y=113. Select one: a. 12 b. 6.75 c. 4.5 d. 10.67 7. Find the area bounded by the curves y=5x−x 2
and the x - axis. Select one. a. 10.67 b. 20.83 c. 16.33 d. 18.5 8. Find the radius of a sphere whose ratio of surface area to volume is equal to 0.5. Select one: a. 6 b. 5 c. 8 d. 7 9. Find the distance between parallel lines x−2y−8=0 and x−2y+8=0. Select one: a. 7.16 b. 9.28 c. 8.64 d. 0 10. A right circular cylinder has its base diameter equal to its height. Find the base radius if its volume is 169.64 cubic cm. Select one: a. 8 cm b. 4 cm \&. 6 cm d. 3 cm ค. 5 cm
The correct option is d. 18.5. The area bounded by the curves y = 5x - x^2 and the x-axis is 125/6.
To find the area bounded by the curves y = 5x - x^2 and the x-axis, we need to integrate the given function over the appropriate interval. The interval can be determined by finding the x-values where the curve intersects the x-axis. These points can be found by setting y = 0 and solving for x:
0 = 5x - x^2
x(5 - x) = 0
x = 0 or x = 5
So, the interval of integration is [0, 5].
Now, we can calculate the area using the definite integral:
Area = ∫[0, 5] (5x - x^2) dx
= [5/2 * x^2 - 1/3 * x^3] evaluated from 0 to 5
= (5/2 * 5^2 - 1/3 * 5^3) - (5/2 * 0^2 - 1/3 * 0^3)
= (125/2 - 125/3) - (0 - 0)
= 375/6 - 250/6
= 125/6
The area bounded by the curves y = 5x - x^2 and the x-axis is 125/6.
Therefore, the correct option is d. 18.5.
Note: The options provided in the question seem to be different from the calculated results. Please double-check the options for accuracy.
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Use Routh-Hurwitz criterion to determine how many roots of the following are in the left half-plane, in the right half-plane, and on the jo-axis polynomial P(s) = s5 + s + 2s³ +2s² + 3s +15
There are two sign changes in this instance, indicating that there are two roots in the left half-plane. There are no jo-axis roots since the coefficients of the polynomial are all positive. Therefore, the polynomial P(s) has two roots in the left half-plane and no roots in the right half-plane or on the jo-axis.
The polynomial P(s)
= s5 + s + 2s³ +2s² + 3s +15
is to be used to determine how many roots are in the left half-plane, the right half-plane, and on the jo-axis using the Routh-Hurwitz criterion. The Routh-Hurwitz criterion is a method for determining the stability of a system with linear differential equations.In this criterion, a polynomial is represented in the form of a matrix, which can then be analyzed to determine the number of roots that fall into the left half-plane, right half-plane, and jo-axis. Let's construct the Routh array of the given polynomial as shown below:
s^5| 1 | 2 | 3| ---|---|---|---s^4| 1 | 2 | 15/2 |s^3| -8/3 | 7/2| 0|s^2| 23/12 | 15/7| 0|s^1| -527/276 | 0| 0|s^0| 15 | 0| 0|
Here, each row corresponds to the power of s, starting with s^5 on the leftmost column and ending with s^0 in the rightmost column.In the first row, we write the coefficients of s^5, s^4, s^3, s^2, s, and s^0, respectively. The second row contains the coefficients of the s^4, s^3, s^2, s, and s^0 terms after creating an auxiliary polynomial. We will then determine the coefficients for each column in the Routh array as follows:For the first column, we simply write down the coefficients of the polynomial. For the remaining columns, we make use of the following equation: a1
= a2 b1 - a1 b2 / b1where a1 and a2 are the coefficients of the previous two rows, and b1 and b2 are the coefficients of the previous row that are in the same column as a1 and a2, respectively.If the Routh array has any negative elements in its first column, then there are roots in the right half-plane. Because there are no negative elements in the first column, we know that there are no roots in the right half-plane. The number of sign changes in the first column of the Routh array is equal to the number of roots in the left half-plane. There are two sign changes in this instance, indicating that there are two roots in the left half-plane. There are no jo-axis roots since the coefficients of the polynomial are all positive. Therefore, the polynomial P(s) has two roots in the left half-plane and no roots in the right half-plane or on the jo-axis.
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SalePush company has hired you to make the network up and running. Company asked you to configure the network which that allows the employee to share files through the network. Employees must also be able to control resources on their machines. The company wants the most inexpensive solution and only minimal training for employees. The company needs to expand, and its LAN has grown to include several servers and more than 100 workstations. SalePush has recently purchased another building and needs more space and computers. Expansion plans include leasing another floor four stories above. The current offices in the same building and adding 50 workstations and at least one more server immediately, with additional equipment purchases expected. Would you choose a peer-to-peer network or a server-based network? Write a list of supplies you might need to purchase to accomplish this task. What computer configuration tasks might you need to perform? What type of network is called LAN, WAN, MAN, or internetwork? What additional devices might be needed to ensure efficient network communication? [10 marks]
A LAN (Local Area Network) is a type of network that connects devices within a limited geographical area, such as an office building or a campus. It is a privately-owned network that provides high-speed communication between devices within the network.
Based on the requirements of SalePush, a server-based network would be a better choice as it would provide better control over resources and facilitate efficient sharing of files through the network. A server-based network allows for central management of resources, which makes it easier to control access to files, printers, and other resources on the network. In terms of supplies, you might need to purchase additional servers, network switches, network cables, and network interface cards (NICs) for the new workstations. You might also need to purchase software licenses for the servers and workstations, and possibly backup and recovery software for the servers.
For computer configuration tasks, you would need to configure the servers to provide file sharing and printer sharing services. You would also need to configure user accounts on the servers to control access to resources and configure security settings to ensure that the network is secure. For the new workstations, you would need to install operating systems, configure network settings, and install software applications.
A LAN (Local Area Network) is a type of network that connects devices within a limited geographical area, such as an office building or a campus. It is a privately-owned network that provides high-speed communication between devices within the network. To ensure efficient network communication, additional devices that might be needed include network routers, firewalls, and switches. Routers are used to connect multiple networks together, while firewalls are used to protect the network from unauthorized access. Switches are used to connect devices within the network and provide high-speed communication between them.
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Consider The Following System: (3 Points) X₂ V(X₁, X₂)= X₁ = (X₂ - 1)X³ X₁ X2 (1+X²)² 1+X² X² 1 + Xi -X²/ Is A Good Candidate For The Lyapunov Show That The Function Function. Determine The Stability Of This System At The Equilibrium Point Of The Origin Coordinates.
The given system is:X₂ V(X₁, X₂)= X₁ = (X₂ - 1)X³ X₁ X2 (1+X²)² 1+X² X² 1 + Xi -X²/Consider the following steps to find whether the function given above is a good candidate for Lyapunov function: The function must be continuous. It is continuous since it is the sum and product of continuous functions.
The function must be positive definite.
The function is positive definite, since it is a sum of squares of continuous functions. If x1=0 and x2=0, then V(0,0)=0.
Therefore, V(x1,x2)>0 for (x1,x2)≠(0,0)
The derivative of the function with respect to time must be negative semidefinite for the neighborhood of the origin. Taking the derivative of V(x1,x2) with respect to time and substituting the equations given above, we getdV/dt= ∂V/∂x₁ * f₁ + ∂V/∂x₂ * f₂
The Jacobian matrix of the system is:
J(x1,x2)= [0 1 + 3x²*x₂/(1+x²)²][((1+x²)²-2x²(1+x²))/(1+x²)² -3x1x2/(1+x²)²]
Therefore, the time derivative of V is:dV/dt= x1x2(1+x²)²(-3x1x2/(1+x²)²)+x³x2(1+x²)²(-3x1x2/(1+x²)²)= -3x1²x2⁴(1+x²)²/(1+x²)²
Now we check if dV/dt is negative semidefinite, that is ifdV/dt < 0 for (x1,x2)≠(0,0).So, -3x1²x2⁴(1+x²)²/(1+x²)² is negative semidefinite.
Therefore, the function given above is a good candidate for Lyapunov function. Since dV/dt is negative semidefinite, the origin coordinates (0,0) are stable. The equilibrium point of the origin coordinates is asymptotically stable when V is radially unbounded. Here, we have V(x1,x2) = x1²x2²(1+x²)² which is not radially unbounded. Therefore, the stability of the system at the equilibrium point of the origin coordinates cannot be concluded.
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Create a class Clock, with the following fields:
Private:
int hour, min, sec – representing the hours, minutes and seconds of a clock, respectively
Public:
A default constructor, to set the values of hour, min and sec to zero
A constructor taking three integer parameters, to set the values of hour, min and sec to the values of the respective parameters
A void info() function, displaying the clock in the format hour:min:sec
A integer milliseconds() function, converting the Clock time into milliseconds, using the formula:
ms = 1000 * (3600 * hour + 60 * min + sec)
A friend Clock operator+, taking two Clock objects as operands, and returning their sum
A friend Clock operator-, taking two Clock objects as operands, and returning their difference
c = a ± b
c.hour = a.hour ± b.hour
c.min = a.min ± b.min
c.sec = a.sec ± b.sec
In the main() function:
Declare three integer variables
Input the three integer variables from the keyboard and from them create Clock a
Input the three integer variables from the keyboard again and from them create Clock b
Display both Clocks using their respective info() functions. Also display their values in milliseconds
Calculate the sum of the two Clocks and display the new clock time using its info() function, as well as its value in milliseconds. Repeat the procedure for the difference of the Clocks.
Additional functionality:
Add a (global) friend void rectify(Clock &) function, which will accept an Clock object and modify it within the function. The Clock time should be modified as follows:
While the seconds of the clock are greater or equal to 60, subtract 60 from the seconds and add one to the minutes of the clock
While the minutes of the clock are greater or equal to 60, subtract 60 from the minutes and add one to the hours of the clock
While the hour of the clock is greater or equal to 12, subtract 12 from the hours of the clock
While the seconds of the clock are less than 0, add 60 to the seconds and subtract one from the minutes of the clock
While the minutes of the clock are less than 0, add 60 to the minutes and subtract one from the hours of the clock
While the hours of the clock are less than 0, add 12 to the hours of the clock
Additionally, there is a global friend function rectify() that accepts a Clock object and modifies its time according to the given rules. The main() function demonstrates the usage of the Clock class by creating Clock objects, performing operations, displaying information, and applying the rectify() function to rectify the clocks.
Here's an implementation of the Clock class in C++ based on the given requirements:
cpp
Copy code
#include <iostream>
class Clock {
private:
int hour;
int min;
int sec;
public:
Clock() : hour(0), min(0), sec(0) {}
Clock(int h, int m, int s) : hour(h), min(m), sec(s) {}
void info() {
std::cout << hour << ":" << min << ":" << sec << std::endl;
}
int milliseconds() {
return 1000 * (3600 * hour + 60 * min + sec);
}
friend Clock operator+(const Clock& a, const Clock& b) {
Clock c;
c.hour = a.hour + b.hour;
c.min = a.min + b.min;
c.sec = a.sec + b.sec;
return c;
}
friend Clock operator-(const Clock& a, const Clock& b) {
Clock c;
c.hour = a.hour - b.hour;
c.min = a.min - b.min;
c.sec = a.sec - b.sec;
return c;
}
friend void rectify(Clock& clock);
};
void rectify(Clock& clock) {
while (clock.sec >= 60) {
clock.sec -= 60;
clock.min += 1;
}
while (clock.min >= 60) {
clock.min -= 60;
clock.hour += 1;
}
while (clock.hour >= 12) {
clock.hour -= 12;
}
while (clock.sec < 0) {
clock.sec += 60;
clock.min -= 1;
}
while (clock.min < 0) {
clock.min += 60;
clock.hour -= 1;
}
while (clock.hour < 0) {
clock.hour += 12;
}
}
int main() {
int h1, m1, s1;
std::cout << "Enter hour, minute, and second for Clock A: ";
std::cin >> h1 >> m1 >> s1;
Clock a(h1, m1, s1);
int h2, m2, s2;
std::cout << "Enter hour, minute, and second for Clock B: ";
std::cin >> h2 >> m2 >> s2;
Clock b(h2, m2, s2);
std::cout << "Clock A: ";
a.info();
std::cout << "Milliseconds: " << a.milliseconds() << std::endl;
std::cout << "Clock B: ";
b.info();
std::cout << "Milliseconds: " << b.milliseconds() << std::endl;
Clock sum = a + b;
std::cout << "Sum of Clock A and B: ";
sum.info();
std::cout << "Milliseconds: " << sum.milliseconds() << std::endl;
Clock diff = a - b;
std::cout << "Difference of Clock A and B: ";
diff.info();
std::cout << "Milliseconds: " << diff.milliseconds() << std::endl;
rectify(a);
rectify(b);
rectify(sum);
rectify(diff);
std::cout << "Rectified Clock A: ";
a.info();
std::cout << "Rectified Clock B: ";
b.info();
return 0;
}
In the Clock class, there are private fields for hour, min, and sec. The class provides a default constructor, a parameterized constructor, and member functions to display the clock time (info()) and calculate the time in milliseconds (milliseconds()). Two friend functions operator+ and operator- are defined to perform addition and subtraction on Clock objects.
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If "I" is the length of the line and angle made with x-axis is x° then departure (D) of the line is
A right-angled triangle formed by the line, where the hypotenuse represents the length (I) and the adjacent side represents the departure (D). If the line does not form a right angle with the x-axis, additional adjustments or transformations may be required to accurately calculate the departure.
The departure (D) of a line can be calculated based on its length (I) and the angle (x°) it makes with the x-axis. The departure represents the horizontal distance between the starting point and the ending point of the line.
To calculate the departure, we use the following trigonometric formula:
D = I * cos(x°)
Here, "cos" represents the cosine function, which relates the adjacent side (departure) to the hypotenuse (length of the line) based on the given angle.
By multiplying the length (I) of the line by the cosine of the angle (x°), we can determine the departure (D) of the line.
Note that this calculation assumes a right-angled triangle formed by the line, where the hypotenuse represents the length (I) and the adjacent side represents the departure (D). If the line does not form a right angle with the x-axis, additional adjustments or transformations may be required to accurately calculate the departure.
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Answer True or False
18. A three-phase variable reluctance stepper motor with 12 stator windings has the windings set 30 degrees apart 19. The two-phase operation of a resolver uses the variation in voltage amplitude to determine shaft position
20. The micro-stepping mode generates small stepping angles through motor current control via PWM of the excitation voltage of the rotor windings
21. A resistance-start motor has a start winding, which is a coil made of many turns of heavy-gauge wire
22. In the capacitor-start induction-run motor, the capacitor generates a current in the start-winding that lags the phase of the voltage input.
23. The centrifugal start switch is usually mounted on the shaft of the AC motor.
24. The performance of the resistance-start induction-run motor and the capacitor-start induction-run motor is basically identical
25. The shaded-pole induction motor is a single-phase motor that has no start and run windings
26. In a three-phase motor, the voltages are 180 degrees out of phase with each other
27. The current required to run the AC motor flows through the thermal overload heater and bimetallic switch contacts
28. In a synchronous motor. The salient poles project outward from the shaft of the motor
29. Detent torque is the maximum load-originating torque that the motor can stand without moving from its position with the stator energized
30. The speed torque and the current-torque curves are linear for the permanent magnetic motor
31. The total encoder error is the sum of the instrument error, quadrature error, interpolation error, and quantization error
32. The coreless motor is chosen over the wound-rotor motor in high torque application
33. The figure of merit is a number that provides and objective method of determining the relative ability of a permanent magnetic motor
34. An optical encoder is a sensor
35. The AC motor is better suited than the DC motor for those applications that require variable speeds
36. On a AC motor plate, the type specifies the motor’s enclosure.
37. Slip is the difference between the speed of the rotating field and the speed of the rotor
38. When selecting a motor, the current rating of the starter must be less than the actual current that the starter will switch
39. In an AC servomotor, the voltage applied to the winding and the voltage applied to the control winding are either in phase or ninety degrees out of phase.
An electric motor known as a stepper turns electrical pulses into exact mechanical movements. The term stepper motor refers to the fact that it is intended to move in distinct steps or increments. The True or False asked in the various questions are as-
18. True: A three-phase variable reluctance stepper motor with 12 stator windings has the windings set 30 degrees apart.
19. False: The two-phase operation of a resolver uses the variation in voltage amplitude to determine shaft position.
20. True: The micro-stepping mode generates small stepping angles through motor current control via PWM of the excitation voltage of the rotor windings.
21. True: A resistance-start motor has a start winding, which is a coil made of many turns of heavy-gauge wire.
22. True: In the capacitor-start induction-run motor, the capacitor generates a current in the start-winding that lags the phase of the voltage input.
23. True: The centrifugal start switch is usually mounted on the shaft of the AC motor.
24. False: The performance of the resistance-start induction-run motor and the capacitor-start induction-run motor is not basically identical.
25. True: The shaded-pole induction motor is a single-phase motor that has no start and run windings.
26. True: In a three-phase motor, the voltages are 120 degrees out of phase with each other.
27. False: The current required to run the AC motor does not flow through the thermal overload heater and bimetallic switch contacts.
28. True: In a synchronous motor, the salient poles project outward from the shaft of the motor.
29. True: Detent torque is the maximum load-originating torque that the motor can stand without moving from its position with the stator energized.
30. False: The speed-torque and the current-torque curves are not linear for the permanent magnetic motor.
31. True: The total encoder error is the sum of the instrument error, quadrature error, interpolation error, and quantization error.
32. False: The coreless motor is not chosen over the wound-rotor motor in high torque applications.
33. True: The figure of merit is a number that provides an objective method of determining the relative ability of a permanent magnetic motor.
34. True: An optical encoder is a sensor.
35. True: The AC motor is better suited than the DC motor for those applications that require variable speeds.
36. True: On an AC motor plate, the type specifies the motor's enclosure.
37. True: Slip is the difference between the speed of the rotating field and the speed of the rotor.
38. True: When selecting a motor, the current rating of the starter must be less than the actual current that the starter will switch.
39. True: In an AC servomotor, the voltage applied to the winding and the voltage applied to the control winding are either in phase or ninety degrees out of phase.
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Question 1 Needs Grading 108 491 980 a. Generate the matrix below.A1 = 559 382 207 845 657 714 b. Generate A2 = 4 x 4 random matrix. c. Change A2 so that all its elements are in the range of 10 to 50. d. Convert A into a 4 x 4 matrix by inserting a new column of Os into the 4th column of A, and also row 4 of Az into the 2nd row of Aj. e. What is the 2D correlation coefficient between A1 and A2? f. What is the difference between the standard deviations of all the elements in A1 and A2?
A1 is a 3 x 3 matrix.A1 = 559 382 207 845 657 714b) A2 is a 4 x 4 matrix of random numbers.A2 = rand(4);c) To obtain random numbers within the range of 10 and 50 in A2, we can multiply rand(4) by 40 and then add 10 to every entry, as shown below.
A2 = rand(4) * 40 + 10;d) To convert A1 into a 4 x 4 matrix by inserting a new column of Os into the 4th column of A and row 4 of Az into the 2nd row of Aj, we can perform the following operations.A1 = [559 382 207; 845 657 714; 0 0 0];Az = [1 2 3 4];Aj = [5 6 7 8];A1(1:end - 1, end + 1) = 0;A1(end + 1, :) = Az;A1(2, :) = Aj;
To find the 2D correlation coefficient between A1 and A2, we can use the corr2 function as shown below.corr2(A1, A2);f) To calculate the standard deviation of all the elements in A1 and A2, we can use the std function. The difference between the standard deviations of A1 and A2 is given by.std(A1(:)) - std(A2(:));
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