Suppose we have two negative charges, one located at the origin and carrying charge −9e, and the other located on the positive x
1 answer:
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Answer:
Yes the statement is true.
Explanation:
Power is defined as the rate at which energy is transferred by an object on account of work done.
Mathematically
An object that does work loses it's energy while an object on which work is done gains energy.
Power is often dependent on the type of energy transfer thus we have Electrical Power, Mechanical Power depending on the type of energy involved in the system.
Concept of power is important since it gives us a measure of how fast energy can be derived to given to a system.
Answer:
The system can be described by a convolution
Explanation:
Thinking process:
If we consider a discrete input to a linear time-invariant system, then the system will be periodic with respect to the period, say N. This therefore, means that the output must also be periodic. The proof is as follows:
The LTI system can be written for the system where:
y (n+N) = ∑
= ∑
From the proof, it turns out that y(y + N) = y(n) for any value of n, then the output will be the periodic with the period N.
Answer:
Here is the complete program:
#include <iostream>
using namespace std;
int main()
{ int x = 1, y = 3;
int X = 2, Y = 4;
cout << "tryIt 2A" <<endl;
cout << x << y << endl;
cout << "x" << "y" << endl;
cout << X << " " << Y << endl;
cout << 2 * x + y << endl;
cout << 2 * X + Y << endl;
//cout << x + 2*y << endl;
cout << "x = ";
cout << x;
cout << " y = ";
cout << y;
return 0;
}
Explanation:
I will explain the code line by line in the comment with each line of code and the output of each cout statement.
- int x = 1, y = 3;
This statement assigns value 1 to integer variable x and 3 to int variable y
- int X = 2, Y = 4;
This statement assigns value 2 to integer variable X and 4 to int variable Y As C++ is a case sensitive language so variable x and y are different from variables X and Y.
- cout << "tryIt 2A" <<endl;
This statement has cout which is used to display output on the screen. So the output displayed by this cout statement is:
tryIt2A
- cout << x << y << endl;
This statement will print the values stored in x and y variables. So output displayed by cout statement here is 1 and 3. As there is not space or next line specified in the statement so output displayed will look like this:
13
- cout << "x" << "y" << endl;
This statement will display x and y but these are not the variable x and y. They are enclosed in double quotation marks so they are treated as strings not variables so the output displayed is:
xy
- cout << X << " " << Y << endl;
This statement will print the values stored in X and Y variables. So output displayed by cout statement here is 2 and 4. As there is space " " specified in the statement so 2 and 4 are displayed with a space between them so the output displayed will look like this:
2 4
- cout << 2 * x + y << endl;
This statement has an arithmetic operation in which 2 is multiplied by the values stored in variable x and then the result is added by value of y. So 2*1 = 2 and 2 + 3 = 5. So the result produced by this cout statement is:
5
- cout << 2 * X + Y << endl;
This will work same as above cout statement but the only difference is that the values of capital X and Y variables are calculated here. So 2 * 2 = 4 and then 4 + 4 = 8. The result produced by this cout statement is:
8
- //cout << x + 2*y << endl;
This is a comment because before this statement // is written which is used for single line comment. So compiler ignores comments and will not compile this statement.
- cout << "x = ";
This will display "x = " as it is not variable but it is treated as a line to be displayed on the screen. So cout statement displays:
x =
- cout << x;
This will print the value stored in x variable as there are no double quotes around x so it is a variable which contains value 1. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:
x = 1
- cout << " y = ";
This will display "y = " as it is not variable but it is treated as a line to be displayed on the screen. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen
x = 1 y =
- cout << y;
This will print the value stored in y variable as there are no double quotes around y so it is a variable which contains value 3. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:
x = 1 y = 3
So the output of the entire program along with the program is attached as screenshot.
