Answer:
<u>Pseudocode:</u>
INPUT velocity
INPUT time
SET velocity = 0.44704 * velocity
SET acceleration = velocity / time
SET acceleration = round(acceleration, 1)
PRINT acceleration
<u>Code:</u>
velocity = float(input("Enter a velocity in miles per hour: "))
time = float(input("Enter a time in seconds: "))
velocity = 0.44704 * velocity
acceleration = velocity / time
acceleration = round(acceleration, 1)
print("The acceleration is {:.2f}".format(acceleration))
Explanation:
*The code is in Python.
Ask the user to enter the velocity and time
Convert the miles per hour to meters per second
Calculate the acceleration using the formula
Round the acceleration to one decimal using round() method
Print the acceleration with two decimal digits
Answer:
Stack
Explanation:
Stack is a linear data structure that follows a particular order in the way an operation is done or sequence a job is completed.
It uses either LIFO ( Last In, First Out) which is also known as first come first served sequence or FILO (First In, Last Out) sequence.
There are several real life examples of which we can use the example of replacing the snack items Sarah brought for the customer.
If Sarah used the LIFO method, it means she replaced the snack items first ontop of the already existing snack items that's why there is a mismatch.
Answer:
// here is code in java.
public class NAMES
{
// main method
public static void main(String[] args)
{
int n=4;
// print the upper half
for(int a=1;a<=n;a++)
{
for(int b=1;b<=n-a;b++)
{
// print the spaces
System.out.print(" ");
}
// print the * of upper half
for(int x=1;x<=a*2-1;x++)
{
// print the *
System.out.print("*");
}
// print newline
System.out.println();
}
// print the lower half
for(int y=n-1;y>0;y--)
{
for(int z=1;z<=n-y;z++)
{
// print the spaces
System.out.print(" ");
}
for(int m=1;m<=y*2-1;m++)
{
// print the *
System.out.print("*");
}
// print newline
System.out.println();
}
}
}
Explanation:
Declare a variable "n" and initialize it with 4. First print the spaces (" ") of the upper half with the help of nested for loop.Then print the "*" of the upper half with for loop. Similarly print the lower half in revers order. This will print the required shape.
Output:
*
***
*****
*******
*****
***
*
Answer:
Check the explanation
Explanation:
1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
Note : Log (m *n) = Log m + Log n : this is property of logarithm
3) f(n) = 
, since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n
for n times, this makes the complexity – n * n = n2
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
