A mass m is attached to the end of a rope of length r = 3 meters. The rope can only be whirled around at speeds of 1, 10, 20, or
1 answer:
Answer:
Explanation:
Let's do this in Python. We know that the formula for the centripetal force caused by whirling the mass is:
where v is the speed (1, 10, 20, 40) and r = 3 m is the rope length.
Then we can use for loop to try calculating the tension force of each speed
def maximum_speed(m):
for speed in [1, 10, 20, 40]:
tension_force = m*speed^2/3
if tension_force > 60:
return speed
return speed
You might be interested in
Answer:
The sample output is below.
The code is attached.
Explanation:
Sample Output:
Original array is: 1 2 457 4 Reversed array is: 4754 2 1
Code to copy:
import java.io.*;
class Main
{
/*Define the function revArray passing the array
and its indices.*/
static void revArray(int arr[], int start, int end)
{
/*Declare an integer variable to store the
first index.*/
int temp;
//Check if array is empty.
if (start >= end)
return;
//Store the first index value.
temp = arr[start];
//Swap the last index to first index.
arr[start] = arr[end];
//Store the temp value to last index.
arr[end] = temp;
/*Call the function recursively to swap
rest of the values.*/
revArray(arr, start+1, end-1);
}
public static void main (String[] args)
{
//Declare and initialize array.
int[] arr = {1,2,4,5,7,4};
/*Declare an integer variable to store the
* length of array.
*/
int count=0;
//Set the length of array to count.
for(int i=0;i<arr.length;i++)
count++;
System.out.println("Original array is:");
///Print the original array.
for(int i=0; i<count;i++)
{
System.out.print(arr[i]+" ");
}
/*Call the function to reverse the values in
* the array itself.
*/
revArray(arr, 0, count-1);
System.out.println("\nReversed array is:");
///Print the reversed array.
for(int i=0; i<count;i++)
{
System.out.print(arr[i]+" ");
}
}
}
Answer:
So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
Explanation:
Previous concepts
Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"
Solution to the problem
For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:
Available = 4096 MB - 512 MB = 3584 MB
Now we can find the maximum simultaneous process than can use with this:
And then we can find the maximum wait I/O that can be tolerated with the following formula:
The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:
So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %