The first thing we are going to do is find the equation of motion:
ωf = ωi + αt
θ = ωi*t + 1/2αt^2
Where:
ωf = final angular velocity
ωi = initial angular velocity
α = Angular acceleration
θ = Revolutions.
t = time.
We have then:
ωf = (7200) * ((2 * pi) / 60) = 753.60 rad / s
ωi = 0
α = 190 rad / s2
Clearing t:
753.60 = 0 + 190*t
t = 753.60 / 190
t = 3.97 s
Then, replacing the time:
θ1 = 0 + (1/2) * (190) * (3.97) ^ 2
θ1 = 1494.51 rad
For (10-3.97) s:
θ2 = ωf * t
θ2 = (753.60 rad / s) * (10-3.97) s
θ2 = 4544,208 rad
Number of final revolutions:
θ1 + θ2 = (1494.51 rad + 4544.208 rad) * (180 / π)
θ1 + θ2 = 961.57 rev
Answer:
the disk has made 961.57 rev 10.0 s after it starts up
The word DoS refers to the Disk Operating System. In the case given above, if it happens that there is a distributed DoS attack, the attacker then sends messages directly to the BOTS. Bots refer to internet or web robots. Hope this answer helps. The answer is the first option.
Answer:
def isdivisible():
maxint=input("Enter the Max Int")
int1=0
int2=0
int1=input("Enter the first Integer")
int2=input("Enter the second Integer")
tup1=(int1, int2)
print(tup1)
i = 1
for i in range(1, int(maxint)-1):
if int(tup1[0])%i==0 & int(tup1[1])%i==0:
print(i)
else:
continue
isdivisible()
1.2 Outputs
First test case:
Enter the Max Int6
Enter the first Integer2
Enter the second Integer8
('2', '8')
1
2
Second test case: returning empty list
Enter the Max Int2
Enter the first Integer13
Enter the second Integer27
('13', '27')
Test case 3:
Enter the Max Int4
Enter the first Integer8
Enter the second Integer10
('8', '10')
1
2
Explanation:
The program is as above, and the three test cases are also mentioned. We have created a tuple out of two input integer, and performed the output as required.
Answer:
critical thinking, judgment and decision making, speaking, active listening, and science
Explanation: just took the test on edge
Answer:
- import java.util.Scanner;
-
- public class Main {
- public static void main(String[] args) {
- Scanner input = new Scanner(System.in);
- String numStr = input.nextLine();
- double num;
- try{
- num = Double.parseDouble(numStr);
- }
- catch(NumberFormatException e){
- num = 0;
- System.out.println("Value entered cannot be converted to a floating point number.");
- }
- }
- }
Explanation:
The solution code is written in Java.
Firstly, we create a Scanner object and prompt user to input a number (Line 5-6). Next, we create a try -catch block and place the parseDouble inside the try block. If the input is invalid (e.g. "abc"), a NumberFormatException error will be thrown and captured and set the num to 0 and display the error message (Line 11 - 13).
