Answer: look left to right twice b/c somebody may try to run the light
Explanation:
Answer:
- import java.util.Scanner;
- import java.util.Random;
- public class Main {
-
- public static void main(String[] args) {
-
- Scanner input = new Scanner(System.in);
- Random rand = new Random();
-
- System.out.print("Enter number of questions you wish to practice: ");
- int n = input.nextInt();
-
- int i = 1;
- while(i <= n){
- int num1 = 1 + rand.nextInt(10);
- int num2 = 1 + rand.nextInt(10);
- int answer = num1 * num2;
- System.out.print("Question " + i + ": " + num1 + " x " + num2 + " = ");
- int response = input.nextInt();
-
- if(response == answer){
- System.out.println("Correct answer!");
- }
- else{
- System.out.println("Wrong answer. It should be " + answer);
- }
- i++;
- }
- }
- }
Explanation:
Firstly, we import Scanner and Random classes (Line 1-2) as we need to get user input for number of questions and generate random number in each question.
Next, create a Scanner and Random object (Line 7-8). Prompt user to input number of question and get the input using the Scanner object (Line 10-11).
Next create a while loop and set the condition to loop for n number of times (based on number of questions) (Line 14). In the loop, use Random object nextInt method to generate two integers between 1 -10 (Line 15-16). Prompt user to input an answer using the Scanner object (Line 18-19). If the response is matched with answer print a correct message (Line 21-22) other wise inform user they write a wrong answer and show the correct answer (Line 24-25).
Answer:
C
Explanation:
Pitot tube:
Pitot tube is a device which is used to measure the velocity of flow by measuring pressure difference between the points.
As we know that stagnation pressure is the summation of dynamic and static pressure.
Stagnation pressure = Static pressure + Dynamic pressure
So
Dynamic pressure = Stagnation pressure - Static pressure
We know that dynamic pressure
On the other hand Pitot tube measure the dynamic pressure.
So option C is correct.
Answer:
T = 20.42 N
Explanation:
given data
standard altitude = 30,000 ft
velocity Ca = 500 mph = 0.4 m/s
inlet areas Aa = 7 ft² = 0.65 m²
exit areas Aj = 4.5 ft² = 0.42 m²
velocity at exit Cj = 1600 ft/s = 487.68 m/s
pressure exit 
j = 640 lb/ft² = 0.3 bar
solution
we get here thrust of the turbojet that is express as
thrust of the turbojet T = Mg × Cj - Ma × Ca + ( j Aj - a Ag ) .............1
here Ma = Mg
Ma = a × Ca Aa = 0.042 kg/s
put value in equation 1 we get
T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )
T = 20.42 N
Answer:
I am a girl want private photos
Explanation:
