Whats the best used for Arch bridge
1 answer:
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Answer:
// Scanner class is imported to allow program
// receive input
import java.util.Scanner;
// RomanNumerals class is defined
public class RomanNumerals {
// main method that signify beginning of program execution
public static void main(String args[]) {
// Scanner object scan is created
// it receive input via keyboard
Scanner scan = new Scanner(System.in);
// Prompt is display asking the user to enter number
System.out.println("Enter your number: ");
// the user input is stored at numberOfOrder
int number = scan.nextInt();
// switch statement which takes number as argument
// the switch statement output the correct roman numeral
// depending on user input
switch(number){
case 1:
System.out.println("I");
break;
case 2:
System.out.println("II");
break;
case 3:
System.out.println("III");
break;
case 4:
System.out.println("IV");
break;
case 5:
System.out.println("V");
break;
case 6:
System.out.println("VI");
break;
case 7:
System.out.println("VII");
break;
case 8:
System.out.println("VIII");
break;
case 9:
System.out.println("IX");
break;
case 10:
System.out.println("X");
break;
// this part is executed if user input is not between 1 to 10
default:
System.out.println("Error. Number must be between 1 - 10.");
}
}
}
Explanation:
The program is well commented. A sample image of program output is attached.
The switch statement takes the user input (number) as argument as it goes through each case block in the switch statement and match with the corresponding case to output the roman version of that number. If the number is greater 10 or less than 1; the default block is executed and it display an error message telling the user that number must be between 1 - 10.
Answer:
(a) = 0.308 m³/s
(b) = 0.732 kg/m³
(c) v₂ = 5.94 m/s.
Explanation:
(a) The volume flow rate is given by the cross sectional area of the pipe × Velocity of flow of air
Diameter of pipe = 28 cm = 0.28 m
The cross sectional area, A, of the pipe = 0.28²/4×π = 0.0616 m²
Volume flow rate = 5 × 0.0616 = 0.308 m³/s
= 0.308 m³/s
(b) From the general gas equation, we have;
p₁v₁ = RT₁ which gives;
p₁/ρ₁ = RT₁
ρ₁ = p₁/(RT₁)
Where:
ρ₁ = Density of the air
p₁ = 200 kPa
T₁ = 20 C =
R = 0.287 kPa·m³/(kg·K)
ρ₁ = 200/(0.287 ×293.15) = 2.377 kg/m³
The mass flow rate = Volume flow rate × Density
The mass flow rate, = 2.377×0.308 = 0.732 kg/m³
= 0.732 kg/m³
(c) The density at exit, ρ₂, is found from the the universal gas equation as follows;
ρ₂ = p₂/(RT₂)
Where:
p₂ = Pressure at exit = 180 kPa
T₂ = Exit temperature = 40°C = 273.15 + 40 = 313.15 K
∴ ρ₂ = 180/(0.287×313.15) = 2.003 kg/m³
= ρ₂×
=
/ρ₂ = 0.732/2.003 = 0.366 m³/s
= v₂ × A
v₂ = /A = 0.366/0.0616 = 5.94 m/s.
v₂ = 5.94 m/s.