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Ad libitum [116K]
2 years ago
6

ABS system is necessary?

Engineering
1 answer:
Monica [59]2 years ago
3 0

Explanation:

I think it helps you

I don't know the answer sorry

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In poor weather, you should __ your following distance
Ratling [72]

Answer:

I think reduce your following distance

5 0
3 years ago
Read 2 more answers
12.50 An air conditioner operating at steady state takes in moist air at 28°C, 1 bar, and 70% relative humidity. The moist air f
Mandarinka [93]

Answer:

Hey smith please see attachments for answer:

Please rate me good.

The attachments will provide you a detailed answer

Explanation:

8 0
3 years ago
java Your program class should be called RomanNumerals Write a program that asks the user to enter a number within the range of
Ann [662]

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.

6 0
3 years ago
Read 2 more answers
A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar
Westkost [7]

Answer:

a) m_{2} = 0.753\,kg, b) Q_{in} = 2122.963\,kJ

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.02726\,\frac{m^{3}}{kg}

u = 1192.94\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.2

State 2 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.06643\,\frac{m^{3}}{kg}

u = 1718.12\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.5

The model for the rigid tank is created by using the First Law of Thermodynamics:

Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}

Initial and final masses are:

m_{1} = \frac{V_{1}}{\nu_{1}}

m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }

m_{1} = 1.834\,kg

m_{2} = \frac{V_{2}}{\nu_{2}}

m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }

m_{2} = 0.753\,kg

a) The final mass within the tank is:

m_{2} = 0.753\,kg

b) The total amount of heat transfer is:

Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}

Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )

Q_{in} = 2122.963\,kJ

5 0
3 years ago
A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where the
Archy [21]

Answer:

Work transfer is - 97.02 KJ. It means that work is given to the system.

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

Explanation:

Given that

m= 3 kg

P₁=2 bar

T=T₁=T₂=30 °C

T=303 K

P₂=2.5 bar

PV=  Constant

This is the isothermal process .

We know that work for isothermal process  given as

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=mRT\ln \dfrac{P_1}{P_2}

For air

R= 0.287 KJ/Kg.K

Now by putting the values

W=mRT\ln \dfrac{P_1}{P_2}

W=5\times 0.287\times 303\ln \dfrac{2}{2.5}

W= - 97.02 KJ

So the work transfer is - 97.02 KJ. It means that work is given to the system.

We know that for ideal gas internal energy is the only function of temperature.The change in internal energy ΔU

ΔU = m Cv ΔT

Here ΔT= 0

So

ΔU =0

From first law of thermodynamics

Q= ΔU +W

ΔU = 0

Q= W

Q= - 97.02 KJ

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

8 0
3 years ago
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