This is it dont anwser this is for my other account

1 import java.util.Scanner; 3 public class EqualityAndRelational { 4 public static void main (String args) args) { int userBonus

Anastaziya [24]

Missing Part of the Question

Complete the expression so that userPoints is assigned with 0 if userBonus is greater than 20 (second branch). Otherwise, userPoints is assigned with 10 (first branch

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

( Your solution goes here)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

Answer;

Replace

( Your solution goes here)

With

if(userBonus>20).

The full program becomes

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

if(userBonus>20)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

7 0

3 years ago

Suppose you wanted to convert an AC voltage to DC. Your AC voltage source has Vrms= 60V. If you use a full wave rectifier direct

vovangra [49]

Answer:

$V_{dc}=84.15\ V$

Explanation:

Given that

Vrms= 60 V

Vf= 0.7 V

We know that peak value of AC voltage given as

$V_{o}=\sqrt{2}\ V_{rms}$

Now by putting the values

$V_{o}=60\sqrt{2}\ V$

The output voltage of the DC current given as

$V_{dc}=V_{o}-V_f$

$V_{dc}=60\sqrt{2}-0.7\ V$

$V_{dc}=84.15\ V$

Therefore output voltage of the DC current is 84.15 V.

7 0

3 years ago

Explain how the ideal vapour compression refrigerator cycle is an improvement upon the Carnot cycle. Use any two of the problems

Grace [21]

answer explanation:

<u>Basically, the Carnot cycle is turning a heat differential into mechanical energy, then using mechanical energy to recreate the heat differential. A heat pump only turns mechanical energy into a heat differential. It exchanges the gas being compressed on an ongoing basis and does not extract mechanical energy from it, this continuous loss of mechanical energy input is what allows continuous generation of a heat differential.</u>

<u>Also, the Carnot cycle is an idealized theoretical model, not a practically achievable engineering goal. It illustrates the essential mechanism by which heat differentials can be turned into mechanical energy and vice versa, it has to ignore the fact of inefficiency in conversion between mechanical energy and heat differential for the sake of illustrating a set of ideal processes. A real refrigeration system has to deal with the question of whether the system is pumping away more heat than it generates itself through inefficiency.</u>

8 0

3 years ago

Define cooling tower "range" as it applies to cooling towers.

Arlecino [84]

Answer:

The answer to the range of a cooling tower is the difference between the temperature of the water going in and the temperature of the water going out of the tower.

Explanation:

Cooling towers are used to cool a stream of water using the air of the environment where it's placed or installed. In theory, the lowest temperature the water can be cooled to would be the wet bulb temperature (temperature of the air going in the cooling tower). The range would be the difference (R=Ti-To) between the temperature of the hot water going in and the cooler water going out.

8 0

3 years ago

I need help on Problem 2.11

Jet001 [13]

Answer:

No, energy as work cannot be transferred to the fluid.

The change in internal energy of the fluid is 0.

Explanation:

Work is dW = -P dV.

The fluid is incompressible, so dV = 0. If there is no change in volume of the fluid, no work can be done. dW = 0.

Internal energy is dE = Q + W.

The work is zero. Since the piston is insulated, the heat transferred is also zero. If Q = 0 and W = 0, then the change in internal energy of the fluid dE is also zero.

8 0

3 years ago