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docker41 [41]
3 years ago
9

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

; int userPoints; userPoints-0; Scanner scnrnew Scanner(System.in); 7 10 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;
Engineering
1 answer:
Anastaziya [24]3 years ago
7 0

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;

}

}

}

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It creates airflow in the engine

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Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rat
attashe74 [19]

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

              COP_{HP}=\frac{T_H}{T_H-T_L}

Therefore, the temperature relationship, T_H=1.15\;T_L

Then, we should apply the values in the COP.

                           =\frac{1.15\;T_L}{1.15-1}

                           =7.67

The number of heat rejected by the heat pump must then be calculated.

                   Q_H=COP_{HP}\times W_{nst}

                          =7.67\times5=38.35

We must then calculate the refrigerant mass flow rate.

                   m=0.264\;kg/s

                   q_H=\frac{Q_H}{m}

                         =\frac{38.35}{0.264}=145.27

The h_g value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

                   T_L=\frac{T_H}{1.15}

                        =\frac{64+273}{1.15}=293.04

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8 0
3 years ago
A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the interna
Arte-miy333 [17]

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

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Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

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Answer:

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The answer is C

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