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Ymorist [56]
2 years ago
10

. Carly's Catering provides meals for parties and special events. In Chapter 2, you wrote an application that prompts the user f

or the number of guests attending an event, displays the company motto with a border, and then displays the price of the event and whether the event is a large one. Now modify the program so that the main() method contains only three executable statements that each call a method as follows:
Engineering
1 answer:
klio [65]2 years ago
6 0

Answer:

Explanation:

public class Event

  public final static int PRICE_PER_GUEST = 35;

  public final static int CUT_OFF = 50;

 

  //Attributes

  private String eventNum;

  private int noOfGuest;

  private int price;

 

  /**

  * param eventNum the eventNum to set

  */

  public void setEventNum(String eventNum)

      this.eventNum = eventNum;

 

 

  /**

  * param noOfGuest the noOfGuest to set

  */

  public void setNoOfGuest(int noOfGuest)

      this.noOfGuest = noOfGuest;

      this.price = this.noOfGuest * PRICE_PER_GUEST;

 

 

  /**

  * return the eventNum

  */

  public String getEventNum()

      return eventNum;

 

 

  /**

  * return the noOfGuest

  */

  public int getNoOfGuest()

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To water his lawn, a homeowner uses two hoses. One connects to the faucet, the other to the end of the first hose to make the ho
Shtirlitz [24]

Answer: to be exact you need 28mm of tubing for that

Explanation:

When the election

8 0
2 years ago
A cylinder with a 0.25 m inner diameter and a 0.43 m outer diameter is internally pressurized to 175 MPa. a) Determine the maxim
Ivan
The answer is (b) the second one
8 0
2 years ago
The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su
VladimirAG [237]

Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

- Temperature of hot-fluid inner surface T_i = 333°C

- The convection coefficient hot-fluid h_i = 7 W/m^2K

- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK  ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

- The convection coefficient ambient air h_o = 7 W/m^2K

Find:

Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

 The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

 - We will calculate the total heat flux for the entire system q:

                       q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

                       R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

                       R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

                       R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

                       q = ( 333 - 69 ) / 0.2900649351

                       q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

                      q = ( T_i - T_2 ) / R_i2

Where,

                      R_i2 = 1 / h_i + L_1 / k_1

                      R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

                      ( T_i - T_2 ) = q*R_i2

                        T_2 = T_i - q*R_i2

Plug the values in:

                        T_2 = 333 - 910.141*0.14357

                        T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

3 0
2 years ago
During an experiment conducted in a room at 25°C, a laboratory assistant measures that a refrigerator that draws 2 kW of power h
zvonat [6]

Answer:

Not reasonable.

Explanation:

To solve this problem it is necessary to take into account the concepts related to the performance of a reversible refrigerator. The coefficient of performance is basically defined as the ratio between the heating or cooling provided and the electricity consumed. The higher coefficients are equivalent to lower operating costs. The coefficient can be greater than 1, because it is a percentage of the output: losses, other than the thermal efficiency ratio: input energy. For a reversible refrigerator the coefficient is given by

COP_{R,rev} = \frac{1}{\frac{T_1}{T_2}-1}

Where,

T_1 =High temperature

T_2 =Low Temperature

With our values previous given we can find it:

T_2 = -30\°C = (-30+273)

T_2 = 243K

T_1 = 25\°C = (25+273)

T_1 = 298K

With these values we can now calculate the coefficient of performance:

COP_{R,rev} = \frac{1}{\frac{298}{243}-1}

COP_{R,rev} = 4.42

At the same time we can calculate the work consumption of the refrigerator, this is

W = \dot{W}\Delta t

Where,

\dot{W} = Required power input

t = time to remove heat from a cool to water medium

W = 2kJ/s * 20 min

W = 2kJ/s * 1200s

W = 2400kJ

In this way we can calculate the coefficient of the refrigerator directly:

COP_R = \frac{Q_L}{W}

Where,

Q = Amoun of heat rejected

COP_R = \frac{30000}{2400}

COP_R = 12.5

Comparing the values of both coefficients we have that the experiments are NOT reasonable, because the coefficient of a refrigerator is high compared to  coefficient of reversible refrigerator.

5 0
2 years ago
In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed
Alina [70]

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8  

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒  0.2\times 0.8

⇒  0.16

On the third attempt the success probability will be:

⇒  0.2\times 0.2\times 0.8

⇒  0.032

So that the success probability will be:

⇒  0.8 + 0.16 + 0.032

⇒  0.992

7 0
2 years ago
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