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kaheart [24]
3 years ago
11

The question is in the following image.

Engineering
1 answer:
Leona [35]3 years ago
3 0

Answer: Statement "E" is true

Explanation:

A - Statement is not true because, though static methods can not directly access any instance variables or methods. But they can access them by using their object reference.

B - Statement is not true because a non static method can also access a static variable.

C - Statement is not true because according to the rules for instance variables in java, it can be marked as final.

D - Statement is not true because though it can be declared as final by initialization of variable mandatory and initialization of class loading, there is no rule that says it must be declared as final.

E - This statement is true because in the constructor, static variables are not associated with each object of the class since it is shared by each object. Therefore, if you initialize static variable into constructor, it means that you are trying to associate with a particular member of class and it's not possible to do so!

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Who can work on a fixed ladder that extends more than 24 feet?
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Answer:

Explanation:

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3 years ago
Implement the following Matlab code:
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3 years ago
Determine the following parameters for the water having quality x=0.7 at 200 kPa:
ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

   Temperature = 120.21°C

c). Specific volume

  $v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

  $v_x=v_f+x(v_g-v_f)$

       $=0.001061+0.7(0.88578-0.001061)$

       $=0.62036 \ m^3/kg$

d). Specific energy (u_x)

    $u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

   $u_x=504.5 + 0.7(2024.6)$

         $=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

   At $h_f = 504.71, \ \ h_{fg} = 2201.6$

   h_x=504.71+(0.7\times 2201.6)

        $= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

  $H=mh_x$

       $= 0.5 \times 2045.83$

       = 1022.91 kJ

7 0
3 years ago
An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/
Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

<u>Step 1:</u> Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

<u>Step 3:</u> The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755  / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min  = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

3 0
2 years ago
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