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vredina [299]
3 years ago
8

Which of the code pieces below should replace the underline?public class Test{public static void main(String[] args){Test test =

new Test();test.setAction(_______);}public void setAction(T1 t){t.m();}}interface T1{public void m();}
Engineering
1 answer:
nikdorinn [45]3 years ago
5 0

Which of the code pieces below should replace the underline?public class Test{public static void main(String[] args){Test test = new Test();test.setAction(_______);}public void setAction(T1 t){t.m();}}interface T1{public void m() is explained below

Explanation:

Update of /cvsroot/pydev/org.python.pydev/src/org/python/pydev/editor/correctionassist

In directory sc8-pr-cvs1.sourceforge.net:/tmp/cvs-serv24839/src/org/python/pydev/editor/correctionassist

Modified Files:

PythonCorrectionProcessor.java  

Log Message:

Index: PythonCorrectionProcessor.java

===================================================================

RCS file: /cvsroot/pydev/org.python.pydev/src/org/python/pydev/editor/correctionassist/PythonCorrectionProcessor.java,v

retrieving revision 1.12

retrieving revision 1.13

diff -C2 -d -r1.12 -r1.13

*** PythonCorrectionProcessor.java 23 Feb 2005 14:29:36 -0000 1.12

--- PythonCorrectionProcessor.java 25 Feb 2005 12:34:28 -0000 1.13

***************

*** 419,423 ****

                 //all that just to change first char to lower case.

                 callName = lowerChar(callName, 0);

!                 if (callName.startsWith("get")){

                     callName = callName.substring(3);

                  callName = lowerChar(callName, 0);

--- 419,423 ----

                 //all that just to change first char to lower case.

                 callName = lowerChar(callName, 0);

!                 if (callName.startsWith("get") && callName.length() > 3){

                     callName = callName.substring(3);

                  callName = lowerChar(callName, 0);

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Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

-The mass of C_o2 = 1 kg  

-The volume of the tank V_tank = 1 m^3  

-The added energy E = 14 W  

-The time of adding energy t = 10 s  

-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

<u>Required  </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

the heat transfer.  

Assumption  

-Quasi-equilibrium process.  

<u>Solution</u>  

(a) The volume and the mass doesn't change then, the specific volume is constant.

 v= V_tank/m ---> 1/1= 1 m^3/Kg  

(b) The added work is defined by.  

W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  

(c) From the first law of thermodynamics.  

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

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