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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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Answer:

Answer for the question:

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2 years ago
(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1
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Answer:

The answer is "4.35 \times 10^{-3}\  mm and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point  

y =  yo + \frac{k}{\sqrt{x}}

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.  

Replacing the equation for each condition:  

y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k

People can get yo = 275 MPa with both equations and k= 15.5 Mpa mm^{\frac{1}{2}}.

To substitute the answer,  

310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\  mm

In point b:

The equation is \sigma y = \sigma 0 + k y d^{\frac{1}{2}}

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75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\

by putting the above value in the formula we get the \sigma y value that is= 157.5 MPa

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Step-by-step explanation:

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