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GR9677 #27
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Problem
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This problem is still being typed. |
Lab Methods }Log Graphs
Log graphs are good for exponential-related phenomenon. Thus (A), (C), and (E) are appropriate, thus eliminated. The stopping potential has a linear relation to the frequency, and thus choice (B) is eliminated. The remaining choice is (D).
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Alternate Solutions |
Izaac
2012-08-21 01:18:44 | One can also simply remember that Bode plots (gain VS  ) are semilog ones, so obviously D is inappropriate . |  |
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Comments |
Memol
2012-09-18 07:17:56 | Can anyone help me with an study reference about these graph stuff? |  | Izaac
2012-08-21 01:18:44 | One can also simply remember that Bode plots (gain VS ) are semilog ones, so obviously D is inappropriate . | | keenanman
2007-10-16 12:38:05 | In choice D, the graph gain vs 1/frequency is linear. The graph gain vs frequency is hyperbolic. |  | eshaghoulian
2007-10-02 04:11:09 | Just to add a little bit as to why log graphs are good for exponential related phenomena, note that a power law in log-log coordinates is a line:
 = log(ax^m) = mlog(ax) = mlog(x)+ mlog(a) \Rightarrow log(y) = mlog(x) + mlog(a)) which is of the form  (since ) is just a constant (like  ) and we identify  with ) and  with ) , as these are our new axes in log-log coordinates). So the exponent in the power law becomes the slope in log-log coordinates. Testing this is a GRE favorite, as it is a major tool in experimental physics.
tachyon788
2009-10-06 11:48:47 |
You have a small math error in your use of logs. The equation should be:
=log(ax^m)=log(a)+log(x^m)=log(a)+mlog(x))
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