DQW

Cheers ! I don't believe there is a way to delete a post entirely.

Note that China and India have respectively 4 times and 3 times the population of the US. On the other hand, they both have much greater fractions of their populations living below the poverty line.

Wrong slash on "\math"

One of my favorites was Martin Gardner's Science Fiction Puzzles. There's some really neat puzzles at rec-puzzles.org

I believe it was either Hardy or Bertrand Russell, but I could be remembering wrong.

If you can find chemical equations for the combustion (burning in oxygen) of diamond/graphite/fullerene that specify the enthalpy change, then you have something. Since the in-plane C-C bonds in graphite (and fullerene) are a little stronger, I'd expect them to have slightly larger anthalpies. Edit : found a link; scroll down on page - http://wine1.sb.fsu.edu/chm1045/notes/Energy/HessLaw/Energy04.htm

http://www.unr.edu/sb204/geology/mas.html (most reactive) Lithium Potassium Calcium Sodium Magnesium Aluminum Manganese Zinc Chromium Iron Lead Copper Mercury Silver Platinum Gold (Least Reactive)

Consider Na (not a transition) metal. Each Na atom has 1 electron in the outermost shell. For Na to react' date=' it needs to have this extra electron ripped out. And after this electron gets pulled out, Na attains a very stable, Noble Gas configuration. [b']The high stability of the end-product and the relative ease of getting there is what makes Na very reactive.[/b] Now consider Fe (a transition metal), which has the configuration [Ar] 3d6 4s2. The best you can do with Fe is pull out 2 electrons to make it [Ar] 3d5 4s1 (both half-filled subshells) or 3 electrons, to make it [Ar] 3d5 (one half-filled subshell). So, you see that for Fe to react, you must (i) pull out more electrons, (ii) and still not reach the "most" statble configuration (only a local minimum). Removing all the 3d and 4s electrons takes too much energy, so Fe can really never attain a Noble Gas configuration. The not-so-great stability of the end-product and the relative difficulty of getting there is what makes Fe not so reactive.

Voila !

If you are allowed to use the derivative of the logarithm (which is no more fundamental than that of the exponent), you have correctly found the answer, dy/dx = y = e^x If you have to prove this from "more fundamental" results, I suggeswt you use the series expansion for e^x and differentiate term by term.

...except that you haven't stated what the question asks for. You have merely provided an expression and asked for an answer. Yes, it's obvious that you are required to simplify the provided expression, but unless you state this explicitly, the question does not exist. Henceforth, please state the question in its entirety.

Wait a minute : chemical equations are completely different from mathematical equations !!! To all posters : please, PLEASE, post you problems/questions EXACTLY as they are given to you. What can you possibly gain by rewording it ?

There are formulae used to calculate various hardness values based on expirimental data. For instance the Vickers hardness depends on the apex angle of the indenter, the applied load and the size of the indentation. You can probably find this online somewhere. On the other hand, if the question wants you to derive (empirically) a relation between say, bond length and bond energy, that would be more interesting.

Perhaps the bond energy ? The C - C single bond has an energy of 348 kJ/mol. Also, the bond energy varies roughly inversely with the bond length, which for the C - C bond is 154 pm. http://www.science.uwaterloo.ca/~cchieh/cact/c120/bondel.html The in-plane bond length in graphite is about 142 pm and hence these sp2 bonds are stronger that the bonds in diamond. But the interplanar spacing is about 335 pm, and the van der waals energy holding the layers together is small (roughly about 10 kJ/mol). Also, to say that "These [pi] bonds allow electrons to travel between atoms with greater ease" is like putting the cart before the horse. It is the delocalized 4th electron (of each C-atom) that gives rise to the distributed (meaning, the positions of the pi bonds are not fixed, just like in benzene) pi bonds in graphite. Correct. The C-atoms in fullerenes are also sp2 hybrid, but being non-planar, the bond angles are smaller than 120 deg.

Oh, but if the units you are using are the least count (or resolution) of your fluxmeter, then you can only measure whole number units of flux.

Not just possible, but a certainty ! Or by chance are you getting confused with the flux through a superconductor in the Abrikosov phase? That will always be an whole number multiple of h/2e.

Sounds like the kind of strategery that the dude in your picture would come up with.

Ah, I see.

Don't know what I was thinking...ignore that. The wall I found myself running against was in understanding completeness (for QM). I got a feel for completeness (of an infinite set of orthogonal functions) in terms of being able to write any continuous function as a linear combination of the basis functions. Beyond that, one runs into Lebesgue integrals

Alternatively, noticing that the Langrangian is not explicitly time-dependent, you can directly write H=T+V. PS : This is a purely classical problem - nothing quantum about it (in fact, I don't see any physical chem in it either).

Sorry for the "extended digression"...this will be the last of it (and it's only for the sake of completeness). I recall arriving at the energy-time uncertainty by thinking about correlation amplitudes (courtesy Sakurai). For a continuous spectrum you can write : [math]C(t) = \langle \alpha,0|\alpha,t \rangle = \int dE |g(E)|^2 \rho(E) exp \left( \frac{-iEt}{\hbar} \right) [/math] Now for a real, physical problem let [imath]|g(E)|^2 \rho(E)|[/imath] be peaked at some E = U. Then writing [math]C(t) = exp \left( \frac{-iUt}{\hbar} \right) \int dE |g(E)|^2 \rho(E) exp \left( \frac{-i(E-U)t}{\hbar} \right) [/math] For large t, the integrand oscillates rapidly unless |E-U| is small compared to [imath]\hbar /t [/imath]. So to not see significant deviations from C(t) = 1, you need [imath]t |E-U| \equiv t~\Delta E < \hbar [/imath]. And this is not true only in the case of a continuous spectrum. Okay, now back to the naked singularity.... It has been shown repeatedly, I believe, (and Hawking has had to eat his words on this) that observable singularities can exist (from the collapse of a scalar field, for instance). So I don't see the validity of the question.

That's not fair, Swansont. The energy-time uncertainty does not come out of QM the same way as the x,p relation. Yes, energy (frequency) and time are canonical conjugate variables (fourier transforms of each other), and this provides an argument for the existence of an energy-time uncertainty relation. You can't do the same thing that you do with position-momentum, because time is not an observable in NRQM. The way I arrive at the HUP relation for x,p is by plugging in for <[x,p]> and <{x,p}> in the Schwarz Inequality and juggling with the math (of hermitian and anti-hermitian operators), but there exist no canonical commutation relations for E,t so I can't do the same with them.

The book's answer is correct. So are your steps up to everything that you've shown. (points are correct; slope is correct) Perhaps you made a mistake substituting (x1,y1) in the final equation. Which point did you use - M or N ? I suggest you recheck this last bit of working, or show what you did.