Somehow not nervous yet, but the problem is I get nervous on the day (as I do in any exams). I hope I don't panic during the test because of time constraints.
Anyway, good luck to everyone!
The ratio of the distance from any point(say P) on a conic (ellipse, hyperbola, circle, parabola) to the foci(S) to the perpendicular distance from the point(P) to the directrices(M).
PS/PM=e.
When you dissolve ethanol in water, ethanol does not chemically react with the water or decompose into smaller molecules.
It is simply mixed with water.
P(x) = x^3 + px^2 + qx + r = 0
1.Sum of roots=sqrt(k)-sqrt(k)+a=-p
a+p=0
2.Product of roots=-ka=-r
ka=r
3.Sum of roots taken 2 at a time=-k-a sqrt(k)+a sqrt(k)=q
k=-q
using 2and 1, ka=r=-qa=-q(-p)=pq
pq=r
LOL Black body radiation is exactly why Planck had to hypothesise about quantisation of energy!
Hence they r related
Don't have time to write explanation now but 'll do if I have time later
Edit: See annabackwards' post below
Although there are some errors there, they are minor and won't...
I think I get it now. Most of this black body stuff makes sense to me. (at least, that's what I think, although complex mathematics associated with this is just well beyond my brain's current capacity.)
Thanks
maybe wrong terminology
I mean the shape of the intensity vs wavelength curve.
Very small wavelength - very low intensity
Whoops, the quote on above post was actually from wikiEducator, not wikipedia
Hmm I see.
Thanks for help!!
second one is from wikipedia. Your and wiki's explanations for near absence of higher frequency seems quite different. Is wiki's also correct?
So after equilibrium has been reached, the temperature of the black body will not increase?
And does the fact that radiation absorption leads to higher temperature mean that black body produces IR radiation when atoms oscillate?
Yep, like a black body box with one small hole
Hmm.. So it's mainly the oscillation of elections but protons do contribute a minute amount?
Thanks anyway! You are the first one to either be able to answer or to be bothered to answer my question.:)