black body radiation in relation with wein,s and stefan,s law?
108893
What specifically do you want to know?  Stefan's law tells us the rate at which energy radiates from a black body radiator, or cavity radiator, at a given temperature, (per unit area of the radiator). It turns out that this amount of radiation is independent of the particular material doing the radiation, and depends only on the temperature. :  Rc = sigma T^4 where sigma is a constant (the Stefan-Boltzman constant).    This is Stefan's law.  The distribution of the energy at each frequency of radiation was also determined experimentally to obey a formula that was independent of the material used.  \n\nRl = (C/L^5) (1/e^(D/LT), where C and D are constants, e is the base of the natural log, and Rl is the spectral radiancy, i.e. the rate of energy radiation (per unit area) at a given wavelength.  This is Wein's law.\n\nMax Plank took this experimentally derived law, and modified it slightly so that it agreed with experiment better.  He then was able to derive his version of Wein's law directly from theory, by assuming that the energy given off by the cavity radiator is emitted by tiny oscillators, each with a characteristic frequency of oscillation.  The energy the oscillators was quantized, that is, it couldn't be just any amount, it had to be energies given by the formula E = n h v, where n is an integer, h is a constant called Planck's constant, and v is the characteristic frequency of the oscillator.  Thus the energy given off by the oscillator also had to be quantized, since the energy of the oscillator always had to be an integer multiple of hv.\n\nPlank's theory could also be used to derive Stefan's law.
202172
It hasn't been observed, if that's what you mean by proven, but it is a widely accepted idea because the theoretical underpinnings are very solid.  I don't think anyone contests the idea much these days.  If we're lucky, when the LHC at CERN comes on line in late 2007, we may be able to make microsingularities and watch them evaporate, giving direct proof of Hawking radiation.
781911
None.\n\nCalories are a measure of the heat content of objects.  If you are referring to the amount of heat radiated by the earth, it would be stated in some unit appropriate to the measure of radiant energy, such as watts or newtons per second.\n\nYou may be able to find an answer by visiting the library or using the internet.
121167
http://hyperphysics.phy-astr.gsu.edu/Hbase/surten.html
88803
Go to this website and it will give you the info you need!\n\nhttp://www.sciencebuddies.org/mentoring/project_ideas/Phys_p009.shtml?from=Home
943578
do a yahoo search on    "black absorption and radiation"
505967
Dark animal fur. Black surfaces are both good absorbers and good emitters of infra-red radiation.
370272
Yes. Radiant energy strikes the matter surface it repel electrons from the matter and it release heat energy.
81056
ja :), that's a good one. E=mc^2 talks about the equivalence between matter and energy; c is just the constant that joins them both ..you won't attain a c ¨velocity¨out of it ; you'll just have to use \n E=mc^2/(sqrt(1-(v/c)^2) ( Is it Lorenz transf?) and I think Xav was pretty clear with his explanation! :)
864425
One source that breaks down these classifications by temperature, or maybe it was only color photometry (I don't recall) is Kenneth Lang's Astrophysical formula:\n\nhttp://www.amazon.com/Astrophysical-Formulae-Radiation-Processes-Astrophysics/dp/3540296921/sr=1-1/qid=1163374728/ref=sr_1_1/102-7735916-8685767?ie=UTF8&s=books\n\nVolume I has some tables I think you will find helpful.  You might need to do a little interpolation...\n\nanother place to look would be the iau website http://www.iau.org . The IAU is the kind of group that would define a standard, though this might be too detail oriented for them to mess with.\n\nYou will be hard pressed to find an equation, however.  You are going to be stuck with a lookup table.
74435
Try reading this, it should answer your question.\nhttp://phyun5.ucr.edu/~wudka/Physics7/Notes_www/node58.html
1118486
Suppose a charged particle moves at a more or less constant velocity v1 followed by a finite time interval t of acceleration resulting in a new velocity v2. The radiation emitted always spans a large range of frequencies which (as a rough approximation) goes from frequencies of about 0 to 1/t, dropping to zero at frequencies much higher than 1/t.\n\nFor the acceleration events in your examples: For an electron passing through a sharp local magetic field, t ~ a/v where a is the size of the region with the magnetic field and v the electron velocity. For a collision with a nucleus or other charged particle t ~ b/v where b is the impact parameter (the closest distance between the straight line trajectory of the electron and the particle it's colliding with).\n\nFor purposes of order of magnitude calculations, the total energy per frequency interval radiated by a particle with charge e (in gaussian cgs units) is roughly:\n\nd Energy/d omega  ~ e^2 * |v2-v1|^2/c^3\n\nfor frequencies in the region 0 < omega < 1/t.\n\nThere is never one single frequency emitted except when there's a sustained pure sinusoidal varying current as in a radio antenna. Of course atomic and nuclear transitions produce photons with precise frequencies (wavelengths) but those are not really the result of accelerating charges in the usual sense.
1026247
No.\n\nIt would violate the second law of thermodynamics.\n\nThe entropy of a black whole is huge. That of the big bang very low.
1040095
heat radiation is possible in a vacuum, thats how the suns energy reaches us
1032932
mass, length, time, current, luminous intensity and solid angle steradians r fundamental. Rest all can b expressed by these alone.\n\nDarren G - missed d last 1.\n\nSo ur ans is option D
1095651
Okay, everyone else so far is WRONG. The sun, like any well-behaved (approximate) blackbody, does absorb light which impinges upon it. A good illustration would be if the sun were part of a binary star system and it's companion star were 20x brighter. When you are positioned such that all of both stars were visible to you, things would be quite bright and you'd be wishing for some SPF 10,000. Now, if the sun moved in front if it's brighter companion (from your point of view), the total amount of light you see from the binary system would be reduced by a factor of around 20 because you would be in the SHADOW of the sun with respect to the brighter star... in other words the sun itself, although it is a light source, would be shading you from it's brighter companion because it does absorb light. I hope this helps.
1113966
You are correct.  The dominant effect with a black shirt in sunlight is absorption of energy.  At night, there is minimal incoming energy, so the dominant effect is radiation of energy.  A dark object under an open sky (without clouds to reflect infrared energy back) can get considerably cooler than the surrounding air.