If 15% of a starlight passes through each 1000 pc of interstellar medium, what % passes through 3000 pc?
371630
the first 1000 blocks 85% of the light and passes 15%.\n\nThat means only 15% of the original light "makes it" to the second 1000.  The second 1000 ALSO blocks 85% of the light and pssses 15%; BUT those numbers apply to the light that "made it", not to the "original" light source.\n\nand so on.\n\nThe total light passing through, converting 15% to decimals (0.15) is:\n\n(0.15) * (0.15) * (0.15).
1103364
each magnitude corresponds to a ratio of 10^(-.2), so you want to multiply the (-.2) by (1.9) and look at 10 to that power. This gives the  fraction of photons that make the trip.\n\nHomework and repeat question.
471913
The daily path of an object on the celestial sphere, including the possible part below the horizon, has a length proportional to the cosine of the declination. Thus the speed of the diurnal motion of a celestial object is this cosine times 15 °/hr = 15'/min = 15"/s, i.e. (compare angular diameter):\n\nup to a Sun or Moon diameter every two minutes \nca. four seconds for the largest planet \n2000 diameters of the largest stars per second
1187636
Light always travels at the same speed in a vacuum, the speed changes ever so slightly when it has to go through a different medium (water e.g.) but its so fast to begin with you wouldn't really notice it. 300,000 kilometers per second is beyond human comprehension, I mean you can only imagine how fast that is, you can't experience it or 'know' it.
983322
frequency = C / wavelength.  c is speed of light
26330
There is an ambiguity in the question!\nIt is not told in which medium it is.\nThe speed of light is 299,792,458 in vaccum & it is almost the same in air.
1074904
This can be proven with a simple experiment.  Light travels at 186,000 mps, if you go to a desk or anything with a drawer that can be pulled out, while the drawer is closed it can be fairly safe to say that it's dark in the drawer.  When you open the drawer you will find that the dark excaped before the light got in.  So, you can safely say that darkness travels faster than the speed of light.
739516
The light intensity is inversely proportional to the SQUARE of the distance.\n\nUse this formula:\n\nEarth sunlite * Earth dist. ^ 2 = Mercury sunlite * Mercury dist. ^ 2\n\n1400 * Earth dist. ^ 2 = Mercury sunlite * 5.7e10 ^ 2\n\nYou use multiplication, not a proportion, because the relationship is inverse. \n\nRadagast is wrong in three ways. First, the proportion is INVERSE, not direct, so we use multiplication, not division. Second, the relationship depends on the square of the distance, not the distance itself. And third, he means similar triangles, not congruent ones.\n\nRadagast's formula will give you lower light intensity on Mercury than on Earth, clearly nonsense.
710893
light my son light.\ndarkness means 0% illumination.As light from a star  travels in all direction it covers all of the universe(and never gets destroyed in vaccum so keeps on travelling) And there are so many stars.\nonly a small part of the space is filled with light absorbing object and black holes.So almost 90% of space is occupied by light.\nso light is the king!!!!
16428
If your question is based on what you can see as well as are actually above us and that you are a young person with pupils that will dilate to ~ 7mm tops AND you are observing from a dark dark site with steady seeing conditions (very little "twinkling" of the observed stars)...I think around 15,000 "bright" (we won't get into "limiting magnitude" as that is a large variable) stars is a safe guess. However, technically due to the fact that the as you look towards the constellation Sagittarius in the summer from the N. hemisphere you are looking towards the center of the Milky Way (our galaxy name). That  splash of "cloud" is the combined light of (as C. Sagan would say) "billions and billions" of stars.\nThat latest estimation of our sister galaxy Andromeda is upwards of 300 billion stars alone (it's number of orbiting stars that are influenced and follow it's spin has just recently been upped quite a bit). The current estimate for the Milky Way is 250 billion...however it has also recently been re-classified as a "barred spiral" and no longer a pinwheel shape like "whirlpool spiral". So...who knows a re-estimation of it's star number content may change.\nIn general most galaxies average out to have between 200 - 300 billion stars.\n\nNow...if we take all the assumptions mentioned above (about your eyes, seeing conditions etc) and we add to the number 15,000 of bright individually distinguishable stars in the Milky Way and the number of stars represented by the glow of the Milky Way and the number of stars in the Andromeda galaxy (which a kids good eyes should be able to see ...no sweat) well then the number you can see goes...WAY UP... Carl Sagans answer is an understatement.
852339
Well, obviously the sun doesn't stop outputting light just because we're cloudy.  We're still getting 1120 W/m^2 through the atmosphere.  It's just being reflected, scattered, or absorbed by clouds, moisture, etc.\n\nSince clouds appear white from space, that says that a big chunk of the sunlight that would otherwise have it the ground winds up sailing off into space.
684422
I know EM can go up to 100,000x (maybe beyond?), and light can go up to 1000x.  I'm not sure what the limit on EM is, but i think it's less than 800,000x.  So #4 looks right.
835254
The sun emits energy at a rate of P = 4x10^26 W. The amount of power density that reaches the icy planey is then this\n\nPd = P/(4*pi*R^2)\n\nwhere R is the distance from the sun to the planet. This is the total flux of energy emitted by the sun spread over the surface area of a sphere of radius R. The albedo (A) of a planet indicates how much of the incident energy is reflected back into space. The absorbed power density is then\n\nPa = (1 - A)*Pd = (1 - A)*P/(4*pi*R^2).\n\nThe absorbed energy heats the planet and also leads to radiation. The radiated power follows the Stefan-Boltzmann law\n\nPr = e*s*T^4\n\nwhere e is the emissivity, s is Stefan's constant, and T is the temperature of the radiating body. Stefan's constant is s = 5.67x10^-8 W/(m^2*K^4), and for a blackbody, the emissivity is one.\n\nAt equilibrium, the absorbed power and the radiated power are equal\n\n(1 - A)*P/(4*pi*R^2) = s*T^4\n\nwhich can be solved for T, the temperature of the planet.\n\nThe temperature of the planet will impart kinetic energy to gas molecules in the air. The speed of the molecules will be distributed with an rms spedd os\n\nVrms = sqrt(3*k*T/m) = sqrt(3*R*T/M)\n\nwhere k is Boltzmann's constant, R is the universal gas constant, m is the mass of the molecule, and M is the molar mass.\n\nFor a planet the escape velocity for an object is given by\n\nVesc = sqrt(2*G*M/r)\n\nwhere G is the universal gravitation constant, M is the mass of the planet, and r is the planet's radius.\n\nIf the speed of the gas molecules is greater than the escape velocity of the planet, then that planet will not be able to retain that gas at the given temperature.
966236
No matter how fast you go or think you are going, light always appears to be going the speed of light.\n\nEven if you are going 0.99999999999 c from the viewpoint of an outside observer, you would still see light traveling at the speed of light around you. Now, light sent by objects moving at 0.9999999999 c towards you (those "ahead" of you) would be severely blue-shifted and light from objects moving away from you at that speed (those "behind" you) would be severely red-shifted.\n\nBut light from any object moving the same speed as you would look perfectly normal. In fact if you covered the windows of your spaceship (it better be a spaceship, since you won't be in any atmosphere for very long), you wouldn't know you were moving at all!
402996
Light travels at 186,000 miles per second, or 300,000 kilometers per second.
416072
It depends on the wavelength of the light and its energy.
721172
A) Speed is distance divided by time, as in kilometers per second or km/s, which can also be expressed as km*s^-1 (kilometers times seconds to the minus one power). The incorrect notation km/s-1 appears to be a mixed up combination of these two ways of expressing speed.\nB) Light speed is 300,000 km/s, so 9,000 is 3% of that.\nC) A Parsec is 30,856,802,500,000 km and a year is 365*24*60*20 = 31,536,000 seconds. Doing the math gives 0.000009198 parsecs per year.\nD) A megaparsec is a million parsecs so 0.000009198 * 4 billion / a million = 0.036792 Mpc/year