Ehryk
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 Apr 29 comment How is 6W equivalent to 40W, as claimed by adverts for LED light bulbs? @BrianKnoblauch wire two '75W equivalent' LED lights to a hair dryer, and use that in your garage or under the hood then! You get 15x the heat output of a 100W light bulb, with more light! Apr 29 comment How is 6W equivalent to 40W, as claimed by adverts for LED light bulbs? @CuriousOne I clarifying that the power ratings aren't overspecced needlessly - though they may be an upper bound not a median for safety reasons, it wasn't a marketing thing to increase sales or anything. For LED lights, I somewhat agree they are marketing these really hard; the other thing they don't mention is that they are more directional, and a '40W' equivalent LED may not provide as much illumination in all directions as a 40W incandescent, and that also isn't mentioned. Apr 29 comment How is 6W equivalent to 40W, as claimed by adverts for LED light bulbs? Incandescent power ratings are important - they need to be matched to an outlet capable of supplying that wattage, or there can be fire risks. Over time, though, the buying public started equating power ratings (not over-specced) with brightness (not correct to do), and so we have these "40W equivalent" for people who still think of brightness in terms of incandescent power ratings. Feb 2 comment Why is 7 TeV considered as a big amount of energy? I'll try to answer that with two assumptions: the 'average' collision rate is held constant for the entire season of "600 million collisions per second" (Source), and the energy released in every collision is ONLY the kinetic energy of stopping the particles. Thus, $600,000,000\ per\ second * 7\ TeV * 2 = 1345.83\ Joules\ per\ second = 1.35\ kW$, or $0.001\%$. This is, of course, absurd, as the 'purpose' of the LHC is more than just 'producing collisions', and the 120MW includes additional processing, cooling, etc. Jul 24 comment How does a planet's size really affect its surface gravity? Great ball park analysis! Good work. Apr 29 comment Is there any disadvantage to sending rockets straight up? Right, but crossing into the Moon's SOI is roughly equivalent to what I meant by crossing the Earth - Moon L1. Apr 28 comment Is there any disadvantage to sending rockets straight up? Launch Site latitude is dependent on the orbit desired and chosen to minimize fuel. For a Polar orbit, the equator is the worst place to launch as you'll have to spend $500 m/s \Delta v$ extra fuel to cancel this effect, and the north or south poles would be ideal. For equatorial orbits the direction of Earth's rotation (including geostationary), indeed the equator is the ideal place to launch and has the minimum fuel requirements, but these 'ideal' locations have to be tempered against the practicality of getting a rocket and launch facility there. Apr 27 comment Is there any disadvantage to sending rockets straight up? Picture posted showing the trajectory for a 'straight up' burn. I might try this in KSP later and post the video. Apr 27 comment Is there any disadvantage to sending rockets straight up? Agreed, I'll look for an Earth-Moon one and I'll draw in a trajectory Apr 24 comment Is it possible to generate energy by the moon orbit? It's possible that you could use the Moon to harvest the energy from Earth's rotation instead, which may end up adding to the moon's orbital energy (causing it to escape Earth's gravitational sphere of influence). Apr 17 comment Why is the singularity of the Big Bang not considered to be the center of the Universe? Or did the original $1m^3$ volume somehow vanish or disappear onto a different dimension? Apr 17 comment Why is the singularity of the Big Bang not considered to be the center of the Universe? Are you saying that the big bang happened in a fourth spatial dimension? If not, I think the balloon analogy falls short. At one point, the universe occupied $1m^3$. Then it expanded to $10m^3$, while fully encapsulating the volume of the original $1m^3$. Why is it not accurate to find 'where' this original $1m^3$ was in relation to the current universe, and call that the center? Apr 8 comment Is there any difference between fusing and smashing particles? I was referring to stellar fusion with H + H => He, and did not specify the isotope. Or were you saying that the 'stable' qualifier is incorrect? Apr 8 comment Is there any difference between fusing and smashing particles? It is not just about stability. Two nuclei => hundreds of stable Hydrogen nuclei is still not fusion. Put simplistically, you're either combining nuclei (Fusion), splitting nuclei with neutrons (Fission). Smashing things together can do either of those, or neither (energy production, misses, breaking nucleii with momentum). Apr 8 comment Is there any difference between fusing and smashing particles? In short, though perhaps not without some exceptions, the 'products' of Fusion will be more massive than its 'reactants'. The products of a Particle collider, if there are any stable products (it could in theory be 100% energy), are less massive than the 'reactants'. Apr 5 comment Why is the sky of the moon always dark? Indeed! I meant 7% / 0.07, not both. Mar 29 comment Why is the sky of the moon always dark? However, where landings happen the duller dust is stirred up and moved; and the angle of incidence to the camera is not perpendicular to the surface. Check out this page with a reproduction on earth with a soup can: www3.telus.net/summa/moonshot/fillit.htm Mar 29 comment Why is the sky of the moon always dark? I can't find many good measurements of lunar surface albedo near landing sites, but it does have an overall average of 0.07% to 0.11% (0.12% including earthshine), comparable to worn asphalt on earth. This would put it quite low compared to snow and deserts, see this chart on the Albedo wiki page. Mar 28 comment Why is the sky of the moon always dark? Rovers have, and they have cameras. Jan 20 comment Would cosmological redshift be present in the following situation? What's a better way? I was hoping to pin it down to this (preposterous) example to nail down specifically what I'm having a hard time understanding; which is the two planets are fully stationary with respect to each other.