# Tag Info

63

When you say "why aren't things being destroyed", you presumably mean "why aren't the chemical bonds that hold objects together being broken". Now, we can determine the energy it takes to break a bond - that's called the "bond energy". Let's take, for example, a carbon-carbon bond, since it's a common one in our bodies. The bond energy of a carbon-carbon ...

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You've calculated the speed of a remote-triggered gun after it fires the bullet, true. However, there's actually nothing about space in your calculation, as @ACuriousMind noted. In theory, a gun fired on Earth could fly off just as fast, at least for a second. What you should use is not $m_\mathrm{gun}$ but $m_\mathrm{gun} + m_\mathrm{person}$. The gun never ...

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Electricity isn't a gas that expands out to shock anything in contact with it. Electricity is a flow from high voltage to low voltage. Touching a charged object is only dangerous if you become a current path--if it uses you to get somewhere. Even if the earth had a net charge, you aren't providing it anywhere to go, so you will not be shocked. It's somewhat ...

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As a very rude guess, fresh snow (see page vi) can have a density of $0.3 g/cm^3$ and be compressed all the way to about the density of ice, $0.9 g/cm^3$. Under perfect conditions you could see a 13 feet uniform deceleration when landing in 20 feet of snow, or about 4 meters. Going from $30 m/s$ to $0m/s$ (as @Sean suggested in comments), you'd have ...

23

Let's work this out from the Stephan Boltzmann law. What color is a fire? If you look at color charts for black body radiators at various temperatures, I estimate it to be about 1000K. (Be careful: some flames are colored by strong emission spectra, making their light very different from a blackbody radiator's color). Glancing around the web from various ...

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For most guns, you can roughly hold them in place while fired. That is, the repulsion will not only "hit" the gun's mass but the astronaut's mass too, not allowing the gun to gain such high speed. With your numbers this leaves at most $$v \approx 0.11~\text{ms}^{-1} = 0.38~\text{kmh}^{-1}$$ for an astronaut + spacesuit + gun with $m=225~\text{kg}$, if no ...

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In the LHC, we are talking about mini black holes of mass around $10^{-24}kg$, so when you talk about $10^{15}-10^{20}kg$ you talk about something in the range from the mass of Deimos (the smallest moon of Mars) up to $1/100$ the mass of the Moon. So we are talking about something really big. The Schwarzschild radius of such a black hole (using the ...

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I'll answer the concrete question, because it's one of those fun ones where the units are all wrong and the scales are just absurd. Does this also mean that if I release a million amperes of current into the earth, every living entity walking barefooted should immediately die? It depends on how long you do it and with how much power. And ...

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This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens. Apparently the solar wind pressure is of the order of a nanoPascal. As I write this ...

18

@Señor O gives a very good answer, but he assumes an ideal deceleration. Based on a viewing of the scene, Anna sinks a little under a meter, while Kristoff doesn't sink more than half a meter. Since they fell about 200 feet (about 60 m), my initial estimate for their impact velocity is (assuming no air resistance): v = \sqrt{2gh} = \sqrt{2*60*9.8} ... 17 This is another chance to use one of my favorite approximations ever! I first offered it as an answer to a question about how deep a platform diver will go into the water. Now is the chance to use it again! Issac Newton developed an expression for the ballistic impact depth of a body into a material. The original idea was expressed for materials of ... 17 Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately: $$P = ... 17 Yes it is, if you can throw it hard enough. Not bothering with things like air resistance etc. (I think this is the least of the plausibility problems) you need to put the spear into a low-Earth orbit, such that the centripetal force is provided by gravitational acceleration.$$ \frac{v^2}{R} = \frac{GM}{R^2}Using R=6400\ km and M= 6\times10^{24}\ ... 16 updated calculations - based on neutrino energy escaping and vapor inhalation risk Your math is close but not quite right. First - the number of tritium atoms. There are 1000/(16+3+3) = 45 moles (as you said) This means there are 45*2*N_A = 5.5 \cdot 10^{25} atoms of Tritium Now the half life is 12.3 years or 4500 days, that is 3.9\cdot 10^8 s. ... 14 Although I don't know anything about this, using some rough estimates I think I can get the right order of magnitude: Volume of graphite in a pencil: 10 cm cylinder of 1 mm thick = 0.314 mm^3 (error: ~factor 2) Maximum surface a pencil can write: 50 km \times 1 mm = 10 m^2 (error: ~factor 5) Thickness of the graphite layer: Volume / Surf. ... 13 I don't think this sounds unreasonable as an estimate at all. Let's check it. One designs a building as a compromise between two competing factors: One needs all of the load bearing materials to be well mildly loaded - working in their linear region so that there is no danger of their undergoing plastic (irreversible) deformation, creeping then ... 12 The observable universe contains about 100 billion galaxies, each containing on average close to a trillion stars. That is a total of about 10^{23} stars. A typical star is like our sun. Sun has a mass of about 2×10^{30} kg, which equates to 10^{57} atoms of hydrogen per star. A total of 10^{23} stars containing 10^{57} atoms each gives us a total ... 11 The horizontal component of running is believed to be fairly negligible for humans. Some research suggests that the limit isn't strength related at all, but design --- in particular, based solely on power, humans could theoretically run up to almost 40 mph. The issue is two fold: first, our limbs are actually too heavy, for big strength (e.g. climbing in ... 11 Physics Land: The physics 101 answer to there kinds of questions goes like this: Assume the cars will lock up their wheels so that I can apply a simple analysis of sliding friction. The frictional force F_f is dependent only on the normal force N between the car and the surface and on the coefficient of friction \mu between the tires and the road. ... 11 Nice theoretical answers (I can certainly appreciate them, I'm a mathematician). But why delve into theory when experiment is available? In this video you can see a skier jump from more than 200 feet and get head first into the snow, without a helmet. The video starts with the aftermath, if you want to see the jump right away fast forward to about 1 ... 11 The fog you are seeing is condensation of atmospheric water, not sublimed CO_2. The water fog is made very near the boiling surface, and then sinks slowly, exactly as it does in rainclouds. Therefore, just because you can see fog gathering on the floor does not mean that the CO_2 is confined there. The CO_2 molecules have a speed, in random ... 11 What is the size/scale of a wood fire that is producing 1kW? Based on what my understandings re energy content of fuels and combustion processes, about 0.5 to 1 cubic inch per minute of typically dry wood in an open fire. Based on an utterly superb 80 page Wood Fuels handbook which I discovered along the way - about the same, rather to my surprise. ... 11 anyway, how likely is it the ice ages could be explained by the earth 'realigning' so that polar regions would migrate over the surface of the earth? How about zero? The geological evidence of the Ice Ages clearly says that, between the ice episodes, the ice did not move. It's just that the polar caps shrank. For instance, the extent of the last ice ... 9 Firstly we are not the best conductors, so current might be having a relatively hard time getting through us. But I believe the real reason is that you also need a high potential difference in order to get current flowing through you. Like lightning which needs a huge potential difference between the clouds and earth (so big that most of times a neutral ... 9 The cosmological estimation of the number of atoms in the observable universe works as follows: one of the Friedmann equations can be written as \dot{a}^2 -\frac{8\pi G}{3}\rho a^2= -kc^2,$where the scale factor$a(t)$describes the expansion of the universe,$\rho$is the total mass density (radiation, baryonic matter, dark matter, and dark energy) ... 9 No. No matter how hard you throw. Since orbits are ellipses, all trajectories meeting the criteria must pass through the ground at one point except for the surface-grazers. Air resistance will not be negligible so there's no point in assuming it will be. The effect of air resistance on any shape other than a lifting body is a drag force straight backwards ... 8 Live on earth is protected from solar wind by the earth's magnetic field. Charged particles from the sun (mostly) penetrate the earth's atmosphere with great velocity. These particles can be trapped by a magnetic field to follow circular path's around the magnetic field lines, thereby losing their energy due to collisions or bremstrahlung. From first ... 7 Dr. Phil Plait has written about this extensively. He has a book (Death from the Skies) with a chapter that deals with this. He has a blog entry about this very subject as well (in addition to a link to one just talking about getting hit by a meteorite). Here is an excerpt: what are the odds of getting killed by one? Turns out, they’re a lot ... 7 You should always find an answer that is a formula, and then only apply significant figures once you get to the one final step of substituting your numbers back into the problem in place of variables. Avoid multiple intermediate steps of substituting numbers at all costs. Not only will this save your pencil a lot of work, but it will also cause your ... 7 I will attempt to answer this question with some basic dynamics and some contact mechanics. There are two special cases here. a) There is sufficient friction to keep the base of the pin A fixed (imparting a reaction impulse$J_A$when hit by the ball, or b) The floor is smooth and the pin will translate and rotate at the same time with$J_A=0\$. There is ...

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