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The mechanism of a speedometer determines the travel speed of the vehicle indirectly by measuring the rotation speed of the transmission output or wheels. One turn of the wheel causes the vehicle to cover a distance equal to the circumference of the wheel, so there is a direct relation between the wheel's angular speed and the vehicle's linear speed; ...

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Yes. With low air pressure, the axle is closer to the road: the radius of rotation is reduced. Some cars have systems that warn you when your tire pressure is low. The way they work is by measuring the rotation rate of the tires. If the rotation rate exceeds a certain limit, the dashboard light glows. With under-pressured tires, your speedometer will ...

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No. Automobile tires to not expand radially to any great extent - the steel belts will keep that from happening. So, the tire radius still determines how far the car travels per rotation. Now, if your tires are slipping on the road, or are slipping with respect to the rims, than yes you have speedometer problems, but you have lots of other problems as ...

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Brionius has the right answer, but there is more to be said. Water at room temperature in air will slowly evaporate. Water at room temperature in a vacuum will boil, as is shown here. So these mini torpedos can prevent damage to chemical bonds. Water molecules are polar. The O's are a little negatively charged. The H's are a little positive. The H's and ...

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In fact, they do!! Watch what happens to an ice cube that is left in the air... trillions of particles of its exterior are torn out of their stable arrangement, and soon they cascade down the sides—a microscopic waterfall! So in this case you are right, but it is just the very exterior surface of an object that is exposed to the air and thus affected ...

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Another way of looking at this is that things that would be destroyed by the environment (be it heat, light, etc) have already been destroyed (like ice on a hot summer day). The things that you see around you are the ones where the bond energy was high enough that they survived.

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Things actually do get destroyed by what those air molecules pick up and throw around. Take look at this example [image from here: http://en.wikipedia.org/wiki/File:Arbol_de_Piedra.jpg ] Just like their bigger sized brothers, it's the load of those mini-torpedos that brings the destruction.

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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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The pressure of falling water can vary by a LOT - it depends in detail on the shape of the interface between the water and the surface it hits. When a perfectly spherical drop of water hits a hard surface, there will actually be a short moment in time when the contact point between the water and the surface travels faster than the speed of sound in water - ...

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Each kilogram of water that falls from 10 m will have the kinetic energy of about 100J. I can't tell you the pressure since you didn't specify the area. Shorter falls will have less energy.

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Let me expand on (and correct a minor error) in what I said in the alluded-to thread. Unfortunately, I do not know the mentioned article (although "The Moon's Twin", published in 1989 in "The Magazine of Fantasy and Science Fiction", discussing the Jupiter/Io system, might be it). Therefore, I can't directly address it, but I think I can say three major ...

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Let us calculate the energies. For the bomb, Google gives $3.8\times 10^{16}\,\text{J}$. For the stone, we use the Einstein equation $$KE=mc^2(\gamma-1)\qquad\gamma\equiv\sqrt{1-\beta^2}^{-1};\quad \beta=v/c$$ Plugging in the numbers, this gives a stone energy of $3.8\times 10^{18}\,\text{J}$. The stone is actually more powerful than the bomb. This is ...

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This is a well known and kenned estimate. The Wikipedia article Observable Universe covers this well; look especially at the sections under headings "Mass of ordinary matter" and "Matter content — number of atoms" for how it is derived. In summary: Cosmological models, especially the Lambda CDM-Model refined from the FLRW Metric, imply a relationship ...

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In general the acceleration of gravity at the surface of a planet depends on both its radius and its mass (density times volume): $$g=\frac{GM}{R^2}= \frac43\pi G\rho R$$ For a given amount of work done by the athlete ($F\Delta x$), height jumped scales roughly with $g$. A vertical jump looks like this: From a standing start you jump up and measure ...

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It is a nice question, however the answer is complicated : I recommend to look at Eddington standard model in http://www.astro.umass.edu/~wqd/astro640/model.pdf, there are also nuerical methods mentioned there.

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The overly simplified (and empirically incorrect) way is to just balance the pressure at the surface $$P_{\text{gravity}} = \frac{Gm^2}{4\pi R^4}$$ And $$P_{\text{radiation}} = \frac{\epsilon \sigma T_{\text{surface}}^4}{c}$$

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You can estimate number of atoms by finding out average molar mass of paper and mass of one sheet of paper. If we assume that paper is mainly composed of cellulose, we can neglect other components as insignificant. Then we find out that cellulose's molar mass is approximately 162.14 g/mol. Mass of one sheet of paper is about 4.5 g. Doing the math you get ...

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So, a piece of paper is made of wood, and wood is some organic substance. I don't know what the chemical formula is, but let's say it's mostly carbon. In fact, let's just pretend it's all carbon, since you only want order of magnitude. Wikipedia tells me a piece of A4 paper weighs about five grams, and then I divide by the atomic mass of carbon and get ...

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What's the difference between where you are and space? Atmosphere (ie air pressure), temperature, gravity, radiation. Think about how each of those affects what happens to someone after they pull the trigger of a gun, on earth. Atmosphere: has no significant impact on the effects of recoil. It will help slow the bullet down but that's not relevant to ...

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Gravity pulls straight down. When you are at an incline, the force you need is lower (you know this from experience: pulling something up a steep slope is harder work than pulling it up a gentle slope). The easy way to do this: If you need a length of rope $L$ to lower something on an incline, and a length $H$ to lower it vertically, then the force you need ...

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I haven't done a calculation yet, but I would use an extrapolation based on the Clausius-Clapeyron formula: $$\frac{dP}{dT} = \frac{L}{T\Delta V}$$ You then take any two known thermodynamic quantities of water and water vapor and linearly extrapolate to that point where the difference is zero. A good choice could be the entropy, the entropy of water vapor ...

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As the other answers have stated, the primary oversight in the original question is the mass of the astronaut/cosmonaut holding the firearm. However, your original number for the mass of the projectile is off by an order of magnitude. Therefore, the original calculation - as well as some of the other samples provided afterward - are all still an order of ...

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The maximum Voltage before a current flow will occur from moon to earth is not that Schwinger limit! It is simply the value when field strength at the moons surface (smaller radius than earth) will be high enough to start field emission. Due to roughness of the moons surface this value is not easily calculated.

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The problem (read error) with your calculation is when you assume the speed of the bullet to $$v_{bullet}=800 m/s$$ This is the speed the bullet gets if the gun is being held relatively still by a person standing on fixed ground. The speed of the bullet comes from the impulse of the high pressure in the piston after the "explosion". Since the impulse is ...

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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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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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Based on the calculation in my earlier answer, we were going to try to charge the earth with $10^{12}C$ and put that charge on the moon. Sending all the charge back in a giant lightning strike would then cause such a rapid change in electric field (not to mention that it dumps all the energy of twelve suns for a tenth of a second...) that it would ...

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I will estimate the potential difference, and later add in the other considerations. I am taking the entire space between the Earth and Moon to be vacuum - so I am totally ignoring the effects of the Earth's atmosphere. The Swinger Limit (http://en.wikipedia.org/wiki/Schwinger_limit) is the largest electric field that can exist before nonlinear effects start ...

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A counterpoint to Schwern's answer (which was instructive, but I believe wrong on some key points - but I will borrow a couple of numbers from it). I think the correct way to pose the question is: If a 300 mA current for 100 ms will kill a human, what should be the rate of change of the electric field around the body to induce that current? Treating ...

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Mass of Sun = 1.989 x 10^30 Kg Mass of 1 lion = 190 Kg Mass of 1 lion in spacesuit = 250 Kg Mass of 1 trillion lions = 190 x 10^12 Kg Mass of 1 trillion lions in spacesuit = 250 x 10^12 Kg Mass of 1 trillion pregnant lioness in spacesuit = 300 X 10^12 Kg Temperature of Sun's surface = 5778 K Any objects would instantly disintegrated at close ...

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There is a concept of "voltage of a step"* in energy industry - if a high voltage power line is leaking into the ground and isn't shut down, then near that point the ground voltage difference over a single human step (when one feet is closer than the other) can be enough to kill a person; that's why it may be dangerous to approach fallen wires after a storm ...

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Adding to other answers here and giving another perspective, the wiki article on Ground says The Earth serves as a (reasonably) constant potential reference against which other potentials can be measured. An electrical ground system should have an appropriate current-carrying capability to serve as an adequate zero-voltage reference level. In electronic ...

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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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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 ...

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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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Start digging at the equator and move all the dirt to the polar regions. This will decrease the moment of inertia of the planet about its spinning axis. Due to the conservation of angular momentum this will result in an increase in angular velocity, akin to a figure skater who retracts her arms while spinning.

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Cover it in mirrors that are highly reflective on one side and painted black on the other. Position the mirrors so that the "faces" are perpendicular to the surface. A sketch is below (I have only shown three mirrors, the idea is that you would cover the planet with them, but they will be most effectively placed close to the equator). The plan is that each ...

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