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49

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 ...


43

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 ...


31

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 ...


21

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 ...


20

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 ...


20

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 ...


19

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 ...


17

@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

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}\ ...


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. ...


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

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 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

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 ...


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

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 ...


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) ...


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

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 ...


6

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 ...


6

There are a few ways I might approach this experimentally: (1) - Strip a pencil down to the cylindrical graphite core (or simply use a mechanical pencil), weigh this core to obtain a value for $m_{core}$, and then counting as you go, draw fixed-length lines on paper using a straight-edge. After some time, weigh the remaining section of the core to determine ...


6

The practical detection limit for HST is about a visual magnitude of 30 - that sort of number was reached in the ultra-deep field. Assuming that the solar sail kept reasonably stationary for the 100 hours or so of required exposure then we could do a calculation based on that. There is absolutely no need to resolve the object in order to detect it. If you ...


6

$$ 110 \text{ hp} = 82 \text{ kW} $$ This is 1000x what the laptop draws. You won't notice. To put that change in perspective, you would see a similar increase in the power (85 W) used by the car if your speed changed from 65.00 mph to 65.02 mph, since $P \propto v^3$ (at high speed, the power goes as the velocity cubed) as per this answer, so $ ...


5

The problem is that humans are complex systems, and one of the characteristics of complex systems is that they show emergent behaviour. By this I mean that the behaviour is not simply related to fundamental properties of the system. People have noticed various trends, for example it has been observed that many mammals live for about the same number of ...


5

The V2 reached a height of 80 km at the peak of its approximately parabolic trajectory. To reach such altitude (ignoring friction) requires a supersonic take-off speed of $$V = \sqrt{\frac{gh}{2}}=630 m/s$$ That's a tough nut to crack with conventional ballistics, although regular firearms have muzzle velocities that are up there - the Swift reaches 1200 ...


5

Is it possible to estimate? Yes. I'll give it a quick try. But the details of whether the planet will be incinerated and so on will make the reality much more complicated. As a ballpark, I think supernovae release about $10^{53}$ erg of energy. Spread over a sphere of, say, 1 AU gives $3.55\times10^{22}$J.m$^{-2}$. This energy isn't all released in one go ...


5

Here's my quantitative attempt at $4.$ and $1.$: The Coandă effect here is the tendency of the airflow to adhere to the surface of the ball. This means that near the surface of the ball, the streamlines are curved with a radius of curvature approximately equal to the radius of the ball $R$; this curvature results in a pressure gradient just as it does in ...



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