Tag Info

Hot answers tagged

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 ...


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. ...


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

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 ...


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 ...


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

$$ 110 \text{ hp} = 82 \text{ kW} $$ 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 $ ...


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 ...


5

TL,DR: it cannot be done because there is no spear strong enough to withstand the acceleration. Long answer: There are two orbits that would give you this result. The first is one where you throw the spear "slightly up" - it would rise out of the atmosphere as drag slows it down, and slowly descend at ever-steeper angle upon re-entry. With the right launch ...


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 ...


5

If the black hole simply swalled matter, and didn't lose any energy, it probably isn't too hard a calculation, just assume the earth is unsupported mass that falls into the BH, which grows in mass as it adds more stuff. The problem, is we know this isn't how it would happen, and some significant fraction of swalled mass will be released as energy, maybe one ...


5

We run an experiment on my A Level Physics course to answer this question. Expose the graphite in the pencil you wish to use at either end. Measure the length of the graphite, its diameter (then calculate its cross-sectional area) and the electrical resistance along its length (either by direct measurement using a multimeter or by passing a current through ...


5

Since I have much better answer from Vagelford -- I'll write my own version. When matter falls on the black hole it gets fractioned and radiates. As far as I know (correct me if I'm wrong) one can estimate the radiated energy as $\simeq 0.05mc^2$. Where $m$ is the mass of the falling matter. The Earth's matter is pulled by the black hole gravitation ...


4

I'll take a go at it - as with the piano tuners in Chicago, I take the approach as if I have "no facts to go on". Your head has a surface area of $4\pi r^2$, the fraction of it which is covered with hair is $\gamma$. The density of hairs per unit area is $\sigma$, and the number of hairs is then $N=4\pi r^2 \gamma \sigma$ Hairs per unit area is obviously ...


4

Let me give the naivest possible estimate, so that people have something to criticize. Assuming that the most of the jet interacts with the ball and is deflected at a substantial angle then the force on the ball is roughly the momentum flow through the pen. In your units this is $\rho_{air} Q^2/(\pi d^2)$. Saying the force to levitate a ball is $1\times ...


4

This question is different from, but related to another question How is it that the Earth's atmosphere is not “blown away”?. In answering that question with respect to solar wind, I remarked that the orbital speed of Earth is 30 km/s while the speed of the solar wind varies between 300 km/s and 800 km/s in a nearly orthogonal directions (fully ...


4

Diehl et al. (2006) used gamma ray observations to map $^{16}$Al in the galaxy. Because $^{16}$Al has a half-life long compared to the expected rate of supernova, but not so long we expect the SN rate in the galaxy to have changed dramatically over that time, it might be an indicator of the recent SN rate. Actually carrying through this calculation relies on ...


4

What is the force required to move a raindrop? The force $F_{(t)}$ that pushes against the drop at time $t$ is $$F_{(t)} = \frac{cW\cdot A\cdot \rho\cdot v_{(t)}^2}{2}$$ where $cW$ is the drag coefficient, $A$ the cross section area of the drop, $\rho$ the density of the air and $v_{(t)}$ the velocity at any given time, but what you might be asking ...



Only top voted, non community-wiki answers of a minimum length are eligible