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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 ...

17

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 ...

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 ...

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

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 ...

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

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 ...

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

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

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 ...

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 ...

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

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 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 ... 3 Supernovae can release several times 10^44 J of energy. This has resulted in the adoption of the foe (10^44 J) as the standard unit of energy in the study of supernovae. The Foe is a unit of energy equal to 10^44 joules. To measure the vast amounts of energy that produces a supernova, the scientists used a unit of energy occasionally called foe was an ... 3 I'll take a slightly different approach to the others. I just got a close haircut (not for science, but why waste a good opportunity right?) and managed to keep something like 90% of the hair. So I can use the fact that$N$hairs of diameter$d$, length$\ell$and density$\rho$have a mass $$M = N \frac{\pi}{4} d^2 \ell \rho.$$ Accounting for the fact ... 3 I just went to a mirror to count the linear hair density of my head. I found that in about$1 cm$there are$15 hairs$, thus the linear hair density is about$\lambda=15 hairs/cm$. So density of hair per unit area is$\sigma=\lambda^2=225 hairs/{cm}^2$And assume that this hair density is roughly constant. I found that it takes about 6 times the area of my ... 3 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 ... 2 The chances are of the order of 1 in 10,000. You can derive this number by assuming the most probable impact as the chance order of magnitude (there's a 1 in 10,000 chance in 2019). All that you wanted to know about near Earth objects, dangerousness and probability of impact is here: http://neo.jpl.nasa.gov/risk/ 2 I'm a first year physics student, so my answer might not be satisfactory - but I hope it will give some insight to the problem. 1) from what I know we need to consider: Drag - which I will address Turbulence - which I know next to nothing about, and therefore I will ignore with hope someone will be able to expand. we need the drag force to be equal to ... 2 Firstly, I assume that we have 300 hairs per square cm on our head. This can be tested by waxing an area of 1cm^2 on your scalp and counting the number of hairs that are removed. Step 2, we must calculate the area of the scalp, and we assume 100 hairs per square cm applies to the whole area of the scalp. I assume the radius of my head is sphere. I ... 1 The odds of a very-high-casualty rate impact in the next decade are even lower than previous answers have stated, since there are ongoing surveys looking for just such dangerous "near Earth objects" and "Earth approachers" and there are no known near-term threats: JPL's Current Impact Risks New survey instruments are coming online now. For instance, the ... 1 Given your tag of 'estimation' I would just use $$a=\frac{v-v_0}{t}$$ plugging in some numbers for terminal velocity$v$and a 'safe' parachute deployed velocity$v_0$. Then assume ~$1\rm{s}$for$t$and we have an a, those wiki links should confirm whether your numbers are good! For fun can I suggest calculating$a$for$v=0$and$t\approx0.5\rm{s}\$ ;-)

1

To a good approximation the deceleration felt by the tourist will be the same as that felt by the parachutist. There may be some elasticity in the link joining the two, but I'd be surprised if this made much difference. As always (is there anything not on Wikipedia) there are a couple of useful articles on Wikipedia. ...

1

I have not done the math but would expect that the radiation from the asphalt as T^4 will favor larger gradients for higher temperatures. I have the impression that air goes something like T^6, so even conduction energy transferred will have larger gradients the hotter it is. Your g is temperature dependent I guess. Edit in response to edit of question. ...

1

I just want to add that oscillations about the center of the Earth are dampened due to the momentum of the entering mass. Figures for volume of the mass continually eaten by the black hole differ by orders of magnitude going by previous posters. But the consumed material as it falls will depend on the cross section of this volume times the radius of the ...

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