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70

The energy of a bullet is around 735 joules (see bullet details here). This is about the same energy that I have when I'm running at about 4.6 m/s. Would you rather be hit by me or the bullet? The bullet kills you because it concentrates all the energy onto a small impact area while my impact area is rather larger (and sadly getting even larger as ...


67

7 TeV is not that much kinetic energy, that has been covered by your question and previous answers. However, in the context of a proton, with a rest mass of $1.672×10^{−27}~\mathrm {kg}$ (very, very little mass), when a single proton has 7 TeV then it is travelling at a specific speed: $$E= mc^2$$ \begin{align}E& = E_0 + \text{KE}\\ \text{KE}&=E- ...


38

Neither of those statements are true. It's an easy approximation to make: a neutron star has all of that 'space' removed from between nucleons --- so we just need to know how big a neutron star of mass equal to the solar system would be. Well, the only significant mass is the sun (jupiter is about 1% the mass of the sun---negligible). If the sun were ...


34

There is a consistent definition, but it involves a couple of arbitrary thresholds, so I doubt you'd consider it rigorous. The construction $X \gg Y$ means that the ratio $\frac{Y}{X}$ is small enough that subleading terms in the series expansion for $f\bigl(\frac{Y}{X}\bigr) - f(0)$ can be neglected, where $f$ is some relevant function involved in the ...


11

$\gg$ doesn't mean anything about a specific order of magnitude. If you are told to assume $Y \gg x$, what that means is if you have a series expansion of the form $$\sum C_n \biggl(\frac{x}{Y}\biggr)^n\tag{1}$$ you can neglect all subleading terms (i.e. all except the first one with a nonzero coefficient). For comparison, a simple statement $Y > x$ does ...


10

If you could compress the mass into that small a space, it would collapse into a black hole, at which point the notion of "size" becomes harder to define, with space-time being so warped. The "event horizon" radius would be about 3 km, if I get the formula correctly. The idea of "there's a lot of space in atoms" comes from computations which state that the ...


10

Generally taking note of these relationships is a precursor to either (a) applying an approximation or (b) using a purturbative or series solution. In case (a) what qualifies is completely a matter of your sensitivity to error. If you are going to throw out terms $\mathcal{O}(C^{\pm 2})$ and require a 1% approximation then $C$ had better differ from 1 by a ...


10

This is perhaps similar to what mbq meant, but I will elaborate. The T-p phase diagram of water tells us, for a given temperature and pressure, what phase we will get if we have a bunch of that substance. If I apply different pressures to a bottle of water, I am moving around in the p-direction of the T-p plane. I am not changing the pressure of the triple ...


9

This is a complement to dmckee's answer by way of an example. Suppose that I want to determine the acceleration due to gravity of an object near the surface of the earth. Let $h$ be the height of the object above the surface, then according to Newton's Law of Gravitation, I get \begin{align} a = \frac{GM}{(R+h)^2} = ...


9

For forces that can be expressed in terms of a quantum field theory, you can compare the size of the coupling constants (which are dimensionless). In short this means that perturbative expansions of weaker forces are well represented by a small number of leading terms because the series converges quickly, while those of stronger forces require more terms (or ...


8

There is a great answer (with references) to this at http://answers.google.com/answers/threadview/id/539329.html which I'll summarize as follows: From http://www.hawaii.edu/suremath/jsand.html the estimate for the grains of sand is 7.5 x 10^18 or 7.5 billion billion. From http://www.faqs.org/faqs/astronomy/faq/part8/section-3.html the estimate for the ...


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 ...


8

a great topic. First, ten gigatesla is only the magnetic field near a magnetar - a special type of neutron star. They were discussed e.g. in this Scientific American article in 2003: http://solomon.as.utexas.edu/~duncan/sciam.pdf Ordinary neutron stars have magnetic fields that are 1000 times weaker than that. It is true that in the magnetar stars, atoms ...


7

The "removing the space" and "atoms are mostly empty" memes for atomic nuclei are interesting, but I do grit my teeth every time I hear this. A description that fits better with me might be "remove the electromagnetic force". Concepts of size and space of particles are based on how they interact using forces. There is no evidence that fundamental particles ...


7

The most common case where this comes up is when you're dealing with a problem where it's helpful to linearize using a Taylor expansion. For example, a decaying exponential $$ e^{-t/t_0} = 1 - \frac{t}{t_0} + \frac12\left(\frac{t}{t_0}\right)^2 - \frac1{3!}\left(\frac{t}{t_0}\right)^3 + \cdots $$ or a smallish logarithm $$ \ln(1+x) = x - \frac{x^2}{2} + ...


6

Let's make some assumptions. First, assume the fish is rigid. Second, let's assume he's not flapping. Third, I guess let's assume it's a male fish since I said "he." We'll also assume this is 2D because we're looking for an approximation. I would approximate the fish as an airfoil. NACA airfoils are a pretty good choice because they are analytically defined ...


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

One notable class of exceptions are what are called "hierarchy problems" in particle physics. For example, if you identify the Planck mass as a fundamental energy scale, you end up with huge dimensionless numbers which don't have an obvious explanation (i.e. ratio of Planck to electroweak scale, etc.). Explaining these large (or small depending on how you ...


6

Hmmm...some back-of-the-envelope calculations: The depth of the air column at sea level is $14\text{ lbs/in}^2 = 2 \times 10^5\text{ g/cm}^2$, so neglecting space-charge effects and assuming minimum ionization the whole way we get about $4 \times 10^5\text{ MeV} = 0.4\text{ TeV}$ energy loss. We are actually above minimum ionization, so we can multiply that ...


5

Consider a spherical drop of water, initial temp 40C, radius 3mm, mass 0.1g To get it down to 0C, you need to remove 4.18 (J/gK) * 0.1 g * 40 K = 17 J then, to freeze it solid, you need to remove latent heat of fusion 333 (J/g) * 0.1 g = 33 J for a total of 50 J. The heat conductivity equation is $H=\frac{\Delta Q}{\Delta t} = k A\frac{\Delta T}{x}$ ...


5

According to http://ga.water.usgs.gov/edu/earthhowmuch.html the total volume of water on the earth is $1.386\times 10^9$km$^3$, which is about $1.4 \times 10^{21}$kg (I'm rounding because I don't know the average temperature and therefore density of the water). According to http://en.wikipedia.org/wiki/Biomass_(ecology)#Global_rate_of_production the annual ...


5

Molecules vibrate with frequencies in the range 10$^{12}$ to 10$^{14}$Hz. Although I don't know of any strict definition, I would take the view that a molecule must hold together for a few vibrations otherwise what you have is a collision not a molecule. That means the lifetime must be greater than 10$^{-14}$ to 10$^{-12}$ seconds, depending on the molecule. ...


5

I think the answer is no. It generally precedes some approximation method with a bounded error, but there are so many approximations methods in physics -- some rigorous, some nonrigorous -- that it's way too presumptuous to give it a rigorous definition. Generally, it means one of several things: If $a\ll b$, expanding in powers of $\frac{a}{b}$ is ...


4

It is a symbol and an idea used in mathematics too. But the important part is just that $B$ is 'ignorable' relative to $A$. This depends on the level of precision that is being used experimentally. If you're working to a precision of 1 part in 100, then $B$ should not effect the answer to that level of precision. If you're working to 1 part in a million, ...


4

These things don't have to be 'smallest' or 'largest'. They are simply (what especially high-energy physicists would agree to be) the most natural units in which to carry out calculations when doing fundamental research. The crux is realizing that things like a 'second' and a 'meter' or a 'kilogram' are purely invented because they are convenient in everyday ...


4

Yes, I just did it. Arced a small spheriod of spit (no phlem) and it hit the ground and rolled instead of splattered. -15 degrees F, light wind in parking garage (not sure if that matters) attempted with numerous quantaties of spit and trajectories but small quantity with upward trajectory is what got the tiny frozen droplet.



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