Hot answers tagged

89

A "perfectly efficient" computer can mean many things, but, for the purposes of this answer, let's take it to mean a reversible computer (explained further as we go). The theoretical lower limit to energy needs in computing is the Landauer Limit, which states that the forgetting of one bit of information requires the input of work amounting $k\,T\,\log 2$ ...


65

Quantum mechanics has almost no bearing on the operation of the brain, except insofar as it explains the existence of matter. You say that signals are carried by electrons, but this is very imprecise. Rather, they are carried by various kinds of chemical signals, including ions. Those signals are released into a warm environment that they interact with over ...


64

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 ...


51

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 ...


44

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 ...


44

Your approach is incorrect. You cannot do this calculation by considering that the energy absorbed by the object is converted into a change in gravitational potential energy. For one thing the object would just get hot and radiate away most of the energy and for another this is a dynamical problem, you have to be able to accelerate the object upwards. What ...


41

The formula you want is called Planck's Law. Copying Wikipedia: The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. $$ ...


40

In the UK a packet of biscuits would typically be 200g and contain about a thousand Calories or 4.2MJ. By contain I mean that if the biscuits were burned in oxygen the energy released would be about 4.2MJ. If a train has a mass $m$ and is moving at a speed $v$ then its kinetic energy is: $$ T = \tfrac{1}{2}mv^2 $$ Equating this with the energy in the ...


37

We tend to think that our modern electronic devices are very energy-efficient so mechanical mainsprings etc. must be enough but they're not. After all, the (Intel i7) microprocessors have over 1 billion transistors per chip and each transistor has to consume some nonzero (and not "totally" negligible) energy, after all, to do an operation and they do ...


34

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 ...


33

This is very rough and based on eyeballing without careful measurements: I've got a four-watt nightlight. I can read by it (not comfortably) at a distance of about a meter. The sphere of radius 1 meter has a surface area of about 12 square meters, so it appears that 1/3 of a watt per square meter will (barely) suffice for reading. The earth gets about ...


25

The Wikipedia article you linked states: Atomic clocks show that a modern day is longer by about 1.7 milliseconds than a century ago If we take this change of 1.7 ms/century and multiply by 2.5 million centuries (250 million years) then we get a change of 4,250 seconds or 1.18 hours. So 250 million years ago the day length would have been 22.82 hours. ...


24

As mentioned by Stephen Mathey in the comments, for each body with mass $M$ and radius $r$, there is a velocity one has to attain to completely escape the body's gravity well. This is the escape velocity $$v_e=\sqrt{\frac{2GM}{r}}$$ where $G$ is Newton's constant of gravity, $M$ is the mass of the body you are escaping from, and $r$ is the distance from the ...


23

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 ...


23

Let's work this out from the Stephan Boltzmann law. What color is a fire? If you look at color charts for black body radiators at various temperatures, I estimate it to be about 1000K. (Be careful: some flames are colored by strong emission spectra, making their light very different from a blackbody radiator's color). Glancing around the web from various ...


23

Are we increasing the gravity of Earth with our population No, we don't increase the total mass of the Earth, because our bodies are made of things that were already on Earth (food, water, air, minerals etc). Do we, other species and things have our own gravity? Yes everything with mass has gravity (and things without but that's complicated) ...


23

Yes - but only in the sense that all macroscopic processes depend on underlying quantum mechanics at the microscopic scale. No - quantum mechanics is not the best model for describing what happens in the brain. In one sense, the behaviour of a neuron is similar to a quantum process, such as (for example) the decay of an electrically excited or radioactive ...


22

One point, the difficulty of seeing colors in dim light is due to properties of the human vision system. Most cameras will not have the same effect and will be able to show vivid colors in even dim light (as long as the light is sufficient for imaging). But as a good guess, with accommodation, you can read (to some extent) under a full moon. The sun ...


21

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 ...


21

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 ...


21

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

The unit of illumination is the lux, lumens per square meter. What is the minimum lux required for reading? How many lux does the Sun provide at distance D? What is the minimum lux required for reading? You can plug all sorts of numbers into this depending on how good your eyes are, how big the print is, and how close you hold it to your face. I'm going ...


18

@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} ...


18

Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately: $$ P = ...


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}\ ...


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

Let's assume you mean that Earth now has the mass of Jupiter (as opposed to actually launching from the literal planet Jupiter - whole different question...). Then: radius of Earth = $6.4 \times 10^6~\text{m}$ mass of Jupiter = $1.9 \times 10^{27}~\text{kg}$ Escape velocity, $v_\text{escape} = \sqrt{\frac{2GM}{r}}$ This gives a value for ...


16

updated calculations - based on neutrino energy escaping and vapor inhalation risk Your math is close but not quite right. First - the number of tritium atoms. There are 1000/(16+3+3) = 45 moles (as you said) This means there are 45*2*$N_A$ = $5.5 \cdot 10^{25}$ atoms of Tritium Now the half life is 12.3 years or 4500 days, that is $3.9\cdot 10^8 $s. ...


16

The simplest formula for the centrifugal acceleration is $$ a = r\omega^2 $$ Here, $r$ is the radius which is 0.25 meters in your case. $\omega$ is the angular velocity which is $2\pi$ times the frequency $f$. Your $f$ is 1500 revolutions per minute which is $1500/60=25$ revolutions per second. In the SI units, we have $$ a = 0.25\times 4\pi^2 \times 25^2 = ...



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