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Another possibility is that a fan would "de-stratify" the air in the room. If the temp sensor was fairly high up a fan would mix the hot and cooler layers of the air in the room making the sensor a bit cooler than it would normally be.


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A fan moves air around. It makes people feel cooler, by causing evaporation of skin moisture (sweat). A fan's motor also gets hotter. Air moving over a thermostat would have no affect (thermostats don't sweat), but the increase in temperature of the fan's motor, would increase the air temperature slightly, causing the air conditioning to work harder. If ...


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The answers so far are OK, but at the risk of being downvoted for humor, let me suggest reading This What-if Comic If you start with an evacuated canister, things will be different from starting with a canister full of atmosphere at STP.


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The idea is that helium weighs less than normal air. To consider a more extreme example, imagine two situations where the milk canister is underwater. In the first situation it is filled with air; in the second, lead. In the case where the milk canister is filled with air, it may even float. It would still want to go down because of gravity, but the ...


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The difference will be : $$\Delta M=V*(\rho_{He}-\rho_{air})$$ where $\rho$ stands for the volumetric mass density and $V$ the volume inside the canister. Therefore $V*\rho_{air}$ is the mass of air inside the canister. $$M_{displayed}=V_{inside}*\rho_{gaz}+V_{canistersteel}*\rho_{steel}-V_{tot}*\rho_{air}$$ The scale measures the difference between the ...


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I recently saw a video of a demonstration by a Japanese researcher who came up with a method that used a pair of high-powered (presumably) infrared laser beams that, where they intersected, heated the air enough to turn it into plasma, creating a pulse of white light. It works, but it's slow, low-resolution, & requires staggering amounts of power. If you ...


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It is possible to hit spots in the air with laser pulses from multiple directions in such a way that air molecules in that spot become ionized and emit light, see the technology discussed in this article, along with this demonstration video. And if you just want a 2D screen rather than a 3D display, then of course you can also just use lasers to project it ...


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Not really; you need to have the laser light pass through particles in a medium. Laser light is made of photons; in order to see the laser, photons must be reflected off of a something to your eyes. You cannot otherwise "see" a photon because photons don't interact via the electromagnetic force - in other words, photons don't emit photons. To see the laser ...


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Although this is really more an engineering question, I will give a few pointers. First - the pressure equation you give relates pressure in a liquid to depth - I think. But I can't quite make out the units you use. Air pressure is "essentially constant" over most volumes - although air pressure does change with height, the density is so low that you have ...


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For practical purposes, we can assume that the disappearance of a 1.2 cubic cm (1.2 ml) object gives a waveform that's very similar to the sudden appearance of a 1.2 ml object, except for the sign of the resulting pressure wave. Now we do have a simple means to create that effect: setting off gunpowder will suddenly produce a lot of gas. You'd need just a ...


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The following is not my own research, but taken from Randall Munroe' wonderful what-if "Glass Half Empty" where he describes a glass of water, bottom half filled with water, top half filled with vacuum (or: nothing). (edited to exclude other two glasses) But what if the empty half of the glass were actually empty—a vacuum? (Even a vacuum arguably isn’t ...


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I suggest a totally different approach. But it's only a partial approach with much guesswork, too. The ear is able to perceive 20 µPa. (at 2 kHz). Of course you could calculate some pressure changes at the closing void, but these actually have nothing to do with the sound pressure at your ear drum. Let's do some energy calculations. 20 µPa at 1 cm² area at ...


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Kudos to the question-asker for thinking about everything they read! :-) I was pleased to note that the author of the previous answer mentioned "audible" means "audible" to the human ear. Note also that "audible" also depends on the frequency a bit...generally-speaking, as humans, for high-frequency sounds we need them a little more intense if we're going ...


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Sound intensity is measured on the dB scale, which is a logarithmic scale of pressure. The "threshold of hearing" is given by the graph below: which tells you (approximately) that 0 dB is about "as low as you go" - the "threshold of hearing". Note that sound signal drops off with distance - we will have to take that into account in what follows. If you ...


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This only a partial answer. Your understanding of acoustics needs to be enhanced a bit: for one thing, an "audible sound" means there are frequencies our ear responds to. This means roughly 20 Hz to 20 000 Hz . Now, if we assume (I know, I know :-) ) that the parchment left a vacuum in its place of some small but specified volume, and we assume a ...


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According to the research paper linked here: http://www.researchgate.net/profile/Weicheng_Cui/publication/222221948_An_overview_of_buckling_and_ultimate_strength_of_spherical_pressure_hull_under_external_pressure/links/53f1a2950cf26b9b7dd0da3c The pressure difference which can be held by a sphere of any particular material is a function of (t/R), where t is ...


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An interesting question, but I think this would be absolutely impossible for the following reason. Vacuum chambers, particularly large ones, need thick strong walls to prevent air pressure making them collapse. The larger the volume the greater the problem. Pressure = force over area - so force = area times pressure - $F=PA$ For area of $10 m^2$, Force ...


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A parachute is a device specifically designed to create viscous friction. Viscous friction generates a force that: is oriented opposite to the velocity; is proportional to (a certain power of [*]) the velocity. So the falling velocity will increase until the drag force (pointing upwards) becomes equal to the weight of the falling object (pointing ...


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It could be possible if the parachute was very large, rigid, shaped like a floating object, and you started descending from the vacuum of space. In this case the parachute would float on top of the atmosphere. It's easier to visualize if you imagine the parachute being a boat and you fell into some water; the boat would float on top of the water and reduce ...


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Without the ability to change the shape of the parachute, no. With the ability, yes - briefly. A modern square parachute acts as a wing, producing enough lift to slow the descent of the vehicle, but it relies on forward momentum to do so and to remain inflated. If the trailing edge of the parachute is pulled down quickly, the air moving under the wing ...


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Yes it is possible. The trick is to have a parachute which is large enough that it's Schwarzschild radius extends down to the object it is lifting. Under such a circumstance, the parachute would stop ALL motion of the object it is lifting. PS I just watched Interstellar :D


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The closest you are going to get is a parachute large enough to slow your descent to the point where you can find lift in rising air and climb away. They exist and are called paragliders! Strictly speaking they are still falling at 1 to 2 metres per second but rely on rising air ( thermals, ridge etc ) to 'fall' slower than a parcel of surrounding air. ...


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It would be possible in theory, but only in a very side-thinking way: if you make a parachute so large it encapsulates the whole Earth, it will in effect act as a balloon and not fall down, due to the internal pressure of the atmosphere. This wouldn't work in practice for obvious reasons, but maybe in Kerbal you might be able to do something like it..


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I will answer "yes" if you think out of the box for a parachute, which is a way for a person ejected from a plane to fall on the earth safely. Theoretically, one might design a parachute with a layer of helium so as to match the parachute and person downward gravitational force at a certain height, possibly 4 km above ground so as to avoid mountains, with ...


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No. All parachutes, whether they are drag-only (round) or airfoil (rectangular) will sink. Some airflow is needed to stay inflated, and that airflow comes from the steady descent. Whether your net descent rate is positive or negative is a different question. It is quite easy to be under a parachute and end up rising (I have done it myself), you just need an ...


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To first order, the speed of sound is not affected by pressure. Pressure waves can be shown to fulfill the D'Alembert wave equation $(c_S^2\,\nabla^2 - \partial_t^2)\psi=0$ where the wavespeed $c_S$ is given by: $$c_S = \sqrt{\frac{K}{\rho}}$$ where $K$ is the bulk modulus of the medium in question and $\rho$ its density. Now, for an ideal gas, the bulk ...


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No the denser sphere pulls away from the lighter sphere immediately. Just think about the instant they're both dropped and they fall some tiny amount of distance. They both run into the same amount of air molecules in the short distance but the net effect of those air molecules is less on the heavier of the two objects because it has more mass and therefor ...


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Emerson & Hedges' Chemical Oceanography and the Marine Carbon Cycle (Google book link, you'll have to scroll to page 87) has a table that gives the value $\beta_0=2.71\times10^{-2}$ for air in seawater at 20$^\circ$C and one atmosphere. This website has a table of values at multiple temperatures from a few different published sources--the value at ...


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According to Engineering Toolbox: the water vapor in air at 90°F and 70% RH is about 9.85 g per pound of dry air. The maximum amount of water vapor at 32°F is about 1.62 g per pound of dry air. So, depending on the air flow around your glass, it can condense a lot of water. Is it possible to reduce humidity of my room (10 feet × 10 feet ) in a ...


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At 75% relative humidity and 90 F, there is abound 26 grams of water per cubic meter of air. If you had a sealed room 10 ft x 10 ft x 10 ft, at 90 F, reducing the humidity from 75% to 25% would require removing 615 grams of water vapor from the air, or 327 grams at 70 F.



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