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May 15

Space launch, part 2

I have asked this question already (how long does it take at Baikonur to prepare a Soyuz rocket for launch), but it was answered only partially (that it takes 2 days from roll-out to launch). So here's the other part of the question: What is the minimum amount of time it takes from the beginning of payload encapsulation and mating to the rocket until roll-out of the rocket to the pad? I know it can't take longer than 5 days (from the document previously shown), but can they do it any faster? 2601:646:8A00:A0B3:441F:C8FF:AED4:31F7 (talk) 11:46, 15 May 2018 (UTC)[reply]

Finding off-target cleavage sites of Cas9 by dCas9-ChIP: biased technique?

This paper discussing CRISPR-Cas9 features this sentence: "Although a number of studies have employed sequence similarity-based off-target search or dCas9-ChIP to predict off-target sites for Cas9, such approaches cannot assess the nuclease activity of Cas9 in a comprehensive and unbiased manner."

My question is: how could dCas9-ChIP be biased? 129.215.47.59 (talk) 13:43, 15 May 2018 (UTC)[reply]

I have not accessed the paper, so this is a terrible answer, but my first thought is that if the dCas9-ChIP relies on binding, then it won't necessarily match up with where Cas9 actually cuts. It might bind in some way where it can't cut at all, or it might have a site where it has somewhat lower affinity for the uncut DNA sequence but somewhat higher affinity for the transition state. Wnt (talk) 14:23, 15 May 2018 (UTC)[reply]

My reading of that sentence is not that the Cas9 is biased, but that the off-target searches are. Fgf10 (talk) 19:26, 15 May 2018 (UTC)[reply]

Half-cell basics review

This isn't a real test question, but if it had been, maybe I'd keep how reduction potentials and half-cells work clearer in my head. My memory has corroded...

A man is wearing a gold bracelet on one wrist and a platinum bracelet on the other. He is coated with sweat and an unspecified but non-corrosive and non-valuable grime from work which is the same on both sides. A perfect voltmeter is used to measure the potential difference between the bracelets. The reading will be:

a) zero, because neither of these metals reacts with anything unless a voltage is applied.
b) zero, because neither metal has ions in solution.
c) 0.32, because Pt2+ + 2e- has reduction potential of 1.2 and Au3+ + 3e- has reduction potential of 1.52.[1]
d) some other number because I fouled that up, for example using the Au+ + e- reduction potential of 1.83.
e) depends on the composition of the grime and whether it can form compounds with gold or platinum.

The real gold here is if we can explain persuasively why some answers aren't applicable. Wnt (talk) 14:39, 15 May 2018 (UTC)[reply]

None of the above, but e) is closest. It has nothing to do with forming compounds with gold or platinum, but on the potential to be oxidized or reduced by gold or platinum. The half-reactions will depend on 1) what ions are present in the grime and what sorts of half reactions they could undergo and 2) the reduction potentials of those reactions. Regardless, the platinum band will be the anode of the reaction, and gold the cathode if any DOES occur, because platinum has a lower Standard electrode potential, and thus would be more likely to be oxidized than the gold. Because of that, there is no reaction which would cause the gold band to be the anode preferentially over the platinum band. However, WHICH reaction would occur, if any, would depend on what complementary reduction half-reaction is happening at the gold band. --Jayron 32 16:10, 15 May 2018 (UTC)[reply]

Platinum atoms have a proton less than gold atoms (Pt/Au atomic numbers: 78/79) and are less electronegative, which equates to being more electropositive or "noble". Their electronegativities may be compared numerically (Pt/Au: Pauling 2.28/2.54, Allen 1.72/1.92). Therefore to the extent that the man's sweat functions as a shared conductive electrolyte between his bracelets, he becomes the frame of a Galvanic cell where his gold bracelet is the cathode. A high impedance voltmeter will consequently show the platinum bracelet has a positive e.m.f. relative to the gold. Over a long time the relatively "less noble!" gold could even develop sacrificial Galvanic corrosion while the platinum is preserved. Since there are umpteen ways an unspecified grime might react preferentially with one bracelet, the answer e) is indicated. If the "gold" bracelet is in fact only gold plated, its shine may not last long. DroneB (talk) 20:07, 15 May 2018 (UTC)[reply]

From Galvanic series "When two metals are submerged in an electrolyte, while also electrically connected by some external conductor, the less noble (base) will experience galvanic corrosion." The bracelets are not connected by an external conductor, so no corrosion. 114.124.149.159 (talk) 10:31, 16 May 2018 (UTC)[reply]

Galvanic corrosion relies on the flow of a current, and so requires a conduction path outside the cell. However an isolated cell can still show an EMF, without there being current flow (other than that through the voltmeter). Andy Dingley (talk) 11:09, 16 May 2018 (UTC)[reply]

Supernova is loud

I believe the supernova is so loud that if there's air between earth and the exploding star for sound to travel through, we would hear it from many light-years away. Even the loudest recorded sound here on earth, the eruption of Krakatoa in 1883 which is heard for thousands of miles, is merely a pin drop compared with supernova. I judge supernova explosions generate thousands or tens of thousands of decibels. How loud exactly can the supernova be? Planet Star 22:56, 15 May 2018 (UTC)[reply]

Human death by sound is estimated to be 198 to 202. I think the lungs burst or something. So above that. Sagittarian Milky Way (talk) 23:18, 15 May 2018 (UTC)[reply]

Under a thousand though, because logarithms. Sagittarian Milky Way (talk) 23:19, 15 May 2018 (UTC)[reply]

How much air is there in the vacuum of space? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:34, 15 May 2018 (UTC)[reply]

Very little. The Moon's surface has 100,000,000,000,000 times thinner than Earth and that's still in the solar wind. Sagittarian Milky Way (talk) 23:55, 15 May 2018 (UTC)[reply]

So, for all practical purposes, a supernova explosion would not be loud. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:59, 15 May 2018 (UTC)[reply]

Above c. 170 decibels (forgot exact number) if the center of a sine wave was at sea level pressure the bottom would have to be below pure vacuum but you can't go lower than complete absence of atoms. The peak to trough value of a shock wave could still be higher than 2 atmospheres but it's debatable if that's still sound. Sagittarian Milky Way (talk) 00:16, 16 May 2018 (UTC)[reply]

If you were on a close planet it'd briefly do things to your eardrums that could be interpreted as sound but since the things are shock waves that'd kill you on contact and neurons cannot transmit anything faster than about 150 meters/second you will never feel the hit on your body much less hear it. Sagittarian Milky Way (talk) 00:29, 16 May 2018 (UTC)[reply]

If space were filled with air, all kept magically at Earth surface temperature and pressure, it would take almost a million years for a sound wave to travel a distance of one Light-year. (Calculating 9.46x10¹⁵/340/3600/24/365). The sound of Krakatoa erupting was widely reported and happened during the days of early experiments with mechanical recording, see History_of_sound_recording#The_acoustic_era_(1877_to_1925), but it is not known to have been knowingly recorded. The estimated energy of the eruption was a nearly infinitesimal fraction 5.6x10^-27 to 8.4x10^-29 of the net energy released in a Supernova (Calculating in joules: 2x10⁸x4.2×10⁹/1.5 to 100 x10⁴⁴), see Orders of magnitude (energy). DroneB (talk) 00:44, 16 May 2018 (UTC)[reply]

But in this alternate magical air-filled universe, the acoustic energy will presumably radiate isotropically, so the energy that reaches Earth is an infinitesimal part of the whole, just like the light energy. -Arch dude (talk) 02:15, 16 May 2018 (UTC)[reply]

This is a science desk, so let's be careful with words like "infinitesimal". In a consistent medium, both sound and light energy are distributed along the surface of an expanding sphere and therefore follow an inverse-square law. In this magical universe filled with (perfectly uniform and dust-free) air, the fraction of the acoustic energy that reaches Earth will be exactly the same fraction as that of the light energy. For example, in comparison to a position 100,000,000 (or 10⁸) kilometers from the supernova (about the Sun–Venus distance), if we are 100 light-years away (or about 10¹⁵ km), then we would receive 10^{2 × (8 − 15)} = 10⁻¹⁴ of both the sound and the light energy. --76.69.47.55 (talk) 06:54, 16 May 2018 (UTC)[reply]

While I agree with your sentiments about technical terms, please note that many words have technical meanings in a technical context but still have general meanings in a general scientific context. "Infinitesimal" is technical in math, but not when used to mean "teensy-weensy". In your sentence, you use the general sense of the terms "consistent", "medium", "energy", "distributed", and "surface", all of which also have technical meanings. -Arch dude (talk) 18:18, 16 May 2018 (UTC)[reply]

... but the usage of those other terms was not inconsistent with the technical meaning, as it was with infinitesimal. If one means extremely small, then one should say so, rather than use terms like infinitesimal or "teensy-weensy". Dbfirs 07:35, 17 May 2018 (UTC)[reply]

If there were air between the earth and any given supernova that has ever been observed from earth, and we ignore all other consequences of such air existing, the supernova's energetic ejection of ionized gas particles would impact upon air particles and transfer kinetic energy to them, with those impacted air particles then in turn impacting nearby air particles and transferring energy to them (according to their mean free path in air). The resulting pressure wave could reach earth. But as it is, with no air between the supernova and earth, the highly energetic ejected particles can just come all the way across space to earth on their own. Once here, having lost essentially no energy due to inelastic collisions, they impact the upper atmosphere, the particles of which can in turn transfer the energy downwards to earth, and once the mean free path shortens enough, this can also form a pressure wave. But that doesn't happen because all observed supernovas have been so far away that the energy for any such pressure wave is insignificant. Having air between earth and the supernova won't abrogate the inverse square law. 202.155.85.18 (talk) 01:55, 17 May 2018 (UTC)[reply]

Are you sure the supernovae of recorded history haven't been deflected before reaching Earth's atmosphere? The stellar winds of the galaxy are slowed to almost nothing by the heliopause but supernovas are much more powerful of course. Sagittarian Milky Way (talk) 02:52, 17 May 2018 (UTC)[reply]

See Local Bubble and Superbubble. Graeme Bartlett (talk) 23:15, 17 May 2018 (UTC)[reply]

The energy released by a supernova "can be as much as 10⁴⁴ joules". [2] The Absolute threshold of hearing is 0.98 pW/m2. Often supernovae are visible for long periods, suggesting a slow release of power, but let's suppose aliens are able to convert the entire force of the explosion into a sound that lasts for 1 second that is able to cross space. Then we need 10^-12J/m² to hear it. This gives us a maximum area of coverage of 10⁵⁶m². Taking area = 4/3 pi r², we get r of about 5 x 10³⁷m. Well, a light year is just 9.4607 x 10¹⁵ m, so that actually allows for a 1-second sound, in concept, to be heard about 5 x 10²¹ light years away, which is indeed quite a reach! Now I'm not sure if there is a source on the entire internet that clearly explains the relationship between sound energy and sound pressure (certainly not on Wikipedia), but the "intensity" of a sound is apparently the square of the sound pressure. [3] So the "decibels" related to the "standard reference sound intensity" of 10^-12 W/m^2 at a distance of, say, 1000 light years would be (5x10¹⁸)^2 = 2.5E+36 = something like 363. But the "db SPL" might be twice that? Anyway, let's try to avoid ticking off people who can weaponize supernovas. Wnt (talk) 13:37, 22 May 2018 (UTC)[reply]

Star Wars predicts this won't happen till farm boys can afford trans-galactic travel. Moore's Law predicts in 2080 a few weeks' wages can buy the FLOPS of 10 billion humans so supernova weapons are easily possible by the 21st century. (technological singularity) Sagittarian Milky Way (talk) 16:32, 22 May 2018 (UTC)[reply]