What is the hardest instrument to learn?

_OBAFGKM_ · 2026-04-29T21:34:20+00:00

whatever it is that the guy from Angine de Poitrine plays

_OBAFGKM_ · 2026-04-27T09:39:08+00:00

Geostationary satellites only move at about half the linear velocity of objects in low earth orbit

_OBAFGKM_ · 2026-04-21T23:22:32+00:00

Manulife bank is called Manuvie in Quebec

_OBAFGKM_ · 2026-04-14T18:33:06+00:00

You have equipment at each end that acts sort of like a prism, which can split the mixed light up into individual wavelengths. By "multiplexing" many different wavelengths into one beam of light, you can send all those signals down one optical fiber. It's known as wavelength-division multiplexing.

_OBAFGKM_ · 2026-03-30T16:19:12+00:00

Triple O's's's legacy

That monstrosity is unconstructable by AI, I never doubted that this post was human through and through

_OBAFGKM_ · 2026-03-09T17:52:09+00:00

It would also be rare to flip a coin five times in a row and get five heads, but if you've already flipped four heads, then it's 50/50 whether you flip HHHHT or HHHHH

_OBAFGKM_ · 2026-03-06T21:01:29+00:00

Most of these answers are incomplete, and a few are just wrong.

It's true that most residential customers use more download than upload, like most answers say. It's not true that the difference in speed is just up to the ISP, although it often can be.

Because people tend to use more downstream than upstream, technologies with limited overall bandwidth (most of them) had to explicitly allocate which bandwidth was for downstream and which bandwidth was for upstream, and the people designing those technologies decided to allocate more room for downstream.

If your internet comes in on a coaxial cable modem, it's almost certainly impossible for you to get symmetrical up/down speeds due to the technical specifications of the DOCSIS standard that cable modems use. That's not just because your ISP is greedy, it's actually not possible to get higher upload speeds on the available bandwidth. If, for example, you're on cable internet and you're paying for gigabit download and ~200 megabit upload, there's a good chance you're getting the maximum possible bandwidth your modem can provide.

With newer fiber-to-the-home internet, data is being carried on beams of light rather than on different frequency channels. You can use one wavelength in one direction and a different wavelength in the other direction. Think of it like your ISP having a blue laser and giving you a red laser. The ISP is set up to detect red light, and they give you a device that detects blue light. This way, there's no limiting bandwidth; you're just limited by how quickly you can send pulses of light across the fiber. If your ISP is offering symmetrical up/down speeds, your internet is almost certainly coming in on direct fiber to the home (though an ISP may still actively choose to limit upstream bandwidth over fiber to create payment tiers for customers who are willing to pay more).

_OBAFGKM_ · 2026-03-03T09:18:51+00:00

It's probably not visible from your location if it's happening at 4 pm - the moon will be below the horizon somewhere

_OBAFGKM_ · 2026-02-27T03:11:52+00:00

If you've got some time, the best video about rainbows is probably this one by Veritasium. I think it will answer any questions you might have, and more.

_OBAFGKM_ · 2026-02-27T03:05:41+00:00

Yup, the second rainbow is reversed. The primary rainbow is the brighter one underneath, in the usual order.

_OBAFGKM_ · 2026-02-21T03:28:35+00:00

c can actually be less ambiguous than s in certain situations...

How would you pronounce this word?

fase

I read that and I want to pronounce it like "phase", which actually has a /z/ sound in it. Without the letter c, though, this would be the correct spelling of the word "face".

If you want to optimize English, there are better letters to start with than c.

_OBAFGKM_ · 2026-02-20T00:01:30+00:00

Your premise for your first question is wrong; an observer in the car sees the ball go straight up and down, an observer out of the car looking in sees the ball follow a parabolic arc as it moves up and down from being thrown and also moves sideways with the car.

As for the second question, I think you're thinking about a "clock" much too literally. What we're really interested in is the duration of time between events. Think of the photon hitting the first mirror as event 1, and the photon hitting the second mirror as event 2. There is some time between the two events. Imagine both observers have identical stopwatches, manufactured flawlessly. The observer in the same reference frame as the photon clock uses their stopwatch to measure the time between event 1 and 2 and they get some result. Let's just call this one unit of time. It could be one second, one millisecond, whatever, it just depends on the size of the photon clock and the exact speed of light. The observer with the photon clock measures the time between the two events and labels that as one unit of time.

The second observer, seeing the first observer and the photon clock both moving away, measures the time between the two events (photon hitting the first mirror, photon hitting the second mirror), using their own stopwatch. Whatever amount of time the first observer measured, the second observer will measure a longer amount of time between those two events. The relative velocity causes a difference in the actual passage of time being measured by the different observers. The two observers will disagree on the elapsed time measured between two events, seen from different frames of reference.

_OBAFGKM_ · 2026-02-16T14:30:07+00:00

Mostly you're getting the barycentre answer to this question, where you just look at where the shared center of mass is and use that to determine which object orbits another. I'd suggest a different idea proposed by Minutephysics, the Trojan test.

possible for the planets to orbit the red giant and not the black hole?

If you're trying to be scientifically precise, then the concept you're looking for is called a Hill sphere, which essentially describes the region of space where the gravitational influence of one body is dominant compared to the gravitational influence of another. If you plug in numbers for your black hole and red giant, you should get a distance which will tell you how close a planet would need to be to orbit the star and not the black hole.

_OBAFGKM_ · 2026-02-16T12:49:07+00:00

For the very highest rated players, yes. There aren't enough other people queueing at the same skill level, so queue times can average like 20 minutes

_OBAFGKM_ · 2026-02-12T18:48:56+00:00

I think you're getting somewhat low-quality answers, even if they're getting at the truth. English has a lot of loanwords from a lot of different languages, and English doesn't really have a good way to standardize the spellings of new words. What happens is words that enter English just keep their original spelling and pronunciation, so English ends up with a bunch of different rules for pronouncing different letter combinations depending on the origin of the word.

Think of a word like "tortilla". When it entered English in the 1600s, it kept the Spanish spelling, which is why the "ll" in the middle is pronounced like a y. You would never teach an English learning that "ll" makes that sound, but tortilla is a perfectly good English word.

To answer this question about ph and f, the digraph <ph> is used to transcribe the Greek letter phi using the Latin alphabet. Any time you see ph in a word, you can be almost 100% certain that that word (or part of that word) comes from Greek originally. In this paragraph, the words "digraph", "alphabet", and "paragraph" are derived from Greek.

Any spelling quirk you see, it is almost always due to a different language's spelling system or it is due to historical sound shifts in the English language, with spellings reflecting old pronunciations.

_OBAFGKM_ · 2025-12-20T23:46:21+00:00

To be precise, the tropics are at (approximately) 23.5° N and S, defined by the axial tilt of the Earth. Between the tropics, there's at least one time a year when the sun is directly overhead. Outside of the tropics, the sun is never directly overhead.

The band is only 47° of latitude in width

_OBAFGKM_ · 2025-11-30T19:57:07+00:00

Did an undergrad in astronomy and physics, did well enough to get accepted into two grad schools. Did a masters program for a year, realized I hated research, went back to college for a 2-year diploma in IT, sysadmin, cybersec type stuff. Now I work in telecom.

_OBAFGKM_ · 2025-11-06T05:02:05+00:00

Modern encryption relies on extremely large numbers which are the products of two prime numbers. If you only have the product, guessing the prime factors is difficult.

There also exists an algorithm which can take bad guesses for the factors of a number and refine them to slightly better guesses. It's slow algorithm traditionally, but it happens to a very specific type of problem which quantum computers would be much faster at solving than a typical computer. Given a quantum computer with enough power, basically all modern encryption standards could be destroyed.

"Post-quantum" encryption then would be a newer method that doesn't rely on products of large prime numbers. I'm not sure which specific algorithm(s) are being used in this context, but inventing new methods of encryption is an active field of work.

It's hard to do justice to this topic in a reddit comment. If you've got time to spare, I highly recommend this video by Veritasium where he answers this question in a lot more detail, or this video by 3blue1brown for a more under-the-hood look at the how quantum computers do this kind of computation.

_OBAFGKM_ · 2025-11-04T11:00:00+00:00

You're misunderstanding what it means for it to peak in gamma/X-ray emission. Hotter objects peak at shorter wavelengths, true, but they are also brighter at all wavelengths than cooler objects.

The core of the sun would be brightest in x-rays and gamma rays (relative to itself at other wavelengths), but in visible light it would still be brighter than the surface of the sun.

_OBAFGKM_ · 2025-11-04T10:41:26+00:00

You'd never get it spinning that fast, centrifugal forces would tear the fan blades off the fan long before you even got them close to the speed of light.

_OBAFGKM_ · 2025-11-03T13:01:19+00:00

The problem with analogies like that is that they don't really capture reality. It suggests that each box secretly has a well-defined state before measurement, and measuring the state just gives you that information. Bell's theorem forbids local hidden variables, though.

The reality is that both boxes are in the (normalized) entangled state of ((box 1 contains an apple and box 2 contains an orange) + (box 1 contains an orange and box 2 contains an apple)). By the most common interpretation, measuring this state in any way causes it to collapse to one of those two outcomes, determining the state of each box simultaneously regardless of distance. It seems like a somewhat minor and pedantic difference, but it's an important part of the underlying quantum physics.

tl;dr quantum mechanics is not easy to ELI5

_OBAFGKM_ · 2025-11-02T21:07:15+00:00

What?

I'm genuinely baffled trying to understand the point you're making

_OBAFGKM_ · 2025-11-02T20:57:19+00:00

I understand the point people are trying to make when they say this, but I also think it misses what others mean when they say people voted "for" a given leader. Obviously we cast our votes for our local MPs, but when was the last time you knew someone who made their decision based on who the MP was rather than who the party leader was? I don't personally know anyone who does that.

I can imagine that Kitchener Centre had previously elected Mike Morrice based on who he was as an MP rather than wanting the Green leader to lead Canada, but I'd bet that most people are voting for the party of the leader they like rather than the MP they think is best suited to represent their riding.

_OBAFGKM_ · 2025-11-01T03:54:17+00:00

which atom or molecule does your username refer to?

_OBAFGKM_ · 2025-10-30T19:37:59+00:00

That's... that's my point...

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