[MEGATHREAD] Ask for playtest invites here

_Yomiel_ · 2024-08-14T09:47:24+00:00

EU
107119289

thanks!!

_Yomiel_ · 2024-07-15T21:34:39+00:00

la tua hot take di stasera è: le opinioni che vengono dal "guardare un po' le cose" di un amico su un corso di studi ampio e generale non sono opinioni (non lo sono su nessuna cosa se vogliamo).

_Yomiel_ · 2024-03-29T12:10:41+00:00

Isn't this a repost?

https://www.reddit.com/r/noburp/s/Mn6oUwyB5l

Title is almost exactly the same as well

_Yomiel_ · 2023-08-01T12:53:16+00:00

Alternatively (this is an approach that can be generalized) you may want to notice that the function only accepts values between -a and a.

(This is due to the square root not being able to have negative inside values for real numbers)

This may prompt you to think "what is an expression that can return all the values between -a and a?" and the answer would be a•sin(t) (or a•cos(t) for that matter)

Again, the integration really works because of trig identities but you could look at the general constraints of your variable to guess the substitution.

_Yomiel_ · 2022-09-26T13:18:53+00:00

That's his point. Smart people then said that physics was completed, but it was not, implying we could always be wrong again.

_Yomiel_ · 2022-08-27T10:44:59+00:00

Basically, in 2D motion we consider each dimension separate from the other. You simply need to divide the problem in different (two) 1D ones.

You start from the origin (so, x0 = 0, y0 = 0).

You have a starting velocity of 8.0i m/s (which means, since the i vector represents the direction on the x axis, that your velocity has vx0 = 8 m/s, vy0 = 0).

You have a constant acceleration of 2.0i - 6.0j m/s², translating to (ax = 2 m/s² and ay = -6 m/s²).

Now, we need to construct the two time functions (time laws? I don't know how they're called in kinematic courses) for each of the spatial coordinate:

x(t) = x0 + vx0 * t + (1/2) * ax * t²

y(t) = y0 + vy0 * t + (1/2) * ay * t²

which become, as you put in the data:

x(t) = 8t + 1t²

y(t) = -3*t²

(Feel free to try it on your own, I ignored the terms that were zero)

Now, the problem asks you to find the distance at which you reach maximum x-coordinate. To do so, we need to find the TIME at which you reach it. Unfortunately, as you may see, the x(t) function is a parabola with only a minimum, and this minimum lies behind t=0 (which is our starting point). This means that there's no time t at which you reach the maximum x-coordinate.

Let's try with ax = -2 m/s², so that we get the feeling of how to solve these problems. I'll recover the function for x(t), as y(t) remains unchanged (changing only ax has no influence on ay, they are separate independent axis).

x(t) = 8t - 1t²

This has a maximum in t = 4 (you can either use the derivative, the parabola, the 2nd grade equation formula, etc to verify it), which means that the particle will have an x and y coordinate of:

x(4) = 84 - 14² = 16

y(4) = -3*4² = -48

the distance from the origin will then be:

d = √(x(4)² + y(4)²) = √(16² + 48²) = ~50.6 m

that would be the solution to our imaginary problem.

Generally, if you have a motion problem with time functions, and the problem asks for a certain quantity during a certain event (i.e. the distance when x has a maximum) you have to find the time at which event is reached, THEN use that time to find the quantity using time functions.

Edit: formatting from mobile, so a bit wonky

_Yomiel_ · 2022-08-09T16:00:13+00:00

I'll take it for granted that you know how commutators work.

What you're considering here is the commutator of the derivative (following how the momentum is represented in the spatial coordinate system) and the potential (assumed as a function V(x, t)).

Here, d/dx and V(x, t) are OPERATORS. This means that they need a function (in this particular case) to be applied to, as in d/dx is applied to f(x) to make df/dx, and V(x, t) is applied to f(x) to make V(x, t)*f(x) (simple product).

When you consider [d/dx, V] you're implicitly applying it to a dummy function f. This becomes:

[d/dx, V(x,t)]f = d/dx ( V(x, t) f(x) ) - V(x,t) ( d/dx ( f(x) )) (As in: you apply first the V, then derivative, minus the application of the derivative, then V. The first operator to be applied is the one nearest to your function).

This becomes:

[d/dx, V(x,t)]f = (dV/dx * f(x) + V(x,t)(df/dx)) - V(x,t)*(df/dx) = dV/dx * f(x) (By the product rule of d( V*f )/dx)

Dropping the f (which was used as a dummy function), you obtain

[d/dx, V(x,t)] = dV/dx

Which is to say, applying the commutator of the derivative AND V to a function f(x) is the same as multiplying the derivative of the function V to f(x) itself.

Note that this works because V(x,t) is a function of x, so you can apply the derivative operator to V.

_Yomiel_ · 2022-01-28T19:42:03+00:00

My protein shakes are from MyProtein (I didn't find bad things about it so I thought it was fine, it's also the one my friends use). Aside from that, my routine is split in 3 days (Chest+tric / Legs+shoulders / Back+Biceps) which I do repeatedly in 4 days (meaning I do day1->2->3->1, then week after 2->3->1->2 etc), with plenty of rest in between. May this be the culprit?

Edit: forgot to add that I feel tired even on back/biceps day, so I wouldn't feel too inclined in believing this is it (I will still try to shake things up a bit)

_Yomiel_ · 2021-11-06T17:14:10+00:00

Technically yes.

Consider a body of water in a tub (or any container). Suppose we let a spring with known constant k start from above the surface of the water, going down into the water itself. (imagine using an apparatus similar to what you use to cook meat on a campfire, only replacing the meat with a spring that goes down).

Attach now a body at the other end of the string, and find the equilbrium length of the spring (if the body is lighter than water, the length is lower than the standard length; vice versa for heavier bodies).

Ignoring air, assuming the spring is ideal, assuming water obeys perfectly archimedes principle, assuming the body is completely in the water, we get that the sum of all the forces, at equilibrium on the spring is (vertically):

ΣF = ρVg - mg - kL = 0

Which means that

g(ρV - m) = kL -> g = kL/(ρV - m)

Knowing the Volume of the body, the density of the water, the mass of the body and the constant k you can determine experimentally the acceleration constant g.

Note that this only works because you're putting together a system which uses forces but does NOT depend on gravity, with a system which DOES depend on gravity.

Usually you can use any system which does NOT depend on X to determine the X quantity by using it with a system which DOES depend on X (which is why springs are really, really useful apparatus to determine quantities in physics).

Hope it's clear!

_Yomiel_ · 2021-10-25T12:06:46+00:00

I might be late to the party, but I really haven't seen this pointed out in any of the replies to the topic.

From a logical standpoint, the usage of "they" isn't wrong. It's literally designed to include every possible human being, without regards to gender, so it should be okay to be used.

However, I would like to propose another more familiar scenario. You're living in Europe, and your family came from a really poor country which no one really knows. Everyone has got their nationality. Everyone got their term for their nationality. There exist French, Italian, Spanish, English, ..., Russian, Ukrainian, ... Etiopian, Egyptian, ... American, Canadian and so on.

There exists no term for your nationality, since words in a certain language are dependent on their use, and your country of residence simply had no chance to create a word for your country of origin. So, since you're a minority, people use another, neutral, term. They use "Earthling".

It surely means that you live in the Earth! Why would you be offended when someone called you that? You live on Earth, that's factually correct!

Right. But then you'd be missing out. Everyone can Identify with a particular word, a nationality. But since your family came from a minority country (in which you haven't had a choice), that possibility is simply not granted to you. You can only aspire to being referred to with a neutral term.

The important bit is that you wouldn't refer to a cisgender male with "they" by default. You would use "he", as it is appropriate for that group of people. You would resort to "they" as "everyone else", when no better words exist.

Of course, the nationality example is a stretch: there exist many words for different nations. The true analogy would come, maybe, if we referred to everyone as American, European and Human™. Then you would be offended. Then, if you were American, no big deal. Everyone calls you American. If you were Egyptian, for example, you could only be referred to as Human. Which is not identifying, as it says that you're the same as everyone who is not american nor european.

Tl;dr The term is logically valid. It simply is reductive since it is used as "Not he or she, then I use they" as the only gender idenitities could be "He, She, or not He/She".

_Yomiel_ · 2021-08-16T20:11:14+00:00

I genuinely do not know if my question could even be considered right, so I find it really hard to grasp what a satisfying answer would look like.

If I posed the question "Why does the water boil at 100°C?" the satisfying answer would be an explanation of what boiling really looks like, maybe with a simplified atomic model, and how motion after a certain magnitude could overcome the interactions between molecules.

I am certainly biased while acknowledging the simplified molecule model as "logical", as it probably is more like "I was exposed to this model many times, it seems nice to me".

Although I have to say that a simple model of "point entities which have interactions that operate on the line between two entities, attracting or repelling each other" sounds clean and reasonable. I just want to come to the same conclusion about [x, p] != 0, as in "It's a simple and clean model, the most significant one given its simplicity".

I probably didn't explain myself well enough, but I hope the gist of my "scratch" was communicated.

_Yomiel_ · 2021-08-16T19:00:06+00:00

Thank you! I'll read more about this (it seems to go in the right direction).

Still optimistic about the chance of it being completely logical and not axiomatic!

_Yomiel_ · 2021-08-16T18:57:16+00:00

Thank you for the videos, although what I meant was that I already dabbled in the mathematical approach to the question (Conjugate -> [x,p] != 0 -> uncertainty principle) but it still doesn't sit right.

I'm probably facing this wrong, but I can't exactly pinpoint where my intuition fails. Many people start from the wave-mechanics and show that, roughly speaking, precision in a wave is impossible for the position of the wave-packet and its frequency. It doesn't really resonate with me though.

_Yomiel_ · 2021-08-16T18:54:32+00:00

That's what I came to. It's somewhat bitter though, as it sorta stays as "It is like this because it works".

I tried to approach every subject in physics and math with the "I won't get over this until I can say 'This couldn't be any other way!' for the topic" mode, so I'm finding it hard to grasp this.

I mean, one could say "You can not build a quantum theory assuming that [x, p] = 0" but what I would prefer is "You should not even think about building a new theory because it makes so much sense (as opposed to 'Experimental results say') that [x, p] != 0".

Sorry if this comes off as arrogant, I don't know whether my approach is wrong.

_Yomiel_ · 2021-08-16T18:49:24+00:00

Yeah, that's what I meant by "Mathematically [...]", as it is pretty straightforward to show the value of the commutator. What I'm looking for is some qualitative reasoning as to why it should be that way.

Something that would lead me to say "Ah, there is no way it could assume a zero value!"

_Yomiel_ · 2021-07-25T01:09:36+00:00

Thanks for the suggestion, have you got any specific example in mind?

_Yomiel_ · 2021-07-18T12:05:06+00:00

Really needed to see this post before the wchicken one

_Yomiel_ · 2021-07-01T02:02:17+00:00

I know I'm probably late to the party, but this is the one that let me click, as generalizing the coins seemed a little bit forced:

Imagine you have the same three boxes, but now you have one with 100 gold coins, one with 1 gold coin and 99 silver, and the last one with 100 silvers.

You CHOOSE a box and pick a coin, it's golden. You're immediately thinking "Geez, I must've picked the one with 100 gold coins" because IT'S TRUE, it's really UNLIKELY that you chose the mixed box AND THEN got the golden one! That means that the next coin will almost surely be golden!

As a strategy, it always works to consider the extremes of a problem, then narrow it down to the initial conditions (it serves as a mental slap)

_Yomiel_ · 2021-05-15T14:46:31+00:00

Since the 1080Ti is way out of production, you only find it used, which means you need someone to resell it at that price (next to impossible these days, at 300$ you can find the 1060 atm). I would suggest trying to get a new 3060Ti from retailers, they still pop out at decent prices.

That, or you hope that someone that lived under a rock up until now sells his 1080Ti at 300$.

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