BioPhysicist

@Dale Post 1) "My teacher said centrifugal force is fake and should never be used. But I am seeing other places where it is used. When can we talk about and use the centrifugal force?".... Post 2) "Chat GPT said centrifugal force is fake and should never be used. But I am seeing other places where it is used. When can we talk about and use the centrifugal force?" .... Post 2 should be closed but post 1 should not?

@Dale AI will have the tendency to be similar to "crackpot sources". However, interactions with AI could also reveal misunderstandings of the person as they try to determine the validity of what the AI is saying, similar to any other source. We should always evaluate posts based on their content, not on the source of the misunderstanding. As you can see, posts can be edited to be removed from the explicit source of the confusion. This is the same attitude as "oh, your question involves a homework question, it must be closed"

@Dale There are more problems with this post, but I edited out the Chat GPT part so the actual issues can be discussed on, like clarity or lack of focus.

@Dale The policy is against directly using AI generated text directly in answers. There isn't anything wrong with a confusion being cited from coming from AI, just as if someone was confused by an incorrect reference, professor, etc. Where questions come from have never been a determining factor of closure

@KDP no one is saying you get free energy. It does take energy to force the water out. It's just that is all it is: energy to force the water out. The work from the pump is done on the water; that energy doesn't some how transfer back to the submarine

I see your error in considering energy. By putting $\Delta U$ on the left as well as $-F_g\Delta h$ you have "double counted" the work done by gravity. $\Delta U=-F_g\Delta h$ by the definition of potential energy. If you are accounting for a change in potential energy, then you do not count the work done by gravity as work done by an external force. If you want to explicitly list the work done by gravity, then you don't put in the potential energy. By the work-energy theorem, we do have $\Delta K=W_\text{net}=W_B+W_g$, or we can sub $\Delta U=-W_g$ to get $\Delta K+\Delta U=W_B$

@KDP PSE doesn't have a way to follow users. In any case, no, haven't said energy isn't conserved, nor that there is any free energy here. At this point it's best to leave the discussion here. I have not resulted to personally attacking you, and my answer isn't crafted in opposition to you. If you have further issues you can start a post on meta about what to do about accepted answers you think are incorrect.

@KDP To make a different limiting case.... imagine a neutrally buoyant "submarine" that only pushes out a droplet of water in order to float upwards. The work to move this droplet of water out is very small. The "submarine" will slowly rise upward, increasing its potential energy over time as the buoyant force acts on it. Once the "submarine" eventually gets close to the surface, is the change in its mechanical energy equal to the small amount of work to pump out the droplet of water?

@fishinear Thanks for helping me refine the answer and make it clearer!

@fishinear You are right, pushing the water out does ultimately increase the potential energy of the surrounding fluid, and the boat rising does decrease the potential energy of the surrounding fluid. But the only case those are equal is if the "submarine" is initially completely filled with water and then expels all of that water to become an empty, thin shell.

@fishinear I added to my answer. Hopefully it clears my position up.

@fishinear I am not conflating anything. The work done by gravity is accounted for in the potential energy. If you want to talk about the work done by the net force, then that accounts just for a change in kinetic energy. That is fine; we can either explicitly account for the work done by gravity, or we can instead call it potential energy. In either case though, the buoyant force does work on the submarine; it points upward and the submarine moves upward. I feel like you are trying to refute a claim I never made

@fishinear I already stated my agreement that the net upward force acting on the submarine and what is inside of it does depend on the water pumped out, and that it does require energy to pump water out. As the submarine rises, the buoyant force does work on it, and that work increase the kinetic and potential energies of the submarine and what is inside of it; I am not sure what you mean by it not being relevant. Also, I am not trying to "feel" anything here.

@KDP An object with holes does not have the same surface on which external hydrostatic pressure acts on. You have now changed the boundary. A boat with a hole sinks because it's top surface now it's being pushed on by water, and the boundary in question surrounds the entire boat material rather than just the bottom of the boat.

@KDP "Consider two identical submarines with equal shell volumes." Equal shell volumes mean equal buoyant force. Full stop. You are conflating buoyant force with net force. This is an extremely common introductory physics misconception where students think that submerged objects of the same volume but different weight experience different buoyant forces

@KDP I'm not sure what you're countering. I have no issues with your example.

@fishinear Yes, if you pump more water out while keeping the enclosed volume the same, the net upward force will be larger since there is a smaller weight force. That doesn't change how much work the buoyant force is doing though. It's not about how I want to "name the forces"; buoyant force doesn't have multiple or subjective meanings.

@mr3 Your "derivation" is slightly flawed, but the buoyant force being equal to displaced fluid is a consequence of the underlying principle of the force just arising from hydrostatic pressure acting on the surfaces of the object. This force happens to be equal to the weight of the fluid displaced by the boundary of the object. Taking in fluid, like by a hole, changes the boundary of the object being considered. It sounds like we are using "displaced fluid differently" but agree on the underlying ideas here.

@DewiMorgan No one ever said that. If you don't want to seriously engage with me, that's fine. The Titanic sinking is not what I'm talking about. If I'm missing something about how submarines work, please let me know. In any case, the main point of my answer is that the buoyant force does work on the submarine after it expels water, and this work is what causes the energy (kinetic + potential) of the submarine to increase as it rises.

@DewiMorgan You need to look up why Archimedes' principle is true, then you will understand. In your boat-hole example the object is not fully submerged, so there are confounding factors at play. If you have a fully submerged object, it being hollow or filled with water will not impact the buoyant force (no holes in the object)

@DewiMorgan You're getting mixed up. The displaced water just depends on the enclosing volume. The reason it would sink from taking in water is because its weight increases, not because the buoyant force changes. If it helps, look up a derivation of Archimedes' principle, that should help you see that all that matters for the buoyant force is the volume of the object, not what is inside of it.

@fishinear That's not correct. That's the net force acting on the submarine, sure, because the buoyant force happens to be equal to the weight of the displaced fluid. But the buoyant force itself is just the force that the fluid exerts on the object, and that does not depend on the weight of the object. This is a typical introductory physics mistake: two objects of the same volume but different weight will have the same buoyant force acting on them if fully submerged.

@fishinear The buoyant force just depends on the volume of the submarine. Unless I am missing how submarines work and they do change their volume in this scenario; I don't know much about submarines so maybe I'm missing that part. Does expelling the water also increase the volume of the submarine?

@StevanV.Saban They are not equal. The work done to move water of mass $m$ at a depth $h$ would be $mgh$. The work done by the buoyant force to lift the submarine a distance $h$ is $\rho Vgh=m_dgh$, where $m_d$ is the mass of water displaced by the submarine.

It's a privilege to get the right answer without having to worry if you did it correctly

"It looks similar, just do the integrals"

@ACuriousMind Oh yeah, those are less complicated, haha. Not a lot of meat to those, although fun in the right context

Maybe because I am trained in physics haha

Hm, I enjoyed Shanker's quantum book

Ooof! Yeah, that will do that haha. I primarily play boardgames solo at this point. Still pretty fun. A lot of good solo-able games out there

Been a while since I chatted here :P CuriousMind, you still into board games?

To be fair the teacher did have some other issues with mathematical rigor that even I had issues with haha. But yeah, delta functions are more acceptable forms of "hush hush"

I had a classmate in graduate school who was very troubled with how physicists handle delta functions.

@VirenderBhardwaj I am sorry, I think there is a misunderstanding here. By "related", I just mean that the post is about a similar topic, a similar system, etc. It is a link for someone who finds this post and wants to see other posts related to it. It even adds the post to the "linked questions" on the right side of the page. "Related" does not mean "this post says the same thing. Why did you post this? It already answers your question." That is called a duplicate, and I would have voted to close this post as a duplicate of the linked post if I did think it answers your question.

@BobD I didn't think you were being critical :) I was agreeing with you, the link does not answer the question. Hopefully the OP can see that I do not think it answers this post, otherwise I would have instead voted to close as a duplicate

@BobD Great, because I never made that claim either. It's very common on PSE to list related questions in the comments

I said related, not duplicate. It's about the same video

Related post: physics.stackexchange.com/q/74707/179151

"your last response is EXACTLY why my question should be reopened." You haven't even asked a question here though

@PatDolan I am not a moderator

If you have issues with how questions are handled and how the site works, check out the meta site

I edited your post to remove unnecessary information. This isn't a conversation/discourse site like Reddit. It's a question and answer site. I kept the heart of your post without the subjective and incorrect information obscuring your main point.

This is a Q&A site. The history of other posts is not relevant here. If you have a question, ask it. If there is a question that has been closed that you think deserves an answer, make a post that is better and follows site guidelines. There is no controversy here

Re v1: no moderators closed the question you linked to

I’m voting to close this question because this isn't a question

@FlatterMann I think the OP is talking about the electrons before they hit the detector, not at the detector

Since posting, have you seen the Veritasium videos? first video, and then second video for the doubters

@Chemistry If energy is conserved then yes... We would always find the same energy across any sphere

The energy of a particle at a given point is changing over time, yes. But on average there is an energy flow outwards, and at any distance you can take that energy flow and divide it by the area to report the intensity

@Chemistry the sound source is emitting energy at a constant power; this isn't a single pressure wave pulse being considered. Regardless, you're trying to do more with the definition than what is happening. If you have energy flow (per time time) through some area, we can look at the area density of that energy. That is intensity. It's just a definition. It's not a statement on the physical mechanics at play; you can use it for many systems. Just like how density $\rho=m/V$ isn't saying anything about the molecules in the matter being looked at.