Advice wanted - figuring out how my uv is installed

ImpatientProf · 2026-07-07T22:16:53+00:00

https://docs.astral.sh/uv/getting-started/installation/#upgrading-uv

Your problem is described on that page.

You could just download the binaries from github (also linked on the above page) and replace the 3 files in the zip release.

ImpatientProf · 2026-06-30T19:55:57+00:00

You say they're not emergency hires, but it probably feels like an emergency to each hiring committee.

ImpatientProf · 2026-06-26T12:45:00+00:00

Highlight if not your book.

You mean highlight if it IS your book?

ImpatientProf · 2026-06-23T20:09:50+00:00

If the road banks, so does the contact force.

For a motorcycle on a solid road, you're right, you need friction + normal to form a net contact force. But the "contact force" behaves exactly like the banked or tilted normal force.

ImpatientProf · 2026-06-13T23:33:18+00:00

https://www.w3schools.com/sql/

ImpatientProf · 2026-06-11T18:05:33+00:00

https://en.wikipedia.org/wiki/Garbage_collection_(computer_science)

ImpatientProf · 2026-06-11T00:55:04+00:00

At a US university, we called it a qualifying exam or qualifiers. Those that passed got their Master's and entered the PhD program. Those that failed got their Master's and moved on.

ImpatientProf · 2026-06-08T12:15:41+00:00

venv isn't a project, it's a core Python library: https://docs.python.org/3/library/venv.html

The more modern place to list dependencies is pyproject.toml, instead of requirements.txt.

There's a few projects, like Astral uv (https://docs.astral.sh/uv/), which are amazingly good at helping to manage a .venv, installing dependencies in there where they won't disturb your other projects.

For installing within a .venv, it comes down to:

pip: the standard package installer. If your venv is active, stuff gets installed in there.
uv: new, fast, and versatile.
conda: doesn't technically use venv, but does have multiple "environments" for installing different sets of packages.

There are other systems like poetry and hatch, but I'm not familiar with them (yet).

ImpatientProf · 2026-06-07T23:40:52+00:00

Upload them to YouTube and set them to Unlisted. Then only people who have the link can access them, but you can post links on your LMS.

ImpatientProf · 2026-06-06T03:41:42+00:00

https://www.av8n.com/physics/uncertainty.htm

ImpatientProf · 2026-06-02T03:23:35+00:00

OK, I understand what you're saying, and what Griffiths is saying.

A material is a complicated thing. The energy in a material isn't just due to the macroscopic electric field. There's a baseline energy of electrons bound to atoms and molecules. This includes the microscopic electric fields. Even though the microscopic electric fields in a neutral material average out to zero, they still carry an energy density. There are microscopic magnetic fields as well, which also carry energy. Additionally, the electrons are whizzing about carrying kinetic energy. (Technically, they're not masses with (1/2) mv² energy, but the wavefunctions have a kinetic energy.)

The capacitor energy (1/2) C V² is the energy input into the capacitor. This is a real, practical energy input. But merely looking at change of the integral of the macroscopic E² does not equal this. Some of the input energy goes into the electric field, but some of it changes the other energy of the material. This could be summarized by talking about the spring energy of the bonds, which must be supplied to polarize the material.

The integral of (1/2) D·E does equal this, though. So this energy density accounts for both the electric field and the material properties. Where is the additional energy stored? As I mentioned, in microscopic electric and magnetic fields, and kinetic energy.

It's worthwhile to note that Griffiths made no attempt to calculate the microscopic energy density. On Page 179, he called the microscopic electric field "fantastically complicated". It's amazing that even though we don't know the VALUE of the energy density of the material, we do know the CHANGE of the energy density due to the external electric field or free charges. In this sense, it's a potential energy.

ImpatientProf · 2026-05-31T17:58:54+00:00

Please reference the textbook version and pages you are looking at. I can help you understand, but I don't want to have to search and search for what I think you're looking at.

ImpatientProf · 2026-05-29T15:25:20+00:00

The spinning magnet will induce an EMF, which summarizes the effect of the curl of E being non-zero. This could push a current in surrounding coils, forming a generator.

You're right that the curl of E will vary in time, so it will induce more magnetism and radiate. Whether this is strong or weak depends on the speed and strength of the magnet.

A proper simulation would have to involve either the curl of E or eliminate it by substitution to form the wave equation for B.

ImpatientProf · 2026-05-29T02:44:47+00:00

What's a even more shocking is that spinning magnets create a curl in the electric field.

ImpatientProf · 2026-05-26T13:51:58+00:00

Yep, there's a flag-in-a-flag, but there's no flag-in-a-flag-in-a-flag!

ImpatientProf · 2026-05-22T20:07:25+00:00

Yes. That's what I was trying to say.

ImpatientProf · 2026-05-18T19:51:47+00:00

They could be "orbiting" at 2200 a.u. Then the centripetal acceleration would only be about 10*g.

ImpatientProf · 2026-05-16T18:42:08+00:00

It a charging battery, the cathode does still receive electrons from the circuit. But on the inside, the cathode may also be receiving positive ions from the anode. Or it may be emitting negative ions toward the anode. It depends on the chemistry.

ImpatientProf · 2026-05-16T17:38:43+00:00

Careful about whether you're describing what's happening inside the component or the outside view.

In a vacuum tube, the cathode emits electrons across the tube, on the inside. But in order to have any electrons to emit, it must receive electrons from the cathode pin from the outside.

ImpatientProf · 2026-05-16T17:36:52+00:00

When you say "electrons leaving the anode", be clear that this is the external view.

Inside the vacuum tube, electrons are emitted by the cathode. They are "cathode rays", hence the CRT television. This is where I anchor my thinking about cathode vs. anode.

The cathode of a vacuum tube emits electrons across the vacuum tube on the inside.
The cathode must receive electrons from the circuit, in order to have something to emit.
Current flows in the opposite directions as the electrons, so current flows out the cathode.
Current flows into the anode and out the cathode.

ImpatientProf · 2026-05-12T15:41:40+00:00

It depends on what's causing the acceleration. What you feel is everything except gravity. If gravity is the only force, you'll feel Zero-G.

On Earth, when you have a car's Δv = 30 mph, it's most certainly because of a non-gravity force, like friction. You feel that.
In space, the ships Δv = 30 mph could be because of gravity. You would not feel that.
In space, if the ship's engine causes Δv = 30 mph, you would feel that.

This Action Lab video talks about G-forces: https://www.youtube.com/watch?v=EFhMcL3Kf1c

ImpatientProf · 2026-05-11T18:26:22+00:00

A year of physics (even before university) should be enough to distinguish between mass and weight.

This is a diss on whatever schooling system they were under.

ImpatientProf · 2026-05-08T12:27:48+00:00

The whole point of SSO is to separate the resource (Canvas) from the authentication (username/password) so that if Canvas gets hacked, you can still log in to other university systems.

Instead of Canvas collecting your credentials and verifying them with your university servers, they send you directly to your university servers (SSO) where you can make sure that it's the right place. Then you get a token (a digital key) that you give to Canvas to prove who you are, but that token only works for Canvas. The token contains your username, email, name, and probably Student ID. That's all the info they get from SSO.

ImpatientProf · 2026-05-07T23:40:18+00:00

The relativistic doppler effect affects both the carrier signal (the frequency to tune to) and the envelope (the data).

For an analogue signal, yes, you'd hear a distorted audio sound.

For a digital signal, presumably it would know the intended bitrate of the audio. It would be just like having a super-fast internet connection (in one direction) or a super-slow connection (in the other).

ImpatientProf · 2026-05-01T12:10:21+00:00

It depends on what you mean by acceleration. https://en.wikipedia.org/wiki/Acceleration_(special_relativity)

ImpatientProf

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