I looked at this problem again the other day and one thing that might be worth mentioning is that it can be solved by using the law of sines instead of the law of cosines. For the law of sines, you can do it in two different ways: you can use the angles of 90 degrees and 22.5 degrees, with...
Very interesting @collinsmark :-) I've had my telescope for 30+ years now, and I've seen the rings of Saturn with it, along with 4 of the moons of Jupiter. In December 2020, I saw both in the same field of view. I've also seen a star pass behind the rings of Saturn, (about 25 years ago)...
The above are two photos of today's eclipse in Chicago. I projected the image from my medium power telescope of the eclipsed sun onto a screen. In the first picture about one third of the sun is blocked by the moon, and about two-thirds in the second picture. We had clear skies in Chicago...
The reason the small aperture is placed in the box with the cavity is that then we have very nearly unity emissivity by Kirchhoff's law, basically independent of the emissivity of the walls of the cavity. The derivation with the cavity thereby makes perfect sense. See also...
See https://www.physicsforums.com/threads/a-simple-trigonometry-problem-put-eight-coins-around-a-central-coin.971465/#post-6176083 for another fairly simple problem that I try to motivate students with. It's a good application of the law of cosines. See also post 4 where the 8 pennies are...
The above is a photo of the 2'7" snowman I made Friday, 3-22-24. @DennisN This photo was taken just after I built the snowman. You can get a good idea of how much snow there was=and see where I rolled the balls to make the grass bare. By the next day, all the snow was gone (in the other...
The above is a photo of a 2'7" snowman that I made Friday 3-22-24. Packing conditions were good, but I had very little snow to work with. The photo was taken today, Saturday 3-23-24. We had almost no snow this March in Chicago. This is the first snowman I was able to make since about a...
The change in angular momentum I think is very small, even if you get every electron to go to the other spin state. Let's try to quantify that: ## mvr=\hbar \approx 1.0 E-34 ## joule-sec. Let's work with 22 lbs. of iron=10 kg=10^4 grams. atom weight=56, so we have about 200 moles ## \approx...
It does appear this thing apparently is not completely simple. My experience with the de Haas- Einstein experiment comes from a textbook Modern Physics by Hugh D. Young. He makes it sound like spin one half electrons in the magnetic field that perhaps make a permanent magnet have their spins...
For "phonons", I also wondered the same thing. I think that could be the case especially for nuclear spins, but nothing came up in a google. The reason may be that for nuclear spins, the substance would need to be cryogenically cooled for it to work.
Even with iron and electron spins, I...