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Yeap, now I can see what's wrong. Thanks everyone!

The question is not directly related to the problem itself, but an odd discovery that when using left end of the stick as axis of rotation versus using the CM as axis of rotation, two different normal force expressions are found. Note that the solutions are for the initial moment when the stick...

it's a tool using overleaf api, not public. Not working on overleaf is obviously the problem.

hmm, just tried what you wrote in the tool, unfortunately it's not compiled correctly. It also didn't work on overleaf. $$ v=v_0+at\tag*{[1]} $$ I am guessing this is a pretty special thing about physics forums that coded around the \$\$ math mode. Thanks for your help though.

Unfortunately I can't see your edit through reply, I am not sure if you are still using align there.

For reasons I stated in post #3: Thanks, I know the proper math mode works. There is a tool I am using that only supports the $ and $$ environment to render equations. That's where I need the right-aligned equation tag.

Sorry guys, again the problem is not with proper math environment like align etc. The problem is with \$ or \$$ math mode.

Thanks, I know the proper math mode works. There is a tool I am using that only supports the \$ and \$$ environment to render equations. That's where I need the right-aligned equation tag.

This has bothered me for a while and I have not found a good solution, it's convenient to write math equations using the \$\$ on the fly and occasionally we want to put a tag for an equation. tag works for the equation environment but not for \$\$. For example I would like to have [1] displayed...

Interesting... My first thought was using matlab ode solver but I can see where your approach is coming from. Thanks!

Thanks for the clarification. What would you suggest how to determine ##h## from here? It looks like we need boundary condition for ##\phi(z=0)## which needs to be determined from Young's equation at the air-glass-water contact point? The other boundary condition ##r(z=0) = r_0## is given. i.e...

I agree with the final equation being the Young's equation in this setup, can you elaborate on how this part ##\sigma\frac{\partial (r\hat{i}_s)}{\partial s}ds d\theta+\sigma\frac{\partial \hat{i}_\theta}{\partial \theta}ds d\theta## is obtained. They are correct, I am wondering if these come...

Definitely would be interested to see how you might approach this with more details. Thanks!

Thanks for the ideas, I wasn't sure if I needed to consider Young's equation and the multi-phase interactions to come up with an reasonable estimation. It looks like people are still (in 2022) trying to work out the meniscus for the water surface inside a cylindrical tube. So this is not a...

There are two difficulties, first ##r(y)## is not known, the surface tension force ##F_{surface \; tension}## is not known either. We can write net surface tension force as ##F_{surface \; tension} = \int_0^H 2 \pi r (\sin \arctan \frac{dy}{dr(y)}) dy ## Is there something else we could use to...

kuruman, this is a more reasonable answer without doing the detailed calculation.

A snow cylinder with an initial radius ##R## rolls without slipping down a tall hill with constant slope. As it rolls, snow sticks onto the cylinder which makes its radius slowly increase. The amount of gathered snow is proportional to the distance the snow cylinder has traveled. Consider the...

That's the same thing I derived earlier, ##- \frac{f_e (f_e - \delta )}{\delta }## How do you determine ##\delta## or ##\epsilon## from here?

That's exactly correct, there will be an adjustable delta between the two foci of the objective and eye piece ##\delta##. Therefore exact calculation will yield a final image distance from eye piece ##- \frac{f_e (f_e - \delta )}{\delta }## Note that this has nothing to do with objective lens...

I found a problem where the question is asking for image distance. The focal lengths of the objective and eyepiece convex lenses are in the ratio 8:1 for aparticular telescope. The telescope is pointed at a building that is 10 km from thetelescope and is 100 m tall. What is the distance to the...

For the 2nd question, looking at their solution, I wonder how they draw the conclusion ##\angle K' Sun = 90## and ##\angle K' S Sun = 180##. I came up with this picture. My K is the same as K' (ecliptic pole).The intersections between the Sun's path (slanted dotted path) and the horizon (H) are...

I looked at this in Stellarium before posting, unfortunately I was mislead by the pictures without extending ecliptic into an line. Now it's obvious.

TL;DR Summary: Astro Olympiad Problem determining the latitude of an observer from a picture taken. Well this question and answer are really confusing. There are no cardinal directions labelled on the picture. However because the Sun and the Moon should move on a circular path, the left side...

I get it now. Thanks!

I am re-reading this post and I am confused what's 'today' versus 'at reception'. Is it proper distance at emission?

Okay in reviewing the textbooks and hogg's paper which is great, I have arrived at a conclusion, that comoving distance between two objects is greater than their proper distance. ##d_C = (1 + z) d_P## I found my conclusion contradicting to my understanding between the two distances in the past...

oops, it looks like I can't edit the posts any more. Thanks for the reminder though.

I found a good stackexchange thread on this https://physics.stackexchange.com/questions/400358/the-difference-between-comoving-and-proper-distances-in-defining-the-observable?newreg=ee254fcb84bb473e9e577b216d1bded0 Is there a formal textbook that I can read to look up this information further...

Interesting read of page 17 https://www.astronomy.ohio-state.edu/weinberg.21/A873/notes3.pdf The formula after simplification seems to agree with the given solution. While the formula still seem counter-intuitive which I am still struggling to coneptualize, the key relationship is the luminous...

So you are saying the two given solutions do not contradict each other where in the first part, you divide ##d_c## by ##1 + z##; while in the second part, you multiply ##d_c## by ##1+z##. I will need to read your posts more carefully. I am relatively familiar with these concepts but the...

Thanks, how to reconcile with your explanation with the solution of another question for this problem? What is the bolometric luminosity of the quasar? The answer given was ##L = I 4 \pi (d_c ( 1 + z))^2 = I 4 \pi ( 4.4 Gpc * ( 1 + 1.5))^2 ## I had thought this made sense but now I am more...

A quasar with a bolometric flux of approximately 10−12 erg s−1 cm−2 is observed at a redshift of 1.5, i.e. its comoving radial distance is about 4.4 Gpc. Assume that the quasar in the previous question is observed to have a companion galaxy which is 5 arcseconds apart. What is the projected...

On the night of December 23rd 24th 2015, an occultation of a bright star by the moon will be visible from Britain to Japan. Given that the moon is in full phase on December 25th, which star does the moon occult? a. Aldebaran (RA 4h 37m, Dec 16o 31’) b. Pollux (RA 7h 45m, Dec 28o 2’) c. Regulus...

Entropy change is correct. For probability, you need to go back and look up your lecture note/book on energy partition function and Boltzmann factor.

NVM, these are all given: Find the altitude and azimuth of the Moon in Helsinki at midnight at the beginning of 1996. The right ascension is α = 2 h 55 min 7 s = 2.9186 h and declination δ = 14◦ 42 = 14.70◦, the sidereal time is Θ = 6 h 19 min 26 s = 6.3239 h and latitude φ = 60.16◦. It's...

The sign convention you will need for this is the object distance. When light comes from the same side as the object the object distance is positive; negative other wise. Images can be treated as objects for secondary lens. Use the above convention to figure out the sign then proceed to use...

The answer key wherever it's coming from doesn't make sense. Remember if the cold reservoir has 0 K, you are supposed to get 100% efficiency which is not happening with the answer key you cited.

Everything looks good to me so far.

I think the velocity component equation is a statement of impulse relationship as well.

Let D be the opposite corner. In the CM frame, A moves towards CM, D moves towards CM as well. The other two corners (let them be B and C) moves away from the corner at ##v##. Then ##v_A \cos \theta/2 = v \sin \theta/2## ##v_{CM} = \frac{j}{4m}## This is where it seems like the problem is...

Did the original problem say anything about the fluid being incompressible?

Check very carefully what you are given at final state.

The use of spherical lens equation should be correct where you let R approach infinity. To me this problem is worded poorly. Based on you said, it seems to be asking for the position of the secondary image (from a first virtual image under water) formed in the mirror above water. You probably...

When I read the question, it seems to suggest the fish is between the air-water interface and a spherical metallic mirror below. What you wrote suggests the mirror is above the air-water interface which itself is above the fish.

It looks like the number 100 is a given standard error(uncertainty) for a single measurement. Perhaps you could use that to find the uncertainty for the difference.

Make sure you are taking care of the x direction relative velocity first; then you can work on the y direction relative velocity. You will need to perform relative velocity calculation twice. Keep in mind, relative motion has an effect on both time (indenpdent of direction) and space (dependent...

This is all very good, one thing I would like to add is to numerically check the relevant quantities after one obtains the symbolic solution. If the period ##t_{osc}## is very small in the sense that heat conducted ##Q = k A \frac{\Delta T}{thickness} t_{osc}## through the soap surface can be...

Very good, that was the detail I overlooked. Thanks.

I do not agree, this is bullocks. We can simply set up position vector of ##\vec A(t)## and ##\vec B(t)## with respect to the fixed center of the carousel, their relative velocity is simply ##\frac{d (A-B)}{dt}## or ##\frac{d (B-A)}{dt}## Since this is a pretty popular book, I am wondering if I...

Case 2 looks good, edit: I think I can see now the high potential below ##L## means it's possible for a closed loop: L->T2->D2->L.