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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.

Perhaps when the transistors are off, the inductor has a emf big enough to create a closed loop the other way, upwards. This can happen if ##emf > Vcc##

I stumbled upon this problem and is fascinated by the result. The lack of spin for the tube is counter intuitive. I like the symmetry argument. It still feels odd that the amount of angular impulse applied for each half the motion with respect to the midpoint of the ring could be exactly equal.

isn't ##\tau_{p1}/\tau_{p1} = 1##? are we confusing something here?

Do you still need help? This seems like a trivial case of length contraction with ##c## replaced by 1. The formula is saying the proper length ##l## is contracted down to ##\gamma l##.

there are 41 macro states total. A macro state is different from each other by its macroscopic properties in this case it would be the total magnetic dipole moment.

I don’t recall doing this in the past, my first reaction reading the title is isn’t this what must follow from conservation of energy.

In addition to the formal Lorentz transformation approach, you could also use length and velocity observed in the ground frame and follow kinematics . Length can be found through length contraction and velocity can be found from relative velocity.

@wrobel I missed the part ##\omega## is parallel to the plane.

@wrobel I am not sure what you mean by the curve traced by contact point. Are you referring to 'a' contact point labelled on the ball. And you are looking at the curve traced out by 'that' contact point. Otherwise isn't the curve traced out by all the contact points at different instants in time...

##R(d-R)## is a parabola opening downwards, I do not see how you would be able to maximize the fraction.

We know that the net force on the table must be zero $$\sum F_i = F$$ We know that the components of the torque with respect to the origin is also 0. $$\sum \tau_x = 0$$ $$\sum \tau_y = 0$$ $$\sum \tau_z = 0$$ But the problem becomes insufficiently constrained that there are only 4 equations...

Solution as stated in the problem description.

I am trying to model stress in simple objects such as a disk, a cube, a rectangular prism or a solid ball under their own weight or additional point mass weight on top to demonstrate such material response to students. I understand the usual process is to set up Mesh, boundary condition, and...

Excellent!

To be clear I deduced ##\omega## at which ##D## spins around ##P## is ##\frac{2 \pi s}{Tr}##.

@etotheipi that's all very good, I have no dispute against what you wrote. @hutchphd no dispute with what you wrote either. I have in fact worked out the problems in many different ways. I guess what bothers me is that I have not seen a direct correction and explanation which step was done...

##D## is on the cone. No doubt I can obtain the correct answer through simple distance and time. What bothers me is where the fallacy in this argument presented is in the initial post?

Thanks, It’s understood angular velocity for the cone is more complicated but can we not use Spin omega for ##v_{P/D}##?

Let the vertex of the cone be ##O##, the contact point on the cone all the way to the right be ##D## touching ground. Then ##v_{\text{D relative to the table}} = v_{D/table} =0## since it rolls without slipping. Due to relative motion $$\vec v_{P/table} = \vec v_{P/D} + \vec v_{D/table} = \vec...

Hey fellow physics enthusiasts, how might the volume of a balloon change as you bring it down deep into the ocean (consider both adiabatic (quick) and equilibrium (slow) descend). Looking for insights what most likely will happen, for simplicity we can start with a thin (##t << R##) elastic...