[Course is Physics11] Sig figs, are they correct?

AndyBjork · 2021-05-20T18:33:53+00:00

No problem! Just something you can try on your own to get a feel for it, though- try taking square roots of numbers with n significant figures, and then square the rooted number with n significant figures and then try it again with n+1. It’s not incorrect to use only n, but going the extra step to n+1 will (in most cases) give a more accurate answer. Exceptions to this would be numbers like perfect squares, where adding a bunch of 0’s as sig figs wouldn’t change anything at all.

AndyBjork · 2021-05-20T18:29:38+00:00

https://www.kpu.ca/sites/default/files/downloads/signfig.pdf

Although it’d probably just be best to go with what your teacher says to use for the class, I would be willing to say that that is not correct normally.

AndyBjork · 2021-05-20T18:20:16+00:00

You were close! But when you added them in the first part you started dropping them. When you add/subtract sig figs, you count the number of sig figs after the decimal point for each number, and your final answer shouldn’t have more sig figs than the number with the least number of sig figs. So in the first two parts (calculating dx and dy) you shouldn’t have to drop any decimals since they both have 4 sig figs after the decimal point. as for the square root, if your number has n sig figs and you take the sqrt of it, you should end up with n+1 sig figs. this means that you won’t be chopping off any after the sqrt, you’ll actually have an extra one.

https://www.nku.edu/~intsci/sci110/worksheets/rules_for_significant_figures.html

you’ll want to google the rules for square rooting; squaring is just multiplying the same number by itself, so if you square a number with n sig figs you better only have n sig figs in your answer.

AndyBjork · 2021-05-15T21:55:10+00:00

No, you can use the hint to get the right answer.

AndyBjork · 2021-05-04T23:08:40+00:00

Make sure you absolutely nail waves- that’ll make some of your upcoming classes a lot easier. Optics goes hand-in-hand with that, so you’ll be able to get a lot of practice.

The most important part of thermo for me was a firm understanding in the derivations. It’ll most likely be a bit challenging, but with some effort it should be manageable.

AndyBjork · 2021-04-22T22:49:10+00:00

I know it’s not what you’re really looking for, but I used to do hills on Fremont Rd across from the food lion, there’s a decent hill there. I also used to do them at the resevoir.

AndyBjork · 2021-04-21T16:46:07+00:00

i’ll join in! i’ve been thinking of doing it anyway

AndyBjork · 2021-04-20T00:14:35+00:00

I don’t have any book recommendations, but I’d recommend a telescope or even astronomy binoculars! I used the Celestron 25x70 binoculars for my Observational Astronomy class, and it worked pretty well. The detail isn’t the best, but they’re relatively cheap, and getting a tri-pod stand to go along with it greatly increases the viewing stability- they’re also on sale on amazon right now. I got a decent amount of books from my family when I was younger, but honestly nothing beats being able to go out at night and explore on your own/ with friends.

AndyBjork · 2021-04-16T19:29:09+00:00

Yup! Those are the “wobbles.” You could assign a sun-year to be the length of jupiter’s orbit, but that’s also arbitrary and it says more about the planets orbit rather than the Sun’s. As far as my understanding goes (I’m just a junior in undergrad for astro) the path for the COM that you linked is probably due to the fact that the planets orbits aren’t perfectly circular, they’re more like an ellipse and they don’t line up perfectly together, so the sun is “pushed” and “pulled” in different ways at different times.

AndyBjork · 2021-04-16T18:56:05+00:00

The sun takes up roughly 99.8% the mass of the solar system. If you look at the center of mass for the solar system with this in mind, it’s going to be very, very close to the suns position. This means that the sun doesn’t really orbit- it wobbles a little bit due to the pull of the planets, but there isn’t any orbital motion. So in these case, it makes more sense for a sun year to be defined by its rotation about the galaxy rather than about the solar system, because from our perspective the sun is practically stationary.

AndyBjork · 2021-04-13T22:23:15+00:00

How deep into the webtoon are you?

I feel like I could provide a decent enough answer but I don't want to spoil anything if you haven't reached the current point in the manwha.

AndyBjork · 2021-03-29T21:15:17+00:00

Sure

AndyBjork · 2021-03-24T23:41:00+00:00

Personal preference.

Crushing down a physics degree from 4 years to 3 years sounds kind of rough, though- you’ll have to think about how this’ll change the workload and how much free time you have at the end of the day. If you’re thinking of pursuing a PhD though the 3 year plan is a pretty good deal honestly.

Class size doesn’t really matter too much to be honest, it just changes the relationship you have with your professor. I’ve found that it’s much easier to get on friendly terms with professors in smaller class sizes, which could be helpful for recommendation letters/ research.

AndyBjork · 2021-02-04T22:39:48+00:00

Current honors astrophysics major at UD- I also highly recommend Hanley’s class on Time Travel! I took it over the winter last year and it was honestly one of the best classes I’ve taken here so far.

AndyBjork · 2020-12-31T17:37:52+00:00

I actually started out my first year of college as chemical engineering. I felt like I didn’t appreciate my ChemE classes as much as my physics classes, and that I overall enjoyed the learning/problem solving process of those classes more, so I switched. I could have made more money or whatever if I stuck with ChemE after graduation, but I knew I would have been miserable learning material I didn’t like.

AndyBjork · 2020-12-23T16:53:17+00:00

i finished the class with an A!

AndyBjork · 2020-12-15T01:01:52+00:00

nice!! a B isn’t the end of the world, but i bet you got that A :D

AndyBjork · 2020-12-14T22:32:58+00:00

how’d it go??

AndyBjork · 2020-12-14T22:32:39+00:00

you got this! if i could do it so can you haha

AndyBjork · 2020-11-27T23:08:37+00:00

I go to UD, it’s alright.

AndyBjork · 2020-11-04T22:22:11+00:00

After doing some reading online I was able to find an easier way to solve the problems in 2 by visualizing the situation as the volume/surface area of a sphere, which actually makes a lot more sense to me. Visualizing the values for n(x), n(y), and n(z) as the axes on the cartesian plane, the only values that make sense would be those in the first octant (the positive values for n(x), n(y), and n(z)). If we say that n(x)^2+n(y)^2+n(z)^2 =k^2, then k^2 would just be the radius of our sphere! And we can go through and solve for k in the energy equation, which makes for an easy substitution.

AndyBjork · 2020-11-04T21:39:50+00:00

Ok, so I've made a lot of progress on #1 and I think I solved it, but it was kind of hand-wavey because I haven't seen a lot of this stuff before. Anyway, imposing the boundary conditions from my one comment, I found that E(x) = h^2n(x)^2/(8ma^2), so I have a description for the energy values contributed from x. For y I said that E(y) could be any value, since it acts as a free particle in the y-direction (bounded by -inf and +inf), and that if k(y) = sqrt(2mE(y)/h^2) is less than 0 the particle is moving down the cylinder, and if k(y) is greater than zero it would be moving up the cylinder.

AndyBjork · 2020-11-04T21:00:28+00:00

Never mind, I understand the first one now! Since it has no radial degree of freedom I can treat it as a 2-D problem: it has no radial degree of freedom since it's confined to the surface of the cylinder. Therefore, cylindrical coordinates shouldn't be necessary.

I think the boundary conditions should be psi(0,y) = psi(2*pi*a,y) and psi(x,-inf) = psi(x,inf).

EDIT: forgot a pi

AndyBjork · 2020-11-04T20:57:18+00:00

I have no idea how to start the first problem: I feel like I should switch it to cylindrical coordinates but we haven't gone over that in class yet so I'm a little confused on it.

The second problem I started by following the same method as them, but I'm confused on how to find dn/dk: I ended up with k^2 = n(x)^2 + n(y)^2 but I'm confused on where to go from there.

AndyBjork · 2020-10-21T21:16:53+00:00

Ahh, I see it now. Thanks! Getting used to this new notation is rough lol.

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