In this particular problem, there is only one unknown quantity. This means that you only needed one equation. If you had used the momentum equation, you would have also gotten the same answer. So at a certain level, you just kind of got lucky.

But the general approach (elastic collision -> write down both conservation equations and then start playing around with the algebra) will work no matter what type of problem is thrown at you. One mistake I've seen people make is that they try to memorize a bunch of different formulas for all the situations shown in the examples used rather than understanding that they all stem from just a few basic concepts/equations.

Momentum: p = mv
Kinetic energy: E = 1/2 * mv^2
Conservation of momentum (always): sum of p_before = sum of p_after
Conservation of energy (elastic): sum of E_before = sum of E_after

That's pretty much how to solve every one of these collision problems.

I don't want to sound like I'm piling on, but after a couple minutes of lecture/reading about conservation of momentum and maybe one example, the above question should be trivial. I would imagine that you gave this a significant shot before posting here. Assuming you are working from a textbook (if so, do you mind indicating which one?), I strongly recommend re-reading any and all material related to questions you are struggling with.

Also, pre-req's are normally pre-req's because what you learn in them is relevant for subsequent courses. To me it sounds like you might be treating Physics like people treat their required Gym credits or something like that, where they are just "freebie" albeit PIA credits.

Also, even if actual Physics concepts won't be used in future courses, the ability to problem solve given incomplete data is almost universally useful, so if nothing else that should be your motivation to do well in this class.

GL

you mean this part:

Nope. I meant this part:

...

Because it's a monic polynomial (the leading coefficient is 1), if there's a rational root (a root that you're going to be able to find by hand), it must divide the constant term (which is 1), and so the only two possibilities are 1 and -1. The numbers 0, 2, and -2 have no chance of being roots.

I'd appreciate some help with this:



I decided to split the volume into two regions and then just add them together. I have the volume of the cone fine, but for the volume of the hemisphere I get:



Which evaluates to ~10.4, but if I use this equation for the volume of a spherical cap, I get ~5.5. I've double checked my answer with an integral calculator and it's fine, is there something wrong with how I set up the integral?

Your dz integral goes from one fixed height to another. This means that your top surface is parallel to your bottom surface. But the top surface of the cone is rounded. This means that your upper limit will have to be derived from the equation of the hemisphere instead of being a constant value.

Random question since I dont post in this part of the forum much. My company is recruiting for a Senior Data Scientist in Chicago. Is there a place to post things like that? Figured it might be of some interest if anyone in that area is looking for a math based job..

Ok that makes sense, but then the integral is very messy with sqrt(8-x^2-y^2) in the limit. I'm not really sure what kind of substitution to make.

I'd just go back and do the whole thing in spherical coordinates. But if you wanted to do it the way you were doing it, you can make a polar/cylindrical coordinate substitution at this point.

Wow ok I missed the really obvious trig sub that makes it much easier. Thanks!

Real life math/physics/whatever question.

We measure the static coefficient of friction of paper as it comes off the paper machine. Let's say we cut five sets of samples across the sheet and measure the COF of each sample, getting five measures of the COF. I'm interested in quantifying the distribution that these sample are picked from. That is, let's say we have data like

a1 a2 a3 a4 a5
b1 b2 b3 b4 b5
c1 c2 c3 c4 c5
d1 d2 d3 d4 d5

Where a, b, c, d are paper manufactured at different times and a1, a2, ... a5 are samples that are tested yielding COF measurements. How can I estimate the standard deviation of the testing procedure?

Taking the standard deviation of the 20 individual measurements is wrong because the process itself will be varying. Can I take the standard deviation of a1, a2,...a5 and the standard deviation of b1, b2,...b5 etc and average the four standard deviations? That also seems wrong.

This is a survey sampling problem. The choice of estimator is going to depend on the process by which you are sampling. You clearly have a two-stage sampling process. First, you sample clusters (a,b,c,d) and then within the clusters you are sampling some portion of them. This leaves some questions about how you are sampling in each stage:
1) I the first stage a simple random sample where every sheet of paper is equally likely to be sampled? Is it some sort of systematic sample where you take a sheet at some fixed interval? Such as taking every 10,000th sheet
2) In the second stage, is the sheet of paper just being chopped up into 5 pieces so the whole thing is being used or is it being chopped up into more samples where only some portion of them are being used?

If you are taking every 10,000th sheet and then using all pieces of it, this is a systematic cluster sample (you are taking sheets systematically and then sampling the whole cluster). If you are taking sheets randomly and then chopping the sheet into many pieces and randomly selecting 5 then it is a 2 stage simple random sample. Given that all of the potential sampling schemes seem pretty straightforward, once you have it fully defined, it shouldn't be too hard to find the right estimator. I don't know these off the top of my head, but I'm sure I have them somewhere in some old lecture notes or a sample survey textbook.

We're taking a sample across the width of the sheet every hour or so and sampling five small sections across the sheet. So it is a 2 stage random sample.

So basically what I want is the distribution of results you would expect from the sampling and testing procedure if there was no variability in the underlying process. So lets say that we take x samples, with each sample x1, x2, ..., xn consisting of a subgroup of five tests x11, x12, ..., x15. My intuition would be to normalize the subgroup data by subtracting the difference between the subgroup mean and the overall process mean. So then every subgroup would have the same mean and you could just calculate the standard deviation of all of the individual measurements.

I want to measure this because right now we take a single sample and measure the friction of the top of the sheet and the top and the bottom of the sheet and the top. Then we let the value of either the top/top or top/bottom COF drive decision making on dosing our frictionizer chemical. So it's worse than using a single value, we're using the minimum of two tests! So I want to show that our underlying process could be perfectly stable but we're letting our sampling and testing uncertainty drive our decision making.

like, here's some simulations of what we would be seeing if we take the minimum of two tests, use one test, average 3 or average 5 based on my estimate of the standard deviation of the sampling and testing procedure even if the underlying process mean didn't change at all.

A general rule of thumb is to write down an equation for your sample observations. When you say you want to determine the distribution of the samples, that is saying that they have systematic components and one or more random components.

Writing down an equation is tantamount to identifying the components you are interested in learning about through sampling. Then from the sample you collect you can estimate the effect of the components. And, from that, of course, you can make decisions on specific samples (i.e., accept or reject as being within the norms).

It sounds like a classic Analysis of Variance problem. Loosely speaking you want to "control" for one factor to discern information (make a decision) on the other factor. If I am following how your samples are obtained, I think this is a fixed effects model application.

Not sure how much anyone here knows about inorganic chemistry, but I would appreciate some insight into this question and if I've explained it properly.

Question: Explain why CsICl2 is more stable than KICl2 in the solid state

My Answer: Given that Cs (cesium) is a larger cationic metal with a higher oxidation state in this case than K (potassium), it will form a more stable ionic lattice with the polyhalide anion ICl2 as the packing efficiency increases.

Secondly, as you go from K to Cs, the enthalpy change should become more positive and MICl2 ---> MCl + ICl would become more unfavorable so the stability of MICl2 increases from potassium to cesium.

Is this a sufficient answer? Could I add more or change anything?

Does anyone here know much about Cayley graphs? I have come up with an axiomatization of complete Cayley graphs which must be in the literature somewhere. I wonder if anyone would know where. Here is the axiomatization:

Let X be a set.

For any relation R on X^2 (i.e. R is a subset of X^4), define the relation (x_1, y_1) R^op (x_2, y_2) if (x_1, x_2) R (y_1, y_2). R^op is also a relation on X^2.

Then a complete Cayley graph is an equivalence relation R on X^2 such that R^op is also an equivalence relation.

A more general question about the literature is that I am looking for a longer introduction to the topic, but in the same spirit as this blog post:

https://terrytao.wordpress.com/2010/...try-of-groups/

What do you mean by a "complete Cayley graph"? When I think of a "complete graph" I think of this:

https://en.wikipedia.org/wiki/Complete_graph

To get a Cayley graph to look like that, I think you just take the generating set to be the set of all elements (except the identity, which creates useless loops). And then (because it's a group) everything is connected to everything else. (If g and h are in a group, there exists an element k such that g = hk.)

I don't know how much I can help you here. I first remember seeing Cayley (di)graphs in Fraleigh's Algebra book. I'd check in Dummit and Foote or Hungerford. The "introduction" will probably be at a slightly higher level than the link, but if you can follow the link you probably can follow the presentation in those textbooks.

I don't know "much" about these things, but hopefully this gives you a start.

Right. The underlying graph of a complete Cayley graph is a complete graph. The point is that Cayley graphs can be characterized in a non-group theoretic way, and I have a characterization of this specific class of Cayley graphs.

The standard source is text by Magnus, Karrass and Solitar. But if you want something with more geometric motivation check out survey by Bowditch, available for free here.

Anyone have any idea why the Fst (fixation index) might vary between different loci within a population?

Say I give you 10 values for p: p1, p2, .... p10 and each number has a value between 0 and 1 (most of them are 0.5 usually but whatever).

Then I tell you to solve for 'p' where p is defined here:



Is this the same thing as if I said "p = p1 + p2 + ... p10 / 10" ?
Which is the same as saying "the average value for p" ?

Because if not, that's fine: math is hard and the world is normal. But if so I want an arrow to the knee.

No. If any of the pi happens to be 0 and at least one of the other pi is positive, then the original expression is going to be zero but the arithmetic mean will be positive.

Is it the same as average if we change "p_i is between 0 and 1" to "0 < p_i < 1" ?

This is actually the question I meant to ask in the first place, sorry.

It isn't the same as the average, however one way you can construct it such that it will be the same is if for every p_i, there is also a p_j that is equal to 1-p_i. If this is the case, then the average will be .5 and that expression will also be equal to .5 because prod(pi) will be equal to prod(1-pi). You said that most of the pi are equal to .5. If they were all .5 then it would clearly be the same.