I'm not that familiar with stellar formation, so you might be right. According to this website:
http://www.josleys.com/show_gallery.php?galid=313
the earth is malleable because of its liquid core and tectonic plates on the surface (quite interestingly, according to that website, if the...
Thanks for your help. I appreciate it. This is not my area either, but I teach lab courses to pay for my tuition, and I wanted to give my students examples of some good applications of centripetal acceleration and I thought what could be a more grand example of centripetal acceleration than the...
Actually, according to Wikipedia:
http://en.wikipedia.org/wiki/Clairaut's_theorem
the gravity is modified by:
g[1+(\frac{5m}{2}-f)\sin^2 \varphi]
where m is the ratio of the centrifugal force to gravity at the equator (which should be really small), and f is proportional to the difference...
Thanks. I appreciate it. I wish I could just slip in a factor of 2 on
m\omega^2(r \cos \varphi) \sin\varphi = -mg \frac{1}{2}\frac{1}{r}\frac{dr}{d\varphi}
and say that the 2 comes from considering mass spread over the entire ellipsoid. That would produce the right answer. But...
If you can get the right answer, let us know! I've been trying for hours and I've given up. I teach a lab course and I'm trying to give my students some information on centripetal acceleration, but I can't figure out how to calculate the bulge - like the original poster I get half the value.
I thought the formula should be:
m\omega^2(r \cos \varphi) \sin\varphi = -mg\frac{1}{r}\frac{dr}{d\varphi}
but maybe I'm wrong.
You need the centripetal force along the tangential direction, so m\omega^2r \cos \varphi gets multiplied by \sin\varphi.
When you integrate:
m\omega^2(r \cos...
Shouldn't your r in the LHS of:
m\omega^2r\sin\varphi = -mg\frac{1}{r}\frac{dr}{d\varphi}
be r*cos(phi), or distance from the rotation axis? If you do that, then you get the original poster's value of half the bulge.
Sure. It's a little bit lengthy though, so it might take some work to read it:
P(m)=\frac{\sqrt{2 \pi n}n^ne^{-n}}{\sqrt{2 \pi m}m^me^{-m}*\sqrt{2 \pi (n-m)}(n-m)^{(n-m)}e^{-(n-m)}}p^m(1-p)^{n-m}=
\frac{n^{n+1}}{\sqrt{2 \pi n}*m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}...
Normally I would just dismiss the formula, but I found it in two different sources (both particle physics sources though). One book talked about the vacuum bubble expansion of the integral:
\int \frac{1}{[k^2-m^2][(k-p)^2-m^2]}=\int \frac{1}{[k^2-m^2]^2}
-\int \frac{p^2}{[k^2-m^2]^3}...
This is probably a dumb question, but I have a book that claims that if you have a function of the momentum squared, f(p2), that:
\frac{d}{dp^2}f=\frac{1}{2d}\frac{\partial }{\partial p_\mu}
\frac{\partial }{\partial p^\mu}f
where the d in the denominator is the number of spacetime...
I want to show that the binomial distribution:
P(m)=\frac{n!}{(n-m)!m!}p^m(1-p)^{n-m}
using Stirling's formula:
n!=n^n e^{-n} \sqrt{2\pi n}
reduces to the normal distribution:
P(m)=\frac{1}{\sqrt{2 \pi n}} \frac{1}{\sqrt{p(1-p)}}
exp[-\frac{1}{2}\frac{(m-np)^2}{np(1-p)}]...
For some information on tides, here's a website:
http://www.lhup.edu/~dsimanek/scenario/tides.htm
For some mathematical detail (just algebra):
http://mb-soft.com/public/tides.html
You only need gravity to explain the tides.
Omega is a constant, equal to the angular velocity of the earth...
I'm pretty confused by the post - I think you might have a lot of misconceptions.
First, tidal forces have nothing to do with centripetal acceleration. Tidal forces are due to gravity.
Second, if you're standing still on the earth, unless you're at the poles, there is a centripetal force on...
I always thought it'd be dangerous to ground the primary side since it's at such high voltage. So maybe isolation transformers allow you to ground the secondary side, rather than, as you say, allowing the primary to float when you have a grounded secondary?
When lightning strikes near power lines, it breaks down the air which shorts the power lines, but why would this have an effect on the load? The short and the load are in parallel, so the load should not be affected. When you have two things in parallel what happens to one branch should not...
What other coupling does a transformer have other than magnetic? I thought that was the definition of transformer, two windings that don't touch but share flux.
That makes sense. So can the definition of an isolation transformer be a transformer that is not tapped, i.e., the transformer has...
You can prove it with some clever thermodynamical arguments, but I always find those to be dissatisfying.
Here's my attempt at explaining it. Imagine an object that is painted white. That means if you shine light on it, the light will reflect off it. Now imagine all the atoms inside the object...
Apologies for digging up this old thread, but I have some questions on some of the responses:
But why doesn't the metallic screen inside the microwave window get burnt, as does aluminum foil when you put it in the microwave?
What you say makes a lot of sense. However, what about the...
Does anyone know what is meant by electrical isolation?
From this schematic:
url=http://upload.wikimedia.org/wikipedia/en/1/12/Swer.gif
I don't see the significance of calling the transformer an "isolation transformer". What is the difference between a transformer and an isolation...
I got it from the link that was given by berekeman in this thread. The very last paragraph says:
I can sort of understand how internet operates: some companies build a network of cables that cross the country, and charge ISPs to connect to this backbone. They sell usage of their lines to ISPs...
Thanks.
I find it interesting that the reason for separating the electric industry into power plants, transmission lines, substations, primary and secondary distribution, is to prevent a monopoly of the entire electric industry. But I find that odd, as I'd imagine that power plants have a...
I don't see how the secondary center conductor (the middle black wire) is connected to neutral/ground. That middle black wire seems to end at a dielectric spacer that separates a ground wire (the wire where you can see the braids) from other wires.
Could someone confirm if what I'm saying about this picture is correct:
http://en.wikipedia.org/wiki/File:Polemount-singlephase-closeup.jpg
http://en.wikipedia.org/wiki/File:Polemount-singlephase-closeup.jpg
There are three thick insulated black wires which are the split phase secondary...
Well it's been awhile since I looked at thermodynamics, but I thought a Debye solid didn't require periodic boundary conditions, just vanishing at the endpoints.
So you have a solid, and the frequencies it can vibrate at correspond to wavelengths that are 2L/n for positive integers n and...
I'm not sure why you would need to decompose it. Assuming a Biot-Savart field, the flux through a circle whose center goes through the current carrying wire is zero, since the magnetic fields circle around the wire, so that they are always tangential to planes perpendicular to the direction of...
I'm not sure I know the answer at the level you want it. This is kind of like the example you see in textbooks of charging a capacitor. If you form a loop around one of the wires leading up to the capacitor, and calculate the line integral of the magnetic field, then you get an answer if you use...
Kinetic energy would not be a scalar because of Galilean boosts. If you're standing still, a tree has very little kinetic energy. If you're in a moving car, the tree is moving very fast. However, the kinetic energy is a scalar in the sense that is remains invariant under rotations. So it doesn't...
I think that's an interesting historical question. The equations for electricity and magnetism, for example, are not the same when you make a Gallileo boost of the form y=x+vt. People tried to save it by introducing an ether fluid, and so when you boost you also have to boost the ether fluid...
You have to make some assumptions with Gallilean boosts. For example, if you have a force that depends on velocity, say a drag force instead of a spring force:
-kx'=mx''
The transformation y=x+vt (where v is velocity, t is time) results in the equation:
-k(y'-v)=my''
which is not -ky'=my''...