Incredibly 'lucky' vacuum metastability event. Laws of physics change but somehow all matter simply doesn't perish or turn into something completely unrecognisable.
Are you sure it wasn't just a theoretical "What if that happened, wouldn't it be cool, and these are the things that we predict would have happened"
[QUOTE=MountainWatcher;31569446]Huge-ass asteroid grazing Earth?[/QUOTE]
are you talking about stopping rotation or just destroying earth
because right now it's not really possible to stop the earths rotation without destroying most of it
[QUOTE=Jo The Shmo;31578043]Are you sure it wasn't just a theoretical "What if that happened, wouldn't it be cool, and these are the things that we predict would have happened"[/QUOTE]
Well they sure put alot of effort on the CGI.
[QUOTE=JohnnyMo1;31568438]Why in the world would the Earth's rotation suddenly stop do you know how much torque that would require[/QUOTE]
oi
if i floored my car n did burnouts the torque would stop the earths spin, fuckan fully sick cunt
[IMG]http://profile.ak.fbcdn.net/hprofile-ak-ash2/195744_209064009130288_2894812_n.jpg[/IMG]
bye earth
(No, but seriously, bye Earth as we know it.)
Doing an online physics test for uni that goes towards my final mark. Got this question:
[quote]A small object A, electrically charged, creates an electric field. At a point P located 0.250m directly north of A, the field has a value of 40.0 N/C directed to the south.
What is the charge of object A?[/quote]
I went and used this equation here [img]http://upload.wikimedia.org/math/9/4/1/9410b5b1c771eb6631badefc6477c652.png[/img] rearranging for Q and got the correct answer.
Now, r-hat is the unit vector, and... I didn't have a fucking clue what my unit vector was, but I went and assumed it was 1 which ended up being correct. Is this because the point I was working with was exactly North of the charge, i.e. along the Y-axis so r and |r| would end up being exactly the same (or the 'triangle' has a hypotenuse the same length as one of the other sides)?
[editline]12th August 2011[/editline]
I guess no matter where you are in relation to the particle it can ALWAYS be considered 'directly up' if you're willing to set the Y-axis at an arbitrary position defined as the direction of the point away from the particle, and hence you can always force the triangle to converge as a line and simply ignore that unit vector term.
A unit vector is length 1 by definition.
[editline]12th August 2011[/editline]
Oh wait I see what you mean.
It wouldn't really matter because you only ever actually need to work with magnitudes in this problem. Since you're given the magnitude of the E-field and the distance, you only ever have to work with scalars. You can get rid of the unit vector entirely and just use
[img]http://www.codecogs.com/gif.latex?E%20=%20\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2}[/img]
Yeah, but now I've come across this question:
[quote]Now add a fourth charged particle, particle 3, with positive charge q3, fixed in the yz-plane at (0, d2, d2). What is the net force F on particle 0 due solely to this charge?
Express your answer (a vector) using k, q0, q3, d2, x-hat, y-hat, and z-hat. Include only the force caused by particle 3.[/quote]
So I HAVE to use unit vectors and I'm not sure what to do here. I've already wasted a few attempts at the question and one in specific really pissed me off; all of the previous questions had said:
[quote]Express your answer (a vector) using any [b]or[/b] all of...[/quote]
The first attempt I wasted was one with x-hat in the answer when it's not required. I decided to give the problem a shot and stuck x-hat into the answer (even though I was pretty certain that was wrong due to the problem only existing in the yz plane) solely because of the wording of the question - and surprise surprise - I was right; that was the wrong way to do it..
I only have one attempt left. I think I have the right answer now but... I dunno. I don't know if I want to risk it and get it wrong (given that even getting partial credit on the question at this point would be handy).
Science thread, what do I learn from when I've learned everything I can from high school physics, but not going to college yet?
What books can I read, etc?
[QUOTE=Collin665;31703542]Science thread, what do I learn from when I've learned everything I can from high school physics, but not going to college yet?
What books can I read, etc?[/QUOTE]
Learn calculus if you haven't already. A great math foundation makes most physics a breeze.
Since there has been alot of talk of prosthetics lately.
I began thinking if the "augmentations" in games such as Deus Ex could even be possible?
My main question is, if metal can mend with flesh, without the body completely rejecting it by creating deadly infections.
I remember this being a problem some years ago, and is it still?
[QUOTE=booster;31990712]Since there has been alot of talk of prosthetics lately.
I began thinking if the "augmentations" in games such as Deus Ex could even be possible?
My main question is, if metal can mend with flesh, without the body completely rejecting it by creating deadly infections.
I remember this being a problem some years ago, and is it still?[/QUOTE]
Well we routinely give people artificial hearts, or artificial valves nowadays, don't we? We can give people insulin pumps. So sure, we can beat rejection - it's not impossible. I'm not sure how well it'd go down when you start trying to replace limbs, though. I'd imagine you'd have to try and somehow force the body not to let scar tissue build up at the join or it'd get all stiff I'd imagine (kind of like arthritis but with tissue build up).
I have a question:
If an electrons positional probability increases as you get closer and closer to the centre of the electron cloud wouldn't the electron MOST PROBABLY be at the VERY CENTRE? And if that is so, why is the electron not most probably in the same place as the nucleus? Ergo, why does all matter not suddenly turn into neutrons as protons and electrons should most probably occupy the same space?
[QUOTE=sltungle;32005080]If an electrons positional probability increases as you get closer and closer to the centre of the electron cloud[/QUOTE]
It doesn't though
The highest probability is at the average orbital radius
[QUOTE=JohnnyMo1;32005788]It doesn't though
The highest probability is at the average orbital radius[/QUOTE]
Explain. What is the 'average orbital radius'? I'm guessing it's simply the median value of all separate electron shells in the atom?
The average radius from the nucleus of the orbital that a given electron would be filling. Just one radius in the case of the hydrogen atom. You can see depictions of the wave function in a hydrogen atom having a maximum probability at the nucleus but these aren't taking all quantum numbers into account. An actual wave function will look something like this, tapering off to zero at zero radius and hitting a maximum at the average orbital radius:
[IMG]http://i56.tinypic.com/9lk389.png[/IMG]
What IS the probability of an electron being at the centre of an atom? Is it in fact 0? I'd find that hard to believe - it'd seem intuitive to think there'd be a relatively large probability of an electron being at the centre of the atom (then again, quantum mechanics is anything but intuitive). DOES the [B]occasional[/B] atom (hydrogen in particular because it's so easy to calculate) spontaneously become a neutron?
[editline]30th August 2011[/editline]
Woops. Answered before I could ask my second question. Thanks!
But then that raises a question: why, intuitively speaking, is an electrons probability of being at the nucleus zero? I mean, electrons are negatively charged, protons are positively charged. Shouldn't the electron want to be at the nucleus more than anything?
My physics lecturer said to me that, when you solve the Schrodinger equation for the hydrogen atom, there's no Z component for angular moment in the electron. Mathematically this explains things to me, but intuitively I'm not satisfied by this explanation.
It is possible but not the norm. Happens when there is an overabundance of protons in a nucleus.
[url]http://en.wikipedia.org/wiki/Electron_capture[/url]
[editline]29th August 2011[/editline]
"Around the elements in the middle of the periodic table, isotopes that are lighter than stable isotopes of the same element tend to decay through electron capture, while isotopes heavier than the stable ones decay by electron emission."
[QUOTE=JohnnyMo1;32005953]It is possible but not the norm. Happens when there is an overabundance of protons in a nucleus.
[url]http://en.wikipedia.org/wiki/Electron_capture[/url]
[editline]29th August 2011[/editline]
"Around the elements in the middle of the periodic table, isotopes that are lighter than stable isotopes of the same element tend to decay through electron capture, while isotopes heavier than the stable ones decay by electron emission."[/QUOTE]
Typical decay, right? (beta negative and beta positive decay). I'm still frustrated that there doesn't seem to be an intuitive, non-mathematical reason for why quantum atoms (as opposed to the tradition Bohr atoms) don't decay, though. I mean the mathematics makes sense; I understand it - I like it. But I want some intuitive reasoning, too. I suppose you can't really ask for intuitive reasoning in QM, though.
I can't really speak to the specifics of QM I haven't taken any formal QM courses. Just a cursory glance over it in modern physics.
But I just got back from the lab and the muon detector has been up and detecting all weekend as planned. :D Getting great data and our decays seem to be centered around the average time of ~2 microseconds. Gonna try to keep it going until next monday if we can. That will be like a nice solid 300 hours worth of data. Nothing like a hard day of lab work. *go to lab, check detector, go home*
This may be a retarded question, because I don't know alot in this area.
But I've heard that in the beginning years of our solar system, jupiter was really close to becoming a star. But it failed.
So I wondered, if given alot more resources at once, lets say a vast ammount of asteroids crash into it. Could this somehow act as a catalyst, thus increasing the fusion within jupiter, and then creating a star.
Or is it simply just billions of years too late for this?
Well, I guess if you slammed a huge heavy ball or shit into it, the gravitational pull of it could simulate a star. But with just loads of energy transferal through friction, the most you could have would be a tiny, little supernova, since the gas wouldn't contract into a star due to it.
This isn't really up there in your leagues, but i'm waiting for a reply back from NASA hopefully telling me if my fuel suggestion is possible. Wish me luck.
Holy shit, what? :v:
Basically, I've been getting into building amateur rockets and after learning how to properly build a liquid propulsion system I wanted to come up with a fuel source different then normal liquid oxygen and liquid hydrogen. I don't know anything about Deuterium but I assume that if you have Deuterium enhanced liquid hydrogen you'd get a much faster impulse than just normal liquid hydrogen and liquid oxygen.
Also in the Delta E = c. m . Delta T equation in thermodynamics, does E represent any and all energy that can raise or decrease temperature? Like if we did the Watt experiment and dropped shit on water, could we plug in the work energy for delta E? Or is it just the energy transmitted better bodies of differing temperature?
Dividing by 0!
[QUOTE=RidingKeys;32031622]Basically, I've been getting into building amateur rockets and after learning how to properly build a liquid propulsion system I wanted to come up with a fuel source different then normal liquid oxygen and liquid hydrogen. I don't know anything about Deuterium but I assume that if you have Deuterium enhanced liquid hydrogen you'd get a much faster impulse than just normal liquid hydrogen and liquid oxygen.[/QUOTE]
Why would you assume that? The only difference between deuterium and regular hydrogen is one neutron. Deuterium is chemically identical to hydrogen, it's just got double the mass.
If anything it's WORSE. Kinetic energy is proportional to the velocity squared, and to mass linearly. If you want to go faster you want to throw your propellent away from you quicker, not make it more heavy.
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