[QUOTE=JohnnyMo1;32656718]Assuming you're very small you wouldn't feel anything.[/QUOTE]
But would you like levitate?
Or could you walk on the walls?
I really have no clue about this kind of stuff.
Actually if you were in a spherical room at the center of the earth you wouldn't need to be small. The gravitational force at any point in the room would be zero.
[editline]6th October 2011[/editline]
Yes. It would be the same as if you were floating in space.
[QUOTE=JohnnyMo1;32656805]Actually if you were in a spherical room at the center of the earth you wouldn't need to be small. The gravitational force at any point in the room would be zero.
[editline]6th October 2011[/editline]
Yes. It would be the same as if you were floating in space.[/QUOTE]
Woah, that is pretty god damn cool.
Also, say if you were inside this sphere, and there was a tube leading from the sphere all the way to the "open ground/crust/whatever".
Would you even be able to get away from the centre without using tools or outisde help?
Or would the gravity always pull you back to the centre?
If the tube was smooth I don't really see how you could do it, but you could rock-climb your way out if it was like a cave or something. I THINK the gravity scales linearly with respect to radius inside a sphere, meaning if you were halfway from the surface to the center gravity would be half as strong as it is on the surface.
[editline]6th October 2011[/editline]
So you could easily climb out part of the way but the further you go the stronger gravity would become as you traveled.
ignoring heat, pressure and friction, would throwing a magnet down a copper coil which tunnels through the Earth generate electricity out of gravitational pe for a rather long time?
I think in that case eddy currents would rob the magnet of energy.
If you were directly in between 2 huge masses, would you feel no gravitational pull from them, or would you feel like you were being ripped apart?
[editline]6th October 2011[/editline]
[img]http://4.asset.soup.io/asset/2426/1828_c77f.gif[/img]
I thought it was just an edited gif in photoshop at first
[QUOTE=Jo The Shmo;32658953]If you were directly in between 2 huge masses, would you feel no gravitational pull from them, or would you feel like you were being ripped apart?[/QUOTE]
Directly in between i.e. I am flat?
If you lie entirely on the plane between them, you won't feel a thing. If you're a normal-sized person, it depends on the strength of the gravitational fields from them. Between two planets an AU apart? You won't feel anything. Between two black holes 10m apart? You're fucked.
Snip
i always wondered what would happen if we weaponised a particle collider or something
oh kinda like this
[url]http://davidszondy.com/future/tesla/teslaray.htm[/url]
DISCUSS FOR SCIENCE!!!
[QUOTE=JohnnyMo1;32657073]I THINK the gravity scales linearly with respect to radius inside a sphere, meaning if you were halfway from the surface to the center gravity would be half as strong as it is on the surface.
[/QUOTE]
Yes, assuming the mass density of the earth is a constant.
[editline]7th October 2011[/editline]
I took my first midterm in Quantum Mechanics today. I don't know how I feel about it...
[img]http://i.imgur.com/NjCls.png[/img]
I don't even
Fucking Mastering Physics.
I think it's about time we cut the creators a new one.
I don't want to watch the last episode of Cosmos.
Because then I'll know that I'll never see anything that good ever again :(
If you don't watch it, you'll never watch it again either.
Last episode was my favorite.
Oh hey, I have to use MasteringPhysics too. I didn't know it was so widespread.
[QUOTE=Turnips5;32714958]Oh hey, I have to use MasteringPhysics too. I didn't know it was so widespread.[/QUOTE]
Yes, it started in MIT. Now it looks like its on its way to taking over the whole world.
just realised I missed the deadline for that first masteringphysics homework section, FUCK
good thing it's not assessed + I just did all of it anyway and got everything mostly right
dammit though, why do some of the sections have like 155 points in them while others have four, it makes NO SENSE
there were like 34 marks in the section I missed and 4 in the one I just did, and they both took me roughly the same time to complete
wtf
I'm doing a physical chem assignment at the moment and I'm fucking stumped.
I'm told to find the reaction rate constant, k, for a first order reaction for which I'm only given the concentration of A, and the reaction rate.
My values are:
t (seconds) 100 200 300 400 500
[A] (mol/L) 0.344 0.314 0.286 0.261 0.238
I plotted the graph, which is linear (which is good). The gradient should be the reaction rate if I'm not mistaken (d[A]/dt), but then from there I'm stumped.
If I go:
d[A]/dt = k[A]
And try to solve for k... I don't get a constant value (which is counter intuitive given that k represents a reaction rate CONSTANT).
d[A]/dt is constant. [A] changes over the course of the reaction. If the left hand side of the equation is constant, and one of the two values on the right hand side ([A]) is changing, then the other value (k in this case) must ALSO be changing to maintain the constant value.
No clue what I'm overlooking but my guess is it's incredibly simple. Any help would be greatly appreciated.
edit : no, hang on, I haven't done physical chemistry in ages
[editline]11th October 2011[/editline]
that is bizarre, I see your problem exactly, I don't know what the deal is
The concentration of A isn't constant, and k definitely is constant, so the rate should definitely change with time, maybe it's not a linear graph as you think
[editline]11th October 2011[/editline]
yeah, that's it, now I remember
form a differential equation -dA/dt = k[A]
from that you can rearrange it to get ln A = - kt + c
you should be able to do it from there, it's an exponential decay
[editline]11th October 2011[/editline]
WHO KNOWS, I'm probably just thick but that's my best guess
[QUOTE=Turnips5;32726964]edit : no, hang on, I haven't done physical chemistry in ages
[editline]11th October 2011[/editline]
that is bizarre, I see your problem exactly, I don't know what the deal is
The concentration of A isn't constant, and k definitely is constant, so the rate should definitely change with time, maybe it's not a linear graph as you think
[editline]11th October 2011[/editline]
yeah, that's it, now I remember
form a differential equation -dA/dt = k[A]
from that you can rearrange it to get ln A = - kt + c
you should be able to do it from there, it's an exponential decay
[editline]11th October 2011[/editline]
WHO KNOWS, I'm probably just thick but that's my best guess[/QUOTE]
this is pretty much exactly it.
you need to plot the natural log of the concentration against time.
just plotting concentration against time should give you an exponential curve for a first order reaction - it's possible your data range is so small you're not seeing much of the reaction lifetime and that's why it appears to be straight.
the last experiment i did involved a lot of kinetics - studying a reaction via IR. some of the reactions were so slow that i had to leave them running for 18 hours to get a useful data set.
here's the gist of it.
this is concentration (actually absorbance but beer-lambert law means for these purpose i can use them interchangeably) against time. as you can see, it's a curve! this a first order migratory insertion reaction.
[img]http://mikeh269.com/filedump/ir.png[/img]
by plotting ln(concentration) against time, this second plot is produced
[img]http://mikeh269.com/filedump/ir_1.png[/img]
in this one - you can see - as turnips made obvious in his post, the gradient is -k
so all you have to do is pop a negative sign in front of the gradient and ta-dah! you have your observed rate constant.
Ugh it's amazing how fast I've forgotten everything related to chemistry. Turn the clock back one year and I'd be able to solve that in minutes (or hours).
Just got my lab book back
"excellent first experiment!"
I was expecting it to come back covered in red pen with "YOU TOOL THIS IS ALL FUCKED UP" written at the end
:D
[QUOTE=Turnips5;32729319]Just got my lab book back
"excellent first experiment!"
I was expecting it to come back covered in red pen with "YOU TOOL THIS IS ALL FUCKED UP" written at the end
:D[/QUOTE]
ooooh. what were you doing?
my lab book is pretty much incredibly boring, the first 20 pages are just tables of numbers. hundreds of numbers.
[QUOTE=mike;32730283]ooooh. what were you doing?
my lab book is pretty much incredibly boring, the first 20 pages are just tables of numbers. hundreds of numbers.[/QUOTE]
Magnetic force experiments with Helmholtz coils, it was pretty confusing but my lab partner and I managed it okay. There were like 5 medium sized experiments to get through, we managed 4 but they don't expect everyone to finish all 5 when it's the first week, plus I totally did a sick error analysis which people seemed to like so time reasonably well spent I guess
You basically had a dipole on a spring in the middle of two Helmholtz coils and did various things to it, culminating in trying to measure its magnetic moment by plotting force against field gradient (they threw a lot of entirely new physics at us in that time, but they gave us all the formulae and everything so it's cool)
Today's lab session involved fucking around with an oscilloscope. It wasn't so much an experiment as it was a fuckfest of knobs and switches, but we got good results for 5/6 of them so...
Your lab book might be boring to you but I actually enjoy getting loads and loads of really good data quickly and plotting a sweet curve at the end of it. That doesn't happen too often to me.
The weirdest thing is that my physics teacher from school last year told us we were all good theoreticians but had shitty experimental technique :v: I still like the theory side of things more, but god damn are they letting it out slow at the moment, I haven't learned a single new piece of maths or physics in the theory lectures yet. :|
[editline]11th October 2011[/editline]
[img]http://i.imgur.com/Ck8Us.png[/img]
thanks, physics administrator, that's real helpful!
[editline]11th October 2011[/editline]
mike, I heard kinetics experiments are some of the most boring things ever
[QUOTE=Turnips5;32732512]Magnetic force experiments with Helmholtz coils, it was pretty confusing but my lab partner and I managed it okay. There were like 5 medium sized experiments to get through, we managed 4 but they don't expect everyone to finish all 5 when it's the first week, plus I totally did a sick error analysis which people seemed to like so time reasonably well spent I guess
You basically had a dipole on a spring in the middle of two Helmholtz coils and did various things to it, culminating in trying to measure its magnetic moment by plotting force against field gradient (they threw a lot of entirely new physics at us in that time, but they gave us all the formulae and everything so it's cool)
Today's lab session involved fucking around with an oscilloscope. It wasn't so much an experiment as it was a fuckfest of knobs and switches, but we got good results for 5/6 of them so...
Your lab book might be boring to you but I actually enjoy getting loads and loads of really good data quickly and plotting a sweet curve at the end of it. That doesn't happen too often to me.
The weirdest thing is that my physics teacher from school last year told us we were all good theoreticians but had shitty experimental technique :v: I still like the theory side of things more, but god damn are they letting it out slow at the moment, I haven't learned a single new piece of maths or physics in the theory lectures yet. :|
[editline]11th October 2011[/editline]
[img]http://i.imgur.com/Ck8Us.png[/img]
thanks, physics administrator, that's real helpful!
[editline]11th October 2011[/editline]
mike, I heard kinetics experiments are some of the most boring things ever[/QUOTE]
they are obscenely boring.
the only thing that made the one i did recently even remotely interesting is that it was quite a fun synthesis to make the compound we were studying.
[QUOTE=mike;32727790]this is pretty much exactly it.
you need to plot the natural log of the concentration against time.
just plotting concentration against time should give you an exponential curve for a first order reaction - it's possible your data range is so small you're not seeing much of the reaction lifetime and that's why it appears to be straight.
the last experiment i did involved a lot of kinetics - studying a reaction via IR. some of the reactions were so slow that i had to leave them running for 18 hours to get a useful data set.
here's the gist of it.
this is concentration (actually absorbance but beer-lambert law means for these purpose i can use them interchangeably) against time. as you can see, it's a curve! this a first order migratory insertion reaction.
[img]http://mikeh269.com/filedump/ir.png[/img]
by plotting ln(concentration) against time, this second plot is produced
[img]http://mikeh269.com/filedump/ir_1.png[/img]
in this one - you can see - as turnips made obvious in his post, the gradient is -k
so all you have to do is pop a negative sign in front of the gradient and ta-dah! you have your observed rate constant.[/QUOTE]
Thanks a million, man. I was looking at the rate laws yesterday in my book and wondering, "why do I need to plot the log of concentration against time to get a linear graph? It's ALREADY linear."
I appreciate the help, dude :) I posted for help on the Physics Forums too but surprisingly that place is fucking useless half of the time. I get faster and better replies here.
[editline]12th October 2011[/editline]
So just to make sure I understand here, we have:
-d[A]/dt = k[A]
From there you can transpose (I know differentials aren't really fractions but they can be treated like them) to get:
1/[A] * d[A] = -k * dt
And then integrating from there you'd get:
ln([A]) + b = -kt + c (where b and c are just constants)
Is that about right?
Why isn't there a constant of integration on the left hand side of the equation provided above (by Turnips5)? Is there a particular reason for that?
[QUOTE=sltungle;32740186]
Why isn't there a constant of integration on the left hand side of the equation provided above (by Turnips5)? Is there a particular reason for that?[/QUOTE]
I've no idea about chemistry, but if you have two constants you can add them together or subtract one from the other to make one new constant, if that's what you're asking :v:
[QUOTE=ThisIsTheOne;32740705]I've no idea about chemistry, but if you have two constants you can add them together or subtract one from the other to make one constant, if that's what you're asking :v:[/QUOTE]
Oh... duh. Wow, I can't believe I missed something that blatantly obvious :v: Massive fail on my behalf.
Geeze, I guess I'm just not in a uni-mood at the moment. Had my last class of first year yesterday so I'm completely out of it :v:
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