anonymous
  • anonymous
A large tank contains 100 liters of a salt solution which has a concentration of 0.25 kilograms per liter. Pure water is added to the solution at a rate of 80 liters per minute. At the same time, the well-mixed solution drains from the tank at 80 liters per minute. Find the amount S of the salt in the solution as a function of t: kilograms At what time is there only 12 kilograms of salt remaining in the tank? t= minutes
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Hi dichalao, Do you still need a solution?
anonymous
  • anonymous
Yes! Or just give me some clues first :)
anonymous
  • anonymous
What i have tried is to set up the first differential equation as : dS/dt= 25/(100+80t)(1-80t)

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More answers

anonymous
  • anonymous
You need to consider the physics of what's going on in a little bit of time,\[\delta t\]
anonymous
  • anonymous
and you need to consider what's actually changing. You're told 80 L/min enters the tank, and 80 L/min leaves the tank, so volume's the same. What's changing is the concentration, or even better, the mass, of salt in the solution.
anonymous
  • anonymous
You need to come up with an expression for the change in mass of salt in the tank as a function of time, then use that to get an expression for concentration and solve the system.
anonymous
  • anonymous
Yea i see the solution is diluting
anonymous
  • anonymous
So, let's consider what happens in that little bit of time, delta_t. No salt is entering the system, since what's coming in is pure water. Salt *is* leaving the system in the outflow. I said we need to consider mass, so, let the mass at time t be\[m(t)\] and that at time t+delta_t be\[m(t+\delta t)\]The difference between the two will be\[\delta m = m( t + \delta t) - m(t)\]
anonymous
  • anonymous
Hello
anonymous
  • anonymous
Hi Lok!!
anonymous
  • anonymous
Yea
anonymous
  • anonymous
The change in mass is the negative of the amount leaving (because these delta quantities are always read (final)-(initial), and since you have less at the end, the change will be negative). That is,\[\delta m = -Q \delta t C(t)\]where Q is the flow rate (liters/min) and C(t) is concentration at time t (this is why delta_t needs to be small, otherwise C(t) won't be valid).
anonymous
  • anonymous
Notice the units on the left-hand side and the right-hand side. LHS is (mass), and right-hand side is (volume)/(time) x (time) x (mass)/(volume). The units equate.
anonymous
  • anonymous
So what we have now is\[\delta m = m(t+ \delta t)-m(t)\]\[-Q \delta t C(t)=m(t+\delta t)-m(t)\]Divide both sides by the constant volume in the container, and by delta_t to get\[-\frac{QC(t)}{V}=\frac{C(t+\delta t)-C(t)}{\delta t}\]
anonymous
  • anonymous
because \[C(t) = \frac{m(t)}{V}, C(t+\delta t)=\frac{m(t+\delta t)}{V}\]
anonymous
  • anonymous
That expression above (above what I *just* wrote) should look familiar. Does it?
anonymous
  • anonymous
Yea
anonymous
  • anonymous
Sorry, my computer's acting up.
anonymous
  • anonymous
You can now take the limit as delta_t approached zero on both sides to get,
anonymous
  • anonymous
\[\lim_{\delta t \rightarrow 0}-\frac{QC(t)}{V}=\lim_{\delta t \rightarrow 0}\frac{C(t+ \delta t)-C(t)}{\delta t}\]\[\rightarrow -\frac{QC(t)}{V}=\frac{dC}{dt}\](the left-hand side is independent of delta_t, so it's just a constant in terms of taking the limit).
anonymous
  • anonymous
You now have a separable differential equation:\[\frac{dC}{C}=-\frac{Q}{V}dt \rightarrow \int\limits_{}{}\frac{dC}{C}=\int\limits_{}{}-\frac{Q}{V}dt\]
anonymous
  • anonymous
Thanks i can take it from here on
anonymous
  • anonymous
\[\log C=-\frac{Q}{V}t+c\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
I have one more question
anonymous
  • anonymous
ok
anonymous
  • anonymous
Suppose taht a_n>0, and lim n->infinity n^2a_n=0. Prove that sigma a_n converges
anonymous
  • anonymous
Since lim n^2a_n->0, so i assume n^2a_n behave like 1/n^p where p >0, if so a_n should be have like (1/n^(2+p), a_n is a p-series, with p >1, so a_n converges. Do u think this approach is reasonable?
anonymous
  • anonymous
You might have to leave it with me. I just got up and need to get ready for uni/work. I'll get back to you.
anonymous
  • anonymous
Sure :) Have a good one :)
anonymous
  • anonymous
Before I go, I want to confirm the sequence is\[n^2a_n\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok
anonymous
  • anonymous
i have a question regarding C(t) is concentration at time t (this is why delta_t needs to be small, otherwise C(t) won't be valid).
anonymous
  • anonymous
Do you mean C(t)*delta_t wont be valid? Or just C(t)
anonymous
  • anonymous
Well, C(t) is the concentration at time t, but that's always changing. We're looking at times t and t+delta_t, but only using C(t) to approximate the amount of mass that's leaving. This is valid in the limit, but if you leave everything as a finite difference (which is what you have before taking the limit), what you're really saying is\[-QC(t)/V \approx m(t+ \delta t)-m(t)\]
anonymous
  • anonymous
But because our increment in time was small, the CHANGE in C(t) over that time is assumed very small too.
anonymous
  • anonymous
So then all you need to do to find the mass is multiply it by volume flow rate.
anonymous
  • anonymous
I don't know how well this is being explained. We could have used differentials instead of delta notation, but I think understanding the approximations before taking limits is useful.
anonymous
  • anonymous
This is a good explanation, i understand everything you did
anonymous
  • anonymous
-Q is -80L/min?
anonymous
  • anonymous
No, Q=-80L/min
anonymous
  • anonymous
I took Q as the magnitude of flow, that's why I introduced the minus sign when setting up the equation.
anonymous
  • anonymous
You should end up with\[C(t)=\frac{1}{4}e^{-\frac{4}{5}(\min^{-1})t}kg/L\]
anonymous
  • anonymous
which makes sense, since as t approaches infinity, C(t) approaches zero...the salt is being drained away.
anonymous
  • anonymous
and at time 0, C(0) = 1/4 kg, which is what you've stated as an initial condition.
anonymous
  • anonymous
1/4 kg/L, I mean.
anonymous
  • anonymous
if i neeed to find the amount S of the salt in the solution as a function of t: i just need to multiply both side by 100? since V is constant?
anonymous
  • anonymous
right side*
anonymous
  • anonymous
Yes, (concentration) x (volume) = (mass) here.
anonymous
  • anonymous
got it~
anonymous
  • anonymous
If the volume flow rates were different either side, you'd have a very different situation to look at.
anonymous
  • anonymous
lol it will be super complicated?
anonymous
  • anonymous
Kind of...
anonymous
  • anonymous
where are you from btw
anonymous
  • anonymous
So, in these types of problems, look at what's actually changing, and try to come up with an expression that focuses on that. Try to formulate something you can turn into a differential equation in the end.
anonymous
  • anonymous
UK citizen in Sydney.
anonymous
  • anonymous
Where are you from? Are you male or female?
anonymous
  • anonymous
I am from US, a kid
anonymous
  • anonymous
:P
anonymous
  • anonymous
So is this school stuff?
anonymous
  • anonymous
yea, i am learning sequence/series and my friend is learning 1st degree differential equation
anonymous
  • anonymous
You are really good at problem solving
anonymous
  • anonymous
Is this for the international baccalaureate?
anonymous
  • anonymous
haha, thanks
anonymous
  • anonymous
international baccalaureate?
anonymous
  • anonymous
hey, I just found the answer to your other problem
anonymous
  • anonymous
haha do you have tips on how to approach a solution?
anonymous
  • anonymous
AMAZING !!
anonymous
  • anonymous
the stuff you're studying is covered in the international baccalaureate - it's an international diploma that can be used to get into universities around the world.
anonymous
  • anonymous
actually, hang on, I'll need to check it. I was looking to apply the Dirichlet test with a series 1/n^2...but I want to see if I can get around one of the assumptions
anonymous
  • anonymous
No it is just normal school work
anonymous
  • anonymous
I'll think about tips on how to approach a solution and get back to you. Really, all you have to do is determine what you're being asked to find, put down the information you have before you and build a bridge.
anonymous
  • anonymous
are you told that \[a_n \ge a_{n+1}>0?\]
anonymous
  • anonymous
No, what i posted is everything given
anonymous
  • anonymous
i wouldn't have thought about setting -Qdelta_t(Ct)=m(delta_t+t)-m(t) haha
anonymous
  • anonymous
Did you ask yourself what was actually changing in the system?
anonymous
  • anonymous
Concentration's harder to come to terms with than mass, since concentration is a ratio of quantities, whereas mass is something we have an intuitive feel for.
anonymous
  • anonymous
experience helps too
anonymous
  • anonymous
yea i know that the amount of salt in the container is lessening and the amount of salts going out is also lessening
anonymous
  • anonymous
How old are you ~
anonymous
  • anonymous
if you dont mind :)
anonymous
  • anonymous
20s
anonymous
  • anonymous
You are a genius i guess ..
anonymous
  • anonymous
like I said, experience helps
anonymous
  • anonymous
You should pass some experiences to me haha
anonymous
  • anonymous
I have been! =D
anonymous
  • anonymous
LOL True.. Are you still trying to find the solution to the problem?
anonymous
  • anonymous
yes, but I should be showering...I have an idea, but not much time to check it out.
anonymous
  • anonymous
take your time buddy, i have to go to class right now.. Thank you for helping out :) see you
anonymous
  • anonymous
ok
anonymous
  • anonymous
Damn, I figured it out just after you left. You can use the limit comparison test.
anonymous
  • anonymous
The version we need says, If \[\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=0\]then\[\sum_{n=0}^{\infty}b_n\]convergent implies\[\sum_{n=0}^{\infty}a_n\]is also convergent.
anonymous
  • anonymous
Now, you need to show that the sum of a_n is convergent, and we can consider for out 'b' series, \[\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}\frac{1}{n^2}\]which is convergent (p-series, p=2>1). Therefore, taking the ratio of a_n and b_n and forming the limit gives,\[\lim_{n \rightarrow \infty}\frac{a_n}{b_n}=\lim_{n \rightarrow \infty}\frac{a_n}{1/n^2}=\lim_{n \rightarrow \infty}n^2a_n=0\]by construction (i.e. by the setup of the problem). By the theorem above, your series is convergent.
anonymous
  • anonymous
HAHA got it !!!!
anonymous
  • anonymous
Yeah. Whenever I see a polynomial in the variable, I think ratio test, limit comparison or Dirichlet first.
anonymous
  • anonymous
A large tank contains 450 liters of a salt solution which has a concentration of 0.65 kilograms per liter. A salt solution which has a concentration of 0.8 kilograms per liter is added to the solution at a rate of 20 liters per minute. At the same time, the solution drains from the tank at 20 liters per minute.
anonymous
  • anonymous
It's going to be similar. Volume in the tank won't change again, which is useful. Try to form a mass balance like we did before.
anonymous
  • anonymous
I learned the limit comparison and ratio test just now haha
anonymous
  • anonymous
Ah..well, that may have been why you couldn't do it ;)
anonymous
  • anonymous
There are a couple of versions, depending on how the limit a_n/b_n behaves. I had to pick the right one.
anonymous
  • anonymous
So this time, there is outflow and inflow of salt.
anonymous
  • anonymous
Yes
anonymous
  • anonymous
What does your mass balance look like?
anonymous
  • anonymous
i am trying to figure it out
anonymous
  • anonymous
There's a general formula for changes of 'stuff' in a system (for further reference):\[\Delta = i-o+ \gamma - \delta\]
anonymous
  • anonymous
Change = in - out + generated - destroyed
anonymous
  • anonymous
This isn't a chemical or nuclear reaction, so nothing is generated or destroyed.
anonymous
  • anonymous
So gamma = delta = 0
anonymous
  • anonymous
inflow=Qdelta_t(C(t)) outflow=-Q(deltat)(C(t+deltat)?
anonymous
  • anonymous
Nearly...
anonymous
  • anonymous
Your inflow isn't changing with time - it's fixed at 0.8kg/L * 20 L/min
anonymous
  • anonymous
First, in a unit of time, delta t, how much salt is entering the system?
anonymous
  • anonymous
0.8(20)(delta t) (delta t in min of time)?
anonymous
  • anonymous
Yes :)
anonymous
  • anonymous
Just set it equal to some constant, a, or something.
anonymous
  • anonymous
You don't have to, but when you have numbers floating around, it can stuff the algebra if you're not careful.
anonymous
  • anonymous
so it will just be a-outflow=m(t+delta t)-m(t) and then everything goes the same way?
anonymous
  • anonymous
Yep
anonymous
  • anonymous
give a round of applause to loki haha
anonymous
  • anonymous
but what about the outflow...what's your expression?
anonymous
  • anonymous
LOL ouchie
anonymous
  • anonymous
might want to save the round of applause for a bit =D
anonymous
  • anonymous
You might want to write your inflow as\[a \delta t Q\]where a is your constant concentration and Q is the inflow rate.
anonymous
  • anonymous
outflow= 20(delta t)(Constantly changing concentration) :(
anonymous
  • anonymous
You might want to write your inflow as\[a \delta t Q\]where a is your constant concentration and Q is the inflow rate.
anonymous
  • anonymous
Your outflow is write. Don't be :(
anonymous
  • anonymous
right
anonymous
  • anonymous
not 'write'...
anonymous
  • anonymous
LOL you need more sleep
anonymous
  • anonymous
i know
anonymous
  • anonymous
concentration = mass over volume mass=(mass initial+mass inflow-mass outlfow)?
anonymous
  • anonymous
Now, what does\[m(t+ \delta t)-m(t)\]equal?
anonymous
  • anonymous
(constant inflow)a - (changing outflow), b delta t(C(t+delta t) ?
anonymous
  • anonymous
lol, you're giving me a headache...I'll write it out
anonymous
  • anonymous
sorry.. i am not floating with you :(
anonymous
  • anonymous
\[m(t+\delta t)-m(t)=a \delta t Q-C(t) \delta t Q=(a-C(t))Q \delta t\]
anonymous
  • anonymous
the change is just *in* - *out*, as per that Delta equation I put up.
anonymous
  • anonymous
give me some time to digest haha
anonymous
  • anonymous
a is constant concentration for inflow, while C(t) is changing concentration
anonymous
  • anonymous
Yep.
anonymous
  • anonymous
delta t is intended to cancel the time component in Q?
anonymous
  • anonymous
Yes :)
anonymous
  • anonymous
got it now i think .. haha
anonymous
  • anonymous
i got to go grab food now...otherwise i have to eat myself tonite .see you ~~thanks !!!
anonymous
  • anonymous
Ok. I'll finish it.
anonymous
  • anonymous
Divide both side by V, the volume of the system to obtain\[C(t+\delta t)-C(t)=\frac{Q}{V}(a-C(t))\delta t\]and divide both sides by delta t\[\frac{C(t+\delta t)-C(t)}{\delta t}=\frac{Q}{V}(a-C(t))\]
anonymous
  • anonymous
Take the limit as delta_t approaches zero on both sides to get\[\frac{dC}{dt}=\frac{Q}{V}(a-C(t)) \rightarrow \frac{dC}{a-C}=\frac{Q}{V}dt \rightarrow \int\limits_{}{}\frac{dC}{a-C}=\int\limits_{}{}\frac{Q}{V}dt\]
anonymous
  • anonymous
You should get\[C(t)=a-C(0)e^{-\frac{Qt}{V}}\]
anonymous
  • anonymous
Oh wait...I forgot about the salt already in there! This is what happens when you do it on the fly. Go eat.
anonymous
  • anonymous
I'll write something later.
anonymous
  • anonymous
Okay, this is much easier when you do it on paper. This site needs something like a whiteboard. I'll write it up and scan.
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Wow! you are the most amazing person in the world haha! So for the first question do i also need to add the original amount of salt to the equation?
anonymous
  • anonymous
You know, I've forgotten the first question already. I don't think so since if you put t=0 into the solution, you should get the initial amount in the tank. Just check.
anonymous
  • anonymous
Oh yea. it is right
anonymous
  • anonymous
Always check the initial and terminating conditions (t=0, t-> infinity) to see if your answer makes sense.
anonymous
  • anonymous
LOL i am too lazy to do so when there is a smart guy in the house :)
anonymous
  • anonymous
was*
anonymous
  • anonymous
Yeah, flattery will get you everywhere.
anonymous
  • anonymous
LOL
anonymous
  • anonymous
I am just fooling around ~~
anonymous
  • anonymous
I know :P
anonymous
  • anonymous
Is there a way that i can use to find square root super fast
anonymous
  • anonymous
What do you mean? By hand?
anonymous
  • anonymous
yea or better in mind
anonymous
  • anonymous
Newton-Raphson method is a numerical method for extracting solutions from equations. It has something called, "quadratic convergence" which makes it very fast. Plus, it's easy to use.
anonymous
  • anonymous
I explained it to someone last week. I'll see if I can find it. If not, the result for square root is \[x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\]
anonymous
  • anonymous
You start with an initial guess, x_0, and the number it spits out on the LHS will be x_1, which will be closer to the solution. But that means that this is a BETTER guess than before, so we can plug the result into the RHS to get an EVEN BETTER guess on the LHS and so on...you repeat for infinity, but in reality, it only takes a couple of cycles to get something close.
anonymous
  • anonymous
Oh, that's only for the square root of 2.
anonymous
  • anonymous
LHS= left hand sum?
anonymous
  • anonymous
For\[y=\sqrt{a}\]it would be\[x_{n+1}=\frac{x_n}{2}+\frac{a}{2x_n}=\frac{1}{2}\left( x_n+\frac{a}{x_n} \right)\]
anonymous
  • anonymous
LHS = left-hand side.
anonymous
  • anonymous
So, start with an initial guess for sqrt(2) of 1.5. Plug 1.5 into the RHIS of the equation and you should get an answer, 1.4166666...
anonymous
  • anonymous
How you got a/(2X_n)
anonymous
  • anonymous
Plug that answer into the RHS again and you get on the LHS: 1.414215686...
anonymous
  • anonymous
Already it's getting close.
anonymous
  • anonymous
is this something similar to newton's method?
anonymous
  • anonymous
Yes, it's the same.
anonymous
  • anonymous
You've used it?
anonymous
  • anonymous
My professor talked a little bit about it last quarter
anonymous
  • anonymous
Oh do you have any cool math tricks that you can show me haha
anonymous
  • anonymous
I'd have to think... Some tricks, you can add zero or multiply by 1 and not change a number...so you can extend this idea when it comes to solving a lot of problems since 0 and 1 can be written in many convenient forms. Example \[\int\limits_{}{}\frac{u}{1+u}du=\int\limits_{}{}\frac{u+0}{1+u}du \int\limits_{}{}\frac{u+1-1}{1+u}du\]\[=\int\limits_{}{}1-\frac{1}{1+u}du=u-\log (1+u) + c\]
anonymous
  • anonymous
O.o magic~
anonymous
  • anonymous
Why multiplying two negatives gives a positive...
anonymous
  • anonymous
cuz ..mathematicians say so LOL
anonymous
  • anonymous
\[-2 \times -3 \]\[= -2 \times -3 +0 \]\[=-2 \times -3 + 2 \times 0 \]\[=-2 \times -3+2 \times (3+-3)\]\[=-2 \times -3 + 2 \times 3 + 2 \times -3\]\[=-3 \times (2+-2)+ 2 \times 3\]\[=-3 \times 0 + 2 \times 3\]\[=0+2\times 3\]\[=2 \times 3\]
anonymous
  • anonymous
Everyone takes it for granted. Hardly anyone ever checks or bothers to figure out why.
anonymous
  • anonymous
Yea this is how education been conducted these days
anonymous
  • anonymous
Were you math major? i hear that only math majors actually get into this kind of proofs, for example proof 1+1=2
anonymous
  • anonymous
yes. pure mathematics and physics.
anonymous
  • anonymous
COOL so you can help me with physics too haha!!
anonymous
  • anonymous
why do you have classes so late?
anonymous
  • anonymous
I had no choice, i picked my class on the last day of classes registration because i didn't have many credits when i entered this school.
anonymous
  • anonymous
Hassle.
anonymous
  • anonymous
did you go to University of Sydney ?
anonymous
  • anonymous
Partly.
anonymous
  • anonymous
Postgrad. stuff at Oxford
anonymous
  • anonymous
yea i had to fight for classes
anonymous
  • anonymous
Isn't it a school though?
anonymous
  • anonymous
LOL that explained why you have 129 fans here haha
anonymous
  • anonymous
Have you heard of University of California, Davis?
anonymous
  • anonymous
Ah, yes.
anonymous
  • anonymous
I don't get the American system.
anonymous
  • anonymous
yea i am attending Davis ..since i couldnt get into UC berkerley
anonymous
  • anonymous
It is annoying and difficult to deal with..
anonymous
  • anonymous
Can't you try and transfer, or do postgraduate stuff...or even try for another course?
anonymous
  • anonymous
I can transfer after my sophmore year , but it is really hard to transfer from one college to another college at the same level
anonymous
  • anonymous
It is much easier if you try to transfer from a city college to a University
anonymous
  • anonymous
Your system's quite weird. It seems almost punitive. In the Commonwealth, it's so much easier. We can transfer easily and get exchange placements.
anonymous
  • anonymous
brb
anonymous
  • anonymous
Yea, i had an internship at a company last summer. One of my colleagues is an immigrant from UK. He told me that the school system here is much more complicated than the one in UK
anonymous
  • anonymous
Hell yeah.
anonymous
  • anonymous
I think it has to do with the fact all universities in the Commonwealth (well, most) are public, and there are links between schools that act as 'feeder schools', although you can choose where you go.
anonymous
  • anonymous
The governments ensure as much as possible there is enough money to allow people places in courses they want (assuming the academics think they can handle the work).
anonymous
  • anonymous
So you actually go to lower division schools, elementary, middle and high school, that help you to get into the university you want?
anonymous
  • anonymous
We have nothing like that here. NOw they are evening cutting classes because of the recent budget cut
anonymous
  • anonymous
No, it's not like that. Because the government funds everything, the standards of all the universities have to be the same, which means most people are happy studying at a university that is close geographically to their home. There's no (as far as I'm aware) discrimination in selection.
anonymous
  • anonymous
If you want to do research, then you have to get yourself into one of the major ones here. At Sydney, I was taught and worked with people who were at Oxford, Cambridge, Yale, etc.
anonymous
  • anonymous
I think there was a mathematician who worked at Stanford too.
anonymous
  • anonymous
Where did you do your undersgraduate
anonymous
  • anonymous
Sydney.
anonymous
  • anonymous
There's a guy in my old physics school there who worked at NASA after graduating. He's back now. I think he discovered something to do with black holes.
anonymous
  • anonymous
What are you going to do when you finish your studies? I get the impression with the US system that you guys study for a general degree and then use it (if you want) to get into a professional school.
anonymous
  • anonymous
OMG. You are surrounded by ...genius man so jealous
anonymous
  • anonymous
I dont know...what i am gonna do.. I am majoring in mechanical engineering.. i dont know what i can do with it
anonymous
  • anonymous
Well here, mechanical engineers make a lot of dosh.
anonymous
  • anonymous
All engineers do.
anonymous
  • anonymous
Maybe you could focus on making moolah?
anonymous
  • anonymous
LOL you mean i design and produce my own moolah?
anonymous
  • anonymous
I mean money ;)
anonymous
  • anonymous
yea that s what i meant
anonymous
  • anonymous
LITTERALLY producing moolah
anonymous
  • anonymous
Sorry, subtleties are lost online.
anonymous
  • anonymous
Why not, your Government does.
anonymous
  • anonymous
*?*
anonymous
  • anonymous
LOL
anonymous
  • anonymous
How about you .what you going to do after post grad
anonymous
  • anonymous
Well, I'm taking a break and teaching mathematics at a grammar school in Sydney. I'll get back into research a bit later.
anonymous
  • anonymous
I C
anonymous
  • anonymous
Do you think the earth's magnetic field is gonna turn 180% around
anonymous
  • anonymous
i asked about the international baccalaureate before because that's part of what I'm teaching.
anonymous
  • anonymous
Re. magnetic field, I think it's supposed to flip, and can do so anytime 'soon', but 'soon' is in astronomical terms, so it could be like, forever.
anonymous
  • anonymous
Oh
anonymous
  • anonymous
What kind of research are you working on
anonymous
  • anonymous
gtg i will ttyl man..Thanks for all the help~:)
anonymous
  • anonymous
No worries :D

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