infinity..?..a professor of mine defined it by this statement which is pretty easy to understand.
"pick a number, infinity is a number always greater than any number so chosen."
so ultimately infinity is a concept number...our way to imagine the end of a number line, or increasing sequence of...
nope..we are heavier during day, you'd have been correct provided there was no MOON, the moons gravitational pull is stronger than the sun's ,even when they're on opposite sides of the planet , for the numerical values, they're on google just feed em in the formulae and get your values.
its not always true that friction is`zero in ideal rolling, but two things are implied here on saying "PURE ROLLING OF A RIGID BODY"
1. the Kinetic friction acting is zero
2. Static friction shall act at the sole point of contact having a finite value decided by the forces acting on the...
the "force" and "equal and opposite force" is called action reaction pair and the body moves because the two forces act on different bodies
eg. if u push a cart u apply "action"/"force" on the cart while, the "reaction"/"equal & opp. force" acts on you...
you must have felt it sometime didn't...
but why should the universe be finite(observable one is no doubt)..? why constrain it to be so, it would be like repeatin speed of light mistake-->"that there's a special referece frame for a special event"
its really really dicey topic, isuggest we for now stay cosied up with multiverse one...
only discontinuous..?
try, f(x) =[1/x]...where...[y] is the greatest integer less than or equal to y or as you would call, floor(y)...f(1/n) leads to a jump discontinuity one you can never fix, and hence an implied non-differentiability.
What parties..?..(elaborate please.)
answer any 1 simple question,
Q.what is the probability that on choosing any two numbers the two are relatively prime(no common factor)
Q.In an automotive factory, the vehicles are numbered serially, you a common person on an average spot the vehicle with...
Agreed..but for a moment let's have the desert isle picture(we don't know anything but what we see about something)
...or better let the person who posed choose, which would be far wiser than to hav a "whose the better wildcat" thing
we both have given our veiws,answers et cetera to a...
I agree Tedjn to a certain extent though, i still stick to my discrete interpretation though, as the removal of caffiene from the body(urination) occurs at the period of one hour, that i consider as "caffiene eleimaination complete", but you have clarified the second model to me which "the Chaz"...
M=\oint\rho \partial V
...it isn't allowing latex preview so there's supposed to be a volume element
\partial V in there in case its missing.
for a constant density case,
M =\rho V
the rest i believe you can figure out yourself...through the equation
@Liparulo,
please make the problem statement more clear to me, and i think you must have arrived at the solution by now,(there is no effect after 14.36 hours)..and its quite the time the thread be closed so it won't be spammed, for any further details on the solution you may contact me by...
ok, i have 1 question.
so when after a coffee/any drink do you immediately, go to the bathroom and spend the rest of your life pissing..? or do you visit the loo after a certain period of time..
look kid, i took in a sane assumption that the person is human..and no caffiene seperating...
the point is that the guy who having the cola is taking it on an hourly basis, so we must calculate it as a discrete-input function...
now we'd require a DE to find the solution only if the guy has a cola dispenser installed in his bathroom, from which he continuosly drinks while excreting...
to end it you notice that, its the sum of a GP, probably your teacher didnt expect you to know that...
c(3) is your answer, then find out after how much time c(3) will decaty below 20mg
c(3)=147.843mg of caffiene(verify answer please)
hence solve by sayin tha after c(3) is achieved you lose k th...
ohk its like this c(n) represent caffiene amt. in ur body,assuming u are taking cola for the first time in your life and consumption starts @n=0, n number of hrs passed since u drunk cola(all values in mg, unless mentioned as something else)
"a" is caffiene supplied per drink, k is the fraction...
Euler maclaurin won't fit easily here, its for sums that can be approximated to integrals, if I'm not wrong,
here the domain of the sum is really small, plus due to the square root makes it worse (see graph of 1/sqrt(x), it apparently ceases to fall/falls very slowly after a certian interval)...
try looking at it like this,
time rate of fall(caffiene conc.)= 13% of present conc.
{-d/dt}[C]= .13C...
{-d/dt}=rate of fall...if u haven't guesed by now.
this approach is although wrong let me tell you since you mentioned that the caffiene elimination occurs on an hourly basis.
it's a...
Ah yes, my apologies ,its that i get a bit excited when a of ideas cross my mind(which it did in this thread), which shows up as you tried to read, spellchecker, well let me handle that part,
well the "multipole theorem" is just an idea I am proposing of how to think of the beginning of a mag...
ahem, but why do you need a godforsaken invisibility cloak, u'd hav to loot the bank to pay for it... :|
you could just use your brains and a bunch good processors to make a lumpsum transfer, study codebreaking, not classical E&M for this...:)
the crux here is that there will be not net change in the system+surroundings energy content,the following transforms occur______
PE---->KE...under strain
PE/KE----> chemical energy(very small part) + heat(non-frictional+frictional)...under strain/after rupture respectively.
a part of non...
the reply isn't genuine its just a theory I am proposing, i am yet to get down to the transform equations(just out of high school need some time)
normal to wire as in if u released an electron u kept beside the wire, you would observe that the e- in a complex way moves towards the wire, but if u...
Still its simple, since the spring is not stretched it will begin to dissolve as a whole rupturing arbitrarily, and slowy, because the energy if the system is maintained constant, there being no external parameters involved (as an assumption). so the spring will dissolve but rupturing very...
I wish to Quote the multipole theorem, from what brainstorm has said,
thenet mag. field is the result of multiple "multipoles" self aligning in a certain direction causing this "magnetic field" due to moving sets of multipoles and think of it thi way,
suppose u keep an electron near the wire and...
Destructive interference(DI) does occur, it will have to but consider your self or your ears how far apart are they even if one of ur ear is at the point of D.I the sound from other ear masks it
and to "hear DI occur" the detector has to be at the exact point, considering human dimensions it is...
oh by the way if you haven't noticed PD across a pure conducter, saubhik is zero, hence the resistance is shorted out(PD across it also zero, if you try looking through it via current law) hencethe current through resistance is also ZERO.
problem solved..:P
can't believe everyone except 2 people...
i beg to disagree with tiny tim,loop law is for the ideal situation(what u n i hav learnt),
so if u want to prove it rigorously, no I am not giving full solutions(it'll kill me) to you conserve energy through different paths or more OBVIOUSLY start by saying that "a point in any circuit is at...
here it is:
\vec{F}_{consevative}= \vec{\nabla} (U)
or for a one dimensional case,
\vec{F}_{consevative}=-\frac{\partial}{\partial x} U
so are you preparing for IIT-JEE or are you through with it..(i gave JEE2010 waiting for results)?
conservative forces have many forms, you can write one of them as
(qualitatively for understanding the orm is excellent, but i don't want to get into a fist fight with nabla operator it makes things worse to understand so don't assume it as general.),
\vec{F}=f(|r|)\hat{r}
when you take the...
Well let's give it a try,i don't know a damn about adiabatic theorem so let's try not to think about any theorem already quoted,but get to standard thermodynamic adiabatic process there your "parameter lambda",i expect it to show up as somehow related to "gamma, the ratio of specific heats" We...
i agree but wouldn't it be better to have a take that symmetries exist for the sake of have least possible value of interaction energy?
(i am no scholar on this topic(just outta high school) rather this is but a random guess from generally obseved phenomenon.)
quit the math people its real simple to explain from a change of frame...
suppose you are the standing in the lab observing currnt flow through a wire,we'll call this POV the lab frame.
let the electrons point-of-veiw be the e-frame
now in the e frame at any point which is at rest for the...
Your attempt is correct but you missed somethingthat for a spring if you take natural length as the datum, the force on change in length is given as:
\vec{F}= -k \vec{x}
and hence work done by a spring against external forces
W_{s}=\int\vec{F}\vec{.dx}
over the required limits
in...
there you go its true.
the spring would first at most probably the point with greatest stress(midpoint) and then PE "stored" woul go inot heat,sound, and KE which is dissipated most of the PE is lost barely any is used for the reaction.
To put in complicated terms..
"consider the electron in the lab frame... when it moves you record a current and a mag. field..
on move with the electron's reference... you can see ONLY a time dependent electric field..
all electrons in the current will generate a superimposed field that...
but did it work...?the solution is the exponential one..f(s)=As^{2}
im just a high school kid i may be wrong after all..?
do i mail u my solution to this one..?
f(s)=As^{2}
D_{2}(e^{As^{2}})= (A+2(As)^{2})e^{As^{2}
"D" is d/dx and D2 is 2nd derivative
so we get our condition that...
ohyeah..!..i got the solution its really simple...
you don't even need 2 do a self torture by the infinite series..
didn't you try inspection..??
as in a function giving back on two diferentials: (depressed quadratic)*(dependant function)..with no -ve coefficients...??..doesn't it seem too...
OK apologies for my vague and torturous answer...xDD
(hey can u help me solve the Schrödinger the ie colatitude and radial equations I am stuck badly..??)
finally u have your answer,Thaakisfox has enlightened us(me at least)..
@mathy_girl
the infinite series is domain independant,flexible, mathod to solve a DE but it is a very painful task
did you try substitutions for V(try thinking a kind of function that coul give this PDE.?my claim is it should be the product of three fuctions..in x,s,t)
wat do u say HallsOfIvy.?
@cipher42..pleasez simplify..the answer is
the bounds sre from -infinity to infinity
and the answer for electric field(gauss's law) is
\vec{E}=\frac{{2k}{\lambda}}{d}
where..
d= distance of point fom centre of the infinite wire
{\lambda}=linear charge density
k= dielectric constant of the medium...
find the electric field..by gauss' law..create a gaussian surface..
OR choose "dl" element at "l" from the centre/midpt of the wire..and find net field (dE)due to the element and integrate it with proper bounds/limits
it will have to be normal to the wire as each point on the infinite line is...