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Comments:
<0> http://www.ichblog.eu/content/view/60/1/
<1> Title: Information Clearing House Blog - The Unthinkable: The US- Israeli Nuclear War o ...
<0> night
<2> night
<3> hmm, if you have two counterspinning rigid bodies in free space with a mechanism that can lock the two together suddenly without friction, balanced in such a way that the resulting motion is none, can the energy be lost?
<4> good question
<5> of course it can
<6> the energy isn't lost
<6> it just goes into deforming the bodies
<6> the kinetic energy is lost
<7> anyone feeling particularly knowledgeable on quantum tunneling? (resonance effects specifically)
<6> I wasn't aware there was resonance involved with tunneling
<7> well,,, there is
<6> uh cool
<7> if say you have a double barrier
<6> delta function barrier?
<5> how did you make it ?
<8> anyone here use pgplot ?
<7> microacg, ***ume U(x) = 0 for all x other than x= 0-20 and 30-50 (arbitrary atomic scale units)
<7> there will be some energies of an incident electron where transmission probability is 1
<7> these are resonant energies
<6> ah, that's more in depth than I studied it
<7> me too
<7> unfortunately i must write a report on it for wednesday
<5> energy is conserved
<5> that's all you need to know :P
<7> that is not
<5> ok
<5> F = ma
<5> momentum is conserved
<5> now that is everything
<5> although really momentum is conserved is covered by F=ma
<6> I don't think it is :/
<9> hey guys i need help with this question
<9> A 900kg boat is traveling at 70km/h when its engine is shut off. The magnitude of the frictional force f of k between boat and water is proportional to the speed v of the boat. Thus f of k = 60v, where v is in meters per second and f of k is in newtons. find the time required for that boat to slow down to 35 kmh in seconds
<9> i've keep getting the wrong answers
<9> anyone ?
<9> i keep getting 15
<9> but it isn't right
<10> Tried energy?
<9> what do you mean by energy?
<11> forget energy
<9> then..
<10> You know how much energy has to be burned off as friction because you know the initial and final energy.
<11> what do you get for F_net(t)?
<9> tis all about newton's second law and property of friction
<11> xtmdster: no, that won't work
<10> Just guessing, I didn't get a number.
<9> f net is 0
<9> its equilibrium , engine is stopped
<11> [set]: its not zero, if it was, the boat would keep coasting
<11> at the same velocity
<9> but the engine of the boat is shut down
<11> so we have the force of friction on the boat, in this case it is the net force
<9> ok h/o
<9> ok
<11> so you have F_net(v)=60v
<9> and how do you get the frictional force
<9> yes
<11> so that gives you an acceleration
<11> so ma=60v
<11> then integrate that with respect to time
<9> what do you mean by that last line
<11> you don't know how to integrate?
<9> yes i do
<9> ok
<11> well time is your variable, in this case
<11> so it makes sense to integrate with respect to it...rather than somthing else...
<9> mm
<11> so a=dv/dt
<11> then m*dv/dt=60*v
<11> so m*dv/v=60*dt
<11> and you can integrate
<11> to get v as a function of time
<9> ooh
<9> thx
<11> [set]: that relation shows up ALOT in physics
<11> [set]: so remember it
<9> I see, will do
<9> i keep getting 7.5 but it wasn't correct
<9> sry for the time interval.. i was eating
<11> what is your v(t)?
<11> and make sure your signs are correct
<9> v(t)= 9.2*7.5
<9> v=9.2 and time is 7.5
<11> .....
<11> well for one it should be an exponential....
<11> % DSolve[ y'[x]==-60/m*y[x],y[x],x ]
<11> stupid mbot...
<11> [set]: what is the integral of 1/x ?
<9> logx
<9> or ln if you like it this way
<11> yeah, so do the integral again...
<9> what was this all about 23:12 JabberWalkie >> % DSolve[ y'[x]==-60/m*y[x],y[x],x ]
<9> is that supposed to spit out the integral?
<11> yea, but its not working
<9> yeah i **** with integral , still have much to learn on it
<9> when im taking calc i guess i'll learn it
<12> anyone around to help with some simple 3d kinmatics?
<0> morning // Jreggelt!
<3> zorlac, try
<12> um
<12> say you hit a ball 188 meters, at a 45 degree angle .9 meters above ground level
<12> and i need to find the initial velocity, how would i go about doing tihs
<3> ***uming no air you have a constant acceleration scenario and the formulae you have for that applies. so it's a matter of examining each and see what that gives you in each and whether that will give you the answer
<3> since there is no air then constant velocity applies to the horizontal speed component
<3> draw on paper, find formulae, think
<3> I could figure it out but you will probably learn less that way
<12> I'm confused as to what formula to use
<3> those involving 'constant acceleration'
<3> draw on paper, find formulae, think
<12> do you know how i would find the velocity of the y axis?
<3> by thinking :)
<12> i've been thinking for like 4 hours
<3> you no doubt have a book which has formulae under the title of 'constant acceleration'
<3> the constant acceleration is of course gravity
<12> yes
<12> the accerlation on y is -10
<3> at the apex vertical speed is 0
<3> s = a t^2 might be involved
<12> thats distance, right?
<3> yep
<3> in a constant acceleration situation
<12> would the y axis have an inital velocity?
<13> the 45 degree angle would suggest yes
<3> the would be the same
<3> they*
<3> x and y
<12> oh i see
<12> if they are the same i guess that means i can set them equal toe ach other...
<3> yes if useful
<3> I think what you might want to do is to formulate a formula for how long it would go as a function of the initial velocity and angle. then equal that to 188
<3> and solve
<12> hmm
<3> imagine you knew the initial velocity and call that V0
<12> yeah
<3> if you had that how far would it go. then equate that to 188
<12> 188 = .5at^2 + V0t
<12> thats correct, right?
<3> don't think so
<12> why not?
<3> s=att only applies to the situation where initial (or end) velocity is 0 so you can only use that for half the way
<3> and only for the y part. the x is a constant velocity situation
<3> you find the time using the y part. then the distance from the x part
<3> s=v*t once you know the time
<3> s=v*t applies to constant velocity situations
<12> finally got it
<3> excellent
<14> http://xkcd.com/c114.html
<3> woe he that would pick on the humble and scared
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