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anonymous
 3 years ago
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anonymous
 3 years ago
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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Delta y=\Delta y_1+\Delta y_2=\frac 12 a(\frac{L}{v_0})^2+L +D \] \[x_1=v_0t\] \[t=\frac{x_1}{v_0}=\frac{L}{v_0}\] \[\Delta y_1=v_0t+\frac 12 at^2\] \[\Delta y_1=v_0\frac{L}{v_0}+\frac 12 a(\frac{L}{v_0})^2\] \[\Delta y_1=\frac 12 a(\frac{L}{v_0})^2+L\] now outside o the acceleration plates \[x_2=v_0t\] \[t=\frac{x_2}{v_0} =\frac D{v_0}\] \[\Delta y_2=v_0(\frac D{v_0})\] Here is where I'm stuck: \[ma=qE\] \[a=\frac {qE}{m}\] \[KE=U\] \[\frac 12 mv_0^2=qV_acc\] \[v_0^2=\frac {2qV_acc}{m}\] now i Have \[\Delta y=\frac{qE}{m}(\frac{Lm}{2qV_acc})^2+L +D\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but my equation should look like \[\Delta y=\Delta y_1+\Delta y_2=\frac{EL}{2V_{acc}}\left( D+\frac L 2\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I probably made a mistake somewhere, because my m's aren't cancelling out....and some other stuff too I guess

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0feel free to write on here fellas:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y_1=v_yt+\frac 12 at^2\] is there a y velocity within the plates?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0please stop posting @CaritaDeAngel

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0can I type today on OS (testing)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0please continue posting @TuringTest

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0Okay let's start with the electron gun ... dw:1362064370315:dw

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0oh goodie, I can :) ok, what is the exact info you are given? just what's in the pic of the CRT you showed me?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0so we are supposed to get the initial xvelocity from the voltage generated by the AC, which you say is 500V. I don't know where you get that number, but if it is correct answer @experimentX 's question

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0yep .. it's called accelerating potential.dw:1362064638627:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not worried about that anymore. do we know that the V_acc is exactly 500V or is there some uncertainty. should i mention that there is some uncertainty

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0this is how you accelerate electrons http://en.wikipedia.org/wiki/File:WaterCooledXrayTube.svg No it doesn't matter for theory. Just for error analysis ..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for my analysis I mean

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would you like to see my graph?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0hmm .. sure!! upload it

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0hmm ... looks like you could that data to calculate errors.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0I"v never been good with error analysis ... but let's see what I can do.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no need to I guess....the xcel program did that already

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0he who?? I know couple of error analysis method like Standard ERror

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0never mind @experimentX I'm running out of time....I'm gonna make changes and submit what I have.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0hmm ... what methods are asked to use?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0he said to just describe the method but don't go into too much detail. I described it already and I just need to tweek a couple of things in my report to make it look better. The equations are fine I guess, I'll just use the final displacement equation without too just some derivation steps. Thanks for you help though. I'm not gonna learn anything new in the next 15 mins

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0okay .. I'll send you the details of derivation if you want ... within next 15 hrs. (optional: if you want)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do I insert a picture in latex?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0wow ... i don't know ... maybe these guys know http://tex.stackexchange.com/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks everyone...I learned my lesson, but it's not like I was playing video games, I've been busy...maybe I should quit my job

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0wow!! ... looks like I am the only free loader at home here.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, you don't work?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0Not since i joined grad school.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I see, anyhow gotta go....thanks though @TuringTest and @experimentX ....I'll work on my time commitments

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0well ... didn't do much except troll around. Maybe next time .. :P

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362138326802:dwdw:1362138537113:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362138832951:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0So until now we have expression for \( v_x\) and \(a \) ( and 'a' is perpendicular to the original path) dw:1362139377040:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362139616525:dwdw:1362139773642:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0Your this expression is incorrect \[ \Delta y_1=v_0\frac{L}{v_0}+\frac 12 a(\frac{L}{v_0})^2 \] \(v_0 \) is zero by default ... no ycomponent.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362142311512:dwdw:1362142342862:dwdw:1362142369122:dwdw:1362142375743:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \Delta y = \Delta y_1 + \Delta y_2 = \frac{EL^2 }{4V_{acc} } + \frac{EDL}{2V_{acc}}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0in my equations read d as L ... rest are same.
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