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\[\Delta y=\Delta y_1+\Delta y_2=\frac 12 a(\frac{L}{v_0})^2+L +D \] \[x_1=v_0t\] \[t=\frac{x_1}{v_0}=\frac{L}{v_0}\] \[\Delta y_1=v_0t+\frac 12 at^2\] \[\Delta y_1=v_0\frac{L}{v_0}+\frac 12 a(\frac{L}{v_0})^2\] \[\Delta y_1=\frac 12 a(\frac{L}{v_0})^2+L\] 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 \[\Delta y=\frac{qE}{m}(\frac{Lm}{2qV_acc})^2+L +D\]
but my equation should look like \[\Delta y=\Delta y_1+\Delta y_2=\frac{EL}{2V_{acc}}\left( D+\frac L 2\right)\]
I probably made a mistake somewhere, because my m's aren't cancelling out....and some other stuff too I guess

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Other answers:

feel free to write on here fellas:)
\[y_1=v_yt+\frac 12 at^2\] is there a y velocity within the plates?
please stop posting @CaritaDeAngel
can I type today on OS (testing)
please continue posting @TuringTest
Okay let's start with the electron gun ... |dw:1362064370315:dw|
oh goodie, I can :) ok, what is the exact info you are given? just what's in the pic of the CRT you showed me?
yes
so we are supposed to get the initial x-velocity 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
yep .. it's called accelerating potential.|dw:1362064638627:dw|
I'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
this is how you accelerate electrons http://en.wikipedia.org/wiki/File:WaterCooledXrayTube.svg No it doesn't matter for theory. Just for error analysis ..
for my analysis I mean
would you like to see my graph?
hmm .. sure!! upload it
hmm ... looks like you could that data to calculate errors.
I"v never been good with error analysis ... but let's see what I can do.
no need to I guess....the xcel program did that already
he who?? I know couple of error analysis method like Standard ERror
never mind @experimentX I'm running out of time....I'm gonna make changes and submit what I have.
hmm ... what methods are asked to use?
he 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
okay .. I'll send you the details of derivation if you want ... within next 15 hrs. (optional: if you want)
how do I insert a picture in latex?
wow ... i don't know ... maybe these guys know http://tex.stackexchange.com/
thanks 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
wow!! ... looks like I am the only free loader at home here.
wait, you don't work?
Not since i joined grad school.
I see, anyhow gotta go....thanks though @TuringTest and @experimentX ....I'll work on my time commitments
well ... didn't do much except troll around. Maybe next time .. :P
|dw:1362138326802:dw||dw:1362138537113:dw|
|dw:1362138832951:dw|
So until now we have expression for \( v_x\) and \(a \) ( and 'a' is perpendicular to the original path) |dw:1362139377040:dw|
|dw:1362139616525:dw||dw:1362139773642:dw|
Your 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 y-component.
|dw:1362142311512:dw||dw:1362142342862:dw||dw:1362142369122:dw||dw:1362142375743:dw|
\[ \Delta y = \Delta y_1 + \Delta y_2 = \frac{EL^2 }{4V_{acc} } + \frac{EDL}{2V_{acc}}\]
in my equations read d as L ... rest are same.

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