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- anonymous

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- jamiebookeater

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- anonymous

\[\Delta y=\Delta y_1+\Delta y_2=\frac 12 a(\frac{L}{v_0})^2+L +D \]
\[x_1=v_0t\]
\[t=\frac{x_1}{v_0}=\frac{L}{v_0}\]
\[\Delta y_1=v_0t+\frac 12 at^2\]
\[\Delta y_1=v_0\frac{L}{v_0}+\frac 12 a(\frac{L}{v_0})^2\]
\[\Delta y_1=\frac 12 a(\frac{L}{v_0})^2+L\]
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

but my equation should look like
\[\Delta y=\Delta y_1+\Delta y_2=\frac{EL}{2V_{acc}}\left( D+\frac L 2\right)\]

- anonymous

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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## More answers

- anonymous

feel free to write on here fellas:)

- anonymous

\[y_1=v_yt+\frac 12 at^2\]
is there a y velocity within the plates?

- anonymous

please stop posting @CaritaDeAngel

- TuringTest

can I type today on OS (testing)

- anonymous

please continue posting @TuringTest

- experimentX

Okay let's start with the electron gun ... |dw:1362064370315:dw|

- TuringTest

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?

- anonymous

yes

- TuringTest

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

- experimentX

yep .. it's called accelerating potential.|dw:1362064638627:dw|

- anonymous

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

- experimentX

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 ..

- anonymous

for my analysis I mean

- anonymous

would you like to see my graph?

- experimentX

hmm .. sure!! upload it

- experimentX

hmm ... looks like you could that data to calculate errors.

- experimentX

I"v never been good with error analysis ... but let's see what I can do.

- anonymous

no need to I guess....the xcel program did that already

- experimentX

he who?? I know couple of error analysis method like Standard ERror

- anonymous

never mind @experimentX I'm running out of time....I'm gonna make changes and submit what I have.

- experimentX

hmm ... what methods are asked to use?

- anonymous

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

- experimentX

okay .. I'll send you the details of derivation if you want ... within next 15 hrs. (optional: if you want)

- anonymous

how do I insert a picture in latex?

- experimentX

wow ... i don't know ...
maybe these guys know
http://tex.stackexchange.com/

- anonymous

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

- experimentX

wow!! ... looks like I am the only free loader at home here.

- anonymous

wait, you don't work?

- experimentX

Not since i joined grad school.

- anonymous

I see, anyhow gotta go....thanks though @TuringTest and @experimentX ....I'll work on my time commitments

- experimentX

well ... didn't do much except troll around. Maybe next time .. :P

- experimentX

|dw:1362138326802:dw||dw:1362138537113:dw|

- experimentX

|dw:1362138832951:dw|

- experimentX

So until now we have expression for \( v_x\) and \(a \) ( and 'a' is perpendicular to the original path)
|dw:1362139377040:dw|

- experimentX

|dw:1362139616525:dw||dw:1362139773642:dw|

- experimentX

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.

- experimentX

|dw:1362142311512:dw||dw:1362142342862:dw||dw:1362142369122:dw||dw:1362142375743:dw|

- experimentX

\[ \Delta y = \Delta y_1 + \Delta y_2 = \frac{EL^2 }{4V_{acc} } + \frac{EDL}{2V_{acc}}\]

- experimentX

in my equations read d as L ... rest are same.

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