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JenniferSmart1Best 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 \[\Delta y=\frac{qE}{m}(\frac{Lm}{2qV_acc})^2+L +D\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
but my equation should look like \[\Delta y=\Delta y_1+\Delta y_2=\frac{EL}{2V_{acc}}\left( D+\frac L 2\right)\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I probably made a mistake somewhere, because my m's aren't cancelling out....and some other stuff too I guess
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
feel free to write on here fellas:)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[y_1=v_yt+\frac 12 at^2\] is there a y velocity within the plates?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
please stop posting @CaritaDeAngel
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
can I type today on OS (testing)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
please continue posting @TuringTest
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Okay let's start with the electron gun ... dw:1362064370315:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
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?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
so 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
yep .. it's called accelerating potential.dw:1362064638627:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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 ..
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
for my analysis I mean
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
would you like to see my graph?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
hmm .. sure!! upload it
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
hmm ... looks like you could that data to calculate errors.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
I"v never been good with error analysis ... but let's see what I can do.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
no need to I guess....the xcel program did that already
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
he who?? I know couple of error analysis method like Standard ERror
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
never mind @experimentX I'm running out of time....I'm gonna make changes and submit what I have.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
hmm ... what methods are asked to use?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.0
how do I insert a picture in latex?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
wow ... i don't know ... maybe these guys know http://tex.stackexchange.com/
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
wow!! ... looks like I am the only free loader at home here.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
wait, you don't work?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Not since i joined grad school.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I see, anyhow gotta go....thanks though @TuringTest and @experimentX ....I'll work on my time commitments
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
well ... didn't do much except troll around. Maybe next time .. :P
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1362138326802:dwdw:1362138537113:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1362138832951:dw
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
dw:1362139616525:dwdw:1362139773642:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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 ycomponent.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1362142311512:dwdw:1362142342862:dwdw:1362142369122:dwdw:1362142375743:dw
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
in my equations read d as L ... rest are same.
 one year ago
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