A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

JenniferSmart1
 one year 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 \[\Delta y=\frac{qE}{m}(\frac{Lm}{2qV_acc})^2+L +D\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0but my equation should look like \[\Delta y=\Delta y_1+\Delta y_2=\frac{EL}{2V_{acc}}\left( D+\frac L 2\right)\]

JenniferSmart1
 one year 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0feel free to write on here fellas:)

JenniferSmart1
 one year 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0please stop posting @CaritaDeAngel

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0please continue posting @TuringTest

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

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

JenniferSmart1
 one year 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
 one year 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 ..

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0for my analysis I mean

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0would you like to see my graph?

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

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

experimentX
 one year 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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0no need to I guess....the xcel program did that already

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

JenniferSmart1
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0hmm ... what methods are asked to use?

JenniferSmart1
 one year 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
 one year 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)

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0how do I insert a picture in latex?

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

JenniferSmart1
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0wow!! ... looks like I am the only free loader at home here.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0wait, you don't work?

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

JenniferSmart1
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0well ... didn't do much except troll around. Maybe next time .. :P

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

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

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

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

experimentX
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0in my equations read d as L ... rest are same.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.