## JenniferSmart1 Group Title question one year ago one year ago

1. JenniferSmart1 Group Title

$\Delta y=\Delta y_1+\Delta y_2=\frac 12 a(\frac{L}{v_0})^2+L +D$ $x_1=v_0t$ $t=\frac{x_1}{v_0}=\frac{L}{v_0}$ $\Delta y_1=v_0t+\frac 12 at^2$ $\Delta y_1=v_0\frac{L}{v_0}+\frac 12 a(\frac{L}{v_0})^2$ $\Delta y_1=\frac 12 a(\frac{L}{v_0})^2+L$ 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 $\Delta y=\frac{qE}{m}(\frac{Lm}{2qV_acc})^2+L +D$

2. JenniferSmart1 Group Title

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

3. JenniferSmart1 Group Title

I probably made a mistake somewhere, because my m's aren't cancelling out....and some other stuff too I guess

4. JenniferSmart1 Group Title

feel free to write on here fellas:)

5. JenniferSmart1 Group Title

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

6. JenniferSmart1 Group Title

7. TuringTest Group Title

can I type today on OS (testing)

8. JenniferSmart1 Group Title

9. experimentX Group Title

10. TuringTest Group Title

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?

11. JenniferSmart1 Group Title

yes

12. TuringTest Group Title

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

13. experimentX Group Title

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

14. JenniferSmart1 Group Title

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

15. experimentX Group Title

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 ..

16. JenniferSmart1 Group Title

for my analysis I mean

17. JenniferSmart1 Group Title

would you like to see my graph?

18. experimentX Group Title

19. experimentX Group Title

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

20. experimentX Group Title

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

21. JenniferSmart1 Group Title

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

22. experimentX Group Title

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

23. JenniferSmart1 Group Title

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

24. experimentX Group Title

hmm ... what methods are asked to use?

25. JenniferSmart1 Group Title

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

26. experimentX Group Title

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

27. JenniferSmart1 Group Title

how do I insert a picture in latex?

28. experimentX Group Title

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

29. JenniferSmart1 Group Title

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

30. experimentX Group Title

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

31. JenniferSmart1 Group Title

wait, you don't work?

32. experimentX Group Title

Not since i joined grad school.

33. JenniferSmart1 Group Title

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

34. experimentX Group Title

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

35. experimentX Group Title

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

36. experimentX Group Title

|dw:1362138832951:dw|

37. experimentX Group Title

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

38. experimentX Group Title

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

39. experimentX Group Title

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.

40. experimentX Group Title

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

41. experimentX Group Title

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

42. experimentX Group Title

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