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JenniferSmart1

  • 2 years ago

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  1. JenniferSmart1
    • 2 years ago
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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 \[\Delta y=\frac{qE}{m}(\frac{Lm}{2qV_acc})^2+L +D\]

  2. JenniferSmart1
    • 2 years ago
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    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
    • 2 years ago
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    I probably made a mistake somewhere, because my m's aren't cancelling out....and some other stuff too I guess

  4. JenniferSmart1
    • 2 years ago
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    feel free to write on here fellas:)

  5. JenniferSmart1
    • 2 years ago
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    \[y_1=v_yt+\frac 12 at^2\] is there a y velocity within the plates?

  6. JenniferSmart1
    • 2 years ago
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    please stop posting @CaritaDeAngel

  7. TuringTest
    • 2 years ago
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    can I type today on OS (testing)

  8. JenniferSmart1
    • 2 years ago
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    please continue posting @TuringTest

  9. experimentX
    • 2 years ago
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    Okay let's start with the electron gun ... |dw:1362064370315:dw|

  10. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    yes

  12. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    yep .. it's called accelerating potential.|dw:1362064638627:dw|

  14. JenniferSmart1
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    for my analysis I mean

  17. JenniferSmart1
    • 2 years ago
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    would you like to see my graph?

  18. experimentX
    • 2 years ago
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    hmm .. sure!! upload it

  19. experimentX
    • 2 years ago
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    hmm ... looks like you could that data to calculate errors.

  20. experimentX
    • 2 years ago
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    I"v never been good with error analysis ... but let's see what I can do.

  21. JenniferSmart1
    • 2 years ago
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    no need to I guess....the xcel program did that already

  22. experimentX
    • 2 years ago
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    he who?? I know couple of error analysis method like Standard ERror

  23. JenniferSmart1
    • 2 years ago
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    never mind @experimentX I'm running out of time....I'm gonna make changes and submit what I have.

  24. experimentX
    • 2 years ago
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    hmm ... what methods are asked to use?

  25. JenniferSmart1
    • 2 years ago
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    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
    • 2 years ago
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    okay .. I'll send you the details of derivation if you want ... within next 15 hrs. (optional: if you want)

  27. JenniferSmart1
    • 2 years ago
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    how do I insert a picture in latex?

  28. experimentX
    • 2 years ago
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    wow ... i don't know ... maybe these guys know http://tex.stackexchange.com/

  29. JenniferSmart1
    • 2 years ago
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    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
    • 2 years ago
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    wow!! ... looks like I am the only free loader at home here.

  31. JenniferSmart1
    • 2 years ago
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    wait, you don't work?

  32. experimentX
    • 2 years ago
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    Not since i joined grad school.

  33. JenniferSmart1
    • 2 years ago
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    I see, anyhow gotta go....thanks though @TuringTest and @experimentX ....I'll work on my time commitments

  34. experimentX
    • 2 years ago
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    well ... didn't do much except troll around. Maybe next time .. :P

  35. experimentX
    • 2 years ago
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    |dw:1362138326802:dw||dw:1362138537113:dw|

  36. experimentX
    • 2 years ago
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    |dw:1362138832951:dw|

  37. experimentX
    • 2 years ago
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    So until now we have expression for \( v_x\) and \(a \) ( and 'a' is perpendicular to the original path) |dw:1362139377040:dw|

  38. experimentX
    • 2 years ago
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    |dw:1362139616525:dw||dw:1362139773642:dw|

  39. experimentX
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1362142311512:dw||dw:1362142342862:dw||dw:1362142369122:dw||dw:1362142375743:dw|

  41. experimentX
    • 2 years ago
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    \[ \Delta y = \Delta y_1 + \Delta y_2 = \frac{EL^2 }{4V_{acc} } + \frac{EDL}{2V_{acc}}\]

  42. experimentX
    • 2 years ago
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    in my equations read d as L ... rest are same.

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