Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JenniferSmart1 Group Title

question

  • one year ago
  • one year ago

  • This Question is Closed
  1. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  2. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  3. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  4. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    feel free to write on here fellas:)

    • one year ago
  5. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  6. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    please stop posting @CaritaDeAngel

    • one year ago
  7. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    can I type today on OS (testing)

    • one year ago
  8. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    please continue posting @TuringTest

    • one year ago
  9. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  10. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  11. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

    • one year ago
  12. TuringTest Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  13. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  14. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  15. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  16. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    for my analysis I mean

    • one year ago
  17. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    would you like to see my graph?

    • one year ago
  18. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm .. sure!! upload it

    • one year ago
  19. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  20. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  21. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  22. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  23. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  24. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm ... what methods are asked to use?

    • one year ago
  25. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  26. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  27. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    how do I insert a picture in latex?

    • one year ago
  28. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  29. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  30. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  31. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    wait, you don't work?

    • one year ago
  32. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Not since i joined grad school.

    • one year ago
  33. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  34. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  35. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  36. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1362138832951:dw|

    • one year ago
  37. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  38. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  39. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 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 y-component.

    • one year ago
  40. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  41. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  42. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
    • Attachments:

See more questions >>>

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.