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JenniferSmart1 Group Title

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  • one year ago
  • one year ago

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  1. sami-21 Group Title
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    wow so many people here to help !!

    • one year ago
  2. JenniferSmart1 Group Title
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    @zepdrix

    • one year ago
  3. JenniferSmart1 Group Title
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    First there is no force acting on the particle, just the velocity in the positive x direction. Then there is a force acting in the y direction forcing the electron upwards, Then in the last scenario there is no force acting on the particle except for the velocity... THere should be a y displacement, because of the vertical acceleration

    • one year ago
  4. JenniferSmart1 Group Title
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    |dw:1361683871212:dw|

    • one year ago
  5. JenniferSmart1 Group Title
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    |dw:1361684180784:dw|

    • one year ago
  6. JenniferSmart1 Group Title
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    A B and C are length...distances. So the electron travels along "c" and feels and acceleration upwards, but when it starts traveling in the distance B it doesn't feel any acceleration but it will continue moving to the right but at an angle upwards. I can't type sorry lol

    • one year ago
  7. JenniferSmart1 Group Title
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    oh so after it leaves c , it continues like this... |dw:1361684734571:dw| with a velocity that has an x and y component...oh that's why it's going at an angle =)

    • one year ago
  8. JenniferSmart1 Group Title
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    so the v_1 is the original velocity in the x direction, which has been consist throughout this path. Hhhhhmmm how do we explain the y component of the velocity?

    • one year ago
  9. zepdrix Group Title
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    Yah why does the wind stop blowing when it reaches B? :c hmm that's confusing me...

    • one year ago
  10. zepdrix Group Title
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    Because it's making me think that it would do this,|dw:1361685064192:dw| I'm probably not interpreting that correctly though :\

    • one year ago
  11. JenniferSmart1 Group Title
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    it's actually going through an electric field for a distance c which is causing it to move upwards. the particle is an electron (negative) and it wants to accelerate towards the positive plate. (opposites attract remember?) but you can think of it as wind blowing and then it stops blowing miraculously :) |dw:1361685140543:dw|

    • one year ago
  12. zepdrix Group Title
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    yah i was just being silly by saying wind :) lol

    • one year ago
  13. JenniferSmart1 Group Title
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    since it acquired a \(v_{net}\) at an angle \(\theta\) while traveling in c, it will continue in that manner after it leaves c because there are no forces acting upon it, and it always had this horizontal velocity.

    • one year ago
  14. zepdrix Group Title
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    Ok fair enough c:

    • one year ago
  15. JenniferSmart1 Group Title
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    LOL! This is how printers work apparently. The ink particle acquires a negative charge and it's path towards the paper is manipulated by an electric field that it travels through. So the stronger the electric field that it travels through (positive and negative plates), the more it deflects

    • one year ago
  16. zepdrix Group Title
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    Oooo interesting! c:

    • one year ago
  17. zepdrix Group Title
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    Hmm your class seems hard :\ I think JenJen has gotten a lil too smart1 for me to help.

    • one year ago
  18. JenniferSmart1 Group Title
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    *you're

    • one year ago
  19. JenniferSmart1 Group Title
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    displacement is \(\Delta y= vt^2+\frac 12 at^2\) we probably have two y displacements.

    • one year ago
  20. JenniferSmart1 Group Title
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    One displacement for when the particle is within that electric field and one for when it exits. Because the first displacement has an acceleration \(\uparrow\) factor and the second displacement doesn't

    • one year ago
  21. zepdrix Group Title
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    \[\large \Delta y= vt+\frac 12 at^2\] I don't think it's a t^2 on the v is it? :O

    • one year ago
  22. JenniferSmart1 Group Title
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    no i don't think so, because velocity is m/s and delta y is in m, so just one s would cancel it out. \[\frac m s\cdot s\]

    • one year ago
  23. JenniferSmart1 Group Title
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    so \[m=\frac ms\cdot s+\frac 12 \frac m{s^2}s^2\] and acceleration is in \(\frac m {s^2}\)

    • one year ago
  24. zepdrix Group Title
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    \[\large \Delta y= vt+\frac 12 at^2 \qquad \rightarrow \qquad \Delta y= \frac{m}{s}(s)+\frac 12 \left(\frac{m}{s^2}\right)s^2\] So \(\Delta y\) equals some distance in meters. Bah you wrote it out already, yer too fast for me! lol

    • one year ago
  25. JenniferSmart1 Group Title
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    lol, yep you got it!

    • one year ago
  26. zepdrix Group Title
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    So doesn't that help show that we dont' want the square on the first t? D:

    • one year ago
  27. JenniferSmart1 Group Title
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    what do you mean?

    • one year ago
  28. zepdrix Group Title
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    Maybe you just made a typo :O\[\huge \Delta y= vt^{\color{orangered}{2}}+\frac 12 at^2\]Why is the orange part there?

    • one year ago
  29. JenniferSmart1 Group Title
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    ooooops!!!! yeah that's a BIG TYPO LOL sorry...yeah that makes it more confusing haha

    • one year ago
  30. JenniferSmart1 Group Title
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    so we have to \(\Delta y\) \[\Delta y_1\] and \[\Delta y_2\] the second \( \Delta y\) doesn't have an acceleration...so we can cancel out the \(\frac 12 at^2\) \[\large \Delta y_2= vt+\cancel{\frac 12 at^2}\]

    • one year ago
  31. JenniferSmart1 Group Title
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    and \(\Delta y_{total}=\Delta y_1+\Delta y_2\)

    • one year ago
  32. JenniferSmart1 Group Title
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    \[\Delta y_{total}=\Delta y_1+\Delta y_2\] \[=vt^2+\frac 12at^2+vt^2\]

    • one year ago
  33. zepdrix Group Title
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    There you go doing it again >:O

    • one year ago
  34. JenniferSmart1 Group Title
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    oh my!!!! LOL!!!! I like squaring my velocities LOL!!!! \[=vt+\frac 12at^2+vt\]

    • one year ago
  35. JenniferSmart1 Group Title
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    THis is why I need you here =D

    • one year ago
  36. JenniferSmart1 Group Title
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    now we need to substitute for the t's because we're not given the time that it takes the particle to travel along that path.

    • one year ago
  37. JenniferSmart1 Group Title
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    Let's see....\(\Delta x=vt+\frac 12at^2\) inside and outside of the electric field? or just one x displacement?|dw:1361686876935:dw|

    • one year ago
  38. JenniferSmart1 Group Title
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    or maybe to 2 \(\Delta x's\) because the speed at which the particle goes within in the field is different form the speed outside of that field? What do you think?

    • one year ago
  39. JenniferSmart1 Group Title
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    i mean the time it takes to travel inside of c could be affected by that vertical acceleration.

    • one year ago
  40. JenniferSmart1 Group Title
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    *two not "to 2"

    • one year ago
  41. zepdrix Group Title
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    I dunno :\ I give up. Pretend I'm still here and keep talking XD lol

    • one year ago
  42. JenniferSmart1 Group Title
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    LOL sure. Thanks a lot btw, it honestly helps a lot. I would have cried myself to sleep already :'(

    • one year ago
  43. JenniferSmart1 Group Title
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    Let's see. \[\Delta x_1=v_xt+\frac 12 a_xt^2\] \[\Delta x_2=v_xt+\frac 12 a_xt^2\] there isn't an acceleration in the x direction at any point through this journey... \[\Delta x_1=v_xt+\cancel{\frac 12 a_xt^2}\] \[\Delta x_2=v_xt+\cancel{\frac 12 a_xt^2}\] and the velocity the x direction isn't changing either. That leaves us with: \[t=\frac{\Delta x}v\] cool now I can just substitute.

    • one year ago
  44. JenniferSmart1 Group Title
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    \[\Delta y_{total}=vt+\frac 12at^2+vt\] \[\Delta y_{total}=2vt+\frac 12at^2\] \[\Delta y_{total}=2vt+\frac 12at^2\] and \[t=\frac{\Delta x}v\] \[\Delta y_{total}=2v\frac{\Delta x}{v}+\frac 12a\frac{\Delta x}{v}^2\]

    • one year ago
  45. JenniferSmart1 Group Title
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    OOopppps!!!!!! \[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\]

    • one year ago
  46. JenniferSmart1 Group Title
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    right zep?

    • one year ago
  47. zepdrix Group Title
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    Why did the acceleration get crossed out during the first displacement?\[\Delta x_1=v_xt+\cancel{\frac 12 a_xt^2}\]

    • one year ago
  48. zepdrix Group Title
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    Oh because this is displacement in the x direction? The acceleration is 0, and the velocity is constant?

    • one year ago
  49. JenniferSmart1 Group Title
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    yes sir! you are correct!

    • one year ago
  50. JenniferSmart1 Group Title
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    \[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\] Let's see, now I have to sub for the acceleration. \[F=ma\] \[F=qE\] q is the charge of one electron? I think? LOL i should know this haha. E is the Electric field

    • one year ago
  51. JenniferSmart1 Group Title
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    no that's wrong!

    • one year ago
  52. JenniferSmart1 Group Title
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    if the electric field is the force per unit charge....then.... that is correct....

    • one year ago
  53. JenniferSmart1 Group Title
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    \[E=\frac F q\]

    • one year ago
  54. JenniferSmart1 Group Title
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    but i need to substitute for the acceleration, because that's not something I can measure. I can only measure the deflection (in mm)

    • one year ago
  55. zepdrix Group Title
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    hmmm true

    • one year ago
  56. JenniferSmart1 Group Title
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    \[F=ma\] \[\frac E q=ma\] \[a=\frac E{mq}\] but there is more....I can't just make that substitution apparently. I need to use potential and kinetic energies instead :-(

    • one year ago
  57. zepdrix Group Title
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    ikr!

    • one year ago
  58. JenniferSmart1 Group Title
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    Potential energy equals kinetic energy \[UE=KE\]

    • one year ago
  59. JenniferSmart1 Group Title
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    Kinetic energy \(KE=\frac 12 mv^2\)

    • one year ago
  60. JenniferSmart1 Group Title
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    and for reasons I don't know \[UE=V\cdot q\] Potential energy \(U\) is supposedly different from electric potential \(V\) Electric potential is the Potential energy per unit charge. \[\frac U q=V\]

    • one year ago
  61. JenniferSmart1 Group Title
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    this allows be to sub for velocity, but not for the acceleration :'-(

    • one year ago
  62. JenniferSmart1 Group Title
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    I'd have to use \[a=\frac E{mq}\] Let's see what happens.... \[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\] \[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)^2\] ugh!!!!!!!!!!!!! \[\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)\] \[\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)\]

    • one year ago
  63. JenniferSmart1 Group Title
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    we made a mistake....there was no velocity in the y at any point, huh? I don't know anymore....time to sleep haha.

    • one year ago
  64. JenniferSmart1 Group Title
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    @JamesJ

    • one year ago
  65. TuringTest Group Title
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    What exactly do you want to know? The vertical displacement over A? in terms of which variables? and do you *have* to use energy formulas to get the answer?

    • one year ago
  66. JenniferSmart1 Group Title
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    @TuringTest My goal is to get the displacement as a function of the electric field It should look like this \[\Delta y=\Delta y_1+\Delta y_2 =\frac{Ec}{2v}\left(B+\frac c 2\right)\]

    • one year ago
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is replying to Can someone tell me what button the professor is hitting...

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