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JenniferSmart1

  • one year ago

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  1. sami-21
    • one year ago
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    wow so many people here to help !!

  2. JenniferSmart1
    • one year ago
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    @zepdrix

  3. JenniferSmart1
    • one year ago
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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

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

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

  6. JenniferSmart1
    • one year ago
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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

  7. JenniferSmart1
    • one year ago
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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 =)

  8. JenniferSmart1
    • one year ago
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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?

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

  10. zepdrix
    • one year ago
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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 :\

  11. JenniferSmart1
    • one year ago
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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|

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

  13. JenniferSmart1
    • one year ago
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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.

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

  15. JenniferSmart1
    • one year ago
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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

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

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

  18. JenniferSmart1
    • one year ago
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    *you're

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

  20. JenniferSmart1
    • one year ago
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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

  21. zepdrix
    • one year ago
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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

  22. JenniferSmart1
    • one year ago
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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. \[\frac m s\cdot s\]

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

  24. zepdrix
    • one year ago
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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

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

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

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

  28. zepdrix
    • one year ago
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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?

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

  30. JenniferSmart1
    • one year ago
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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 \(\frac 12 at^2\) \[\large \Delta y_2= vt+\cancel{\frac 12 at^2}\]

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

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

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

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

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

  36. JenniferSmart1
    • one year ago
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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.

  37. JenniferSmart1
    • one year ago
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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|

  38. JenniferSmart1
    • one year ago
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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?

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

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

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

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

  43. JenniferSmart1
    • one year ago
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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.

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

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

  46. JenniferSmart1
    • one year ago
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    right zep?

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

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

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

  50. JenniferSmart1
    • one year ago
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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

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

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

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

  54. JenniferSmart1
    • one year ago
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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)

  55. zepdrix
    • one year ago
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    hmmm true

  56. JenniferSmart1
    • one year ago
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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 :-(

  57. zepdrix
    • one year ago
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    ikr!

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

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

  60. JenniferSmart1
    • one year ago
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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. \[\frac U q=V\]

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

  62. JenniferSmart1
    • one year ago
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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!!!!!!!!!!!!! \[\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)\] \[\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)\]

  63. JenniferSmart1
    • one year ago
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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.

  64. JenniferSmart1
    • one year ago
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    @JamesJ

  65. TuringTest
    • one year ago
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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?

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

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