anonymous
  • anonymous
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Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
wow so many people here to help !!
anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
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

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anonymous
  • anonymous
|dw:1361683871212:dw|
anonymous
  • anonymous
|dw:1361684180784:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
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 =)
anonymous
  • anonymous
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?
zepdrix
  • zepdrix
Yah why does the wind stop blowing when it reaches B? :c hmm that's confusing me...
zepdrix
  • zepdrix
Because it's making me think that it would do this,|dw:1361685064192:dw| I'm probably not interpreting that correctly though :\
anonymous
  • anonymous
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|
zepdrix
  • zepdrix
yah i was just being silly by saying wind :) lol
anonymous
  • anonymous
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.
zepdrix
  • zepdrix
Ok fair enough c:
anonymous
  • anonymous
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
zepdrix
  • zepdrix
Oooo interesting! c:
zepdrix
  • zepdrix
Hmm your class seems hard :\ I think JenJen has gotten a lil too smart1 for me to help.
anonymous
  • anonymous
*you're
anonymous
  • anonymous
displacement is \(\Delta y= vt^2+\frac 12 at^2\) we probably have two y displacements.
anonymous
  • anonymous
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
zepdrix
  • zepdrix
\[\large \Delta y= vt+\frac 12 at^2\] I don't think it's a t^2 on the v is it? :O
anonymous
  • anonymous
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. \[\frac m s\cdot s\]
anonymous
  • anonymous
so \[m=\frac ms\cdot s+\frac 12 \frac m{s^2}s^2\] and acceleration is in \(\frac m {s^2}\)
zepdrix
  • zepdrix
\[\large \Delta y= vt+\frac 12 at^2 \qquad \rightarrow \qquad \Delta y= \frac{m}{s}(s)+\frac 12 \left(\frac{m}{s^2}\right)s^2\] So \(\Delta y\) equals some distance in meters. Bah you wrote it out already, yer too fast for me! lol
anonymous
  • anonymous
lol, yep you got it!
zepdrix
  • zepdrix
So doesn't that help show that we dont' want the square on the first t? D:
anonymous
  • anonymous
what do you mean?
zepdrix
  • zepdrix
Maybe you just made a typo :O\[\huge \Delta y= vt^{\color{orangered}{2}}+\frac 12 at^2\]Why is the orange part there?
anonymous
  • anonymous
ooooops!!!! yeah that's a BIG TYPO LOL sorry...yeah that makes it more confusing haha
anonymous
  • anonymous
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 \(\frac 12 at^2\) \[\large \Delta y_2= vt+\cancel{\frac 12 at^2}\]
anonymous
  • anonymous
and \(\Delta y_{total}=\Delta y_1+\Delta y_2\)
anonymous
  • anonymous
\[\Delta y_{total}=\Delta y_1+\Delta y_2\] \[=vt^2+\frac 12at^2+vt^2\]
zepdrix
  • zepdrix
There you go doing it again >:O
anonymous
  • anonymous
oh my!!!! LOL!!!! I like squaring my velocities LOL!!!! \[=vt+\frac 12at^2+vt\]
anonymous
  • anonymous
THis is why I need you here =D
anonymous
  • anonymous
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.
anonymous
  • anonymous
Let's see....\(\Delta x=vt+\frac 12at^2\) inside and outside of the electric field? or just one x displacement?|dw:1361686876935:dw|
anonymous
  • anonymous
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?
anonymous
  • anonymous
i mean the time it takes to travel inside of c could be affected by that vertical acceleration.
anonymous
  • anonymous
*two not "to 2"
zepdrix
  • zepdrix
I dunno :\ I give up. Pretend I'm still here and keep talking XD lol
anonymous
  • anonymous
LOL sure. Thanks a lot btw, it honestly helps a lot. I would have cried myself to sleep already :'(
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\Delta y_{total}=vt+\frac 12at^2+vt\] \[\Delta y_{total}=2vt+\frac 12at^2\] \[\Delta y_{total}=2vt+\frac 12at^2\] and \[t=\frac{\Delta x}v\] \[\Delta y_{total}=2v\frac{\Delta x}{v}+\frac 12a\frac{\Delta x}{v}^2\]
anonymous
  • anonymous
OOopppps!!!!!! \[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\]
anonymous
  • anonymous
right zep?
zepdrix
  • zepdrix
Why did the acceleration get crossed out during the first displacement?\[\Delta x_1=v_xt+\cancel{\frac 12 a_xt^2}\]
zepdrix
  • zepdrix
Oh because this is displacement in the x direction? The acceleration is 0, and the velocity is constant?
anonymous
  • anonymous
yes sir! you are correct!
anonymous
  • anonymous
\[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\] 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
anonymous
  • anonymous
no that's wrong!
anonymous
  • anonymous
if the electric field is the force per unit charge....then.... that is correct....
anonymous
  • anonymous
\[E=\frac F q\]
anonymous
  • anonymous
but i need to substitute for the acceleration, because that's not something I can measure. I can only measure the deflection (in mm)
zepdrix
  • zepdrix
hmmm true
anonymous
  • anonymous
\[F=ma\] \[\frac E q=ma\] \[a=\frac E{mq}\] but there is more....I can't just make that substitution apparently. I need to use potential and kinetic energies instead :-(
zepdrix
  • zepdrix
ikr!
anonymous
  • anonymous
Potential energy equals kinetic energy \[UE=KE\]
anonymous
  • anonymous
Kinetic energy \(KE=\frac 12 mv^2\)
anonymous
  • anonymous
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. \[\frac U q=V\]
anonymous
  • anonymous
this allows be to sub for velocity, but not for the acceleration :'-(
anonymous
  • anonymous
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!!!!!!!!!!!!! \[\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)\] \[\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)\]
anonymous
  • anonymous
we made a mistake....there was no velocity in the y at any point, huh? I don't know anymore....time to sleep haha.
anonymous
  • anonymous
@JamesJ
TuringTest
  • TuringTest
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?
anonymous
  • anonymous
@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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