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anonymous
 3 years ago
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anonymous
 3 years ago
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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow so many people here to help !!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1361683871212:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1361684180784:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.1Yah why does the wind stop blowing when it reaches B? :c hmm that's confusing me...

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Because it's making me think that it would do this,dw:1361685064192:dw I'm probably not interpreting that correctly though :\

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it'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
 3 years ago
Best ResponseYou've already chosen the best response.1yah i was just being silly by saying wind :) lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOL! 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
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm your class seems hard :\ I think JenJen has gotten a lil too smart1 for me to help.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0displacement is \(\Delta y= vt^2+\frac 12 at^2\) we probably have two y displacements.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0One 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
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large \Delta y= vt+\frac 12 at^2\] I don't think it's a t^2 on the v is it? :O

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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
 3 years ago
Best ResponseYou've already chosen the best response.0so \[m=\frac ms\cdot s+\frac 12 \frac m{s^2}s^2\] and acceleration is in \(\frac m {s^2}\)

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large \Delta y= vt+\frac 12 at^2 \qquad \rightarrow \qquad \Delta y= \frac{m}{s}(s)+\frac 12 \left(\frac{m}{s^2}\right)s^2\] So \(\Delta y\) equals some distance in meters. Bah you wrote it out already, yer too fast for me! lol

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1So doesn't that help show that we dont' want the square on the first t? D:

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Maybe you just made a typo :O\[\huge \Delta y= vt^{\color{orangered}{2}}+\frac 12 at^2\]Why is the orange part there?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ooooops!!!! yeah that's a BIG TYPO LOL sorry...yeah that makes it more confusing haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.0and \(\Delta y_{total}=\Delta y_1+\Delta y_2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Delta y_{total}=\Delta y_1+\Delta y_2\] \[=vt^2+\frac 12at^2+vt^2\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1There you go doing it again >:O

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh my!!!! LOL!!!! I like squaring my velocities LOL!!!! \[=vt+\frac 12at^2+vt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0THis is why I need you here =D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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
 3 years ago
Best ResponseYou've already chosen the best response.0Let's see....\(\Delta x=vt+\frac 12at^2\) inside and outside of the electric field? or just one x displacement?dw:1361686876935:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or 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
 3 years ago
Best ResponseYou've already chosen the best response.0i mean the time it takes to travel inside of c could be affected by that vertical acceleration.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1I dunno :\ I give up. Pretend I'm still here and keep talking XD lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOL sure. Thanks a lot btw, it honestly helps a lot. I would have cried myself to sleep already :'(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let'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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Delta y_{total}=vt+\frac 12at^2+vt\] \[\Delta y_{total}=2vt+\frac 12at^2\] \[\Delta y_{total}=2vt+\frac 12at^2\] and \[t=\frac{\Delta x}v\] \[\Delta y_{total}=2v\frac{\Delta x}{v}+\frac 12a\frac{\Delta x}{v}^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OOopppps!!!!!! \[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Why did the acceleration get crossed out during the first displacement?\[\Delta x_1=v_xt+\cancel{\frac 12 a_xt^2}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Oh because this is displacement in the x direction? The acceleration is 0, and the velocity is constant?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes sir! you are correct!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\] 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
 3 years ago
Best ResponseYou've already chosen the best response.0if the electric field is the force per unit charge....then.... that is correct....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but i need to substitute for the acceleration, because that's not something I can measure. I can only measure the deflection (in mm)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[F=ma\] \[\frac E q=ma\] \[a=\frac E{mq}\] but there is more....I can't just make that substitution apparently. I need to use potential and kinetic energies instead :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Potential energy equals kinetic energy \[UE=KE\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Kinetic energy \(KE=\frac 12 mv^2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and 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
 3 years ago
Best ResponseYou've already chosen the best response.0this allows be to sub for velocity, but not for the acceleration :'(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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
 3 years ago
Best ResponseYou've already chosen the best response.0we made a mistake....there was no velocity in the y at any point, huh? I don't know anymore....time to sleep haha.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0What 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
 3 years ago
Best ResponseYou've already chosen the best response.0@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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