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sami21
 one year ago
Best ResponseYou've already chosen the best response.0wow so many people here to help !!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1First 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1361683871212:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1361684180784:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1A 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1oh 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 =)

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year 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
 one year 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 :\

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1it'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
 one year ago
Best ResponseYou've already chosen the best response.1yah i was just being silly by saying wind :) lol

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1since 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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1LOL! 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
 one year 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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1displacement is \(\Delta y= vt^2+\frac 12 at^2\) we probably have two y displacements.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1One 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
 one year 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1no i don't think so, because velocity is m/s and delta y is in m, so just one s would cancel it out. \[\frac m s\cdot s\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1so \[m=\frac ms\cdot s+\frac 12 \frac m{s^2}s^2\] and acceleration is in \(\frac m {s^2}\)

zepdrix
 one year 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1lol, yep you got it!

zepdrix
 one year 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:

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1what do you mean?

zepdrix
 one year 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1ooooops!!!! yeah that's a BIG TYPO LOL sorry...yeah that makes it more confusing haha

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1so 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 \(\frac 12 at^2\) \[\large \Delta y_2= vt+\cancel{\frac 12 at^2}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1and \(\Delta y_{total}=\Delta y_1+\Delta y_2\)

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\Delta y_{total}=\Delta y_1+\Delta y_2\] \[=vt^2+\frac 12at^2+vt^2\]

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1oh my!!!! LOL!!!! I like squaring my velocities LOL!!!! \[=vt+\frac 12at^2+vt\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1THis is why I need you here =D

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1now 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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Let's see....\(\Delta x=vt+\frac 12at^2\) inside and outside of the electric field? or just one x displacement?dw:1361686876935:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1or 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1i mean the time it takes to travel inside of c could be affected by that vertical acceleration.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1*two not "to 2"

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1LOL sure. Thanks a lot btw, it honestly helps a lot. I would have cried myself to sleep already :'(

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Let'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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\Delta y_{total}=vt+\frac 12at^2+vt\] \[\Delta y_{total}=2vt+\frac 12at^2\] \[\Delta y_{total}=2vt+\frac 12at^2\] and \[t=\frac{\Delta x}v\] \[\Delta y_{total}=2v\frac{\Delta x}{v}+\frac 12a\frac{\Delta x}{v}^2\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1OOopppps!!!!!! \[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\]

zepdrix
 one year 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
 one year 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1yes sir! you are correct!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\Delta y_{total}=2v_y\frac{\Delta x}{v_x}+\frac 12a\left(\frac{\Delta x}{v_x}\right)^2\] 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1no that's wrong!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1if the electric field is the force per unit charge....then.... that is correct....

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[E=\frac F q\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1but i need to substitute for the acceleration, because that's not something I can measure. I can only measure the deflection (in mm)

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[F=ma\] \[\frac E q=ma\] \[a=\frac E{mq}\] but there is more....I can't just make that substitution apparently. I need to use potential and kinetic energies instead :(

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Potential energy equals kinetic energy \[UE=KE\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Kinetic energy \(KE=\frac 12 mv^2\)

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1and 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. \[\frac U q=V\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1this allows be to sub for velocity, but not for the acceleration :'(

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1I'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!!!!!!!!!!!!! \[\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)\] \[\Delta y_{total}=\frac{\Delta x}{v_x}\left(2v_y+\frac 12\frac{E}{mq}\left(\frac{\Delta x}{v_x}\right)\right)\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1we made a mistake....there was no velocity in the y at any point, huh? I don't know anymore....time to sleep haha.

TuringTest
 one year 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1@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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