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anonymous
 one year ago
two horizontal plates erated by 4.00cm , potetinal differece 120V horizaonal beem of eletrons with speed 6.50*10^6 , the electron enter 1.00 above the negative plate , the horizontal distance tht the electrom travels before striking the postive plate?
anonymous
 one year ago
two horizontal plates erated by 4.00cm , potetinal differece 120V horizaonal beem of eletrons with speed 6.50*10^6 , the electron enter 1.00 above the negative plate , the horizontal distance tht the electrom travels before striking the postive plate?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0draw it first and you will get a lot of help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437369186841:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 @e.mccormick @Luigi0210

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What would be the electric field between the two parallel plates?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that how u start?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, you can use the forces to do the problem. Then, F=ma. Then find where the electron intersects on of the plates.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they didnt geive the accelration

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you know the forces, you can find the acceleration.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do we have to find it ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to find the distane?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what formila are we goona conclude to ? in the end ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, we'll need to.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Usually, in these types of problems, we'll ignore gravity. Find the electric field. F = qE=ma ==> a = qE/m We can then use: \[y(t) = \frac{1}{2}at^2 + v_ot + y_o\] to find when the electron meets one of the plates

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so for acc i get 2.85*10^17

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not sure if correct too mnay calcs maybe screwd

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then, when you have t for that, you can find x There's no acceleration in the x direction, so you can just use: x(t) = v*t The v used here is the one given in the problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not getting the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first i found time , then found d but wrong ansswr1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get a different acceleration. Be careful with units.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do u get as an answerr?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.05.276x10^14 is the acceleration I'm getting

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0even if i use ur acc i am getting wrong answr

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Show me what youre doing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first speed didved by acc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then i get the time , so i times by speed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, can't do it quite that way. \[E = \frac{V}{d}\]gives the E field between the plates. V is the voltage between the plates, d is the separation between them (be careful of units! We want this in meters!). As shown above, we can find the acceleration using \[a = \frac{qE}{m} = \frac{V}{d}\frac{q}{m}\] You can look up the value for q/m online. It's very commonly used. After that, we'll find the time of flight. The electron starts 1cm above the bottom, negative plate. It will travel upwards to the positive plate, which sits at a height of 4cm. \[y(t) = \frac{1}{2}at^2 +v_ot + y_o\] From the information given, we see that vo is zero (no initial velocity in the y direction), and y_o = .01m (1cm). We want to know what t will be when y(t) = 4. That is: \[4 = \frac{1}{2}at^2+1\] Only one step after you find t. What do you get for t?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Bleh. 4 should be .04, and 1 should be .01 in the last equation. Units.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can convert to whatever units you'd like.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok yea got it thanks , could u help me with soe theory questions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437372625139:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which way will the electron accelerate?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437372687094:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can think about this two ways. We can use F = qE = ma, or we can think about which way electric fields point between charges. Which way does the field point between a negative and positive charge?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If we say that the direction the field is pointing in is the positive direction, then F = qE will be negative, because q is negative (charge of the electron). This means it would move in the direction opposite of the electric field.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh i was think why the answrr is 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0beacuse i was 100% sure the direction would be 2 and then i looked at the answer it was 4 i ws soo confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its an electron so would make sence since they are nagative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if it was a proton the answr would be 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0their are 2 netual metal objcts x and y on a wooden table

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a negativly charged rod is brought near object y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437373183331:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the answer is x received a negative charge by induction and y received a positive charge by induction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont get what they mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right. So, the negative charges in the rod will repel other negative charges. Since the objects are metal, and thus conductive, their electrons can move relatively freely. This means that the electrons in the metal will move away from the rod. Since the rod is near object y, they'll move away to object x. This will make object x have a net negative charge. The lack of electrons on object y will give it a net positive charge. This displacement of charge is called induction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am reading from my notes about induction and it says it causes the chrages to become opposite

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what ur saying is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the rod was postive then y would be negative and also x ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean x will be negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If the rod were positive, the y would become negative and x would be positive.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0some of the other answrrs are by conductions whats the differenr

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the boxes were not neutral then it would be conductions right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand your questions. If the objects weren't conductors, then the charges wouldn't be able to move freely. If they were perfect insulators, they couldn't move at all, and thus no charge would be induced.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like the correct answer is induction the wrong answer is conduction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, the phenomenon of the rod inducing a charge on the boxes is called induction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what would make it conduction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the boxes werent neutral?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like lets say box x and y were postive then the answer would be conduction right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Conduction is just the metal carrying the charge across a potential. Like a current moving down a wire.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a posively charger sphere near a metal rod . both are insulated stands and the rod is grounded

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437374305785:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the disturbution of the charge on the rod is ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is somewhat similar to what we just did. What will happen when the sphere approaches the rod?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0umm x will become neagative?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's what I'd say. Charge would be drawn up through the ground wire toward the sphere.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0beacuse they are grounded ? like can't move right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Grounded just means that it can exchange charges with a very large, external source. So, the positive charge of the sphere will cause electrons to accumulate on the rod, mostly near the sphere.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to show that an objct is negativly charged we need to ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it should be positve right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the answer is negative?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The positive sphere will cause the rod to become negatively charged.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thats it for tonight

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do u come here often ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0occasionally. I'm kind of erratic about when I'm on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i am going to sleep learned so much today need to settle in the brain

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hope to see u again (kinda hard to meet people here that are good at physics lol)
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