anonymous
  • anonymous
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?
Physics
katieb
  • katieb
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IrishBoy123
  • IrishBoy123
draw it first and you will get a lot of help
anonymous
  • anonymous
|dw:1437369186841:dw|
anonymous
  • anonymous

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anonymous
  • anonymous
What would be the electric field between the two parallel plates?
anonymous
  • anonymous
i did fnet=Fe+Fg
anonymous
  • anonymous
is that how u start?
anonymous
  • anonymous
Yep, you can use the forces to do the problem. Then, F=ma. Then find where the electron intersects on of the plates.
anonymous
  • anonymous
they didnt geive the accelration
anonymous
  • anonymous
If you know the forces, you can find the acceleration.
anonymous
  • anonymous
do we have to find it ?
anonymous
  • anonymous
to find the distane?
anonymous
  • anonymous
what formila are we goona conclude to ? in the end ?
anonymous
  • anonymous
Yeah, we'll need to.
anonymous
  • anonymous
Usually, 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
  • anonymous
ok so for acc i get 2.85*10^17
anonymous
  • anonymous
not sure if correct too mnay calcs maybe screwd
anonymous
  • anonymous
Then, 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
  • anonymous
not getting the answer
anonymous
  • anonymous
first i found time , then found d but wrong ansswr1
anonymous
  • anonymous
I get a different acceleration. Be careful with units.
anonymous
  • anonymous
whatu get/
anonymous
  • anonymous
what do u get as an answerr?
anonymous
  • anonymous
5.276x10^14 is the acceleration I'm getting
anonymous
  • anonymous
m/s^2
anonymous
  • anonymous
even if i use ur acc i am getting wrong answr
anonymous
  • anonymous
the answer is 6.94
anonymous
  • anonymous
Show me what youre doing
anonymous
  • anonymous
first speed didved by acc
anonymous
  • anonymous
then i get the time , so i times by speed
anonymous
  • anonymous
to find d
anonymous
  • anonymous
No, 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
  • anonymous
Bleh. 4 should be .04, and 1 should be .01 in the last equation. Units.
anonymous
  • anonymous
the answer is in cm
anonymous
  • anonymous
You can convert to whatever units you'd like.
anonymous
  • anonymous
ok yea got it thanks , could u help me with soe theory questions
anonymous
  • anonymous
I'll try :)
anonymous
  • anonymous
|dw:1437372625139:dw|
anonymous
  • anonymous
which way will the electron accelerate?
anonymous
  • anonymous
|dw:1437372687094:dw|
anonymous
  • anonymous
You 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
  • anonymous
2?
anonymous
  • anonymous
If 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
  • anonymous
so 4
anonymous
  • anonymous
ohh i was think why the answrr is 4
anonymous
  • anonymous
beacuse i was 100% sure the direction would be 2 and then i looked at the answer it was 4 i ws soo confused
anonymous
  • anonymous
its an electron so would make sence since they are nagative
anonymous
  • anonymous
Exactly :)
anonymous
  • anonymous
so if it was a proton the answr would be 2?
anonymous
  • anonymous
Yep!
anonymous
  • anonymous
their is one more
anonymous
  • anonymous
their are 2 netual metal objcts x and y on a wooden table
anonymous
  • anonymous
a negativly charged rod is brought near object y
anonymous
  • anonymous
|dw:1437373183331:dw|
anonymous
  • anonymous
so the answer is x received a negative charge by induction and y received a positive charge by induction
anonymous
  • anonymous
i dont get what they mean
anonymous
  • anonymous
Right. 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
  • anonymous
i am reading from my notes about induction and it says it causes the chrages to become opposite
anonymous
  • anonymous
so what ur saying is
anonymous
  • anonymous
in the other way
anonymous
  • anonymous
if the rod was postive then y would be negative and also x ?
anonymous
  • anonymous
i mean x will be negative
anonymous
  • anonymous
If the rod were positive, the y would become negative and x would be positive.
anonymous
  • anonymous
oh yea
anonymous
  • anonymous
some of the other answrrs are by conductions whats the differenr
anonymous
  • anonymous
if the boxes were not neutral then it would be conductions right?
anonymous
  • anonymous
I 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
  • anonymous
like the correct answer is induction the wrong answer is conduction
anonymous
  • anonymous
Oh, the phenomenon of the rod inducing a charge on the boxes is called induction.
anonymous
  • anonymous
what would make it conduction
anonymous
  • anonymous
if the boxes werent neutral?
anonymous
  • anonymous
like lets say box x and y were postive then the answer would be conduction right
anonymous
  • anonymous
Conduction is just the metal carrying the charge across a potential. Like a current moving down a wire.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
their is one more
anonymous
  • anonymous
a posively charger sphere near a metal rod . both are insulated stands and the rod is grounded
anonymous
  • anonymous
|dw:1437374305785:dw|
anonymous
  • anonymous
the disturbution of the charge on the rod is ?
anonymous
  • anonymous
This is somewhat similar to what we just did. What will happen when the sphere approaches the rod?
anonymous
  • anonymous
umm x will become neagative?
anonymous
  • anonymous
That's what I'd say. Charge would be drawn up through the ground wire toward the sphere.
anonymous
  • anonymous
how do u knw tht?
anonymous
  • anonymous
beacuse they are grounded ? like can't move right?
anonymous
  • anonymous
Grounded 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
  • anonymous
to show that an objct is negativly charged we need to ?
anonymous
  • anonymous
it should be positve right?
anonymous
  • anonymous
but the answer is negative?
anonymous
  • anonymous
The positive sphere will cause the rod to become negatively charged.
anonymous
  • anonymous
ok thats it for tonight
anonymous
  • anonymous
do u come here often ?
anonymous
  • anonymous
around this time ?
anonymous
  • anonymous
occasionally. I'm kind of erratic about when I'm on
anonymous
  • anonymous
ok i am going to sleep learned so much today need to settle in the brain
anonymous
  • anonymous
thanks alot!
anonymous
  • anonymous
hope to see u again (kinda hard to meet people here that are good at physics lol)

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