Recent content by meme177

! I got threw off with all the sixes! Thanx a bunch!

An electron with speed 2.00×10^7m/s is traveling parallel to a uniform electric field of magnitude 1.16×10^4N/C . #1). How far will the electron travel before it stops? f=ma=Eq=-eE a=-eE/m I got a=-2.81x10^15 Then I used v(f)^2=v(i)^2+2a(x-x(i)) Solved for x... x=-(V(i))^2/2(a) x=.0712m #2)...

I got it! ... thanks guys!

Here is how it looks

no, the units are already there (degrees) counterclockwise from the +x axis

Mastering Physics is still counting the angles wrong :/

Yes, so θ1= -45, θ2= -135, θ3= 135, θ4=45 ? Am I doing the angles wrt positive x-axis right?

with the forces all pointing to the middle, they cancel (no movement) Now I have θ1= -45 θ2= -135 θ3= 45 θ4=135 Is this correct, now?

okay, I agree.

I think I see it! charge #2 is 45 ...Charge#1 (45+90=135)...charge#4 (180+45=225)...charge#3 (45+90=135) ??

Charge #2 where I calculated the y and x component of the force for question #1

θ1,θ2,θ3,θ4 I have tried these answers (in order) 225,225,225,225=WRONG (but then I was like NO this is impossible) Then I tried 135,45,315,225=WRONG I don't know what to do anymore

1. The problem Two negative and two positive point charges (magnitude Q = 3.16mC ) are placed on opposite corners of a square (side lengths a = 0.105m ) #1. Determine the magnitude of the force on each charge. I GOT... F1,F2,F3,F4 =7.44x10^6N (which was marked correct) I am having...