Recent content by xxabr

Ohhhh. Okay, I got it. Thank you

Homework Statement It can be assumed that the Earth, to a good approximation, moves in a circular orbit about the sun. Determine the speed of the Earth.Homework Equations v2 = GMearth/rThe Attempt at a Solution mearth= 5.98x1024 R of Earth's orbit= 1.49x1011 G= 6.67x10-11 v2=...

Oops yeah, I made a mistake. h is 6.63x10-34Js. And I got it. Thank yoou

Homework Statement A photoelectric effect experiment uses a light of frequency 8x1014Hz and a metal with a work function of 1.2eV. A 1.5V stop potential is also applied. What is the maximum kinetic energy of the emmited electrons? Homework Equations I was thinking about using Ek= hf-W...

Uh... No, I don't think so. At least my textbook doesn't say that.. :confused:

Oh! Okay, thanks (: So... V= kq/r = (9x109)(3.2x10-19)/ (4x10-2) = 7.2x-8V ?

Homework Statement Find the electric potential difference at a point 4.0x10-2m away from an alpha particle. Homework Equations V= kq/r The Attempt at a Solution k= 9x109 r= 4.0x10-2 I'm not quite sure how to get this answer considering I don't have any charge and there's nothing...

Wow, yeah, that is easier. Thank you soo much. (:

Oh okay. I actually made a mistake it's not 2-x it's (2-x)2. I think I got the answer. I did: Fe1= kq1q3/r2 = (9x109)(1.6x10-5)(3.0x10-6)/x2 =0.432/x2 Fe2= kq2q3/r2 = (9x109)(6.4x10-5)(3.0x10-6)/(2-x)2 = 1.728/4-4x+x2 Then I made them equal each other and cross multiplied...

Oh, okay. That makes more sense. Sorry, if I don't it this right away. This unit is kind of hard for me right now. How do I find the net force for x? :/

Alright so, If I use E=kq/r^2 for 50x10^-6C ... R would be 0.45m and for -10x10^-6C ... R would be 0.30m?

Homework Statement Two spheres, with charges 1.6x10^-5C and 6.4x10^-5C are 2m apart. (The charges have the same sign). In the middle of the two spheres, should a third sphere,, of opposite charge 3.0x10^-6C, be placed if the third sphere has zero net electrical force? (1.6x10^-5) --------...

Homework Statement Find the electric field at the point given, x. (+50uC)-------0.45m---- (-10uC) -----0.30m----- ( x ) Homework Equations E= kq/r^2 Maybe, E= Fe/q and Fe= kq1q2/r^2 The Attempt at a Solution q1= 50x10^-6 q2= -10x10^-6 q3= ? I have no idea...