Recent content by MedEx

It does! Thanks. so now i found the frequency and the angular frequency. any word on the phase angle or current amplitude?

Do you know what the equation I would need for that is?

Using the givens I found ωd=862 rad/s, and with some help I got tanϕ=-1 and Z=56.5685 ohms. But I can't figure out any of the other variables. I'm not even exactly sure what "driving frequency" is and I can't find an equation for it.

I've been using ##B_y## as .0229 this whole time! no wonder none of the math made sense. Okay so I made 2 mistakes in my last one. The first was that I had .0229 as positive and the second was that I wrote in the value of ##B_x## where I should have put in the value for ##B_z##. I got ##v_y## as...

so that's ## 4.22e-17 = 1.6e-19(.0137v_y - 38.472) ## multiply both sides by 1.6e-19 and you get ## 263.75 = .0137v_y - 38.472 ## add 38.472 to both sides and divdie by .0137 and you get ## v_y ## = 22060. Then you relate ## v_y ## to ##v_x## with that means that ## v_x= .598253v_y## and...

I just don't really know how I would relate the two of them outside of how I'm using them to try to get the cross product, and that isn't working. Sorry.

I can put ## .02192 v_y-.03664 v_x=0 ## that means ## .02192 v_y=.03664 v_x ## and therefore ## v_y=1.67153 v_x ## which obviously doesn't work for the numbers got because my ## v_y ## was negative and not 1.67153 times my ## v_x ## value So we definitely know I'm wrong, but I already knew that

Homework Statement A proton moves through a uniform magnetic field given by B=(13.7i−22.9j+23.5k) mT. at time t1 v=(vxi +vyj+1.68k)km/s and the magnetic force is given by F= (4.22e-17i + 2.52e-17j). What are vx and vy Homework Equations F=q(v x B) The Attempt at a Solution First I converted...

https://edugen.wileyplus.com/edugen/courses/crs11476/art/qb/qu/c27/pict_27_16.gif Sorry about that. The figures show up on my screen but never on anyone else's i guess

1. Homework Statement The ideal battery in Figure (a) has emf = 7.7 V. Plot 1 in Figure (b) gives the electric potential difference V that can appear across resistor 1 of the circuit versus the current i in that resistor. The scale of the V axis is set by Vs = 18.9 V, and the scale of the i...

Yup that worked. Thanks to everyone c:

What do you get when you try to do it that way? I'm really not very good at calc so that wouldn't be my go to.

Well I thought it could, but now I'm nervous.

Did you get 24.781? I might have just took the original 83 number and flipped the sign. alright 24.7808+29.2864=54.0672 54.0672*8.85e-12 is 4.78495e-10C

okay so x is heading backwards into the negative direction. so the back face should be (1.65*-3.2+2.86)(cos 180deg)(10.24) = 83.3536 the total net flux is then 83.3536+29.2864 = 112.64 112.64*8.85e-12= 9.96864e-10C for my net charge? That doesn't seem like it works either