Recent content by Logan Land

d/dx ln(sqrt x^2+y^2+z^2) = x/(x^2+y^2+z^2) d/dy = y/(x^2+y^2+z^2) and d/dz = z/(x^2+y^2+z^2) so then would it become (x+y+z)/(x^2+y^2+z^2)? which is r*/r^2?

i tried to follow how I attempted the next question, gradient f if f=1/r (d/dx)xbar+(d/dy)ybar+(d/dz)zbar * (1/r) (-x/r^3)xbar+(-y/r^3)ybar+(-z/r^3)zbar = -(x,y,z)/r^3 =-rbar/r^3thats how I tried but I guess was wrong?

(ATTEMPT) gradient(ln||^r||) =((d/dx)x*+(d/dy)y*+(d/dz)z*)(ln||r||) =(x/r^2)x*+(y/r^2)y*+(z/r^2)z* =(x,y,z)/r^2 =(^r)/r^2 correct?

To solve the gradient f when f = ln |r| do I start with differentiating each x,y,z term of the vector?Like ln|x| ln|y|...etc.

The current runs from the negative side of the battery to the postive correct? So first should I try and replace what I can with equivalent components to clean up the circuit? or should try try and go ahead and use mesh analysis

Assume that this circuit has been connected as shown for a very long time. Re-draw the schematic and label the currents and voltages in and across all components including the batteries. All components are “ideal.”

Yes I see that, strange. anyhow so my voltage should be small correct?

the voltage I get is a huge number how can it be so big? edit: oh wait shouldn't it be Vc=V0e^(-t/rc)? a negative sign in the t/rc

then 100e^(1/(2500*(6.666*10^-6)))?

since the capacitors are in series can I just replace the 3 with a single using (1/(1/c1)+(1/c2))?

these formulas?

Two capacitors of value 10, 20 microfarads are connected in series. Then charged with a 100 volt battery. If the capacitors is disconnected from the battery, and then connected to a 2500 ohm resistor, what is the voltage across the capacitors after 1 second?

How do I evaluate this integral?

During lab I obtained the data in the attached photo. It then asked for me to plot the data and get an estimated FWHMP as well as a "Q" First question is what does FWHMP and Q stand for? Second, where on my graph would I estimate these?

yes I copied it correctly. Here is an attachment of the sample exam we were given for our test on Thursday