Recent content by Interesting

In your io equation, it's not Vo/8k. You have a voltage missing. Vo/8K assumes that the voltage over your 8K resistor is Vo, but there's a voltage that isn't ground connected to your 8K resistor.

http://imgur.com/vF77U Make a loop is the same as saying sum all the voltages in the loop. This is a simply voltage divider, and it will give you the voltage that's present at the + terminal. This is equal to the voltage at the - terminal. The gain of you op amp is a division of two resistors...

For power generating cycles, I think that's it.

If you make a loop around the + terminal, it will give you the voltage. Then you can figure out the gain (R2/R1). Then you can figure out the output voltage, and then get io.

Homework Statement The magnetic field inside a cylindrical solenoid of area 4 cm^2 is 0.15 T along the axis of the solenoid. What is the magnetic flux through a disk of radius 3 cm placed perpendicular to the solenoid axis? Homework Equations \phi = \zeta B * dA The Attempt at a...

Homework Statement L1 L2 x= 1+t x= 1+2s y= 1+6t y= 5+15s z= 2t z= -2+6s My professor said to find a normal vector and project the lines, but I'm new to this calculus and all these words are just a fuzzy cloud over my head. Homework Equations Projection of c...

Doesn't the even and oddity of a function also have to do with the Taylor Polynomial? For instance, cos(x) is even because the first term (f(0)) is 1, the second term f ' (0) = 0, f ''(0) = -1, so it's Taylor is 1-x^2/2!+x^4/4! ??

Alright, freshman at the University of Minnesota so I have a while to toss this around, but I would eventually like to own my own business someday. So, that leads me to my question. I am going to get a degree in Biomedical Engineering (the twin cities area is heavily populated with medical...

Ohhh good call! I got a final answer of 12pi, which is simple enough to make me think it's the correct one! Thanks Dick!

Okay, just to check in, the solution to the square root should be (x2+1)/(x2-1) (providing my algebra is correct, which it most likely isn't). Then I multiply that solution by (x2/2 - 1/2) (by multiplying the x into dy). Then I get two integrals, and then I long divide and get a really long...

Alright I'm still stuck. so dx/dy= 1/((x/2-1/(2x)) which is the same as 2/x - 2x, so (2/x-2x)(2/x-2x)= 4/x2-8+4x2, take a 4 out and you get 1/x2-2x2+1, get a common denominator and you get (x4-2x2+1)/(x2), factors to (x2-1)(x2-1)/(x2) + 1. I know this is going to be really obvious, but I can...

Thanks for the help, I don't completely follow you though. You can't explicitly write a formula for x(y). But you don't have to. You have to integrate 2*pi*x(y)*sqrt(1+x'(y)^2)*dy. dy=(x/2-1/(2x))*dx. Use that to find dx/dy and convert the integration over dy to an integration over dx. You...

Homework Statement The curve is rotated about the y axis, find the area of the resulting surface. y=(1/4)X2-.5ln|x| 1<_X<_2 Homework Equations S=2(pi)(f(x))\sqrt{}1+f'(x)^2 The Attempt at a Solution Alright I'm not entirely sure where to even begin. Since I'm rotating about the Y-axis I know...