Recent content by Alpha Russ Omega

Ah, I see. My book described that as well but the problem given already had the Is pointing into the + side of Vs so I figured I would just follow the original drawing. You are correct, the Vs did come out to be negative due to the odd sign placement. Thank you for your help!

Oh, ok. I think I'm starting to understand this. I redrew the set-up by labeling more currents and labeling other voltages. Did I set-up this drawing correctly? http://img405.imageshack.us/img405/4074/new6wi9.png Would this set-up work? Is=I2+I3 "left loop" Vs+I2*R2+Is*R1=0 "right loop"...

Hello. This is a strange problem that I ran into in my homework. I did not use one of the resistors in the second loop. I don't think this is correct and I am really confused. I have no idea how else I could set this up to solve for the needed variables. Find Vs and Is, if V2 = 20V, R1 = 6K, R2...

I think there is a product that sounds like what you are describing. http://www.valleydelivery.com/index.php?main_page=product_info&products_id=147&zenid=19bu1q4b34qv43gopq73oav0p1

There is something like this: Light is reflected in through fiber optic cabling. http://www.treehugger.com/files/2005/06/sunlight_direct.php and http://www.treehugger.com/files/2005/04/sunlight_in_a_c_1.php and http://www.sunlight-direct.com/products.html I'll take anything over florescent...

I think I got it figured out now. Thanks guys! :-)

PSC stands for Passive Sign Convention. The following link describes it in detail: http://morley.eng.ua.edu/E125-2PassiveConv.pdf With the image that my teacher gave me, it's kind of hard for me to tell if it follows it or not. Maybe I'm not looking at it correctly...

Hello: I need help determining if I and V assignments follow the passive sign convention. Here is the image of the circuit: http://img178.imageshack.us/img178/9572/pic1xu2.gif I know that in order for it to follow PSC the current I needs to flow towards +V. I'm not sure if that is...

Alrighty, I got it figured out. Thanks! I did the step-by-step thing and that worked out nicely.

Converging lens, object, and mirror Sorry for double posting, but I'm stuck on a similar one. This one is even more confusing to me. :frown: "An object is placed 1.00 m. in front of a converging lens, of focal length 0.54 m., which is 2.00 m. in front of a plane mirror. How far from the lens...

Ah... So the image distance must be b-a. Thus: [1/(a+b)]+[1/(b-a)]=1/f Then I would probably just solve for a. Thanks for your help!

Hello! I'm stuck on this problem... A thin flat plate of partially reflecting glass is a distance (b) from a convex mirror. A point source of light (S) is placed a distance (a) in front of the plate so that its image in the partially reflecting plate coincides with its image in the mirror...

Aha! I was trying to use the wrong equations. It's figured out now. Thank you! :-)

Homework Statement A sound wave of frequency 1250 Hz., propagating through air at 0 degrees Celsius, has a pressure amplitude of 15.0 Pa. What is the maximum particle speed (in meters/second)? Homework Equations v = frequency x lambda = (2 x pi x frequency)(lambda / 2 x pi) = omega / k =...

Thank you for the hint. That helped me solve the problem. :smile: