Recent content by fogel1497

Thanks for your replies, mesh analysis is mandated for this problem. If i annotate my solution such that the 1 ohm resistor has the differences of the two currents crossing it rather then just the single current, is the remainder of my analysis and procedure accurate?

Homework Statement Please see the attached file for the problem statement. I am attempting to find the current across the .25 Ohm resistor. Homework Equations Using kirchhoffs voltage law I write the following equations: (1 Ohm)(2 amps) + (.25 Ohm)(2 - M1) = 0 The Attempt at a...

Thanks for your reply, i'd like some verification by someone on my physics if anyone can help.

Homework Statement An AC circuit consists of an alternative emf of 1 V connected to a resistor of 500 Ohms, an inductance of 0.4 mH, and two capacitors connected in parallel of 50 pF each, We want to find the resonance frequency of this circuit, the maximum power dissipated by the...

It doesn't exist? So when i do that method if I find i have thetas in my answer then the limit does not exist?

Homework Statement Find the limit of: (x^2+y^2+2xy)/(x^2+y^2)Homework Equations x = r*cos(theta) y= r*sin(theta) The Attempt at a Solution So what I did was change to polar coordinates. Then it simplifies to: (r^2 + 2r^2cos(theta)sin(theta) )/r^2 Factoring out an r^2 from everything you...

So by what your saying: Tension (directed downwards) + Gravity (also directed downwards) = Centripetal Force Therefore: 1. Centripetal Force = Tension + Gravity 2. (mass sattelite)(v^2)/(radius sattelite) = Tension + (G)(Mass sattelite)(Mass earth)/((radius sattelite)^2) 3. Sub in v =...

What about this for C: (mass sat)(w^2)(R sat) = (mass sat)(w^2)(R sat - R earth) + [ (G)(Mass Earth)(Mass Sat)/(R sat)^2 ] What this expression says is that the centripetal force of the sattelite must be equal to the tension in the rope plus the force of gravity on the sattelite. The radius...

Can anyone shed some light on C?

I have the same challenge problem in my physics class. Your choice for the anchor point is correct, but for the wrong reason. You place it at the equator not because of the value of g, but because if you place it anywhere else on the planet it wouldn't be in the same place at all times. Its...