Recent content by arkturus

Homework Statement If a 70kg man is running 3 m/s horizontally, how many watts is he consuming? Homework Equations W = N-m/s N = mass*acceleration The Attempt at a Solution I would assume that in order to get the force in N one would use the gravitational constant of the Earth, but...

Ah thanks, simple enough. VeffIeff = V/sqrt(2) * I/sqrt(2) which simplifies to VI/2. Much appreciated.

I don't have a question about a specific HW problem, just a general questions. When calculating complex power, I understand it's S = Veff*Ieff(conjugate). However, S can also be calculate as S = 1/2 * V * I. Why is this? I feel I'm missing something simple, but I'm pretty lost because...

Thanks. I'm now down to: 1 0 0 0 1 1 0 0 0 It looks like x3 is a free variable, so x3 = t. In that case, x2 = -t and x1 = 0. Good to go?

Homework Statement Given the matrix A = [1 0 0 -2 1 3 1 1 -1] Find an invertable matrix X and a diagonal matrix D such that A = XDX^-1 Homework Equations A = XDX^-1The Attempt at a Solution I've found that the eigenvalues are -2, 2...

Homework Statement Determine the inverse of the matrix: -x 1 0 0 1 -x 0 0 0 0 -x 1 0 0 1 -x Homework Equations Augmented matrix method The Attempt at a Solution The augmented matrix would be the matrix above with the identity matrix alongside it. I'm unsure how to manipulate...

Thanks a lot everyone

I like your method, but I'm trying to do it through the current divider law. The correct answer is Ix = ((16||20) / (16||20 + 80)) * 10A I understand where the 10A comes from, that's just the total current. However, I don't understand why they are using 16||20 as Rtotal and 16||20 + 80...

Homework Statement I'm just trying to calculate Ix using the current divider law. http://i273.photobucket.com/albums/jj224/illway17/circuitS.jpg Homework Equations Ix = I_total * (R_total / R_x) The Attempt at a Solution I've had an issue with the voltage & current divider...

Thanks a lot guys.

Homework Statement In last weeks lab, I was asked to find the resistance of two unknown resistors. I hooked them up in series with a 10V source and measured the current across each one. The ammeter read 5.77 mA at Rx and .014 mA at Ry. Shouldn't the ammeter have read the same value at both...

Thanks a lot, you were a big help

Ah I got it, R should be replace with +12 volts. That way there is a net voltage of 0 throughout the circuit thus power at the top and bottom resistors must be 0?

Ah, got it. I was thinking in terms of P = I*V, but P = V^2/R works too. I think my issue with the problem is the phrasing of "dissipating power". Does that mean that power will be zero? I'm assuming the voltages would have to be 0 in order for power to be zero.

I'm honestly not sure what the voltage must be through a resistor in order for it to not dissipate power. Is there a relationship I'm missing?