Recent content by Marc Rindermann

I don't understand it either. If ##s_1## and ##s_2## are the lengths of both springs then ##a = s_1 + s_2##. Now, if one spring stretches by ##x## then the other spring must be compressed by the same amount, ## (s_1 + x) + (s_2 - x) = a## And if ##V_1 = \frac12 k x^2## then ##V_2 = \frac12 k...

I've come here to post this one, even though I thought it might have already been posted

our university is transitioning to online delivery. However, physics labs, as of Friday last week, are still going ahead as normal. So I'll go into work unless labs are cancelled.

it means that you evaluate the differentiation and keep ##T## constant. Similarly you could find ##\left(\frac{\partial P}{\partial T}\right)_V## where you would treat ##V## as a constant

The Langrangian is an equation of the generalised coordinates ##q## and velocities ##\dot q##. To get from the Lagrangian to the Hamiltonian you perform a Legendre Transform and the Hamiltonian becomes a function of ##q## and ##p = \frac{\partial L}{\partial \dot q}##. So you need to consider...

what I learned only recently and find very interesting is the fact that even after the accident in unit 4 they continued construction on units 5 & 6 for two years until 1988 and operation of units 2, 1, 3 until 1991, 1996, 2000

somebody put it up on https://www.calvin.edu/~pribeiro/courses/engr315/EMFT_Book.pdf.

you want to know what the mass ##M## must be to balance the beam on the blue pivot point. That means the sum of torques must be zero or, since the the torques act on either side of the pivot point, they must be equal in magnitude. So you have one mass at ##3m## to the right of the pivot point...

unless the beam is attached to the blue structure it won't supply any support at all. if you look at your png again, you can see that the beam is lifted left of the the blue point. The beam is lifted left of the blue point and lowered on the right of the blue point. so the beam rotates around...

in whatever way you define your domain it is obvious that $$f(x) = \frac{x^2 - 1}{x - 1}$$ is not defined at ##x = 1##. Don't know what book the question comes from, however, I would take the position without having any further information that the function is defined everywhere where it can be...

MATLAB, Mathematica, Maple run natively on Linux. I guess because they were all developed in an academic environment and predate Windows by at least a decade.

How easy? very easy! the only real compatibility issues there are is that games usually are not (yet) ported to linux. but steam is doing a great service to promote games on linux as well. other compatibility issues might occur when using proprietary file types like any of the windows office...

personally I wouldn't include Halliday and Resnick. In my opinion the only criterium this book matches is its volume.

it could certainly be done in more than one step. However, now that you have the time it takes to get up to the highest point of her jump you'd need to calculate how high she's up in the air. you need the information to calculate the time it takes to fall from this height, given that the speed...

In other words, you have calculated the time after which she is at the highest point. Now you have to add the time it takes to fall down into the pool. she won't be at rest when she hits the water. she'll be at her highest speed just before she hits the water.