Recent content by Afo

Thank you ergospherical and ergospherical. I have cleared all of my doubts. I believe that the sentences (in the book) are badly worded, and not because of the authors' knowledge. After reinterpreting it, I found out what they are trying to say.

I think this is consistent after reading other books

Correct me If I am wrong. I think the book meant to say that the entropy change of a closed system is ##\Delta S = \int_{i}^{f} \frac{dQ}{T}## while the second law of thermodynamics states that the entropy change of an isolated system is ##0## for reversible processes. While for irreversible...

Hmm.. I suspect that the book is wrong. See N. Virgo's comment there: https://physics.stackexchange.com/questions/50160/entropy-change-during-reversible-processes

Homework Statement:: Why is the entropy of a closed system constant in a reversible process, and not related by ##\Delta S = \int_{i}^{f}\frac{dQ}{T}## (See below for the question in more details) Relevant Equations:: ##\Delta S = \int_{i}^{f}\frac{dQ}{T}## I am reading chapter 24 of Physics...

If the ball rolls (which seems to be the case in your question), you must account for rotational kinetic energy. The slide must not be smooth (in order for the ball to roll), and the ball must roll without slipping in order for the mechanical energy to be conserved. (So that the work done by the...

:doh:Ok, I got it! ##\rho(V) = \frac{m}{V}## is a function of volume and ##m## is a constant. Thus ##\frac{d}{dV} \rho(V) = \frac{d}{dV} \frac{m}{V} = m\frac{d}{dV} \frac{1}{V} = -\frac{m}{V^2}##

I am still in high school so this is what I know: 1. We can view ##\frac{d}{dx}## as an operator so ##\frac{d}{dx}f(x) = \frac{df}{dx}## is the derivative of ##f## w.r.t. ##x##. Another way is to view it in terms of differentials: 2. ##dx## and ##dy## are called differentials. The...

Okay, I know why this is false. It doesn't work since ##\frac{a}{b} = \frac{c}{d}## doesn't imply ## \frac{a}{b^2} = \frac{c}{d^2}.## If that was true, that would mean ##|dV| = |V|## which is obviously false.

Like this? ##d\rho = \frac{\Delta \rho}{\Delta V}dV## but ##\frac{\Delta \rho}{\Delta V} = -\frac{m}{V^2}##

Is it like this? dm/dv^2 = - m /(V^2) so dp = dm/dv = dv * (-m/(V^2))

Homework Statement:: This is from 5 ed, Physics 1Halliday, Resnick, and Krane. page 428 about sound waves I have highlighted the equation that I don't understand. How did the author get it? I understand how they get from the middle side to the RHS of the equation, but I don't understand how...

Ok, thanks I got it.

Why is the total energy energy equal to the sum of translational kinetic energy and rotational kinetic energy? I understand the derivation KE = 1/2 I w^2 for a rigid object rotating around an axis: sum 0.5 * m_n * (v_T)^2 = sum 0.5 * m_n * (wr_n)^2 = 0.5 * w^2 * sum m_n r_n^2 = 0.5 * I * w^2...

Oh never mind. I got it. I kinda misunderstood what a chimney is. Here's the picture explaining everything: