if $\vec L=I\vec ω$ holds good in all the cases then directions of angular momentum and angular velocity must be parallel always which is not true in some of the situations. So under what conditions does the relation $\vec L=I\vec ω$ holds good ?
While studying rotational mechanics, I came across a section where it mentioned that angular momentum may not necessarily be parallel to angular velocity. My thoughts were as follows:  Angular momentum ($L$) has the relation $L=I\omega$ where $\omega$ is angular velocity and $I$ is the moment of inertia, so following this relation, it seems they should be in the same direction. Why are they not?
A guy scammed me, but he gave me a bank account number & routing number. Can I use that to take out what he owes me? What would happen if every person in the world gathered at a minimum possible area and jumped all at once? Would that have any effect on Earth? How to get BSD FTP and Telnet back in 10.13 (High Sierra)? There are  with not having ftp and telnet present in 10.13 now as per Apple's dictate.  Has anyone found a solid way to bring BSD ftp and telnet back to macOS vs using inetutils? Completion handlers and return values Share Wireless connection with Wired Ethernet Port I am monitoring the number of pages indexed by Googlebot for one of my websites .  I noticed that since last week, the number of pages has dropped by about 60 or so. Now to be fair, I have not added that much over the last week. (Actually I don't think I added any new content.) But at the very least, I would have thought that the number of indexed pages would be static (if I was not adding anything new.)  I don't understand why the numbers have dropped.  Does anyone have an official explanation for this?. I have checked the Google Webmaster Tools, and have not come across anything that addresses this issue directly. I'm getting frequent inaccurate location as the same location, can I make this stop? Let $m$ be Lebesgue measure and $a \in R$. Suppose that $f : R \to R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. This is a corollary to a proof in Bass, but I don't understand why it follows from the proof he gives. I follow everything up until the last statement. Why is it that proving that the integral is $0$ for all Borel measurable sets guarantees it is true for all Lebesgue measurable sets? Am I missing something vital about the relationship between Lebesgue measurable sets and Borel measurable sets? Statement: Let $m$ be Lebesgue measure and $a \in\mathbb R$. Suppose that $f : \mathbb R \to\mathbb R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. Proof: We prove the conclusion first for intervals, then for unions of intervals, then for open sets, then for Borel measurable sets. For any interval $(c,d)$, we have $$\int_c^df = \int_a^d f - \int^c_a f = 0.$$ By linearity, if $G$ is the finite union of disjoint open intervals, then $\int_G f = 0$. Now, if $G$ is open, then $G$ is the countable union of disjoint, open intervals, $G = \bigcup^n_{i=1}I_i$. We have $$\int_G f = \int \lim_{n\to\infty} f \cdot \chi_{\bigcup^n_{i=1} I_i}$$ and since $f_n$ is integrable for all $n$, and $|f_n| \le f$ for all $n$, by the dominated convergence theorem, $$\int_G f = \lim_{n\to\infty} \int_{\bigcup^n_{i=1}I_i} f =0.$$ Now, if $G_n$ is a decreasing sequence of open sets converging to $H$, since $|f \cdot \chi_{G_n}| \le f \cdot \chi_{G_1}$ for all $n$, we again have by the DCT that $\int_H f = 0 = \lim_{n\to\infty} \int_{G_n}f = 0$. Finally, if $G$ is a Borel measurable set, then we know that there exists a sequence $G_n$ of decreasing, open sets, that converge to some set $H$, where $H \setminus G$ is a null set. Thus, $$\int_G f = \int_H f = 0.$$ We have found that for every Borel measurable set $G$, $\int_G f =0$. Since $f$ is real valued and measurable, $f = 0$ a.e. CNAME for top of domain? Is it possible to set a CNAME record at the top of a domain? (i.e. @ CNAME www, @ CNAME foobar.com., etc.)  My ISP says that it's only possible to use CNAME's for subdomains but I've read somewhere else that is should be possible even if not recommended. I heard Andrew Ng (in a video I unfortunately can't find anymore) talk about how the understanding of local minima in deep learning problems has changed in the sense that they are now regarded as less problematic because in high-dimensional spaces (encountered in deep learning) critical points are more likely to be saddle points or plateaus rather than local minima.   I've seen papers (e.g. ) that discuss assumptions under which "every local minimum is a global minimum". These assumptions are all rather technical, but from what I understand they tend to impose a structure on the neural network that make it somewhat linear.  Is it a valid claim that, in deep learning (incl. nonlinear architectures), plateaus are more likely than local minima? And if so, is there a (possibly mathematical) intuition behind it?   Is there anything particular about deep learning and saddle points? Adjust rounded rectangle radii without creating a new one in Photoshop C6 When I create new rounded rectangles, I see the radii and other information in the properties, but after I work on something else and go back to that layer, the information does not show. How do I adjust the radii of the rounded rectangles without deleting and creating a new one? How much money did Charlie Harper have? This is quite tricky to figure out, and I am not looking for a specific number, but for a approximation.  As you know, Charlie (nearly) always had a lot of money to spend, but a few hints suggest that he actually did not have money at all (talking about absolute, not cash flow)   he went bancrupt after a few months without income he told Alan that there is no money, and everything just a house of cards he worries about Bertha giving away a few clothes he has three mortgages on his house when he died, there was no inheritage to Alan or anyone, and they had to sell the house ...   However, considering the lifestyle and spendings, Charlie must have had a substantial amount available.   driving the latest foreign expensive cars owning a malibu beachhouse paying the property tax for said house discussing retirement at 45 spending endless money on booze, betting and women ...   So all things considered, did he have money, or was he just a scam? I am a beginner with Fourier series and I have to evaluate the sum   $$\sum_{n  =1}^{\infty}{\sin\left(n\right) \over n}$$  I don't know which function I have to take to evaluate the fourier series ... Someone can give me a hint ?  Thanks in advance! When is a CDATA section necessary within a script tag? Are CDATA tags ever necessary in script tags and if so when?  In other words, when and where is this:  &lt;script type="text/javascript"&gt; //&lt;![CDATA[ ...code... //]]&gt; &lt;/script&gt;   preferable to this:  &lt;script type="text/javascript"&gt; ...code... &lt;/script&gt; A question occurred to me when browsing SE this evening, just curious.  What specifies the inside of a 3D construct? If I have a hollow sphere, what's to say that the world isn't the inside, and the center of the sphere isn't the outside. Picture the Asylum in the Hitchhiker's Guide to the Galaxy series, if you have read that.  Additionally, if the inside of a 3D object is defined as what is smaller than the outside, what happens when there is a a sphere with a volume equivalent to 51% of the universe? $\mathbb{R}$ with the lower limit topology is not second-countable I am trying to prove that $\mathbb{R}$ with the lower limit topology is not second-countable.  To do this, I'm trying to form an uncountable union $A$ of disjoint, half-open intervals of the form $[a, b)$, $a &lt; b$. Is this possible? I think this would imply the $A$ is open but no countable union of basis elements could coincide with $A$ therefore making the real numbers with the lower limit topology not second-countable.  I think there must exist something like $A$ described above but I am having trouble visualizing it and coming up with a formula to represent it.  Maybe there is some other way to show it is not second-countable. 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