Action and reaction forces during collision How can the action and reaction force be same.? And if so then how the colliding objects could further move as there net force is Zero.  Since F1 = - F2, that means that the force act on both sides and as there magnitude is same then undoubtedly there is no such net or derived force for which one could move further.
Given Newton's third law, why are things capable of moving? Given Newton's third law, why is there motion at all? Should not all forces even themselves out, so nothing moves at all? When I push a table using my finger, the table applies the same force onto my finger like my finger does on the table just with an opposing direction, nothing happens except that I feel the opposing force. But why can I push a box on a table by applying force ($F=ma$) on one side, obviously outbalancing the force the box has on my finger and at the same time outbalancing the friction the box has on the table? I obviously have the greater mass and acceleration as for example the matchbox on the table and thusly I can move it, but shouldn't the third law prevent that from even happening? Shouldn't the matchbox just accommodate to said force and applying the same force to me in opposing direction?
Finding $\int_0^\infty \frac{\cos(ax)-\cos(bx)}{x}dx$ I am practicing calculus section of GRE Math Subject test, and I can't figure out a way to do the following integral: $$\int_0^\infty \frac{\cos (ax) - \cos (bx)}{x}dx.$$  I absolutely have no clue how to do this. Can someone show me the solution explicitly? Thank you. How does a uniform prior lead to the same estimates from maximum likelihood and mode of posterior? I am studying different point estimate methods and read that when using MAP vs ML estimates, when we use a "uniform prior", the estimates are identical. Can somebody explain what a "uniform" prior is and give some (simple) examples of when the MAP and ML estimators would be the same? Unable to mount NTFS external hard drive Cleanest way to toggle a boolean variable in Java? Is there a better way to negate a boolean in Java than a simple if-else?  if (theBoolean) {     theBoolean = false; } else {     theBoolean = true; } Printing an SE page in Firefox shows only the first page I used to print Q&amp;A pages to keep a base of PDFs. I used either the 'print edit' addon or simply the print function of Firefox, which both allow to print a webpage in a file.  Usually the printed file for a webpage extends on 2 or 3 pages. But I'm not able to print more than one page on this site expressly anymore. No matter the method. I just get the top of each discussion. Check existence of input argument in a Bash shell script When polynomials f(x) and f'(x) are relatively prime, f(x) has no repeated roots. Why? The problem is to show that a polynomial $f(x) \in F[x]$ (F is a field) has no repeated roots if and only if f(x) and f'(x) (the derivative of f(x)) are relatively prime. I've managed to prove one direction of this equivalence (if there are no repeated roots, f(x) and f'(x) are relatively prime), but the other one gives me headaches... can anyone help out? Unintentional Lies and Misinformation Show that $A\in\mathbb{C}_n$ is normal $\iff$ $tr(A*A) = \sum_{i = 1}^n|\lambda_i|^2$, where $\lambda_1,...,\lambda_n$ are the eigenvalues of $A$. If $X,Y$ are equivalence relations, so is $X \times Y$ This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:  Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$. To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$. In naive set theory, I believe ∅ = {∅} = {{∅}} is correct, but just wanted to make sure that I understood this correctly.   ∅ is an empty set, so having an empty set as an element of a set that contains nothing else is pretty much the same thing as e.g. marking the number 2 as 2.000000, right? Doesn't the symbol ∅ intrinsically imply that this is an empty set which contains it self? Finding probability of other child also being a boy A man visits a couple who have two children.One of them, a boy, comes in to the room. Find the probability p that other is also a boy  a) 1/3 b) 2/3 c) 1/2 d) 3/4  Correct Ans: a) 1/3  However I am getting answer as 1/2.My solution is as follows.  The couple has two children, hence the sample space is as follows.  B- Boy G - Girl  BB BG GG ( BG and GB mean the same thing as Girl,Boy), but it is already known that one is boy.So the sample space reduces to  BB BG  So now p is 1/2  Is there some mistake is my approach ? How to prevent an email from accidentally being forwarded? Is it possible to prevent an email from being forwarded, or to have sensitive information automatically removed when reply-all/forward is pressed?  For example, a Google Calendar invitation includes a link to directly accept/reject the event. However, if someone clicks reply-all, and they keep the HTML formatting, the link will be shared with everyone else who received the invitation, and they can change the user's RSVP. (Specifically, the email is sent only to you, but has a "To:" header showing all invitees, so the problem remains.)  What can be done to ensure that the recipient does not accidentally share the sensitive link? Would it be possible with some MIME trickery or with a semi-standard header? Or is the best practice to refrain from including such "direct action" links when someone might be tempted to reply-all? How to choose a circle for default stream in Google Plus? How can I choose a specific circle for the default stream in Google+ ? Many extensions don't work, and are not comfortable if using another PC. For example:  "Last week, I found out that NASA stands for 'National Aeronautics and Space Administration.'"  or  "Last week, I found out that NASA stood for 'National Aeronautics and Space Administration.'" In Knuth's book "The Art Of Computer Programming Vol.1" there is a description about Euclid's algorithm to find the greatest common divisor of m and n.   And there is a phrase.     $m = qn+r$.  If $r = 0$, then $m$ is a multiple of $n$, and clearly in such a case $n$ is the greatest common divisor of $m$ and $n$. If $r \neq 0$, note that any number which divides both $m$ and $n$ must divide $m -qn =r$ , and any number which divides both $n$ and $r$ must divide $qn+r =m$.   I don't understand why any divisor of $m$ and $n$ must divide $m-qn =r$. Is it possible to do a multiline pattern match using sed, awk or grep? Take for example, I would like to get all the lines between { and }  So it should be able to match    1. {}  2. {.....}  3. {.....      .....}  Initially the question used &lt;p&gt; as an example. Edited the question to use { and}. Is the Stack Exchange engine that powers Stack Overflow available to download or install internally for an enterprise or company?  I think Stack Exchange's engine is very great and could be very cool to use for internal enterprise patterns and practices, like the engine of Wikipedia. Why couldn't Doctor Strange use his sling ring in space?