Let's suppose we have a spaceship with  the exact speed of light. If a traveller takes this spaceship to go to proxima centauri (approximately 4 years light away from Earth) and come back, we (as observers on Earth) will see the ship coming back after approximately 8 years.  But how much time would have passed for the traveller on the ship? How can be this calculated with a formula?
Suppose I wanted to travel to one of the recently discovered potentially Earth-like planets such as Kepler 186f that is 490 light years away. Assuming I had a powerful rocket and enough fuel, how long would it take me?
How does one go about and try to find all subclasses of a given class (or all implementors of a given interface) in Java? As of now, I have a method to do this, but I find it quite inefficient (to say the least). The method is:    Get a list of all class names that exist on the class path Load each class and test to see if it is a subclass or implementor of the desired class or interface   In Eclipse, there is a nice feature called the Type Hierarchy that manages to show this quite efficiently. How does one go about and do it programmatically? I want to ask a question about modular arithmetic. I know, that modular multiplicative inverse exists only if modulo and integer are relatively prime. I want to know, are there any ways of division in modular arithmetic, if modulo and integer aren't relatively prime? I tried to find info about that, but failed. I understand the proof that  $$Var(aX+bY) = a^2Var(X) +b^2Var(Y) + 2abCov(X,Y), $$  but I don't understand how to prove the generalization to arbitrary linear combinations.  Let $a_i$ be scalars for $i\in {1,\dots ,n}$ so we have a vector $\underline a$, and $\underline X = X_i,\dots ,X_n$ be a vector of correlated random variables. Then $$ Var(a_1X_1 + \dots a_nX_n) = \sum_{i=1}^n a_i^2 \sigma_i^2 + 2 \sum_{i=1}^n \sum_{j&gt;i}^n a_i a_j \text{ Cov}(X_i,X_j)$$  How do we prove this? I imagine there are proofs in the summation notation and in vector notation? What we've gelost — why doesn't English use the prefix "ge-"? The Germanic languages that I'm familiar with all use a prefix similar to ge- on past participles:  German: Ich habe mir den Fuß gebrochen. Dutch: Ik heb mijn voet gebroken.  But English doesn't do this at all:  English: I've broken my foot.  Where did this prefix come from? Did English ever have the ge- past participle prefix? If so, why was it lost? "no public key available" while upgrading using update-manager How to create a database directory structure for MySQL? I was following this tutorial:    So far, I installed mysql-server mysql-client libmysqlclient-dev, and I was suppose to tell MySQL to create its database directory structure where it will store its information using sudo mysql_install_db.  $ sudo mysql_install_db  [WARNING] mysql_install_db is deprecated. Please consider switching to mysqld --initialize [ERROR]   The data directory needs to be specified.   I thought running mysqld --initialize would solve the issue:  $ mysqld --initialize   mysqld: Can't create directory '/var/lib/mysql/' (Errcode: 17 - File exists) [Warning] TIMESTAMP with implicit DEFAULT value is deprecated. Please use --explicit_defaults_for_timestamp server option (see documentation for more details). [ERROR] Aborting   How do I tell MySQL to create its database directory structure where it will store its information? When I double-click on a script in Nautilus to run it, the script just opens in my text editor with no option to run it. Using Nautilus, how do I run executable text files and/or scripts? How to dismiss the Windows 8.1 Upgrade banner? In the beginning of the series, it's mentioned that Muggle-born wizards/witches are identified due to inadvertent spell-casting. However, it seems like later in the book, disarmed wizards/witches are completely helpless.   What is the explanation for why some spells can be cast without a wand and is there an in-universe explanation for why this isn't done? Also, are there spells that can be cast non-verbally (with/without a wand)? The question is $$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$ So my solution went like this:  $$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$ Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$. Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).  Now my question is: Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$ So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$. Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning? Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution  I realised my mistake after some hours of pondering over it. (In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this:  $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time} $. It was just a coincidence that the value I calculated was divisible by $1979$. The correct way, indeed, would be through the link supplied Where can I rent a decent mountain bike in Zürich? I am going for a 4 day trip in Zürich and I would like to bike quite a lot. Though biking is quite advertised, I failed to find a decent bike rental shop in the city. Would highly appreciate some tips! How can I safely prototype a mains power circuit? I'm building a device that has a mains power component, which I need to prototype before building to make sure everything works. I know I need to make sure wiring—and insulation!—is thick enough to cope with the higher current, and that I need to make sure there's enough space between exposed components so there's no arcing. I'll also be stashing whatever I prototype in a nice, insulated box before applying power, and I'll be include a small-value (250mA) fuse just in case. I'll be considering everything double-insulated rather than earthed (plastic enclosure, and there's nothing I'll need to touch while it's operating).  My plan is to solder (through-hole) components directly to each other, then insulate everything (with heatshrink tubing or electrical tape) as much as possible, then use screw-type cable joiners to connect the power cable.  I know I can't use traditional prototyping tools such as breadboards and stripboard for mains voltages, but what can I use to make prototyping at mains voltages easier?  (To be clear, 'mains voltage' in my area means 230V AC +10/-6% at 50Hz; GPOs are generally rated at 10A.) Is there a limit to a total number of months a holder of UK multiple entry visitor visa can spend in UK per calendar year? A relative of mine has received a multiple-entry UK visa for 2 years. Is there a particular limit on how many months in total per year she can stay in UK? The way I understand the current rule is that such visitors can enter UK as many times as they want as long as they do not stay for longer than 6 months per single visit. However, does this mean that if she stayed with us for 5 months and went back home for 1-2 months, she could come and stay for another 5 months? I heard that it is advisable to spend only 180 day per calendar year in UK and not more, but I could not find that rule in any official UK Home Office documents. Thanks in advance for any info about this. I'm planning to use ZFS on my system (HP ML370 G5, Smart Array P400, 8 SAS disk). I want ZFS to manage all disks individually, so it can utilize better scheduling (i.e. I want to use software RAID feature in ZFS).  The problem is, I can't find a way to disable RAID feature on the RAID controller. Right now, the controller aggregates all of the disks into one big RAID-5 volume. So ZFS can't see individual disk.  Is there any way to acomplish this setup? Any solutions to Veritasium's hydrodynamic levitation? So this Youtube video has been around for a while now  . Also this  may be related.  I just wanted to know if there is any standard hydrodynamic terms to explain it. Derek clearly states it is not about Bernoulli principle, as in this case the ball rotation is important.  Has any recent paper/videos discussed what happens if you do the same experiment with hydrophobic spheres? Can it be predicted with laminar flow theories or does it need turbulence?  Too many questions have been posed, but not so many responses on the internet yet.  Edit: I said Derek stated "it wasn't Coandă effect", yet what he says is that it is not "just Bernoulli's principle". Coandă effect is the typical hairdryer ping-pong ball levitation which is explained by Bernoulli's principle.   Edit2: rephrasing.  Additional comment: There seems to be two explanations "water adhere to the sphere and is ejected conserving horizontal momentum" and "Magnus effect" (ME). The later is not so obvious to me as the system is in equilibrium and there are two fluids, water and air, (does ME means air pushes the ball horizontally?).   Edit3: DESCRIPTION To avoid rot links and people that do not want to see the video, I'll try to explain the phenomena. You have a sphere and a water jet. The water jet is vertical (opposite to gravity), and touches the sphere almost tangentially. The ball not only rotates due to the water jet but also stays at constant height from the ground. The angular velocity can be defined in the direction of the cross product between the position of the contact point between the jet and the sphere (origin in the center of the sphere) and the direction of the jet. Some water drops fly away at the top of the sphere away from the jet.  Edit4: New question, would this phenomenon occur if there was no air? Does a passive Perception (Wisdom) check add WIS mod + Perception skill? I'm new to D&amp;D. There seems to be a discrepancy between the 5e Starter Set Rulebook and the associated character sheets. The Rulebook says that a passive wisdom (percep) score is "10 + the creature's wisdom modifier, as well as any bonuses."  In the example they have a 1st-level character (with a proficiency bonus of +2) Wisdom of 15 (+2) and a  proficiency in Perception, he or she has a passive Wisdom (Perception) of 14 (10+2+2).  But the Human Fighter Folk hero character has a WIS mod of +1, and a +3 written beside perception, and a passive Wis percep of 13 (not 14?).  The Dwarven Cleric has Wis mod of +3 and a +3 written beside perception, and only a 13 passive wisdom.  I guess this has to do with proficiency in the skill, but I'm confused. Can someone explain these instances and maybe also when proficiency is used, vs when just the modifier or just the skill without proficiency? Where can I find a reference to the format of the comment field (field 5) of the /etc/passwd file? In the /etc/passwd file on my system, the "comment" field, field 5, is inconsistent in its contents.  I thought that I could extract it to get the full name of the user.  fullname=`awk -F: '$1 == name {print $5}' name=$LOGNAME /etc/passwd`   However this returns with $fullname containing a name with 0, 3, or 4 commas following.  Exploring the man page (man 5 passwd) provides no details of this field other than describing it as "user name or comment field."  Perhaps there is additional information that is stored along with the user name? The change in energy of an object can be determined by the work equation, where work is the change in energy:  $$ W = F \cdot d $$  I conceptualize the transfer of energy as simply a series of small "packets" of energy being transferred at every Planck length. These small "packets" of energy add up to the total energy transferred (i.e. work). I'm not sure if this conceptualization is correct, so correct me if I am wrong.  However, it makes me wonder why the amount of energy transferred is dependent on distance and not time.  $$ m_1 = 10~kg \\ m_2 = 20~kg \\ W_1 = (10~N)\cdot(5~m) = 50~J \\ W_2 = (10~N)\cdot(5~m) = 50~J \\ W_1 = W_2 \\ t_1 \neq t_2 $$  If I apply a constant force on an object, why isn't the energy transferred at a constant rate with respect to time? The energy transfer rate varies dependent on how long it takes to cover the set distance.  In other words: why is the energy transferred consistent per unit of distance, and not per unit of time? In my cssh config file, I have:  terminal_font=6x13   this default font is too small for me.   man cssh says, I can specify other font size with the -f argument, while using "standard X font notation".  is a wikipedia page containing some other fonts types. Unfortunately, none of these work for me:  $ cssh host1 host2 -f "7x14" Fatal: Unrecognised font used (7x14). Please amend $HOME/.clusterssh/config with a valid font (see man page).   Why does "7x14" not work?  How can I use bigger fonts with cssh ? How many syllables are in the word 'hour'? Does the word 'hour' have 1 or 2 syllables?