Recent content by Yayness

I'll try to explain it shortly. The path of the clock B forms a circle. Let this circle be at rest relative to the inertial frame and let the radius be r. I added a non-inertial frame of reference where B is at rest. This reference frame does not rotate around its own axis, so left will always...

Let a clock A be at rest in an inertial frame and let a clock B rotate around it with constant velocity and constant distance from A. Let v be the velocity of B relative to the inertial frame. Both clocks count how long it takes for B to rotate once. According to clock A, the time it takes is...

Let A and B be two points that are at rest relative to an inertial frame S. Let's say the distance between them is 1·109 m. In the beginning you are at rest relative to S, but during exactly 1 second (counted by your wristwatch), you have managed to accelerate up to the velocity v=0.99c. Let's...

I mean two objects where one of them is at rest in one reference frame, and the other one is at rest in another reference frame. If one object uses a force on the other one, will the value of this force be different in the two reference frames?

I am quite new to relativity, but I have read about it and tried to derive different formulas. (So far, I have derived formulas for length contradiction, time dilation and velocity addition for instance.) I read about relativity on the internet and I saw this formula...

One reference frame A uses a force F on another reference frame B. Both A and B counts the time it takes before A stops using the force F on B. They end up with different results. Will B feel a different value of the force coming from A? Will F*t be different in the two reference frames? (Where...

Useful hint: All the prime factors of a square number above 1, can always be sorted in pairs where every pair contains two equal prime factors. Only square numbers have this ability. For instance: 12² = (2·2·3)² = 2·2 times 2·2 times 3·3 That means if a square number is divisible by a prime...

Are you sure you wrote it correctly? The only two cases when \sqrt{2(2r+r\sqrt 3)}=r\sqrt{2r+\sqrt 3}, are when r=0, or when r=\frac{\sqrt{35+16\sqrt 3}-\sqrt 3}{4}. Did you mean \sqrt{r(2r^2+\sqrt 3 r)}=r\sqrt{2r+\sqrt 3} ?

Oh, yeah I know what you mean. That's weird. :bugeye: Perhaps some people just try it a few times because they believe they'll win anyway since the probability of winning is so high.

If you bet on 35 numbers, you win if 1 of them is correct. But you should still not try this, because the few times you lose, you lose a lot more euros than what you gain. If you play for long enough, you'll lose all your money.

Ok, I suppose this is what you mean: There are 37 numbers in total. If you bet on one number (probability of winning is 1/37), you get 35 euros if you win, and you lose 1 euro if you lose. If you bet on 35 numbers (probability of winning is 35/37), you get 1 euro if you win, and you lose 35...

Hehe ok, thanks! :smile: Hmm interesting, that means it'll work for n=13 as well. Anyway, I believe it would work for any n-gon if you allow any a'th root (and if the following method is allowed). One could just take the the n'th root of i. This number to the power of 4, would then be...

JThompson is right. If you wish to combind \binom{n}{k} and \binom{n}{n-k} to get \binom{n}{k}+\binom{n}{n-k}=2\binom{n}{k}, you need to stop in the middle, or you will double the whole thing and get 2·2n=2n+1. If n is odd, you get...

It is, but it's not recommended if you want to calculate really large amounts of decimals in π. Even though it goes quickly towards π, you need to calculate a lot more decimals in \sqrt 2 than what the number of correct decimals in π will be. Let's say you calculate k decimals in \sqrt 2, then...

Ok, we'll allow imaginary numbers. Then, if n is a number on the form you mentioned in the quote below, the n-gon is "constructible". But that's not the only numbers it would work for. The solutions to 64x6-112x4+56x2-7=0 (or if u=x2, then 64u3-112u2+56u-7=0) will be the real parts of the...