i'll do my scientific best to command your fleet

Mar 29

[video]

[video]

[video]

Mar 15

(Source: twinklepowderysnow, via dwukropek)

Feb 25

[video]

pipolaki replied to your post: Nie będzie mojego zderzenia z żelbetową ścianą…
Przerażasz mnie

Siebie też, jestem ostatnią osobą, którą bym posądziła o optymistyczną postawę z gatunku yay go get it gurl! :D Ale taaaaaki fajny semestr mi się szykuje, trudny, ale fajny. A jak u Ciebie?

[video]

Nie będzie mojego zderzenia z żelbetową ścianą wymagań i oczekiwań stawianych przede mną w tym semestrze. Nie. To semestr rozbije się o mnie. O mój uśmiech, poczucie humoru, wygadanie, odwagę, determinację i ogólną likeability. Jestem zmotywowana jak nigdy, czekałam na powrót na uczelnię z niecierpliwością, a nie ze strachem, i póki co wszystko idzie zgodnie z planem. Jeśli tak zostanie do końca tygodnia, potem będzie już tylko z górki. To znaczy nie będzie w sensie wysiłku, jaki będę musiała włożyć, żeby robić wszystko systematycznie i na satysfakcjonującym mnie (mnie, nie kogoś, komu ubzdurałam sobie, że muszę zaimponować) poziomie, ale w każdym innym tak.

Będzie ciężko? Ok. Otóż drogi semestrze, rzucam się na ciebie ze wszystkim co mam, więc nie masz szans. Robienie ciekawych rzeczy z ciekawymi ludźmi pod okiem ciekawych ; ) prowadzących brzmi jak dobry plan na najbliższe 4 miesiące.

A każdemu lektorowi na egzaminach z angielskiego albo każdej nauczycielce z mojego starego liceum, którzy po usłyszeniu odpowiedzi na pytanie, co studiuję i jaki drugi kierunek, odpowiadają z sarkazmem ‘oh, so fascinating’/’arcyciekawe’ (nawet, jeżeli pan lektor jest przeuroczy i zaczyna od ‘what would such a delicate creature do in a nuclear power plant’, a potem stwierdza, że jestem C1, a pani nauczycielka to chyba najważniejsza osoba, którą spotkałam w swoim nastoletnim życiu, no i nauczyła mnie matmy), przesyłam gorące pozdrowienia.

Feb 17

1ucasvb: Better late than never. Happy Valentine’s Day! Here’s a heart’s sine function.

1ucasvb:

Better late than never. Happy Valentine’s Day!

Here’s a heart’s sine function.

1ucasvb: The familiar trigonometric functions can be geometrically derived from a circle. But what if, instead of the circle, we used a regular polygon? In this animation, we see what the “polygonal sine” looks like for the square and the hexagon. The polygon is such that the inscribed circle has radius 1. (There’s a very neat reason for this.) Since these polygons are not perfectly symmetrical like the circle, the function will depend on the orientation of the polygon. More on this subject and derivations of the functions can be found in this other post Now you can also listen to what these waves sound like This technique is general for any polar curve. Here’s a heart’s sine function, for instance

1ucasvb:

The familiar trigonometric functions can be geometrically derived from a circle. But what if, instead of the circle, we used a regular polygon? In this animation, we see what the “polygonal sine” looks like for the square and the hexagon. The polygon is such that the inscribed circle has radius 1. (There’s a very neat reason for this.) Since these polygons are not perfectly symmetrical like the circle, the function will depend on the orientation of the polygon. More on this subject and derivations of the functions can be found in this other post

Now you can also listen to what these waves sound like

This technique is general for any polar curve. Here’s a heart’s sine function, for instance

anndruyan: Electric Field Due to a Uniformly Charged Disk  Perhaps one of the most difficult concepts in general physics is uniformly charged surfaces. Really, I can’t think of anything more rigorous than E&M and complex integration. Usually with these problem sets we are given the task of finding the E field at a distance and/or the E field when two charges are in the system. So when given a uniformly charged disk and asked to find the E field at a distant point here’s how we do it: First we need to divide the disk into flat rings so we can calculate the E field at our point P by adding up all the rings. Each ring will be out at a distance y and will have a radial width of dy.  Next we’ll set up the E field equation E=kQ/r2, but write it as dE=kdQ/√(x2 + y2). Now we come to our first tricky part of E field integration; the dQ. When we do these problems we must remember our charge densities - linear, surface, volume; lambda, sigma, and rho respectively.  Linear - lambda, 1 dimension : dQ = λdx Surface - sigma, 2 dimensions : dQ = σdA Volume - rho, 3 dimensions : dQ = ρdV Here, we use sigma and use the relationship dQ = σdA = σ2πydy. All that means is that our charge, dq, is equal to the surface charge density, σ, times the area of the charged concentric ring.   Now that we have our charge set we can look at the x and y components of the E field. If you look at dEy you’ll see that as you go around the ring there is another y component that cancels each other out. Therefore we only focus on dEx since the vector sum of dEy will add to 0.  Our dEx = kdQ cosθ /(x2 + y2). We use cosθ because on our free-body diagram dEx is the adjacent-hypotenuse component of dE. Further substitute the dEx equation by using x/√(x2 + y2) for cosθ  and 1/4πε for k. Only thing left to worry about is your trig identities of integration. Take the integral of dEx from 0 to R and you will find the E field at point P. 

anndruyan:

Electric Field Due to a Uniformly Charged Disk 

Perhaps one of the most difficult concepts in general physics is uniformly charged surfaces. Really, I can’t think of anything more rigorous than E&M and complex integration.

Usually with these problem sets we are given the task of finding the E field at a distance and/or the E field when two charges are in the system.

So when given a uniformly charged disk and asked to find the E field at a distant point here’s how we do it:

First we need to divide the disk into flat rings so we can calculate the E field at our point P by adding up all the rings. Each ring will be out at a distance y and will have a radial width of dy. 

Next we’ll set up the E field equation E=kQ/r2, but write it as dE=kdQ/√(x2 + y2).

Now we come to our first tricky part of E field integration; the dQ. When we do these problems we must remember our charge densities - linear, surface, volume; lambda, sigma, and rho respectively. 

Here, we use sigma and use the relationship dQ = σdA = σ2πydy. All that means is that our charge, dq, is equal to the surface charge density, σ, times the area of the charged concentric ring.  

Now that we have our charge set we can look at the x and y components of the E field. If you look at dEy you’ll see that as you go around the ring there is another y component that cancels each other out. Therefore we only focus on dEx since the vector sum of dEy will add to 0. 

Our dEx = kdQ cosθ /(x2 + y2). We use cosθ because on our free-body diagram dEx is the adjacent-hypotenuse component of dE. Further substitute the dEx equation by using x/√(x2 + y2) for cosθ  and 1/4πε for k.

Only thing left to worry about is your trig identities of integration. Take the integral of dEx from 0 to R and you will find the E field at point P. 

Feb 13

[video]

Feb 02

jamesnord: New York from Brooklyn. 

jamesnord:

New York from Brooklyn. 

(via scallawag)

Jan 27

[video]