Recent content by physeven

I think brhmechanic answered your first 2 questions but to answer your third one: dx on it's own is called a differential. For example, for f(x) = y, dy = f'(x)*dx, and dy is the differential of y. Realize here the the differential dy is a function of f'(x) and dx, so basically one could say...

for your answer to be true, dz/d(theta) has to equal zero. differentiating: dz/d(theta)=0-(A-B)sin(2∅)+2Ccos(2∅) if dz/d∅=0, then divide by cos(2∅) on both sides and you'l get 0=-(A-B)tan(2∅)+2C solve for tan and you'll get tan(2∅)=(2C)/(A-B)

Will not do anything to my knowledge. It might affect the star's atmosphere infinitesimally, but don't take that word seriously. If you do the math you might get an exact value, but I believe nothing will happen to the star.

you have to understand the hypothesized particle first: the graviton must have a spin of 2 and be massless. to answer your question: Just what is it that the gravitron is supposedly interacting with? the graviton, as aforementioned, is a massless particle with a spin of 2. therefore it cannot...

i highly doubt you can understand the depth of quantum mechanics without knowing the math behind it, which I'm sure that calculus AB or BC offers. However, another point of inquiry. Why are you taking physics, medicinal chemistry and quantum mechanics? seems like a rather odd combination to me...

i don't recall the derivation in a detailed manner, but if you consider time as a fourth dimension when trying to derive the lorentz factor for time dilation using a light clock you will get -c^2t^2. I assume it works similarly for the lorentz factor for length contraction, the lorentz factor...

F=ma?

you can use simple mechanics. consider the electron a point charge moving with a uniform drift velocity and i presume you will get the solution. just don't expect me to get it for you though.

what book are you using??

are you assuming an inelastic or an elastic collision?

i think what he's trying to say is: since an unstable neutron decays into an electron and a proton, an electron and an antineutrino is, how come the decay does not occur in the nucleus, there being no electrons inside the nucleus, only protons and neutrons. This is because the electron is...

Well i guess its funny now since we know better. but i can imagine myself laughing at the theory back then. I MYSELF would be the laughing stock... but fortunately i was born in a period where they are the laughing stock and not i. good point there.

try out chaos. a really interesting field for me. i have a couple of books on chaos that i read in my spare time.

c. c will always be c.

well i guess what i should have said from the beginning was 'put in terms of pi/2, pi/4 and pi/6.' that would've saved us some reply notifications :P.