Iskalni zadetki

{\displaystyle [00,\beta ]_{|\beta }=\left(g^{\alpha \beta }\left[00,\alpha \right]\right)_{|\beta }=-{\frac {\kappa '\rho _{0}}{2}}\!\,.} Christoffelov simbol...

cos ⁡ ( α θ ) , {\displaystyle x=\rho \cos(\alpha \theta )\!\,,} y = ρ sin ⁡ ( α θ ) , {\displaystyle y=\rho \sin(\alpha \theta )\!\,,} z = θ , {\displaystyle...

\rho }: [α]λT=αl×ρ{\displaystyle [\alpha ]_{\lambda }^{T}={\frac {\alpha }{l\times \rho }}}. V enačbi za raztopine je gostota ρ{\displaystyle \rho }...

tsv=(π4)(kρc)[Tsv−ToP]{\displaystyle t_{sv}=\left({\frac {\pi }{4}}\right)\left(k\rho c\right)\left[{\frac {T_{sv}-T_{o}}{P}}\right]} k = toplotna prevodnost [W·m-1·K-1]...

{\displaystyle r\equiv \rho ={\sqrt {\frac {\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}}=s\mathrm {tg} \,{\frac {\alpha }{2}}\mathrm {tg} \,{\frac...

morske vode je: ρ=C(h)+β(h)S−α(T,h)−γ(T,h)(35−S)T{\displaystyle \rho =C(h)+\beta (h)S-\alpha (T,h)-\gamma (T,h)(35-S)T} Pri tem je gostota ρ izražena v kilogram/m³...

_{\alpha ,\beta =0}^{3}T^{\alpha \beta }S_{\alpha \beta }=\sum _{\alpha =0}^{3}\sum _{\beta =0}^{3}T^{\alpha \beta }S_{\alpha \beta }\equiv T^{\alpha \beta...

{\displaystyle \chi ^{2}={\frac {1}{N-M-1}}\sum {\frac {(\rho _{meritve}-\rho _{izracunan})^{2}}{\delta \rho ^{2}}},} kjer je N {\displaystyle N} število izmerkov...

{\displaystyle T^{\alpha \beta }{}_{;\beta }\,=0\,} . Enačbi včasih se poenostavi z zamenjavama: ρ → ρ − Λ c 2 8 π κ , {\displaystyle \rho \rightarrow \rho -{\frac...

p}{\partial z}}={\frac {\mathrm {d} p}{\mathrm {d} z}}=-\rho g=-\sigma \!\,,} kjer je: ρ {\displaystyle \rho \!\,} – gostota kapljevine, g {\displaystyle g\!\...

{\displaystyle \;_{2}F_{1}(\alpha ,\beta ,\gamma ,x)=a_{o}\left(1+{\frac {\alpha \beta }{1!\,\gamma }}x+{\frac {\alpha (\alpha +1)\beta (\beta +1)}{2!\,\gamma...

sicer: 2 π α = r e 2 6 R ρ , {\displaystyle {\frac {2\pi }{\alpha }}={\frac {r_{\rm {e}}^{2}}{6R\rho }}\!\,,} kjer je r e {\displaystyle r_{\rm {e}}\!\,}  elektromagnetni...

+ ρ {\displaystyle \sin \theta ={\frac {\rho }{r+\rho }}} in ρ = r sin ⁡ θ 1 − sin ⁡ θ {\displaystyle \rho ={\frac {r\sin \theta }{1-\sin \theta }}}...

polje: ∇ 2 ϕ = 4 π κ ρ . {\displaystyle {\nabla }^{2}\phi =4\pi \kappa \rho \!\,.} v n-razsežnem Gaussovem gravitacijskem zakonu: ∇ ⋅ g → = − 2 π n /...

[M^{\mu \nu },M^{\rho \sigma }]=M^{\mu \sigma }\eta ^{\nu \rho }-M^{\nu \sigma }\eta ^{\mu \rho }+M^{\nu \rho }\eta ^{\mu \sigma }-M^{\mu \rho }\eta ^{\nu \sigma...

_{n=0}^{\infty }{\frac {1}{R^{n+1}}}\int _{V}r^{n}\,P_{n}(\cos \theta )\,\rho (\mathbf {r} )\,d^{3}\mathbf {r} \!\,.} Pri tem je R radij-vektor od koordinatnega...

{I}{r^{2}}}\cos \alpha } kjer je I{\displaystyle I\!} svetilnost izvora v kandelah r{\displaystyle r\,} razdalja do izvora α{\displaystyle \alpha \,} kot, pod...

\varepsilon _{\alpha \mu \rho }\varepsilon _{\alpha \nu \sigma }=(d-2)!(\delta _{\mu \nu }\delta _{\rho \sigma }-\delta _{\mu \sigma }\delta _{\nu \rho })\!\,...

{\displaystyle P=\int I\,\mathrm {d} \Omega =2\pi \left[1-\cos \left({\frac {\alpha }{2}}\right)\right]I\,\!,} kjer je α vpadni kot. absolutni izsev navidezni...

{v} }{\partial t}}+(\mathbf {v} \cdot \nabla )\mathbf {v} =-{\frac {1}{\rho }}\nabla p+\nu \Delta \mathbf {v} +\mathbf {f} ({\boldsymbol {x}},t)\!\,...