Iskalni zadetki

Φ2=Φ+1 {\displaystyle \Phi ^{2}=\Phi +1\ \!\,} z značilnostima: Φ−1=1Φ,oziromaΦ=1+1Φ{\displaystyle \Phi -1={\frac {1}{\Phi }},\qquad {\hbox{oziroma}}\qquad \Phi =1+{\frac...

( 1 ) , {\displaystyle r=f(\theta )=\phi ^{\theta \over \pi /2}\qquad (1)\!\,,} kjer sta ϕ {\displaystyle \phi \,} število zlatega reza (zlato razmerje)...

\mathbf {\hat {s}} }{\partial \phi }}=-\sin \phi \mathbf {\hat {x}} +\cos \phi \mathbf {\hat {y}} ={\boldsymbol {\hat {\phi }}}\!\,,} ∂ ϕ ^ ∂ ϕ = − cos ⁡...

) + C {\displaystyle \int \phi (x)\,dx=\Phi (x)+C} ∫ x ϕ ( x ) d x = − ϕ ( x ) + C {\displaystyle \int x\phi (x)\,dx=-\phi (x)+C} ∫ x 2 ϕ ( x ) d x =...

{\displaystyle (\pm 1,\pm 1,\pm \Phi ^{3}\!\,,} ( ± Φ 2 , ± Φ , ± 2 Φ ) , {\displaystyle (\pm \Phi ^{2},\pm \Phi ,\pm 2\Phi )\!\,,} ( ± ( 2 + Φ ) , 0 , ±...

{\displaystyle a={\frac {R}{\Phi }}\!\,,} kjer sta R {\displaystyle R\,\!} polmer očrtane krožnice, Φ {\displaystyle \Phi \,\!} pa število zlatega reza:...

{\frac {E_{x}}{E_{0x}}}=\cos(kz-\omega t+\phi _{1})=\cos(kz-\omega t)\cos \phi _{1}-\sin(kz-\omega t)\sin \phi _{1}} E y E 0 y = cos ⁡ ( k z − ω t ) cos...

{\displaystyle \nabla ^{2}\phi ={\partial ^{2}\phi \over \partial x^{2}}+{\partial ^{2}\phi \over \partial y^{2}}+{\partial ^{2}\phi \over \partial z^{2}}=0\...

Φ ( m , n − 1 ) − Φ ( [ m p n ] , n − 1 ) . {\displaystyle \Phi (m,n)=\Phi (m,n-1)-\Phi \left(\left[{\frac {m}{p_{n}}}\right],n-1\right)\!\,.} Če za...

\Phi _{1}} skozi tuljavo. Tuljave so nasajene na feromagnetnem jedru, na stebrih jedra. Vmes je še tuljavnik. Magnetni pretok Φ 1 {\displaystyle \Phi _{1}}...

e i ( ϕ 1 + ϕ 2 ) {\displaystyle r_{1}e^{i\phi _{1}}\cdot r_{2}e^{i\phi _{2}}=r_{1}r_{2}e^{i(\phi _{1}+\phi _{2})}\!\,} in: r 1 e i ϕ 1 r 2 e i ϕ 2 = r...

r ϕ = Q 4 π ε 0 r {\displaystyle U=\Delta \phi =\;\phi _{2}-\phi _{1}\qquad \qquad kjer\qquad \qquad \phi ={\frac {Q}{4\pi \varepsilon _{0}r}}} Mednarodni...

) ) = ρ ( r → ) , {\displaystyle \nabla ^{2}\phi ({\vec {\mathbf {r} }})\equiv \nabla \cdot (\nabla \phi ({\vec {\mathbf {r} }}))=\rho ({\vec {\mathbf...

L=∂μϕ∂μϕ−V(ϕ).{\displaystyle {\mathcal {L}}=\partial ^{\mu }\phi \partial _{\mu }\phi -V(\phi )\!\,.} Potencialni člen (V(φ)) je odgovoren za pojavitev zloma...

\cos \phi &r\,\cos \theta \,\cos \phi &-r\,\sin \theta \,\sin \phi \\\sin \theta \,\sin \phi &r\,\cos \theta \,\sin \phi &r\,\sin \theta \,\cos \phi \\\cos...

{8\pi }{\phi }}T_{ab}+{\frac {\omega }{\phi ^{2}}}(\partial _{a}\phi \partial _{b}\phi -{\frac {1}{2}}g_{ab}\partial _{c}\phi \partial ^{c}\phi )+{\frac...

t 0 ) {\displaystyle \phi =\phi _{0}+\omega (t-t_{0})} Kvocient ω = ϕ − ϕ 0 t − t 0 {\displaystyle \omega ={\frac {\phi -\phi _{0}}{t-t_{0}}}} imenujemo...

|}}\Delta {\widehat {\sigma }}=\arccos {\big (}\sin \phi _{s}\sin \phi _{f}+\cos \phi _{s}\cos \phi _{f}\cos \Delta \lambda {\big )}.\;\!} . Razdalja d...

{\begin{bmatrix}{\frac {e^{\phi }}{1+\beta _{0}}}&-{\frac {e^{\phi }\beta _{0}}{1+\beta _{0}}}&0&0\\-{\frac {e^{\phi }\beta _{0}}{1+\beta _{0}}}&{\frac {e^{\phi }}{1+\beta...

površino S s ploskovnim integralom: Φ m = ∫ S B → ⋅ d S → , {\displaystyle \Phi _{\mathrm {m} }=\int _{S}{\vec {\mathbf {B} }}\cdot \mathrm {d} {\vec {\mathbf...