ħ = h2πγ = 1√(1 − v2/c2)∇×E = −∂B∂tS = −kB Tr(ρ ln ρ)∇·E = ρε₀E2 = (pc)2 + (mc2)2−dEdx = 4πnz2mec2β2(e24πε₀)2× [ln 2mec2β2I(1 − β2) − β2]λ = hpBν(T) = 2hν3c21ehν/kT − 1E = hν∇·B = 0iħ ∂ρ∂t = [Ĥ, ρ]h = 6.626 × 10−34 J·sc = 1√(μ₀ε₀)E = mc2∇×B = μ₀J + μ₀ε₀∂E∂tB = aVA − aSA2/3− aCZ(Z − 1)A1/3 − aA(A − 2Z)2A ± δGμν + Λgμν = 8πGc4 Tμν
Sign in
Grade 10 Project SBA
PHYSICS · ICT
Ask the site owner for the username and password.