Black-body radiation as a photon gas consider a radiation cavity of volume v at temperature t. bose treats the cavity radiation as an ideal gas of photons in thermal equilibrium. a photon gas is an ideal boson gas with zero chemical potential because their number is not conserved. the energy of a photon is proportional to the frequency of the radiation: e = hw and the density of states in w space is: d(w) = w2 t (a) write down the energy density spectrum u(w, t) as a function of w and t (or b). (b) show that the total energy density w(t) scales as u (t) = ato and find b. (c) let wmax be the angular frequency at which the energy density u(w, t) reaches its maximum, shows that wmat t.

The independent variable would be the fertilizer as it is what the dependent variable (the growth of grass) depends on.

Answershe should not have multiplied the sulfur atoms by the subscript 4

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge