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Q: What is h called in the equation?

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Free Energy

Not necessarily. Remember: q = hA(dT) h can change depending on these values, however remember the Nusselt equation used to find h. This Nusselt equation varies depending on the Reynolds number, Prandtl number, and sometimes (usually during free convection) the Raleigh number. The Prandtl and Raleigh numbers have terms in them which are functions of the material chosen.

Characteristic impedance (Z0) is defined as E/H ratio of {E,H} field. It depends on dielectric permettivity (epsilon), magnetic permettivity (mu) and geometry of region in which {E,H} propagates. For free space, it's easy to believe that geometry coefficient is 1 and in the end, you get -> Z0= square root (mu 0 / epsilon 0) = 120 pi, where subscript 0 means mu and epsilon referred to free space and pi=3.14... If you want to demonstrate that, you have to solve Maxwell's equation, imposing the condition of uniform plane wave travelling into free space, so you'll get an Helmholtz equation for Coulomb electric potential phi (you have to apply Lorentz's gauge condition and you'll get laplacian(phi) + k^2 phi = 0, where k=2*pi*frequency/c0 is called wave number). You solve this equation and put it into the equation linking magnetic potential vector (A) and phi. At this point, you can solve Maxwell equations and get E,H values and modulus ratio (Z0).

There are two degrees to a quadratic equation, as the x2 term is present. General form of a quadratic equation: Ax2+Bx+C

The moment of inertia of a cone about its central axis, start with the standard Intertia equation I = integral r^2 dm dm = rho dV (rho is density) (dV is basically volume) dV = r dr dtheta dx not going to prove that here but you will see in the integral that this does indeed form the volume. integral will be refered to as int from here on. This now forms the triple integral I = rho int(0 to H) int(0 to 2pi) int(0 to r) r^3 dr dtheta dx solving the integral leaves I = rho int(0 to H) int(0-2pi) 1/4 r^4 dtheta dx solving the second integral leaves I = rho int(0 to H) 1/2 pi r^4 dx ok so now you have to sub in the equation for r (the radius) of the cone r = (R/H)x this is the radius at the base divided by the height of the cone multiplied by the distance along the x axis. this equation gives you r at any point this gives you I = rho int(0 to H) 1/2 pi [(R/H)x]^4 dx time to do some housekeeping and take all the constants outside the integral I = (rho pi R^4) / (2 H^4) int(0 to H) x^4 dx this can now be solved and simplified to I = (rho pi R^4 H) / 10 At this stage your solution is complete, however you can tidy up the equation by taking out the mass term. m = (rho pi H R^2) / 3 split the Inertia term up to serperate out the mass term I = [(rho pi H R^2) / 3]*[ (3R^2)/10 ] this is now the complete solution in terms of mass I = (3mR^2)/10 I hope this manages to help some poor unfortunate student who gets set this question.

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