I've written out the half reaction
8e- + 9H+ + SO42- = HS- + 4H2O
and I know the logK = 4.25 (that's the constant mentioned in the prompt)
I've written out the equilibrium statement of 10^4.25 = ([x^1/8]*[H2O^1/2])/([10x^1/8]*[e-]*[(10^-8.2)^9/8]
However, from there, it seems like I have two...
I've attached a figure I've made; I know I'm to assume the Earth is a perfect sphere in this case. Assuming the 103 degrees is measured as latitude, I've calculated the distance in kilometers (Xp in the second equation above) to be 1.1453e4 km. I know I need u = p at the turning point, but not...
@haruspex
alpha is the p-wave velocity in km/s.
Some depths and velocities occur twice because they're the depths to the layer and the thickness of layers can be reoccurring, and since the velocities of p-waves in these layers depends on many factors, they may repeat as well. Table 1 is...
@BvU Here's another screengrab of the spreadsheet with more detailed units and formulas. Still confused. I don't use p because 1/p (i.e., velocity) is the relevant value I need - 1/p is the value I am trying to find depth for.
I've created an excel spreadsheet with the given model in addition to calculating radius of the layer by subtracting depth from 6371. I've calculated Zf.
I've also found what I think is vs by doing alpha * (radius column/6371), but that could be wrong. I know I need to find where 20/6371 =...
I was told to solve the second equation above for x to get l2 and l2, but that only gets me those in terms of other unknowns. I'm assuming I just need to solve t for my knowns, but I keep getting caught up by my unknowns. i.e., solving for x gives x = (l2*L*v2)/(l1*v1+l2*v2). Please note "l" is...
I drew this diagram, it looks like theta is just 90 for the incident edge angles? I don't know an equation for focal distance, the lensmaker's equation has two distance variables and I don't know them
Would I be looking for an equation containing dZ to plug things into since that's the change in height? I don't know if I have ds or dX, though. I understand Snell's law in general but not the more advanced scenarios, I suppose.
I appreciate the diagram! In this case, don't I still need the refracted angle of the two outer rays? If the ray parameter has to stay constant, how can 4*the incident angle = 2*outer refracted angle while the same is true for the middle angle despite the two being different?
In that case, I don't know the distance traveled in the lower layer either so doesn't that just introduce another unknown?
Would the equation dz/ds = (1-p^2/u^2)^1/2 where p is just theta*slowness and use U = 1/2? Although in this case theta still seems like 0 for 3/5 of the incident rays.
So far all I can work out is that the angle of incidence of the outer two and inner two rays is zero degrees, however, I can't work out how to get started on the problem. I feel like I need to use vertical slowness rather than the normal snell's law since I'm working with a dZ rather than a dX...