physics:

# Drag force for air

Drag: The drag force from air is usually written as Fd = 0.5⋅ρ⋅v²⋅A⋅C, where ρ is the density of air, v the speed, A the cross sectional area, and C a drag coefficient. Getting C is the hard part. The Reynolds number Re describes whether inertia dominates over friction. It can be thought as inertia/friction. Specifically, Re = ρ⋅v⋅D/μ, where D is the effective diameter of the object, and μ the dynamic viscosity. We can analyse two regimes: 1) high Reynolds number, where inertia dominates over friction, like in air, where C is more or less constant; 2) low Reynolds number, where friction dominates (like a protein in a cell), where C ∝ 1/Re, meaning that one v is cancelled and Fd ∝ v.


# Averaging speeds

A person cycles from A to B, the first half at speed v1=20 and the second half at speed v2=30. The overall average speed is the harmonic mean, v = 2 ∕ (1∕20 + 1∕30) = 24. Now, instead of splitting the distance in halves, the person covers the first third (fraction a=1/3) at speed v1=20 and the remaining distance (fraction 1-a=2/3) at speed v2=30. The harmonic mean no longer works. Instead, 1∕v = a∕v1 + (1-a)/v2, so v = 180 ∕ 7 ≈ 25.71. The first example, with the harmonic mean, could be rewritten as 1∕v = 0.5∕v1 + 0.5∕v2.


# Adding rates

Say a worker A paints a wall in 2 h while a worker B takes 3 h. Rate rA = 1∕2 (walls per hour), while rB = 1∕3. The total rate is simply r = rA + rB = 5∕6 = 1∕1.2 walls per hour. This means it would take 1.2 h for a wall to be painted by both workers together.


#How a refrigerator works

A liquid from a closed circuit is compressed by a pump, and since its pressure goes up at constant volume, its temperature increases. Then, this gas is circulated along an external radiator, a heat exchanger, so that its temperature will equal that of the external air. During this path along the coil, the liquid evaporates. Its pressure will still be high. The fluid continues its circuit and goes through an expansion valve, where it is allowed to go back to regular pressure, but this process will cool it. After that, it goes through an inner radiator, another coil path, where it will absorb heat from the inside of the refrigerator and condense again into liquid form, just before going back to the compressor for a new cycle. The compression and the expansion are approximately adiabatic (with no heat exchange).


# Syphon effect

Take a pipe and fill it with a fluid, no air gaps. One edge at a higher pressure than the other. Then, no matter the path of the pipe (even if it goes up and then down!), the fluid will flow from high to low pressure.


# Carnot's theorem

Consider a triangle with sides a,b,c and vertices A,B,C (ordered counterclockwise). Now, choose an inner point P and draw the (⟂) distances from it to the sides (the trilinear coordinates of the triangle), to find the pedal points Pa,Pb,Pc that define the pedal triangle. These points divide a,b,c in a special way. First, go counterclockwise from each vertex of the triangle to a pedal point, and get APc,BPa,CPb. The same now clockwise and get APb,CPa,BPc. Then, Carnot's theorem says that the sum of squares of the former set equals the sum of squares of the latter: APc²+BPa²+CPb² = APb²+CPa²+BPc². The inverse holds: if the latter equation is valid, then Pa,Pb,Pc form a pedal triangle.


# Malus' law

If linearly polarised light with irradiance I0 is passed through a polariser at an angle a, the outgoing irradiance I is such that (I/I0) = cos²(a). This means that if a = pi/2, I = 0. It also means that, if we want light polarised at pi/2 wrt the incident light with maximum intensity, the best approach is to place as many polarisers in the middle as we can, so that the angle between two consecutive ones is minimised. If the light undergoes n polarisations from angle 0 o pi/2, we get (I/I0) = (cos(π/(2n))^(2n). For n=0,1,2,... we get (I/I0)≈0,0.25,0.42,0.53,0.61,0.66,..., 1. So basically, a smooth transition would not waste any irradiance. The square of the cosine comes from I being proportional to the square of the electric field: I ∝ E².