Are smaller soap bubbles more accelerated by wind? If you blow a bunch of soap bubbles outside, and a gust of wind hits them, will the bigger ones be more or less accelerated by the wind than the smaller ones?
Intuitively, and maybe from remembered experience, I expect the small ones to be accelerated more. But I'm not sure.
Based on theory I would think they'd be affected about the same, because:
$$a = \frac{F}{m}$$
$F$ proportional to cross sectional area, which is proportional to $r^2$
$m$ proportional to surface area, which is also proportional to $r^2$
I'm assuming that bigger bubbles tend to have about the same thickness as smaller ones, but I don't know that.
Similarly, damping due to air friction should affect big and small bubbles about the same, right?
 A: Yes this is right. Friction is also proportional to cross-section which is r^2. So the acceleration should be approximately invariant. (The friction here is nothing but the wind in the opposite direction)
Here is another argument that makes the result much more clear:
Since these things are floating in the air, the mass of the bubble is irrelevant (ie the density is approximately that of air).
Second, any of these bubbles are very large compared to the microscopic properties of air.
Therefore we can treat the air as a continuous medium, and you have a completely scale-invariant situation where the size of the sphere have no influence.
A: F=ma.  Which means $a = \frac{F}{m}$.  To find out any variations in 'a' versus bubble size, we need to find how F/m varies with bubble size.
The force on the bubble due to wind is somewhere between $k r^2 v $ and $k r^2 v^2 $.  In either case, for any given wind velocity of interest, it is proportional to $r^2$: $F = k_1 r^2$.
The mass of the bubble is made of two components: The mass of the bubble film (MF) and the mass of the enclosed air (MA).  The mass of the film (MF) is proprtional to bubble surface area which is proportional to $r^2$: $MF = k_2 r^2$.  The mass of the entrapped air (MA) is proportional to bubble volume which is proportional to $r^3$: $MA = k_3 r^3$.  So...
$$F/m = \frac{k_1 r^2 }{MF + MA} = \frac{k_1 r^2 }{ k_2 r^2+k_3r^3} = \frac{k_1}{k_2+k_3 r}$$
So, acceleration due to wind is proportional to $\frac{k_1/k_3}{(k_2/k_3)+r}$, which means acceleration is INVERSELY proportional to $(k2/k3)+r$.  So, as bubble size goes up acceleration of the bubble goes down: The velocities of bigger bubbles are less affected by the wind.
A: This is a response to Georg: 
The condition of applicability of the Stokes formula $F_d=6\pi\mu Rv$ is:
$$Re=\frac{\rho_aRv}{\mu}<<1$$ $\mu$ is the dynamic viscosity of air  
From $mg=6\pi\mu Rv$;  $m=4\pi R^2l\rho_w$ we get:
$$v=\frac{2\rho_wglR}{3\mu}$$ and $$Re=\frac{2\rho_a\rho_wglR^2}{3\mu^2}<<1$$ and
$$R<<\sqrt{\frac{3\mu^2}{2\rho_a\rho_wgl}}$$ Now $\mu=1.8*10^{-5}$ Pa*s, $\rho_a=1.2$ kg/m^3, $\rho_w=10^3$kg/m^3, $g=10$m/s^2, $l=10^{-7}$m  
This gives: $$R<<0.1cm$$ We're not talking about such bubbles, i think.  
A: This is not a trivial question and it makes sense to consider at first the simplest case where a bubble falls freely in static air to find the bubble's terminal velocity.
$R$-radius of a bubble
$\rho_w$-density of water
$\rho_a$-density of air
$l$-thickness of the bubble 
The force on a bubble due to gravity:  
$$F_g=4\pi R^2l\rho_wg$$ Because it is difficult to decide when the actual drag force is proportional to $v$ or $v^2$ then to avoid Reynolds numbers lets choose a different way.  
Let $v$ be the terminal velocity of the bubble. If the bubble falls distance $h$ then it transmits to the air mass $m=\pi R^2h\rho_a$ amount of energy $E=F_dh=\frac{mv^2}{2}$. From this drag force:  
$$F_d=\pi R^2\rho_a\frac{v^2}{2}$$ So actual drag force is proportional rather to $v^2$, i think.  
Finally from $F_g=F_d$ we get the terminal velocity of freely falling bubble:  
$$v=\sqrt{\frac{8\rho_wgl}{\rho_a}}$$ Thus as a first approximation, terminal velocity of a bubble does not depend of it's radius but depends of it's thickness. So in general, answer to the question in title, i think, is: No
