Two equal drops are falling through air with a steady velocity of 5 cm/s. If the drops coalesce, then the new terminal velocity will be.

The problem is based on the concept of terminal velocity. Terminal velocity is the maximum constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration. This speed happens when the force due to gravity equals the drag force.

The terminal velocity (v) of a droplet is given by the equation:

v = sqrt((2mg) / (ρACd))

where: m = mass of the droplet g = acceleration due to gravity ρ = density of the medium (air in this case) A = cross-sectional area of the droplet Cd = drag coefficient

When two equal drops coalesce, the volume of the new drop will be twice the volume of one drop, and since the volume of a sphere (which we can assume the drops to be) is proportional to the cube of the radius (V = 4/3πr³), the radius of the new drop will be cuberoot(2) times the radius of the original drop.

The cross-sectional area A is proportional to the square of the radius, so the new A will be (cuberoot(2))^2 = 2^(2/3) times the original A.

The mass m of the new drop will be twice the mass of the original drop, since it's made up of two drops.

Substituting these into the equation for v, we get:

v_new = sqrt((22mg) / (ρ2^(2/3)ACd)) = sqrt(2) * sqrt((2mg) / (ρACd)) = sqrt(2) * v_original

So the new terminal velocity will be sqrt(2) times the original terminal velocity, which is approximately 1.414 times the original terminal velocity.

Given that the original terminal velocity was 5 cm/s, the new terminal velocity will be 1.414 * 5 cm/s = 7.07 cm/s.