Radius of a Raindrop: Finding Mass & Volume

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SUMMARY

The discussion focuses on calculating the radius of a raindrop using Stokes' law, given specific parameters such as the viscosity of air (1.8x10-5 Pa s) and the density of water (1x103 kg m-3). The terminal velocity of the raindrop is stated as 7.05 m/s. The correct formula to determine the radius is r = sqrt((9ηv)/(2g(ρ-σ))), where the density of air is approximately 1.225 kg/m3. The discussion emphasizes that buoyancy is negligible in this context, confirming that the net force is primarily due to weight.

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IDK10
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Homework Statement


Using the viscosity of air as 1.8x10-5Pa s, and the density of water as 1x103kg m-3, find the radius of a raindrop traveling at a terminal velocity of 7.05ms-1. Assume Stokes law can be used.

The problem is finding the density of air, I could use F = 6rvπη but I don't know F, because it says in a previous question to assume that in D + U = W, to assume U is negligible, and I don't know the weight, because I don't know the mass or volume of the raindrop.

Homework Equations


r = sqrt((9ηv)/(2g(ρ-σ)))

The Attempt at a Solution


Without a density of air, I get 9.31x10-5m, but that would be wrong since the density of air is not being included.
 
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Density of air is a small correction, so you can use the ideal gas law to find its approximate value.
 
IDK10 said:

The Attempt at a Solution


Without a density of air, I get 9.31x10-5m, but that would be wrong since the density of air is not being included.
Google "density of air" .(It is about 1.225 kg/m3)
And check your numerical result.
 
In your judgment, is buoyancy going to contribute significantly to the net force on the raindrop?
 
If by buoyancy you mean upthrust, then no, a previous part of the question says it is neglibible. So its just F = W.
 
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IDK10 said:

The Attempt at a Solution


Without a density of air, I get 9.31x10-5m, but that would be wrong since the density of air is not being included.
It is not correct, but not because of neglecting the density of air.
 

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