Radius of a Raindrop: Finding Mass & Volume

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Homework Help Overview

The problem involves calculating the radius of a raindrop using Stokes' law, given parameters such as the viscosity of air, density of water, and terminal velocity. Participants are exploring the implications of missing information, particularly the density of air, in their calculations.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of Stokes' law and the impact of air density on the calculations. There are attempts to derive the radius of the raindrop, with some questioning the significance of buoyancy in the context of the problem.

Discussion Status

Some participants have provided guidance on how to approximate the density of air using the ideal gas law. There is ongoing exploration of the assumptions regarding buoyancy and its relevance to the net force acting on the raindrop.

Contextual Notes

Participants note that previous parts of the question suggest certain assumptions, such as neglecting buoyancy, which may affect the interpretation of the forces involved.

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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