Radius of a Raindrop: Finding Mass & Volume

In summary: The main issue is that you are using the wrong formula. In summary, the problem is to find the radius of a raindrop traveling at a terminal velocity of 7.05ms-1 using the given values for the viscosity of air and the density of water. The formula to use is r = sqrt((9ηv)/(2g(ρ-σ))), not F = 6rvπη. Additionally, the density of air can be approximated using the ideal gas law, and buoyancy is negligible in this scenario.
  • #1
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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  • #2
Density of air is a small correction, so you can use the ideal gas law to find its approximate value.
 
  • #3
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.
 
  • #4
In your judgment, is buoyancy going to contribute significantly to the net force on the raindrop?
 
  • #5
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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  • #6
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.
 

1. What is the radius of a raindrop?

The radius of a raindrop can vary greatly depending on factors such as the speed of the fall, the air temperature, and the humidity level. On average, raindrops have a radius between 0.5 and 4 millimeters.

2. Why is it important to find the mass and volume of a raindrop?

Knowing the mass and volume of a raindrop is important for understanding the dynamics of the water cycle and predicting weather patterns. It can also provide valuable information for agriculture, as rainfall is a crucial factor for crop growth.

3. How can you measure the mass and volume of a raindrop?

To measure the mass of a raindrop, you can use a scale that is sensitive enough to measure small amounts. To measure the volume, you can use a graduated cylinder or a micropipette to collect the raindrop and measure the amount of water it displaces.

4. What factors can affect the accuracy of the radius measurement?

The accuracy of the radius measurement can be affected by environmental factors such as wind, which can distort the shape of the raindrop. The method used to measure the radius can also affect the accuracy, as some methods may be more precise than others.

5. How does the radius of a raindrop affect its falling speed?

The radius of a raindrop does affect its falling speed, but not in a linear way. As the radius increases, so does the air resistance, which slows down the fall of the raindrop. However, at a certain point, the raindrop will reach its terminal velocity and will not fall any faster regardless of its size.

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