Solving Cloud Physics Problems - Brad's Request for Help

Therefore, the number of drops can be calculated using the equation: N = (3 * A * h) / (4 * pi * r^3)Where N is the number of drops, A is the area to be covered, h is the depth of rainfall, and r is the radius of each drop. In order to calculate N, we need to know the values of A, h, and r. The given values are A = 1 * 10^21, h = 0.5 cm, and r = 0.5 * 10^-4m. Plugging these values into the equation gives us:N = (3 * (1 * 10^21) * (0.5 cm
  • #1
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Hi I have those 2 problems:

A typical cloud contains droplets of water with an average radius of .5 * 10^-4m.
how many droplets are needed for a cloud that provides a rainfall of .5 cm.
( the answer must be 1*10^21 droplets)


for this second problem I don't know how to find two equations:

A force F acting on a body of mass m a distance r from some origin has a magnitude of F=(A*m*exp(h*r))/r^4, where A and h are both constants.
Given that the force has dimensions kilogram-meter per seconds squared, what are the dimensions of both A and h.

Please can I have some help or suggestions for those problems?

Thank you very much

brad
 
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  • #2
(a) [tex]N = \frac{3 A h}{4 \pi r^3}[/tex]

where A is the area to be covered, h = 0.5 cm is the depth and r is the radius of each drop.

(b) Hint: the argument of the exponential must be dimensionless - and - the dimensions of mA must be that of the force.
 
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  • #3
Yeah (b) shouldn't be a problem...

But what is (a)??

Granted I'm in high school but I've never encountered that kind of physics... where does it come from?
 
  • #4
It is college physics freshman
but I am sure U can do it :smile:

I have no idea how Tide find the solution for the first problem ( N=...)
I will try to work on it but if I don't find something I will ask him how he did it.
Thanks

brad

HiPPiE said:
Yeah (b) shouldn't be a problem...

But what is (a)??

Granted I'm in high school but I've never encountered that kind of physics... where does it come from?
 
  • #5
The total volume of the drops (number of drops times the volume of each drop) is equal to the volume of the fallen rain (depth of rainfall times the area covered by the rain).
 

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