Answer to Question #159222 in Mechanics | Relativity for Jordi Rodela

Question

One cubic centimeter of a typical cumulus cloud contains 110 water drops, which have a typical radius of 10 μm. (a) How many cubic meters of water are in a cylindrical cumulus cloud of height 3.1 km and radius 1.1 km? (b) How many 1-liter pop bottles would that water fill? (c) Water has a density of 1000 kg/m³. How much mass does the water in the cloud have?

Answer 1

$\text{let }V_{cm^3}\text{ volume of water in }1cm^3$

$V_{cm^3}=V_{drop}n$

$n= 110$

$\text{we assume that a drop is a sphere}$

$V_{drop}=\frac{4}{3}\pi{R^3}$

$R =10 μm= 10*10^{-6}m= 10^{-5}m=10^{-3}cm$  

$V_{drop}=\frac{4}{3}\pi{10^{-9}}cm^3$

$V_{cm^3}=\frac{440}{3}\pi{10^{-9}}cm^3$

$k = \frac{V_{cm^3}}{1cm^3}=\frac{440}{3}\pi{10^{-9}}\approx4.6*10^{-7}$

$k -\text{coefficient when multiplied by which total volume}$

$\text{we get the volume of water}$

$a)V_{cloud}=\pi{R^2}h$

$R=1.1km=1100m;h=3.1km=3100m$

$V_{cloud}=\pi*{1100^2}*3100\approx1.18*10^{10}m^3$

$V{water}=k*V{cloud}$

$V{water}=4.6*10^{-7}*1.18*10^{10}=5.43*10^3m^3=5430m^3$

$b) 1 -liter = 1dm^3=10^{-3}m^3$

$\text{count bottles}=\frac{V{watter}}{10^{-3}}=\frac{5.43*10^3}{10^{-3}}=5.43*10^6=5430000$

$c) m =\rho*V$

$m = 5430*1000=5430000kg$

Answer:a)5430 m^3 b)5430000 c)5430000 kg