To solve the problem it is necessary to apply the concepts related to Conservation of linear Moment.
The expression that defines the linear momentum is expressed as
P=mv
Where,
m=mass
v= velocity
According to our data we have to
v=10m/s
d=0.05m
[tex]A=13*10^6m^2[/tex]
Volume [tex](V) = A*d = (15*10^6)(0.03) = 3.9*10^5m^3[/tex]
t = 3hours=10800s
[tex]\rho = 1000kg/m^3[/tex]
From the given data we can calculate the volume of rain for 5 seconds
[tex]V' = \frac{V}{t}*\Delta t_{total}[/tex]
Where,
[tex]\Delta t_{total}[/tex] It is the period of time we want to calculate total rainfall, that is
[tex]V' = \frac{3.9*10^5}{10800}*5[/tex]
[tex]V' = 1.805*10^2m^3[/tex]
Through water density we can now calculate the mass that fell during the 5 seconds:
[tex]m' = V'*\rho[/tex]
[tex]m' = 1.805*10^2*1000[/tex]
[tex]m' = 1.805*10^5m^2[/tex]
Now applying the prevailing equation given we have to
[tex]P=m'v[/tex]
[tex]P = (1.805*10^5)(10)[/tex]
[tex]P = 1.805*10^6 Kg.m/s[/tex]
Therefore the momentum of the rain that falls in five seconds is [tex]1.805*10^6 Kg.m/s[/tex]
