To solve this problem we will use the concepts related to the flow rate, which describes the volumetric amount of a fluid that travels a point in a given time. Mathematically it can be expressed as
Q = AV
Where
A= Area
V = Volume
Our values are given as
[tex]A = 15*8 = 120m^2[/tex]
V = 2.5m/s
Replacing we have to
[tex]Q = 120*2.5[/tex]
[tex]Q = 300m^3/s[/tex]
At this point we know that 1 m ^ 3 of water is equivalent to 1000Kg (Its value in density at normal conditions)
[tex]Q = 300\frac{m^3}{s} (\frac{1000kg}{1m^3})[/tex]
[tex]Q= 3*10^5kg/s[/tex]
