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

V = 2.5m/s
Replacing we have to


At this point we know that 1 m ^ 3 of water is equivalent to 1000Kg (Its value in density at normal conditions)

