To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,
Q = V*A
Where,
A= Cross-sectional Area
V = Velocity
The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,


Our values are given as,


Re-arrange the equation to find the first ratio of rates we have:



The second ratio of rates is



Energy flows with kinetic energy
Answer:
17. h = l − l cos θ
18. 1.40 m
Explanation:
Let's call d the height of the triangle. We can then say:
h = l − d
Using trig, we can write d in terms of l and θ:
d = l cos θ
h = l − l cos θ
If l = 6 m and l cos θ = 40°:
h = 6 − 6 cos 40
h ≈ 1.40