Answer :
Answer:
The depth of the resulting stream is 3.8 meters.
Explanation:
Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:
[tex]\dot V_{1} + \dot V_{2} = \dot V_{3}[/tex] (1)
All volume flows are measured in cubic meters per second.
Dimensionally speaking, we can determine the depth of the resulting stream ([tex]h_{3}[/tex]), in meters, by expanding (1) in this manner:
[tex]w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}[/tex]
[tex]h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}[/tex] (2)
[tex]v_{1}, v_{2}[/tex] - Speed of the merging streams, in meters per second.
[tex]h_{1}, h_{2}[/tex] - Depth of the merging streams, in meters.
[tex]w_{1}, w_{2}[/tex] - Width of the merging streams, in meters.
[tex]w_{3}[/tex] - Width of the resulting stream, in meters.
[tex]v_{3}[/tex] - Speed of the resulting stream, in meters per second.
If we know that [tex]w_{1} = 8.3\,m[/tex], [tex]h_{1} = 3.2\,m[/tex], [tex]v_{1} = 2.2\,\frac{m}{s}[/tex], [tex]w_{2} = 6.8\,m[/tex], [tex]h_{2} = 3.2\,m[/tex], [tex]v_{2} = 2.4\,\frac{m}{s}[/tex], [tex]w_{3} = 10.4\,m[/tex] and [tex]v_{3} = 2.8\,\frac{m}{s}[/tex], then the depth of the resulting stream is:
[tex]h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}[/tex]
[tex]h_{3} = 3.8\,m[/tex]
The depth of the resulting stream is 3.8 meters.