Answer:
the tension in the wire = [tex]3.12 \ N |_{y= 0.9 \ m[/tex]
Explanation:
Parameters given include:
The height of the large tank = 1 m
Diameter D = 0.75 m
diameter of the nozzle where the water exit d = 15 mm = 0.015 m
The flow speed at the exit of the tank [tex]\mathbf{V = \sqrt{2gy} }[/tex]
The first diagram shown in the attached file depicts and illustrate the sketch of the large tank that is fixed on the cart.
Now; using the x component of the momentum equation from the diagram; we have;
[tex]{ \bar F_{xx} } + { \bar F_{bx} } = \frac{\delta }{\delta t}\int\limits_{cv} \bar u pd. V + \int\limits_{cs} \bar u pV. A[/tex]
For steady flow:
[tex]\frac{\delta }{\delta t}= 0[/tex]
So:
T = u{║ρV₁A₁║}
T = ρV₁²A₁
where:
[tex]\mathbf{V = \sqrt{2gy} }[/tex]
T =ρgy[tex]\frac{\pi d^2}{2}[/tex]
replacing y= 0.9 m
The tension of the wire is:
T = 999 × 9.81 × 0.9 × [tex]\frac{\pi *0.015^2}{2}[/tex]
T = [tex]3.12 \ N |_{y= 0.9 \ m[/tex]
Hence, the tension in the wire = [tex]3.12 \ N |_{y= 0.9 \ m[/tex]
The schematic graphical representation showing the plot of the tension in the wire as a function of water depth for: 0 < y < 0.9 can be found in the document file attached below.
