Answer:
The pressure inside the hose 7000 Pa to the nearest 1000 Pa.
[tex]h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}[/tex]
[tex]V_{1} =V_{2}[/tex]
The pressure at the site of the puncture is [tex]P₂ =7161.3 Pa[/tex]
Explanation:
According to Poiseuille's law, [tex]P_{1} - P_{2} = \frac{128μLQ}{πD^{4} }[/tex]
Where [tex]P_{1}[/tex] is the pressure at a point [tex]1[/tex] before the leak, [tex]P_{2}[/tex] is the pressure at the point of the leak [tex]2[/tex], μ = dynamic viscosity, L = the distance between points [tex]1[/tex] and [tex]2[/tex], Q = flow rate, D = the diameter of the garden hose.
Also, from the equation [tex]P =ρgh[/tex], the equations [tex]h₁ = \frac{P₁} {ρg}[/tex] and [tex]h₂ = \frac{P₂} {ρg}[/tex] can be derived.
Combining Poseuille's law with the above, we get [tex]h₁ - ρgh₂ = \frac{128μLQ}{πD^{4} }[/tex]
[tex]h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}[/tex]
[tex]V =\frac{Q}{A}[/tex]
Since the hose has a uniform diameter, the nozzle at the end is closed and neither point [tex]1[/tex] nor [tex]2[/tex] lie after the puncture,
[tex]V_{1} =V_{2}[/tex]
The pressure at the site of the puncture [tex]P₂ =ρgh₂[/tex]
[tex]P₂ =1kgm⁻³ ×9.81ms⁻¹×0.73m[/tex]
[tex]P₂ =7161.3 Pa[/tex]
