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- https://phys.libretexts.org/Courses/Prince_Georges_Community_College/General_Physics_I%3A_Classical_Mechanics/51%3A_Fluid_Statics/51.05%3A_Pascals_LawAnother important principle in fluid statics is Pascal's law. It states that when a pressure change is applied to a fluid (as with a piston, for example), the pressure change is transmitted undiminish...Another important principle in fluid statics is Pascal's law. It states that when a pressure change is applied to a fluid (as with a piston, for example), the pressure change is transmitted undiminished throughout the fluid and to the walls of the container. In other words, there's nothing special happening in the direction of movement of the piston; the pressure change will be "felt" equally throughout the fluid.
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Dourmashkin)/28%3A_Fluid_Dynamics/28.05%3A_Worked_Examples-_Bernoullis_EquationThe pipe is connected to a narrower pipe leading to the second floor that has an inside diameter 2.5 cm . What is the pressure and speed of the water in the narrower pipe at a point that is a height 5...The pipe is connected to a narrower pipe leading to the second floor that has an inside diameter 2.5 cm . What is the pressure and speed of the water in the narrower pipe at a point that is a height 5.0 m above the level where the pipe enters the house? Let’s choose three points, point 1 at the top of the water in the tower, point 2 where the water just enters the house, and point 3 in the narrow pipe at a height \(h_{2}=5.0 \mathrm{m}\) above the level where the pipe enters the house.
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Dourmashkin)/27%3A_Static_Fluids/27.04%3A_Pascals_Law_-_Pressure_as_a_Function_of_Depth_in_a_Fluid_of_Uniform_Density_in_a_Uniform_Gravitational_FieldNewton’s Second Law applied to the fluid element is then \[(P(r)-P(r+d r)) A=-(\rho A d r) r \omega^{2} \nonumber \] We can rewrite Eq. (27.4.9) as \[\frac{P(r+d r)-P(r)}{d r}=\rho r \omega^{2} \nonum...Newton’s Second Law applied to the fluid element is then \[(P(r)-P(r+d r)) A=-(\rho A d r) r \omega^{2} \nonumber \] We can rewrite Eq. (27.4.9) as \[\frac{P(r+d r)-P(r)}{d r}=\rho r \omega^{2} \nonumber \] and take the limit dr → 0 resulting in \[\frac{d P}{d r}=\rho r \omega^{2} \nonumber \] We can integrate Eq. (27.4.11) between an arbitrary distance r from the rotation axis and the open-end located at \(R_{O}\), where the pressure \(P\left(r_{0}\right)=1 \mathrm{atm}\). \[\int_{P(r)}^{P\lef…
