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Q75PE

Expert-verifiedFound in: Page 398

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Pressure in the spinal fluid is measured as shown in Figure **\[{\rm{11}}{\rm{.43}}\]**. If the pressure in the spinal fluid is **\[{\rm{10}}{\rm{.0 mm Hg}}\]**: (a) What is the reading of the water manometer in cm water? (b) What is the reading if the person sits up, placing the top of the fluid **\[{\rm{60 cm}}\]** above the tap? The fluid density is **\[{\rm{1}}{\rm{.05 g/mL}}\]**.**

(a) The height of the water is \[{\rm{13}}{\rm{.6 cm}}\] of water.

(b) The new pressure will be \[{\rm{76}}{\rm{.6 cm}}\] of water.

**Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.**

The density of the stomach fluid is \[{\rm{1}}{\rm{.05 g/mL}}\].

The density in kg/m^{3} = \[{\rm{1}}{\rm{.05 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ kg/}}{{\rm{m}}^{\rm{3}}}\].

(a) By putting all the value into the equation we get:

\[\begin{array}{l}{P_{Hg}} = {P_w}\\{h_{Hg}}{\rho _{Hg}}g = {h_w}{\rho _w}g\\{h_w} = \frac{{{h_{Hg}}{\rho _{Hg}}}}{{{\rho _w}}}\\{h_w} = \frac{{1 \times 13600}}{{1000}}\end{array}\]

\[{h_w} = 13.6\,cm\,of\,{H_2}O\]

Hence, the height of the water is \[{\rm{13}}{\rm{.6 cm}}\] of water.

(b) The additional pressure of the height of the fluid when the new pressure will be the sum of the pressure when they are lying down and the pressure when they are standing up.

\(\begin{array}{l}{P_n} = {P_l} + {P_s}\\{P_n} = {P_l} + {h_s}{\rho _s}g\end{array}\)

Here again we need to convert the pressure into cm of water so:

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