Solve with explanation (2) In the following Figure below, determine the gauge pressure at point A.Density of mercury = 13.6 g/cm3 = 13,593 kg/maDensity of water = 1.0 g/cm3 = 1,000 kg/m³1 |PageOpen to atmosphere, Pah=38cm- Mercury (13,593 kg/m')Waterh= 76 cmBand C are at thesame elevation

Solution: As given that density if mercury = 13593 kg/m3 desnity if water = 1000 kg/m3

(2) In the following Figure below, determine the gauge pressure at point A. Density of mercury = 13.6 g/cm3 = 13,593 kg/m³ Density of water = 1.0 g/cm3 = 1,000 kg/m3 1| P -Open to atmosphere, Pa 38cm - Mercury (13,593 kg/m') Water h= 76 cm B and Care at the same elevation

(2) In the following Figure below, determine the gauge pressure at point A. Density of mercury = 13.6 g/cm3 = 13,593 kg/m³ Density of water = 1.0 g/cm3 = 1,000 kg/m3 1| P -Open to atmosphere, Pa 38cm - Mercury (13,593 kg/m') Water h= 76 cm B and Care at the same elevation

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter14: Fluid Mechanics

Section: Chapter Questions

Problem 13CQ: Does atmospheric pressure add to the gas pressure in a rigid tank? In a toy balloon? When, in...

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Solve with explanation

(2) In the following Figure below, determine the gauge pressure at point A.
Density of mercury = 13.6 g/cm3 = 13,593 kg/ma
Density of water = 1.0 g/cm3 = 1,000 kg/m³
1 |Page
Open to atmosphere, Pa
h=
38cm
- Mercury (13,593 kg/m')
Water
h= 76 cm
Band C are at the
same elevation

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