Lecture24

# Lecture24 - P = P ρ g h Let’s show that disk of fluid of...

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Unformatted text preview: P = P + ρ g h Let’s show that: disk of fluid of mass dm ρ = m/V m = ρ V dm = ρ dV The disk is in equilibrium F 1 F 2 If we think in terms of pressure F 1 = PA F 2 = (P+dP) A dy Equilibrium-dm g + PA – (P+ dP) A =0-dm g + PA – PA - dPA =0 F net = 0 F net = F 1 + F 2 + dm g dmg Variation of pressure in a fluid- dm g - dPA =0 Substituting dm = ρ dV = ρ A dy- ρ A dy g - dP A =0- g ρ dy = dP y A y B P A P B Note : ρ is constant Liquids are incompressible!-g ρ (y B – y A ) = P B- P A-g ρ h = P B- P A P B = P A + ρ gh y A y B y IMP : Does not depend on the shape of the container h - depth below the surface Equivalent to object moving with constant a v = v + gt P = P 0 + ρ gh In general: A h P P = P 0 + ρ gh P A = P B = P C = P D Pressure same in each point with same height! What happens if the tube contains different immiscible liquids? P P A = P + ρ gh What this equation implies for an open connected container? P A = P + ρ gh A ρ 1 ρ 2 A B P B = P + ρ gh B The densities are different P A = P B At the interface the pressure is the same Torricelli Barometer Barometers use this principle to measure pressure P = P 0 + ρ gh P P=0 A, B same height needs to have same pressure A B P A = P P B = ρ g h h ρ Hg = 13,595 10 3 Kg/m 3 Filled with mercury (Hg) g=9.8 m/s 2 h=0.76 m P = 1.013 10 5 Pa = 1 atm P = ρ g h DEMO Case 1) Vacuum Case 2)...
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Lecture24 - P = P ρ g h Let’s show that disk of fluid of...

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