5.6 Questions
07 October 2011 11:53 Collins, p.107 [cid:image001.jpg@01CC88F0.11386960] Note: ρfresh water = 1,000kg/m3; g = 10N/kg ps. 1,000mbar = 1 bar = 100,000Pa! ANSWERS Collins, p107 Q5 pressure difference = height × density × g ∆p = h × ρ × g ∆p = pressure of the fluid (N/m2 or Pa) = 250kPa - 100kPa = 150kPa (x1000) = 150 000 h = height FROM SURFACE LEVEL (m) = ? ρ = density of the fluid (kg/m3) = 1000 g = gravitational field strength (N/kg) = 10 height => h = 150 000/ (1000 x 10) = 15m OR NOTE IN FRESH WATER: EVERY 10m descend = +100kPaIf he was diving in water that was slightly denser than fresh water then we see from the equation that the pressure difference would be divided by a greater number, therefore the answer would be less meaning that the diver would dive less below the surface. Q6 pressure difference = height × density × g ∆p = h × ρ × g ∆p = pressure of the fluid (N/m2 or Pa) = ? h = height FROM SURFACE LEVEL (m) = 50 ρ = density of the fluid (kg/m3) = 0.42g/cm3 = 0.00042kg/cm3 => (1cm3 = 1/100^3) = 0.00042 x 100^3 = 420 OR NOTE: 1 g/cm3 = 1000kg/m3 g = gravitational field strength (N/kg) = 1.4 NOTE: 1 g/cm3 = 1000kg/m3 ∆p =50 x 420 x 1.4 = 29400Pa OR 29.4kPa Total pressure (atmospheric pressure on Titan = 1600mbar (1 mbar = 0.1kPa = 100Pa) => 1600 x 0.1 = 160kPa TP = 160 +29.4 = 189.4kPa
