|Density (ρ)||mass / volume|
|Characteristic property of a pure substance.|
Below air density values (ρ0) at various temperatures:
|Specific Gravity (SG)||unitless||( ρ of substance ) / ( ρ of water at 4° C)|
|Pressure||Pressure = ( Force on fluid ) / ( Surface area of fluid )|
|Standard Pressure (P0)||atm||Standard pressure (atmospheric pressure) is the pressure the atmosphere exerts on a surface. 1 atm = 1.013 · 105 Pa = 14.69 psi|
|Water Pressure at Depth|
P = F / A = ρ A h g / A = ρ g h
The pressure at a given depth is not dependent on area, just on the depth; generally, the pressure at a given depth of a liquid will be uniform. Note that this is just the water pressure. If the container is open to the atmosphere, then total pressure will be:P = P0 + ρ g h
|Pressure on Surface||½ ρ g l ( y22 - y12 )The pressure a vertical surface such as an aquarium viewing window is determined by the above equation, where y2 is the upper edge's depth below water and y1 is the lower edge's depth below water. Note that because the surface is vertical, atmospheric pressure does not play a role.|
|Pressure at Elevation|
P = P0 e- y ρ0 g / P0For air pressure at an elevation y above sea level.
|Gauge Pressure||Gauges measure pressure relative to atmospheric pressure. For example, a tire gauge may read off 29 psi as the tire pressure. In fact, the absolute tire pressure is atmospheric pressure P0 (14.69 psi) plus the gauge reading of pressure PG (29 psi) for a total absolute tire pressure of 43.69 psi.|
P = P0 = PG
Specific Gravity = ( ρ of substance ) / ( ρ of water at 4° C) = ( mass of substance / volume of bottle ) / ( mass of water / volume of bottle ) = ( mass of substance ) / ( mass of water )
Mass of substance = 80 g - 20 g = 60 grams Mass of water = 100 g - 20 g = 80 grams 60 grams / 80 grams = 0.75 is the specific gravity.
Phead = Pfeet + ρ h g h = 1.5 m ρ = density of blood ≈ 1050 kg / m3
Phead - Pfeet = ρ h g = 1050 · 1.5 · 9.8 = 15435 Pa 15435 Pa · 760 / 101300 = 116 mm-Hg
Pout = Pin P = F / A Fout / Aout = Fin / Ain Fout / Fin = Aout / Ain
FB = ρfluid V g = mfluid g
FB = buoyant force ρfluid = density of fluid mfluid = mass of fluid displaced V = Volume of fluid displaced, equal to the volume of the fully submerged object displacing the fluid.
|Buoyant Force (FB)||Archimedes' Principle addresses buoyant force. This is the phenomenon whereby an object weighs less underwater because the pressure at the bottom of the object is greater than at the top of the object. Thus, the water exerts a net force upward against gravity.|
Pascal's Principle: Pin = Fout / Aout
The problem provides pressure and diameter, so it is apparent that Pascal's Principle will provide an answer.
Aout = π r2 = π · ( ½ · .20 )2 = 0.0314159265 m2 Fout = mcar · g = mcar · 9.8 m s-1 Pin = 15.0 atm · 101325 Pascals / atm = 1519875 Pascals
1519875 = mcar · 9.8 / 0.0314159265 mcar = 4872.2736 kg 4870 kg is the maximum weight that the hydraulic can lift.
FB = ρfluid V g
FB = ( 10.00 - 8.00 ) · g = 2.00 · g ρfluid = 1000 kg m-3 ρobject = ? V = ?
2.00 · g = 1000 · V · g 2.00 = 1000 · V V = 2 · 10-3 m3 → volume of water displaced, and of object.
Density of object → 10.00 kg / ( 2 · 10-3 m3 ) = 5.00 · 103 kg m-3 = 5000 kg m-3
FB = ρfluid V g
ρoil = 900 kg m-3 ρwater = 1000 kg m-3 Vcube = .1003 m3 = 10-3 m3 Vcube in oil = 10-3 m3 · 20% = 10-3 m3 · .20 = .0002 Vcube in water = 10-3 m3 · 80% = 10-3 m3 · .80 = .0008
FB in water = 1000 · .0008 · 9.8 = 7.84 FB in oil = 900 · .0002 · 9.8 = 1.764 FB total on cube = 7.84 + 1.764 = 9.604 Newtons is buoyant force on the cube
Mass of cube = 9.604 N / 9.8 m s-2 = 0.98 kg is mass of the cube
FB = ρfluid V g
ρfluid = 1000 kg m-3 FB = ? V = ? SGwood = 0.50 ρwood = SGwood · 1000 kg m-3 = 500 kg m-3
Notice that the tribemembers are just getting their feet wet. Therefore, the raft must be entirely submerged. Volume of log ≈ Volume for a cylinder = π r2 l = π · .252 · 4 = 3.14159265 m3
Recall that the upward (buoyant) force must break even with the raft's and the peoples' mass. FB = ( mraft + mtroops ) · g = ( Vraft · ρwood + N · 40 ) · g = ( 1570.79633 + N · 40 ) g Where N is the number of people on the raft.
( 1570.79633 + N · 40 ) g = 1000 · 3.14159265 · g 1570.79633 + N · 40 = 3141.59265 N · 55 = 1570.79633 N = 28.5599333 → The raft can support 28 people.
This problem is worded in a complicated manner, and with a variety of different units. The first step is to take the material and make it simple, in SI units.
Atmospheric pressure = 1013.25 mbar = 101325 Pa Tank pressure reading = 30.0 cm → In mm-Hg, this would simply be 30.0 cm · 10 mm cm-1 = 300 mm-Hg → Converting Torr (mm-Hg) to Pascals = 39996.7105 Pa
The problem posed is to find the absolute pressure in the tank. Absolute pressure is gauge (or relative) pressure + atmospheric pressure.
101325 Pa + 39996.7105 Pa = 141321.71 Pa → 141000 Pa
This problem is simple, but the presentation is complicated due to wordiness and varying units. Take the material and simplify it all into SI units.
Atmospheric pressure = 1013.25 mbar = 101325 Pa Tank pressure reading = -7.0 cm → In mm-Hg, this would simply be -7.0 cm · 10 mm cm-1 = -70 mm-Hg → Converting Torr (mm-Hg) to Pascals = -9332.56579 Pa
With the above information, it is straightforward to figure out the absolute pressure in the tank. Absolute pressure is gauge (or relative) pressure + atmospheric pressure.
101325 Pa + ( -9332.56579 Pa ) = 91992.4342 Pa → 92000 Pa