ap physics chp 11
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AP Physics Chp 11. Density (mass density) ρ =m/V Solids and liquids are fairly constant at our range of temp/pressure but gases vary. Pressure P = F/A the force is perpendicular to the surface area but pressure is not a vector 1 Pa = 1N/m 1 atm = 101,325 Pa bar = 10 5 Pa - PowerPoint PPT PresentationTRANSCRIPT
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AP Physics Chp 11
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• Density (mass density)
• ρ=m/V
• Solids and liquids are fairly constant at our range of temp/pressure but gases vary.
![Page 3: AP Physics Chp 11](https://reader034.vdocuments.mx/reader034/viewer/2022051020/568165a2550346895dd8818c/html5/thumbnails/3.jpg)
• Pressure
• P = F/A the force is perpendicular to the surface area but pressure is not a vector
• 1 Pa = 1N/m 1 atm = 101,325 Pa• bar = 105 Pa • The english system used lb/sqin = psi
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• The pressure we experience now is due to the fluid above us being pulled down by gravity.
• 14.7 lb/si or 101,325 Pa is the standard at sea level
• The higher up you go the lower the pressure gets because of less fluid above you
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• Thus in a static fluid
• P2 = P1 + ρgh
• ρ is the density and h is the depth in m
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• We use this in constructing a barometer (measure air pressure).
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• As well as manometers which measure other fluid pressure.
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• Gauge pressure is the pressure compared to atmospheric pressure
• Absolute pressure is the actual pressure
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• How high can a submersible pump raise water if it has a maximum pressure of 110 kPa?
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• P2 = P1 + ρgh
• 110000 = 101325 Pa + (1000kg/m3)9.8m/s2(h)
• h = 0.89 m
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• How much force would it have at the pump if the pipe was 2.3 cm in diameter? How much at the maximum height with the same size pipe?
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• At the pump
• P = F/A • P = 110 kPa or 110000 Pa • A = πr2 r = 2.3cm/2 = 1.15cm = 0.0115m• A = π(0.0115m)2 = 0.00042m2
• F = PA = (110000 Pa)(0.00042m2) = 46 N
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• At the maximum hieght
• P = F/A • P = 101325 Pa • A = πr2 r = 2.3cm/2 = 1.15cm = 0.0115m• A = π(0.0115m)2 = 0.00042m2
• F = PA = (101325 Pa)(0.00042m2) = 43 N
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• Why can’t a surface mounted pump pull water up out of a well that is 12 m deep?
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• P2 = P1 + ρgh
• 101325 Pa = 0 Pa + (1000kg/m3)(9.8m/s2)(h)
• h = 10 m
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• Pascal’s Principle
• Any pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and enclosing walls.
• P1 = P2 if the level of the fluid is equal at each point otherwise we add ρgh as needed
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• F1/A1 = F2/A2 useful in hydraulic jacks
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• Arcimedes’ Principle
• Fluids apply a buoyant force on any object partially or totally submerged in it with a magnitude equal to the weight of the displaced fluid.
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• Fb = Wfluid
• Fb = ρgV
• Why do supertankers float?
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• The supertanker’s design displaces enough water/salt water such that Fg = Fb
• Why will filling a ship to its limit in Boston create problems when it tries to go to Panama?
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• As fluid enters a tube that is open on both ends the mass of the fluid going in has to be equal to the mass coming out.
• A1v1 = A2v2 continuity equation
• How does this explain the effect on kinking a hose?
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• Bernoulli’s Equation
• P1 + 1/2pv12 + pgh1 = P2 + 1/2pv2
2 + pgh2
• When the fluid isn’t moving this gives the pressure equation P = Po + pgh
• This equation is used often to explain the affects of moving air like wind over a roof.
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• What is the pressure above the roof sheeting if the wind is blowing 32 m/s and there is no wind inside your attic space. Assume the height difference is negligible. Density of air is 1.29 kg/m3.
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• P1 + 1/2pv12 + pgh1 = P2 + 1/2pv2
2 + pgh2
• P1 + ½(1.29 kg/m3)(32m/s)2 = 101325 Pa
• P1 = 100004 Pa