Physics III: Fluids and Thermodynamics Flashcards
As an ideal fluid flowing in a narrow pipe passes from a region of cross-sectional diameter d to a region of cross-sectional diameter d / 2, the flow speed of the fluid will:
A. decrease by a factor of 4.
B. decrease by a factor of 2.
C. increase by a factor of 2.
D. increase by a factor of 4.
D. increase by a factor of 4.
If the diameter of the flow tube decreases by a factor of 2, the cross-sectional area decreases by a factor of 22 = 4. Since the Continuity Equation implies that flow speed is inversely proportional to the cross-sectional area of the flow tube, so the flow speed will increase by a factor of 4.
vasoconstriction increases/decreases blood flow?
decreases
Elastic properties relate to the tendency of a material to maintain its shape and not deform when a force is applied to it. A material such as steel will experience a smaller deformation than rubber when a force is applied to the materials and is therefore said to have higher elastic properties because their rigidity makes them act like springs that control how quickly its particles return to their original positions. Steel is a rigid/elastic material while rubber deforms easily and is a more flexible/less elastic material.
Therefore, sound can travel faster through mediums with higher elastic properties (like steel) than it can through solids like rubber, which have lower elastic properties.
Pgauge = ρgh
equation for hydrostatic pressure in units of Pa (Pa is N/m<span>2</span>)
P=pressure of gauge (ie the pressure of the system due to water above object regardless of external pressure such as surface gas pressure and atmosphere)
ρ = densityfluid (mass/volume in kg/m3)
g= acceleration d/t gravity = 10 N/kg
(instead of m/s2 because it’s in fluid)
h=height between object and top of fluid (ie depth)
^essentially you are determining the pressure due to the water mass of water above the object
T/F: density of a liquid remains constant
true.
This means that the hydrostatic gauge pressure will increase proportionally with both density and depth
Difference between hydrostatic gauge pressure an total hydrostatic pressure?
hydrostatic gauge pressure: only expresses the pressure within the system and is proportional to depth
total hydrostatic pressure is Patm + Pgauge and is not proportional to depth.
what variations can you have in a gas system that affects thermodynamics
what is the value of density of water
1000kg/m3
or, 1g/cm3
what is the unit of pressure for fluids
the pascal, Pa= 1N/m2
e.g. sea level: 1 atm= 100kPa (kilopascals)
T/F: hydrostatic gauge pressure does not depend on area of fluid in the system
true.
only depth matters (ie height from surface)
calculation would be the same whether youre in swimming pool or in an ocean
_________rather than net force determines whether a body moves or remains at rest in a fluid
pressure difference
equation for density
ρ=m/v
how do you determine specific gravity?
sp. gr. = ρ/ρH2O
where ρH2O is 1000kg/m3
since the units kg/m3 cancel out, sp. grav. is unitless
you can also use specific gravity to find density of a substance by rearranging the equation
*a substance with a density that is n times the density of water will have a specific gravity equal to n
bouyant force is defined as
the net force exerted on an object that is partially or fully submerged due to the pressure difference between the area of water above vs area of water below the object
what is the magnitude of the bouyant force?
equal to the weight of the fluid displaced by the object (Archimedes principle)
this translates to the equation: FBouy=ρfluidVsubg
(from F=ma)
ρfluidVsub = mass of displaced fluid
Vsub =volume of the portion of the object that is submerged, also equal to the volume of displaced fluid
T/F: the pressure on a submerged object is greater pushing up than pushing down on the object
true
and the pressure increases with depth of submersion, from Pgauge=ρgh
this is simply due to the fact that the pressure is greater at the location in the fluid that corresponds with bottom of object
hydraulic jack scenario
two pistons rest above two cylinders of different cross-sectional areas, connected by a pipe/passageway.
Pascal’s law: the force exerted on the smaller cylinder equals the force exerted upwards on the other cross-sectional area.
what is the relationship between force and area in a hydraulic jack
the pressure increase on the body of fluid is
F1 / A1
by Pascal’s law, F1 / A1=F2 / A2
⇒ F2 inversely proportional to A1