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Flashcards in Fluids, Gases & Heat Deck (133)
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1
Define:

a fluid

A fluid is a phase of matter that is capable of yielding to pressure and changing its shape to fit a container.

On the AP exam, fluids come in two forms: liquids and gases which are in motion. Static gases are technically fluids, but they are non-ideal ones and are not tested as such.

2

What are the properties of an ideal fluid?

An ideal fluid:

  • is incompressible
  • is non-viscous (no friction)
  • changes shape to fit its container
  • flows without turbulence

On the AP Physics exam, all fluids are assumed to be ideal unless otherwise noted.

3

Water is flowing through a level pipe. What can be said about the nature of the flow?

Since water can be assumed to be an ideal fluid, the flow will be frictionless and non-turbulent. The speed of the flow will be constant at all points along the pipe, and the pressure on the walls of the pipe will be constant at all points.

4
Define:

Density (ρ) of a substance

A substance's density (ρ) is the mass of a unit volume of that substance.

The SI units of density are kg/m3.

5

What are some commonly-used units of density?

g/cm3, g/cc, g/mL, kg/L, and kg/m3 are all commonly-used measures of density.

The first four are equivalent in magnitude, while the fifth differs by a factor of 1000.

6
What is the density of water in:

  • g/cm3
  • g/mL
  • kg/L
  • kg/m3

The density of water is:

  • 1 g/cm3
  • 1 g/mL
  • 1 kg/L
  • 1000 kg/m3

This is the one density you should memorize for the AP Physics exam.

7
Define:

specific gravity of a substance

A substance's specific gravity is that substance's density, divided by the density of water.

The magnitude of a substance's specific gravity is identical to the substance's density, as expressed in g/mL, but specific gravity is a unitless quantity.

8
Define:

buoyancy

Buoyancy is the tendency of an object to weigh less when partially or fully submerged in a liquid.

Buoyancy is caused by the liquid displaced by the object. The liquid pushes up on the object with a force equal to the weight of the liquid displaced.

9

How is the buoyancy force of a submerged object calculated?

The buoyancy force of a submerged object is simply the weight of the liquid displaced by the portion of the object that is submerged.

FB = ρliq * Vsub* g

Where:

  • FB = the buoyancy force pushing up on the object
  • ρliq = the density of the liquid
  • Vsub = the volume of the object that is submerged in the liquid
  • g = the gravitational acceleration (commonly 10 m/s2)

10

An object with a volume of 1.5 L is fully submerged in water. What is the buoyancy force on the object?

The buoyancy force on the object is 15 N.

Using the definition of buoyancy force:

FB = ρliq * Vsub * g
FB = (1 kg/L) * (1.5 L) * (10 m/s2)
FB = 15 N

11

Equivalent objects are submerged in ethyl alcohol (ρ = 0.79 kg/L) and mercury (ρ = 5.43 kg/L). Which one experiences the larger buoyancy force?

The object submerged in mercury will experience the larger buoyancy force.

Remember, the buoyancy force is equal to ρliq * Vsub * g. Since the objects are identical, the submerged volume is equal in each case, and presumably gravity is constant, so the fluid with the larger density will have the higher buoyancy force.

12

If a solid object is dropped into a liquid, under what conditions will it float?

The object will float if its density is less than or equal to the density of the surrounding liquid. This is known as the buoyancy rule.

13

A cube of styrofoam, with a density of 0.5 g/cm3, is dropped into water. How much of the cube is submerged in the water when the cube begins to float?

One half of the cube will be submerged in the water.

The proportion of an object that is under the surface of a liquid when the object floats is simply equal to ρobjliq, where ρobj is the object's density and ρliq is the liquid's density.

In this case, the object is one-half the density of the liquid, and thus one-half of it will be submerged when it floats.

Note that if ρobj > ρliq, this fraction is greater than 1, and the object sinks to the bottom.

14
Define:

the effective weight of an object submerged in a liquid

The effective weight of an object submerged in a fluid is equal to the object's weight on dry land minus the buoyancy force due to the liquid displaced by the object's submersion.

Weff = mobjg - (ρliq * Vobj * g)

Where:

  • mobj = the object's mass
  • ρliq = the liquid's density
  • Vobj = the object's volume
  • g = 10 m/s2

15

An aluminum ball with a density of 3 g/cm3 has a mass of 9 kg. What is its effective weight when it is submerged in water?

The ball's effective weight is 60 N

With a mass of 9 kg and a density of 3 g/cm3, the ball's volume must be 3000 cm3, or 3 L. Therefore, it displaces 3 L of water when it is submerged. The buoyancy force is simply the weight of 3 L of water:

FB = ρliq * Vsub * g
= (1 kg/L) (3 L) (10 m/s2) = 30 N

With a mass of 9 kg, the ball's dry land weight is 90 N. Subtracting the buoyancy force leaves the final answer, 60 N.

16

A plastic ball with a density of 2 g/cm3 has a mass of 6 kg. What is a shortcut to calculating the effective weight of the ball when it is submerged in water?

The shortcut is that the proportion of the object's weight that is cancelled due to buoyancy is exactly equal to the ratio of the liquid's density to the object's density.

In this case, the liquid, water, has a density one-half that of the object. Thus, the proportion of the object's weight cancelled by buoyancy is one-half the object's dry land weight.

With a mass of 6 kg, the object's dry land weight is 60 N. One half (30 N) of that is cancelled by buoyancy, so the remaining effective weight is 60 - 30 = 30 N.

17
Define:

Pascal's Law of fluid pressures

Pascal's Law states that a fluid will carry pressure undiminished; that is, pressure exerted on any part of a fluid will be carried equally to all the walls of the container.

Pascal's Law is most commonly tested on the AP Physics exam using hydraulic lifts.

18

What does Pascal's Law predict about the pressure on plates 1 and 2 in the diagram below?

Pascal's Law predicts that the pressures will be equal.

The force F1 will be exerted on the fluid behind plate 1 as a pressure P1. The fluid will exert that pressure, undiminished, on all the walls it is contacting, including plate 2.

19

How does the force F2 change if plate 2 in the hydraulic lift pictured below doubles in area?

The force F2 doubles.

According to Pascal's Law, the pressures on plates 1 and 2 must be equivalent. From the definition of pressure, therefore:

P1 = F1/A1 = F2/A2 = P2

So force and area are directly proportional, and doubling A2 will double F2 as well.

20

How does the distance D2 compare to the distance D1 if plate 2 in the hydraulic lift pictured is double the area of plate 1?

D2 = ½D1.

Since the liquid between the plates in incompressible, the volume displaced by plate 1 must be equal to the volume displaced by plate 2:

V1 = A1D1 = A2D2 = V2

So distance and area are inversely proportional, and a plate of double the area will move half the distance.

21
Define:

hydrostatic pressure on a submerged object

Hydrostatic pressure is the pressure exerted on a submerged object by the liquid in which the object is submerged.

Hydrostatic pressure increases proportionately with distance beneath the surface of the liquid.

22

What is the pressure exerted on an object submerged a distance z below the surface of a liquid?

Hydrostatic pressure is calculated:

PH = ρliq * g * z

Where:

PH = the hydrostatic pressure
ρliq = the liquid's density
g = 10 m/s2
z = distance beneath the liquid's surface

23

What is the difference between the hydrostatic pressure at a depth of 20 and 50 cm beneath the surface of water in a large, round-bottomed flask?

The pressure 50 cm beneath the surface is 2.5x the pressure 20 cm beneath the surface.

Since the hydrostatic pressure is

PH = ρliq * g * z

pressure is proportional to depth. 

Note that the pressure is independent of vessel shape; the only variables are liquid density and depth beneath the surface.

24

Vessel 1 is filled with water, while vessel 2 is filled with an unknown liquid. The pressure 10 cm beneath the surface of the liquid in vessel 2 is 3 times the pressure the same depth beneath the surface of the water in vessel 1. What is the density of the unknown liquid?

The unknown liquid's density is 3 g/cm3.

Since the hydrostatic pressure is PH = ρliq * g * z, the pressure is proportional with liquid density. If the pressure at an equal depth is higher, the liquid's density must also be higher, by the same proportion.

25

What are the properties of ideal fluid as it flows through a pipe?

Ideal fluids undergo laminar flow, a form of motion in which the fluid is flowing at the same speed at all points in the pipe.

This is opposed to turbulent flow, where the fluid can have eddies, vortices, and local disruption to the fluid's speed and direction of flow.

26

How does the flow of a viscous fluid through a pipe differ from the flow of an ideal fluid?

Viscous fluid undergoes Poiseuille flow, where the fluid near the walls of the pipe flows much more slowly due to viscous interaction with the walls.

Above a certain fluid speed and pipe diameter, viscous fluids begin to flow turbulently as well.

27

Under what conditions does fluid flow become turbulent?

Turbulent flow occurs in non-ideal fluids when the cross-sectional area of the fluid flow becomes large, and when the fluid velocity increases.

For a given system, there are threshold values of cross-sectional area and velocity, but the specifics won't be tested on the AP Physics exam.

28

How does the speed of a fluid flowing through a pipe change, as the diameter of the pipe decreases?

As pipe diameter decreases, fluid flowing through the pipe begins to flow faster.

This is akin to placing your thumb partially over the end of a water hose, which causes the water to flow through faster, creating a jet of water.

29

Water is flowing through a pipe, and the cross-sectional area of the pipe decreases by one-half. How does the fluid velocity change as the pipe area decreases?

The fluid velocity doubles as the area reduces by half.

The equation to describe the relationship between cross-sectional area and velocity is the continuity equation:

A1v1 = A2v2

Where:

A = the cross-sectional area of the pipe
v = the fluid velocity

30

Water is flowing through a round pipe, and the radius of the pipe increases by a factor of 3. How does the fluid velocity change as the pipe radius increases?

The fluid velocity decreases by a factor of 9.

According to the continuity equation A1v1 = A2v2, there is an inverse proportionality between cross-sectional area and velocity.

However, cross-sectional area of a round pipe is proportional to the radius squared, so if the radius triples, the area goes up by a factor of 9, and therefore the velocity decreases by the same amount.