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Chapter 13 States of Matter. Fluid - A material flows and have no definite shape of their own. Pascal’s Principle: The change in pressure applied at any point in a confined fluid is transmitted undiminished throughout the fluid. Examples Toothpaste Hydraulics. How much can be lifted?. - PowerPoint PPT Presentation
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Chapter 13States of Matter
Fluid - A material flows and have no definite shape of their own.
Pascal’s Principle: The change in pressure applied at any point in a confined fluid is transmitted undiminished throughout the fluid.
Examples•Toothpaste•Hydraulics
2
2
1
1
A
F
A
For
A
FP
How much can be lifted?F1/A1 =F2/A2 F2 = F1A2/A1
F2 = (20 N)(.1 m2)/(.05 m2) = 40 N
If the little piston moves 1 meter, how far does the big one move?
V1=V2 A1H1=A2H2
H2 = A1H1/A2 H2= (.05 m2)(1m)/(.1 m2)
H2= .5 m
Chapter 13States of Matter
Swimming under pressure
P F
A
mg
ABut d = m/v or m = dv
P F
A
mg
A
dvg
A
dAhg
Ahdg
P = hdg
Chapter 13States of Matter
Taking P = hdg and multiplying both sides by A gives
PA = Ahdg or F = vdg
Where F = vdg is the buoyant force
Chapter 13States of Matter Archimedes’ Principle - An object
immersed in a fluid has an upward force on it equal to the weight of the fluid displaced by the object.
1. A body sinks if the weight of the fluid it displaces is less than the weight of the body.
2. A submerged body remains in equilibrium if the weight of the fluid it displaces exactly equals its own weight.
3. A body floats if it displaces a weight greater than that of its own weight
ABC
Chapter 13 States of Matter
A block of wood has a volume of 100 cm3 and a mass of 85 grams.
a. Will it float in water water = 1000 kg/m3 ?b. Will it float in gas gas = 700 kg/m3
d= m/v = 85g/100 cm3 = .85 g/cm3= 850 kg/m3
YESNO
Chapter 13States of Matter
What is the weight of a rock submerged in water if the rock weighs30 newtons and has a volume of .002 m3?
V = .002 m3
W = 30 Nwater = 1000 kg/m3
mg
vdgFnet = Weight - buoyant force
Fnet = mg - vdg
Fnet = 30 N - (.002 m3)(1000 kg/m3)(9.8 m/s2)
Fnet = 30 N - 19.6 N = 10.4 N
The acceleration of the rock will be a = F/m
A = 10.4 N/3.06 kg = 3.2 m/s2
Chapter 13States of Matter
What the maximum weight a helium balloon of volume 2 m3 can support in air?
V = 2 m3
air = 1.2 kg/m3
helium = .177 kg/m3Fnet = Weight - buoyant force
Fnet = mg - vdg
Fnet = (2m3)(.177kg/m3)(9.8m/s2)-(2m3)(1.2 kg/m3)(9.8 m/s2)
Fnet = 3.462 N - 23.52 Nmg
vdg
It can support 20 N
Chapter 13States of Matter
Bernouilli’s Principle For the horizontal flow of a fluid through atube, the sum of the kinetic energy per unit volume and the pressureis a constant.
Chapter 13States of Matter Cohesion: The force of attraction
between like particles. Adhesion: The force of attraction
between unlike particles. Capillary action: The rise of a liquid in a
narrow tube because the adhesive force is stronger than the cohesive force.
Volatile Liquid: A liquid that evaporates quickly.
Chapter 13States of Matter Cohesion •Adhesion
Chapter 13 States of Matter
Surface Tension: The tendency of the surface of a liquid tocontract to the
smallestpossible area
Chapter 13States of Matter
Solid Liquid Gas
Melting
CondensationFreezing
Sublimation
Supercooled
Vaporization
Chapter 13States of Matter
Thermal Expansion: The increase in length or volume of asubstance when heated.
Volume expansion V2 = V1+βV1(T2-T1)
Linear expansion L2 = L1+αL1(T2-T1) Chart Pg 317
Chapter 13States of Matter
Linear expansion L2 = L1+αL1(T2-T1)
A iron bar is 3 m long at 21ºC. What is the length of the barat 100º C?
L2 = 3 m+(12 x 10 –6 (ºC-1)(3 m)(100ºC- 21ºC)
L2 = 3 m +.002844 m = 3.002844 m