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Physics Warm Up: Agenda Copy these assignemts into your
binderDecember 2WarmUp: Agenda/ Review of buoyant forceInClass:
Archimedes WorksheetHomework: Read and take notes p358-362DUE NEXT
CLASSDecember 3-4WarmUp: Specific Heat of CopperLab: Specific
HeatHomework: Answer q 5p374 DUE NEXT CLASSDecember 5-6WarmUp: Heat
of FusionLab: Heat of FusionHomework: When the air temperature is
22.2C many people find it a comfortable temperature, yet the same
people often find a swimming in 22.2C water too cold to be
comfortable. Use specific heat to explain the reason for the
difference in sensation of temperature of the air and the water.
Explain it to somebody and bring in a written copy of your
explanation. WHEN YOU FINISH WRITING DOWN THE HOMEWORK, PLEASE GET
OUT YOUR BUOYANCY HOMEWORK FROM LAST WEEK, QUESTIONS 1 AND 2 P324
SO THAT WE CAN GO OVER THE QUESTIONSBuoyant force p324There was
some difficulty with problems 1 and 2 of the homework last week. We
didnt go over it in class.Question 1: A piece of metal weighs 50.0N
in Air. 36.0N in water, and 41.0N in oil. Find the density ofa. The
metalTwo pages back are the equations used in the sample
problemFnet = (fVf oVo) gwhere Vnet is the apparent weight, and the
subscripts f and o stand for The fluid and the object
respectivelyBecause the object is submerged, the volumes are equal
and we can simplifyFnet = (f o) VgFB f VgFg(object) o Vg=Use the
ratio (you can find this on p322 if you forgot where it comes from)
and solve for the density of the object (o).FB Fg(object) f= o
Substitute values from the question14.0N 50.0N 1.00g/cm3=
3.57g/cm3= 3.57x103kg/m3Buoyant force p324There was some difficulty
with problems 1 and 2 of the homework last week. We didnt go over
it in class.Question 1: A piece of metal weighs 50.0N in Air. 36.0N
in water, and 41.0N in oil. Find the density ofThen B asks you to
find the density of the oil.FB f VgFg(object) o Vg=We can use the
same simple ratio, but now solve for the density of the fluid, (f)=
640x103kg/m3o FB Fg(object)f =Substitute values from the
question3.57x103kg/m3 9.0Nf =50.0NArchimedes PrincipleQuestion 2
from that homework is easy if you just remember that the buoyant
force acting on a submerged object is equal to the weight (force)
of the fluid it displaces.Verify it:On the demo table is a can with
a spout. Fill the with just enough water to run out of the
spout.Weigh the small beaker. When the spout stops dripping, put
the beaker under it.Add the 100g weight to the water and collect
the displaced water. Weigh the beaker with the water and calculate
the weight of the displaced water. Record it on your paper.
Should equal the difference in weight
The mass of this water
Weigh the 100g weight in air by hanging it from the hook on the
balance.Fill the 250mL beaker and place it on the arm of the
balance.Weigh the 100g weight in water and calculate the
difference.The difference is the buoyant force.It should be very
close to the weight of the water displaced.Buoyant force p3242. An
empty rubber balloon has a mass of 0.0120kg. The balloon is filled
with helium at 0C, 1atm pressure and a density of 0.181kg/m3. The
filled balloon has a radius of 0.500m.a. What is the buoyant force
acting on the balloonThe buoyant force acting on the balloon is
equal to the weight of the air it displaces (p320).The mass of the
displaced air is the volume of the balloon times the density of
the. Multiplying that by acceleration due to gravity gives its
weight.Fg(air)= v f g
Fg(air)=0.524m3 1.29kg/m3 9.8 m/s2
Fg(air)= 6.62kgm/s2 = 6.62NV= 4/3 r3
V= 4/3 (0.5m)
V= .524m3
B. The net force will be the difference between the balloons
weight and the buoyant force. The weight of the balloon is the
density of the helium times the volume of the balloon added to the
mass of the empty balloon.B. Fg(balloon)=9.8 m/s2 (0.524m3
)(0.181kg/m3)+9.8 m/s2(0.0120kg)Fg(balloon)=1.05NFnet= Fg(air) -
Fg(balloon) Fnet= 6.62N- 1.05NFnet= 5.57N
Pressure & Pascals LawPressure is force per area. p =
f/aLiquid Pressure = (depth)(density) p = hUnits of pressure:
pascal = n/m2 Atmosphere, mm of Hg (in manometer) kg/cm2 (mass
units).
Common pressures:1 atm = 100 kPa 760 mm of Hg 10 m of H2O 1
kg/cm2
Blaise Pascal Tap water pressure = 4 atm 400 kPa 4 kg/cm2 Car
tire 2 atm6Pressure contdTotal Force = (pressure)(area) TF =
pA.7Two Sample ProblemsFind the pressure needed to push water to
the top of Tower 2, a height of 8.0 meters. (Use mass units). The
density of water is 1.0 g/cm3.
Find the Total Force on the filled school dam whosedimensions
are 5.0m by 2.0m. The average depth is 1.0m.
MainmachineryCastle attack
Solutions are on the next page8SolutionsFind the pressure needed
to push water to the top of Tower 2, a height of 8.0 meters. (Use
mass units). The density of water is 1.0 g/cm3.Find the Total Force
on the filled school dam whose dimensions are 5.0m by 2.0m. The
average depth is 1.0m.
1234p = h = (1.0m)(100cm/m)(1.0g/cm3 2 ) = 100 g/cm2 p = h=
(8.0m)(100cm/m)(1.0g/cm3 2 ) = 800 g/cm2 1
2
341234A = LW = (5.0m)(2.0m) = 10.0 m21234f = pA =
(100g/cm2)(10.0m2)(104cm2/m2) * = 1 X 107 g = 1 X 104 kg or 10
tons!Note: 1kg = 1000g1 metric ton = 1000kg* There are 104 cm2 in a
m2.9Pascals LawThe pressure on a confined fluid is transmitted in
all directions.
Pascals Vases show pressuredepends only on depth &
density.
10Pascal Pressure in Rocket (1)pascalrocket.mov
11Hieros Fountain Demo
h2 is higherthan h1, so thepressure is greaterin the right
systemwhich pushes thewater up into thefountain.Try it! Add a about
50mL of water to the funnel of the fountain on the demo
table!12Hydraulic Lift: Demo: SyringesThe pressure is transmitted
undiminished in all directions.
Try this with different sized syringes.13Hydraulic BrakesThe
applied pressure to the master cylinder is transmitted equally to
all four brake pistons.
14Hydraulics lifts a House! (2)hydraulics.movOh, Pascal! Thanks
to Mark Shisler
15An Uplifting Experience: Demo
A strong rubber balloon inflated beneath a car or truckcan lift
15 metric tonnes of load. The pressure is low, but the surface area
is large. Total Force = (press)(area).Demo: A garbage bagblown up
by a vacuumcleaner can lift a massiveperson.16