q

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