PRE AP CHAPTER 9 PHYSICS I CHAPTERS 18, 19, 20

MATTER  MATTER HAS FOUR STATES  SOLID  DEFINITE VOLUME AND SHAPE  LIQUID  DEFINITE VOLUME AND INDEFINITE SHAPE  GAS  INDEFINITE VOLUME AND SHAPE  PLASMA  INDEFINITE VOLUME AND SHAPE  DOES NOT EXIST ON EARTH NATURALLY

WHAT IS A FLUID?  MATERIALS THAT :  FLOW  ALTER THEIR SHAPE AS THEY FLOW SOLIDS ARE NOT CONSIDERED FLUIDS. WHY?

FLUID DEFINITION  A NONSOLID STATE OF MATTER IN WHICH THE ATOMS OR MOLECULES ARE FREE TO MOVE PAST EACH OTHER, AS IN A GAS OR A LIQUID.

LIQUIDS  HAVE A DEFINITE VOLUME  TAKE THE SHAPE OF THEIR CONTAINER.

GASES  INDEFINITE VOLUME  FILL THEIR CONTAINER ON MATTER THE SIZE.

DENSITY  THE MASS PER UNIT VOLUME OF A SUBSTANCE.  EX: WATER 1 g/cm 3  THE MEASURE OF HOW MUCH THERE IS OF A QUANTITY IN A GIVEN AMOUNT OF SPACE.

MASS DENSITY  REPRESENTED BY THE GREEEK LETTER ρ (RHO)  EQUATION FOR DENSITY ρ = m/v DENSITY = MASS/VOLUME SI UNIT – kg/m 3

DENSITY OF WATER 1 g/cm X 10 3 kg/m3 AT 4 DEGREES CELCIUS

EFFECT OF PRESSURE  DENSITIES FOR SOLIDS AND LIQUIDS DO NOT CHANGE WITH PRESSURE, BECAUSE THEY HAVE A DEFINITE VOLUME.

GAS DENSITIES  DENSITIES FOR GASES ARE EFFECTED BY PRESSURE. WHY?  BECAUSE GASES HAVE AN INDEFINITE VOLUME.  BECAUSE OF THIS THERE ARE NO STANDARD VALUES FOR DENSITIES IN GASES

GAS DENSITIES  DENSITIES LISTED FOR GASES IN TABLES ARE THE VALUES OF THE DENSITY AT A STATED TEMPERATURE ANDPRESSURE.  STANDARD TEMPERATURE AND PRESSURE IS CONSIDERED TO BE:  0° C AND 1 ATM OF PRESSURE

BOUYANT FORCES  A FORCE THAT ACTS UPWARD ON AN OBJECT SUBMERGED IN A LIQUID OR FLOATING ON THELIQUID’S SURFACE.  EXAMPLE OF BOUYANT FORCE  FLOATING ON AN AIRMATTRESS  A BOAT

APPARENT WEIGHT  BECAUSE THE BUOYANT FORCE ACTS IN THE OPPOSITE DIRECTION OF FORCE DUE TO GRAVITY, OBJECTS SUBMERGED APPEAR TO WEIGH LESS IN WATER THAN THEY DO IN AIR.  THIS IS CALLED THEIR APPARENT WEIGHT

FLUID DISPLACEMENT  WHEN AN OBJECT IS LOWERED INTO A FLUID THE OBJECT WILL DISPLACE ITS VOLUME  DO YOU REMEMBER HOW TO MEASURE THE VOLUME OF AN IRREGULARLY SHAPED SOLID?  IT IS THE SAME PRINCIPLE  Demonstration here

ARCHIMEDES  GREEK MATHEMATICIAN WHO WAS BORN IN SYRACUSE, SICILY  LEGEND OF THE CROWN  “EUREKA! ” MEANING I’VE FOUND IT!

ARCHIMEDES’ PRINCIPLE  ANY OBJECT COMPLETELY OR PARTIALLY SUBMERGED IN A FLUID EXPERIENCES AN UPWARD BUOYANT FORCE EQUAL IN MAGNITUDE TO THE WEIGHT OF THE FLUID DISPLACED BY THE OBJECT.  F B =Fg (displaced fluid) = m f g  MAGNITUDE OF BUOYANT FORCE = WEIGHT OF FLUID DISPLACED  M f – MASS OF DISPLACED FLUID

WILL IT FLOAT?  OBJECTS WILL FLOAT OR SINK DEPENDING UPON THE NET FORCE ACTING ON THE OBJECT.  IF THE NET FORCE IS UPWARD IT WILL FLOAT.  IF THE NET FORCE IS DOWNWARD IT WILL SINK.  F NET = F B – Fg (object)

F NET = F B – Fg (object)  F B = m f g (MASS OF DISPLACED  FLUID X G)  F g = m O g (MASS OF OBJECT X G)  F NET = m f g - m O g

F NET = m f g - m O g  REMEMBER: ρ = m/v  SO…. ρv = m  THEREFORE….WE CAN SAY….  F NET = (ρ f v f - ρ O v O ) g  ( REMEMBER THE FLUID QUANTITIES REFER TO THE DISPLACED FLUID)

FLOATING OBJECTS  IN FLOATING OBJECTS THE BOUYANT FORCE IS EQUAL TO THE WEIGHT OF THE OBJECT.  F B = m O g  **ARCHEMEDES PRINCIPLE IS NOT REQUIRED TO FIND THE BUOYANT FORCE ON A FLOATING OBJECT IF THE WEIGHT OF THE OBJECT IS KNOWN

DENSITY DETERMINES DEPTH  THE DENSITY OF AN OBJECT DETERMINES THE DEPTH AT WHICH IT FLOATS.  WHEN AN OBJECT IS FLOATING THE NET FORCE EQUALS…..WHAT?  ZERO!

WE CAN DERIVE A RATIO  F NET = 0 = (ρ f v f - ρ O v O ) g  ρ f = v O  ρ O v f  FOR AN OBJECT TO FLOAT, THE OBJECT’S DENSITY CAN NEVER BE GREATER THAN THE DENSITY OF THE FLUID IN WHICH THE OBJECTS FLOATS.

FLOATING  EQUAL DENSITIES PRODUCES A FLOATING THAT IS TOTALLY SUBMERGED BUT DOES NOT SINK.

CHANGING BOUYANCY  BOUYANCY CAN BE CHANGED BY ALTERING THE DENSITY OF AN OBJECT.  CARTISIAN DIVER DEMO HERE

EXAMPLES?  SWIM BLADDER OF A FISH.  BALLAST TANKS OF A SUBMARINE.  THE HUMAN BRAIN IS IMMERSED IN A FLUID OF DENSITY 1007 KG/M 3, WHICH IS SLIGHTLY LESS THAN THE AVERAGE DENSITY OF THE BRAIN (1040 KG/M 3 ) WHAT BENEFIT IS THIS?

QUESTION  WHY DOES A SHIP SINK?  BECAUSE ITS DENSITY IS GREATER THAN THE DENSITY OF THE WATER.

APPARENT WEIGHT DEPENDS UPON DENSITY  AS VOLUME DECREASES THE AMOUNT 0F WATER DISPLACED DECREASES UNTIL IT BECOMES TOTALLY SUBMERGED.  WHEN SUBMERGED THE VOLUME DISPLACED = VOLUME OF THE OBJECT AND SO IF THE DENSITY OF THE OBJECT IS GREATER THAN WATER IT WILL SINK

EQUATION VIEW  FLOATING OBJECT  F NET = (ρ f v f - ρ O v O ) g  SUBMERGED OBJECT  F NET = (ρ f - ρ O ) Vg  SO WE CAN CONSIDER  F G = ρ O V gF G = ρ O  F B = ρ f V g F B = ρ f  WHICH IS OFTEN USED TO SOLVE BUOYANCY PROBLEMS.

QUESTION  ASTRONAUTS SOMETIMES TRAIN UNDERWATER TO SIMULATE CONDITIONS IN SPACE. EXPLAIN WHY.  THE BUOYANT FORECE CAUSES THE NET FORCE ON THE ASTRONAUT TO BE CLOSE TO ZERO.

QUESTION  A STUDENT CLAIMS THAT IF THE STRENGTH OF EARTH’S GRAVITY DOUBLE, PEOPLE WOULD BE UNABLE TO FLOAT ON WATER. DO YOU AGREE OR DISAGREE WITH THIS STATEMENT? WHY?  DISAGREE, BECAUSE THE BUOYANT FORCE IS ALSO PROPORTIONAL TO g.

QUESTION  EXPLAIN WHY BALLOONISTS USE HELIUM INSTEAD OF PURE OXYGEN IN BALLOONS.  HELIUM IS LESS DENSE THAN AIR AND THEREFORE FLOATS IN AIR.

PROBLEM  CALCULATE THE ACTUAL WEIGHT, THE BOUYANT FORCE, AND THE APPARENT WEIGHT OF A 5.00 X M 3 IRON BALL FLOATING AT REST IN MERCURY.  (DENSITY OF Hg = 13.6 X 10 3 kg/m 3 )  DENSITY OF Fe = 7.86 X 10 3 kg/m 3 )

ANSWER  F B = ρ f V g  ρ FE = 7.86 X 10 3 kg/m 3  V = 5.00 X M 3  G = 9.81  F B = 7.86 X 10 3 kg/m 3 X 5.00 X M 3 X 9.81  FB = 3.86 N  Fg = 3.86 N  APPARENT WEIGHT = F NET = ZERO

PRESSURE  PRESSURE IS THE MEASURE OF HOW MUCH FORCE IS APPLIED OVER A GIVEN AREA.

EQUATION  PRESSURE = FORCE/AREA  P = F/A  SI UNIT FOR PRESSURE IS PASCAL (Pa) = 1 N/m 2.  10 5 Pa = 1 atm  EX: ABSOLUTE AIR PRESSURE INSIDE A CAR TIRE IS ~ 3 atm OR 3 X 10 5 Pa

EXAMPLES OF PRESSURE  CENTER OF THE SUN2 X  CENTER OF EARTH4 X  BOTTOM OF THE PACIFIC 6 X 10 7  AT SEA LEVEL1.01 X 10 5  10 KM ABOVE SEA LEVEL2.8 X 10 4  BEST VACUUM IN A LAB.1 X

TRANSMITION OF PRESSURE  BLAISE PASCAL NOTICED A TREND WHEN PRESSURE IN APPLIED TO A FLUID.

PASCAL’S PRINCIPLE  IF THE PRESSURE IN A FLUID IS INCREASES AT ANY POINT IN A CONTAINER, THE PRESSURE INCREASES AT ALL POINTS INSIDE THE CONTAINER BY EXACTLY THE SAME AMOUNT.

EQUATION  PRESSURE INCREASE = F1 = F2  A1 A2  PRESSURE INCREASE = A SMALL FORCE F 1 APPLIED TO A SMALL PISTON OF AREA A 1 CAUSES A PRESSURE INCREASE IN A FLUID. THIS INCREASE IS TRANSMITTED TO A LARGER PISTON OF AREA A 2 AND THE FLUID EXERTS A FORCE F 2 ON THIS PISTON.

REARRANGE  PRESSURE INCREASE = F 1 = F 2  A 1 A 2  SOLVE FOR F 2  F 2 = A 2 X F 1  A 1  WE CAN SEE THAT F 2 IS LARGER THAN THE INPUT FORCE BY A FACTOR EQUAL TO THE RATIO OF THE TWO AREAS.

DON’T BE CONFUSED!  WHILE AN INCREASE IN PRESSURE IS TRANSMITTED EQUALLY THROUGHOUT A FLUID, THE TOTAL PRESSURE AT DIFFERENT POINTS IN THE FLUID MAY VARY WITH DEPTH.

PRACTICE  THE SMALL PISTON OF A HYDRAULIC LIFT HAS AN AREA OF 0.20 M 2. A CAR WEIGHING 1.20 X10 4 N SITS ON A RACK MOUNTED ON THELARGE PISTON. THE LARGE PISTON HAS AN AREA OF 0.90 M 2. HOW LARGE A FORCE MUST BE APPLIED TO THE SMALL PISTON TO SUPPORT THE CAR?

ANSWER  A 1 =.20 M 2 F 1 =  A 2 =.90 M 2 F 2 = 1.20 X 10 4 N  F 1 = F 2  A 1 A 2  F 1 = A 1 X F 2 F 1 =(.20 /.90) 1.20 X 10 4 N  A 2 F 1 =2.7 X 10 3 N

PRESSURE VARIES WITH DEPTH  FLUID EXERTS PRESSURE IN ALL DIRECTIONS.  PRESSURE AGAINST THE SIDES OF A BOX ALSO INCREASES WITH DEPTH.  EX: ON A SUBMARINE, THE PRESSURE OF AN ENTIRE COLUMN OF WATER IS ON THE SUBMARINE AS WELL AS THE PRESSURE OF THE ATMOSPHERE ABOVE.

EQUATION GAUGE PRESSURE  GAUGE PRESSURE IS NOT THE TOTAL PRESSURE AT A DEPTH.  WARNING … THIS IS ONLY GOOD IF THE DENSITY IS THE SAME THROUGHOUT THE LIQUID.  P = F = MG = ρVg = ρAhg= ρhg  A A A A  P = ρhg

EQUATION ABSOLUTE PRESSURE  ABSOLUTE PRESSURE IS THE TOTAL PRESSURE AT A DEPTH  P = P o + ρhg  ABSOLUTE PRESSURE = ATMOSPHERIC PRESSURE + DENSITY X FREE-FALL ACCELERATION X DEPTH)

PRESSURE FROM ABOVE  ATMOSPHERIC PRESSURE IS THE WEIGHT OF THE AIR ON THE LAYERS OF AIR BELOW.  ATMOSPHERIC PRESSURE IS ~ EQUAL TO 40,000 LBS. OR 200,000 N OF FORCE ON OUR BODIES.  WHY DON’T OUR BODIES COLLAPSE?

ANSWER  OUR BODIES ARE FILLED WITH FLUIDS PUSHING OUT AT THE SAME PRESSURE OF THE ATMOSPHERE.  OUR BODIES ARE IN EQUILIBRIUM  THE FORCE PUSHING IN EQUALS THE FORCE PUSHING OUT.

BAROMETERS  INSTRUMENT THAT IS USED TO MEASURE ATMOSPHEREIC PRESSURE.

KINETIC THEORY OF GASES  GAS PARTICLES WILL CONSTANTLY COLLIDE WITH ONE ANOTHER.  KINETIC ENERGY INCREASES WITH INCREASE IN TEMPERATURE.  THE HIGHER TEMPERATURE IN THE GAS THE FASTER THE MOLECULES MOVE.

TEMPERATURE IN A GAS  TEMPERATURE IS THE MEASURE OF THE AVERAGE KINETIC ENERGY OF THE PARTICLES IN A SUBSTANCE.  SI UNITS FOR TEMPERATURE ARE KELVINS AND DEGREES CELSIUS.  ROOM TEMPERATURE IS ABOAUT 293 K OR 20 DEGREES CELSIUS.

FLUID FLOW  CONTINUITY EQUATION  RESULTS FROM MASS CONSERVATION  M1=M2  BECAUSE p = M/V WE CAN SAY THAT  M = pv  So WE CAN SUBSTITUTE AND SAY  p1v1 = p2v2