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PRE AP CHAPTER 9 PHYSICS I CHAPTERS 18, 19, 20
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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
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WHAT IS A FLUID? MATERIALS THAT : FLOW ALTER THEIR SHAPE AS THEY FLOW SOLIDS ARE NOT CONSIDERED FLUIDS. WHY?
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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.
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LIQUIDS HAVE A DEFINITE VOLUME TAKE THE SHAPE OF THEIR CONTAINER.
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GASES INDEFINITE VOLUME FILL THEIR CONTAINER ON MATTER THE SIZE.
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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.
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MASS DENSITY REPRESENTED BY THE GREEEK LETTER ρ (RHO) EQUATION FOR DENSITY ρ = m/v DENSITY = MASS/VOLUME SI UNIT – kg/m 3
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DENSITY OF WATER 1 g/cm 3 1.00 X 10 3 kg/m3 AT 4 DEGREES CELCIUS
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EFFECT OF PRESSURE DENSITIES FOR SOLIDS AND LIQUIDS DO NOT CHANGE WITH PRESSURE, BECAUSE THEY HAVE A DEFINITE VOLUME.
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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
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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
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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
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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
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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
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ARCHIMEDES GREEK MATHEMATICIAN WHO WAS BORN IN SYRACUSE, SICILY LEGEND OF THE CROWN “EUREKA! ” MEANING I’VE FOUND IT!
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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
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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)
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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
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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)
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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
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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!
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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.
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FLOATING EQUAL DENSITIES PRODUCES A FLOATING THAT IS TOTALLY SUBMERGED BUT DOES NOT SINK.
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CHANGING BOUYANCY BOUYANCY CAN BE CHANGED BY ALTERING THE DENSITY OF AN OBJECT. CARTISIAN DIVER DEMO HERE
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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?
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QUESTION WHY DOES A SHIP SINK? BECAUSE ITS DENSITY IS GREATER THAN THE DENSITY OF THE WATER.
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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
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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.
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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.
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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.
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QUESTION EXPLAIN WHY BALLOONISTS USE HELIUM INSTEAD OF PURE OXYGEN IN BALLOONS. HELIUM IS LESS DENSE THAN AIR AND THEREFORE FLOATS IN AIR.
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PROBLEM CALCULATE THE ACTUAL WEIGHT, THE BOUYANT FORCE, AND THE APPARENT WEIGHT OF A 5.00 X 10 -5 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 )
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ANSWER F B = ρ f V g ρ FE = 7.86 X 10 3 kg/m 3 V = 5.00 X 10 -5 M 3 G = 9.81 F B = 7.86 X 10 3 kg/m 3 X 5.00 X 10 -5 M 3 X 9.81 FB = 3.86 N Fg = 3.86 N APPARENT WEIGHT = F NET = ZERO
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PRESSURE PRESSURE IS THE MEASURE OF HOW MUCH FORCE IS APPLIED OVER A GIVEN AREA.
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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
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EXAMPLES OF PRESSURE CENTER OF THE SUN2 X 10 16 CENTER OF EARTH4 X 10 11 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 10 -12
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TRANSMITION OF PRESSURE BLAISE PASCAL NOTICED A TREND WHEN PRESSURE IN APPLIED TO A FLUID.
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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.
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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.
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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.
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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.
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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?
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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
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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.
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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
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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)
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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?
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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.
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BAROMETERS INSTRUMENT THAT IS USED TO MEASURE ATMOSPHEREIC PRESSURE.
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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.
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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.
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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
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