1Physics 1010:The Physics of Everyday LifeTODAY• Exam-3 review2General Test Information• 7:30 Tonight (11/14) in this room (G1B20).• Closed book.• One 3x5 note card with own notes written on it.• Calculators are allowed.• Worth 40 points total.• Multiple choice plus one essay question.3Last Extra Credit• Will be posted Thursday• Will be due Thursday after Tanksgiving• Worth 10 pts• Will be group project again (so startpicking a group)4Concept Overview• Thermodynamics• Electric and magnetic forces• Circuits and transformersExam is CUMULATIVE, but emphasis will be placed on concepts we covered since Exam-2:5Thermodynamics• Temperature is a measure of the molecules’ averagekinetic energy• Kelvin Temperature scale is absolute. There is no suchthing as a negative Kelvin.• Stefan-Boltzmann Law; power radiated by black body• σ = 5.67*10-8 J K-4 m-2 s-16Black Body Radiation• All bodies radiate according to their temperature• A “black” body is a body that absorbes all radiationincident upon it• Total energy radiated per unit area given by Stefan-Boltzmann Law• P = power radiated per unit area = = energy radiated per unit time per unit area7A black body is easy to construct:Consider a a large cavity with a tiny hole drilled into it. Of all the light that goes through the hole, only an insignificantly small amount gets reflected back out. The rest is bounced around the interior of the cavity so many times, that it is finally totally absorbed. The cavity is then a "Black Body".8Total Solar Power OutputThe radius of the sun is 696,000km. Its surfacetemperature is 5800K, what is the total powerradiated by the sun?The surface area of a sphere is given by A=4πR2Stefan constant: σ = 5.67*10-8 J K-4 m-2 s-1The total power output of a star is call the Luminosity of the starHere Rs is the radius of the star, so 4πRs2 is total power radiated by the star9Total Power Intercepted by EarthThe Earth orbits the sun at a distance of 150,000,000km, and its radius is 6400km.What fraction of the sun’s power is incident on the Earth?πRE2/4πRO2 =(1/4)(6400km/150000000km)2=4.55*10-10At the Earth’s distance from the sun, the total solar power is spread over thearea of a sphere of radius 150,000,000km, but out of all that area, the Earth covers only a circle of radius 6400kmHere RE is the rdius of the Earth, Ro the radius of the Earth’s orbit10Total Power Radiated by EarthThe Earth absorbs 70% of the solar powerincident upon its surface. What is the totalpower radiated by Earth?Conservation of Energy: at equilibrium, the Earth has to radiate as much power as it receives from the sun, or else it would either heat up or cool downPout=0.7*Pin=0.7*(4.55*10-10)*(3.83*1026 W)=1.22*1017 W11“Naked” Earth TemperatureWhat would the temperature of the Earth be ifit has to radiate 1.22*1017W, and it had nogreenhouse effect? Tkelvin = Tcelsius + 273Here P is the total power radiated by Earth. RE the radius of the Earth,so 4πRE2 is the total area of the Earth12Greenhouse Effect• Sun (5800 oK) radiates mostly at optical wavelengths• Earth (~300 oK) radiates mostly at infraredwavelengths• Greenhouse gasses in atmosphere absorb some IRwavelengths, let others pass to space• Greenhouse gasses re-radiate absorbed energy inrandom direction, so half goes back to the ground.• Overall effect:τ is the greenhouse strength13Greenhouse Effect• For one-layer atmosphere,the incindent power, Is, isabsorbed by the ground.• The ground then radiatesback Ig• Some of Ig gets absorbed bythe atmosphere, and some isallowed to escape• The absorbed power gets re-radiated in a randomdirection, so on average half goes to space andhalf back to the ground• So now we have Is+Ia absorbed by ground, so theground has to have a higher T.14Electrostatics221rqkqFc=• Two kinds of charge – positive and negative. Likecharges repel; opposite charges attract.• Force decreases with distance – when thedistance is doubled, the force is one-fourth asstrong.15Magnetic Forces• Very similar to electric force – like poles repel, oppositepoles attract.• Isolated poles obey an inverse square law – just likeelectric charges.• At large distances, force between two magnets decaysquicker than the inverse square of their separation.• Currents produce magnetic fields – use right-hand-rule.• Lenz’s Law: Current induced by a changing magnetic fieldproduces a field that opposes the change.16Circuits• Voltage: The work done per unit charge.• Series: One element connected after another. Thecurrent stays the same throughout the circuit, but thevoltage changes across each element.• Parallel: Each element is connected directly to thebattery. The current branches along each path, but thevoltage is the same across every branch.IRV =RVRIIVP22===qVEPE =17Transformers• Use Lenz’s law to step up or step downvoltage.• The change in voltage depends on thenumber of coils.currentNorth pole18Transformers• In a coil, the direction of the magneticfield is given by the right-hand rule.currentNorth poleNorth pole is:A UpB DownC Sideways19Transformers• In a coil, the direction of the magneticfield is given by the right-hand rule.currentNorth poleNorth pole is:A UpB DownC SidewaysSo the magnetic field in a transformer changes direction 60 times per second20In the circuit below, the voltage of the battery is 12V and all lightbulbs have resistance 1 Ohm.+-ABCDEFThe voltage betweenA and G is:a. 3.0 Vb. 6.0 Vc. 12. Vd. 18. VG21In the circuit below, the voltage of the battery is 12V and all lightbulbs have resistance 1 Ohm.+-ABCDEFThe voltage betweenA and G is:a. 3.0 Vb. 6.0 Vc. 12. Vd. 18. VGThe current from A to BIs the same for both bulbs.The resistances are the same, so the voltage across each isthe same, so the voltage across each is 6V.22In the circuit below, the voltage of the battery is 12V and all lightbulbs have resistance 1 Ohm.+-ABCDEFThe current at A is:a. 3.0 Ab. 6.0 Ac. 12. Ad. 18. AG23In the circuit below, the voltage of the battery is 12V and all lightbulbs have resistance 1 Ohm.+-ABCDEFThe current at A is:a. 3.0 Ab. 6.0 Ac. 12. Ad. 18. AGThe current from A to Bss the same for both bulbs.The resistances are the same, so the voltage across each isthe same, so the voltage across each is 6V.I = V/R = 6V / 1 Ohm = 6A24In the circuit below, the voltage of the battery is 12V and all lightbulbs have resistance 1 Ohm.+-ABCDEFThe current at F is:a. 3.0 Ab. 6.0 Ac. 12.