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Final Exam Review Units: Mechanics Dynamics Work, Power, Energy, Thermodynamics Electricity & Magnetism Waves, Optics, Modern Physics UNIT 1- Mechanics • Measurements • Vectors • Kinematics Measurements • A precise measurement means that you used an appropriate tool to measure something as closely as possible. • For Example: A Bathroom Scale can’t be used to measure the atomic mass of an atom. Measurements • However a tape measure can be used to measure the length of a pencil to the closest half centimeter… Measurements • Example: Using the ruler, what is the most precise measurement you can get for the paper clip? (How many decimal places can you go?) 0cm 1cm 2cm Measurements • Example: Using the ruler, what is the most precise measurement you can get for the paper clip? (How many decimal places can you go?) 0cm 1cm 2cm 0.85 cm Vectors • Vectors are number that have a magnitude and a direction. • Like walking 10m North then 5m East. • They can be put together to make a triangle. Vector Addition • Vector Addition means basically finding out what the hypotenuse of the triangle looks like. • Imagine the two vectors. They would look like: 10m North 5m East Vector Addition • What would their resultant (hypotenuse of the triangle) look like? Vector Addition • What would their resultant (hypotenuse of the triangle) look like? Velocity vs. Time Graphs • In a velocity vs. time graph we can calculate the acceleration of an object by finding the slope of the graph. Velocity vs Time 25 Velocity (m/s) 20 15 10 5 0 0 0.5 1 1.5 Time (s) 2 2.5 Velocity vs. Time Graphs • What is the acceleration of this object? Velocity vs Time 25 Velocity (m/s) 20 15 10 5 0 0 0.5 1 1.5 Time (s) 2 2.5 Velocity vs. Time Graphs m y2 y1 x2 x1 Velocity vs Time 25 Velocity (m/s) 20 (1.5, 15) 15 10 (0.5, 5) 5 0 0 0.5 1 1.5 Time (s) 2 2.5 Velocity vs. Time Graphs m (15m / s 5m / s ) a (1.5s 0.5s ) 10m / s a 10m / s 2 1. 0 s Velocity vs Time y2 y1 x2 x1 25 Velocity (m/s) 20 (1.5, 15) 15 10 (0.5, 5) 5 0 0 0.5 1 1.5 Time (s) 2 2.5 Kinematics • One of the kinematics formulas that we used was: v f vi 2ax 2 2 Kinematics • If a stroller, initially at rest, is pushed so that it has a velocity of 5m/s after travelling 12m what is the stroller’s acceleration? Kinematics • If a stroller, initially at rest, is pushed so that it has a velocity of 5m/s after travelling 12m what is the stroller’s acceleration? vi=0m/s vf=5m/s Δx=12m a=? Kinematics • If a stroller, initially at rest, is pushed so that it has a velocity of 5m/s after travelling 12m what is the stroller’s acceleration? vi=0m/s vf=5m/s Δx=12m a=? v f vi 2ax 2 2 (5) (0) 2a (12m) 2 2 25 24a 1.04m / s a 2 Acceleration of Gravity • Gravity effects all objects the same. g=9.8m/s2 (constant) Who hits the ground first? Acceleration of Gravity • What if two similar objects are dropped but one gets a horizontal velocity? Acceleration of Gravity • They hit at the same time because the downward acceleration is the same for both. g g Using Acceleration of Gravity • With another kinematics formula and “g” we can determine how far an object will fall over time… • Second Kinematics Formula: 1 2 y vi t at 2 Using Acceleration of Gravity • Example: If a tennis ball is dropped from a roof, how far will it travel in 4s? Using Acceleration of Gravity • Example: If a tennis ball is dropped from a roof, how far will it travel in 4s? vi=0m/s a=9.8m/s2 t=2s Δy=? 1 2 y vi t at 2 Using Acceleration of Gravity • Example: If a tennis ball is dropped from a roof, how far will it travel in 4s? vi=0m/s a=9.8m/s2 t=2s Δy=? 1 2 y vi t at 2 1 y (0m / s )( 4s ) (9.8m / s 2 )( 4s ) 2 2 1 y (9.8)(16) 2 y 78.4m Acceleration of Gravity • When an astronaut is in orbit, they feel weightless. Why? Acceleration of Gravity • The earth’s gravity keeps pulling him around and around instead of going right past. Acceleration of Gravity • When gravity pulls something, it makes it fall. Acceleration of Gravity • …So in a way the astronaut is ALWAYS falling, as long as he’s in orbit. Acceleration of Gravity • Astronauts feel weightless because they are in constant free fall. Unit 2- Dynamics • • • • Newton’s Laws Forces Impulse Momentum Net Force • Consider the block with two ropes attached. 15kg Net Force • Consider the block with two ropes attached. • What is the net Force on the crate? 15N 10N 15kg Net Force • Consider the block with two ropes attached. • What is the net Force on the crate? 15N 10N 15kg 5N to the Left. 15N-10N=5N Net Force • What is the Acceleration of the create? 15N 10N 15kg 5N to the Left. Net Force • What is the Acceleration of the create? F=ma F=5N m=15kg 15N 10N 15kg Net Force • What is the Acceleration of the create? F=ma F=5N m=15kg 15N 10N F ma 5 N (15kg) a 5N a 15kg 0.33m / s 2 a 15kg The Atwood Machine • Newton’s Second Law F=ma can be used to calculate the acceleration of a system of pulleys. • Consider the following situation: The Atwood Machine • What is the acceleration of the system? (No friction here) 2kg 3kg The Atwood Machine • Step 1- What is the force on the system? 2kg 3kg The Atwood Machine • Step 1- What is the force on the system? 2kg mg 3kg The Atwood Machine • Step 1- What is the force on the system? F=mg = (3kg)(9.8m/s2)= 29.4N 2kg mg 3kg The Atwood Machine • Step 2-What is the acceleration of the system? F= 29.4N 2kg mg 3kg The Atwood Machine • Step 2-What is the acceleration of the system? F= 29.4N m=2kg+3kg (total mass) 2kg mg 3kg The Atwood Machine • Step 2-What is the acceleration of the system? F= 29.4N m=5kg F ma 2kg mg 3kg The Atwood Machine • Step 2-What is the acceleration of the system? F= 29.4N m=5kg F ma 29.4 N (5kg)a 5.88m / s a 2 2kg mg 3kg Momentum and Impulse • If the momentum (p=mv) of an object is changed, we can find out the force needed to do it. • Impulse formula: m(v f vi ) Ft Momentum and Impulse • A 0.025kg golf ball is accelerated from 0m/s to 110m/s is 0.1seconds. What is the force applied to do so? Momentum and Impulse • A 0.025kg golf ball is accelerated from 0m/s to 110m/s is 0.1seconds. What is the force applied to do so? m= 0.025kg vi=0m/s vf=110m/s t=.1s m(v f vi ) Ft 0.025kg(110m / s 0m / s) F (.1s) 2.75 F (.1s) 27.5 N F Conservation of momentum • Remember momentum is indestructible. It can be transferred or shared but never lost. • Consider these two situations: • Who has more momentum? More velocity? Conservation of momentum • What is the speed of the boy/board after he jumps on the skateboard? The boy’s mass is 50kg, the board is 1kg. v=5m/s v=? Conservation of momentum • What is the speed of the boy/board after he jumps on the skateboard? The boy’s mass is 50kg, the board is 1kg. v=5m/s Step 1 p=mv p=(50kg)(5m/s) p=250 v=? Conservation of momentum • What is the speed of the boy/board after he jumps on the skateboard? The boy’s mass is 50kg, the board is 1kg. v=5m/s Step 1 p=mv p=(50kg)(5m/s) p=250 Step 2 250=(50kg+1kg)v 250=51v 4.90m/s=v Unit 3- Work, Power, Energy • • • • Friction Power Conservation of Energy Temperature Change Sliding down a ramp • Imagine an object sliding down a ramp… Sliding down a ramp • If the object has an initial speed of 10m/s, but only 8m/s at the bottom, what is responsible? 10m/s 8m/s Sliding down a ramp • If the object has an initial speed of 10m/s, but only 8m/s at the bottom, what is responsible? 10m/s FRICTION!! 8m/s Work & Power • Work is the force needed to move an object some distance. Formula: W=Fd • Power is how fast work gets done. Formula: P=W/t Work and Power • How much power is generated by a weightlifter picking up a 50kg weight a distance of 1m in 0.5s? Work and Power • How much power is generated by a weightlifter picking up a 50kg weight a distance of 1m in 0.5s? Step 1: Calculate Work. 50kg W=Fd Work and Power • How much power is generated by a weightlifter picking up a 50kg weight a distance of 1m in 0.5s? Step 1: Calculate Work. 50kg W=Fd W=? F=mg F=(50kg)(9.8m/s2) F=490N Work and Power • How much power is generated by a weightlifter picking up a 50kg weight a distance of 1m in 0.5s? Step 1: Calculate Work. 50kg W=Fd W=(490N)(1m) F=mg F=(50kg)(9.8m/s2) F=490N Work and Power • How much power is generated by a weightlifter picking up a 50kg weight a distance of 1m in 0.5s? Step 1: Calculate Work. 50kg W=Fd W=(490N)(1m) F=mg W=490J F=(50kg)(9.8m/s2) F=490N Work and Power • How much power is generated by a weightlifter picking up a 50kg weight a distance of 1m in 0.5s? Step 1: Calculate Work. Step 2: Find Power W=Fd P=W/t W=(490N)(1m) W=490J Work and Power • How much power is generated by a weightlifter picking up a 50kg weight a distance of 1m in 0.5s? Step 1: Calculate Work. Step 2: Find Power W=Fd P=W/t W=(490N)(1m) P=(490J)/(.5s) W=490J P=980W PE, KE , and Range • Consider an object swinging from a cliff: m=2kg A (h=50m) B (h=20m) PE, KE , and Range • We will need these two formulas: m=2kg PE=mgh A (h=50m) KE=1/2mv2 B (h=20m) PE, KE , and Range • What is the change in PE (ΔPE) between A and B?: m=2kg A (h=50m) B (h=20m) PE, KE , and Range • What is the change in PE (ΔPE) between A and B?: m=2kg A (h=50m) B (h=20m) PEa=mgh =(2kg)(9.8)(50m) =980J PEb=mgh =(2kg)(9.8)(20m) =392J PE, KE , and Range • What is the change in PE (ΔPE) between A and B?: m=2kg A (h=50m) PE=980J ΔPE=PEf-PEi ΔPE=392J-980J= -588J B (h=20m) PE=392J PE, KE , and Range • What is the KE at point B? m=2kg A (h=50m) PE=980J B (h=20m) PE=392J PE, KE , and Range • What is the KE at point B? m=2kg A (h=50m) PE=980J B (h=20m) PE=392J The total energy is equal to PE at A. So Etotal= 980J PE, KE , and Range • What is the KE at point B? m=2kg A (h=50m) PE=980J B (h=20m) PE=392J KE=588J Total Energy is Etotal= 980J. At B, PE is 392J, so KE is: 980J-392J = 588J PE, KE , and Range • Point B is the bottom of the rope swing. How fast will he be going and in what direction? A (h=50m) PE=980J m=2kg B (h=20m) PE=392J KE=588J PE, KE , and Range • Find velocity at point B using KE formula. KE=1/2mv2 A (h=50m) PE=980J B (h=20m) PE=392J KE=588J PE, KE , and Range • Find velocity at point B using KE formula. KE=1/2mv2 A (h=50m) PE=980J B (h=20m) PE=392J KE=588J v=24.25m/s 588J=1/2(2kg)v2 588=v2 24.25m/s =v PE, KE , and Range • Since he is at the very bottom of the swing, the velocity of 24.25m/s is straight to the right. A (h=50m) PE=980J m=2kg 24.25m/s B (h=20m) PE=392J KE=588J v=24.25m/s PE, KE , and Range • If the string breaks, how long will it take to hit the ground? !!! A (h=50m) PE=980J m=2kg B (h=20m) PE=392J KE=588J v=24.25m/s PE, KE , and Range • As it’s falling down: vi=0m/s Δy=20m a=9.8m/s2 !!! A (h=50m) PE=980J m=2kg B (h=20m) PE=392J KE=588J v=24.25m/s t=? PE, KE , and Range • As it’s falling down: vi=0m/s Δy=20m A (h=50m) PE=980J a=9.8m/s2 1 2 y vi t at 2 (Same equation from Unit 1) B (h=20m) PE=392J KE=588J v=24.25m/s t=? PE, KE , and Range • As it’s falling down: vi=0m/s Δy=20m A (h=50m) PE=980J B (h=20m) PE=392J KE=588J v=24.25m/s a=9.8m/s2 1 2 y vi t at 2 2 20 (0)t 1 / 2(9.8)t 20 4.9t 2 t 4.08 2.02s t=? PE, KE , and Range • After the string breaks the guy will go flying. How far away from here will he land? A (h=50m) PE=980J B (h=20m) PE=392J KE=588J v=24.25m/s !!! Δx=? PE, KE , and Range • After the string breaks the guy will go flying. How far away from here will he land? v=24.25m/s !!! A (h=50m) t=2.02s PE=980J Δx=? B (h=20m) PE=392J KE=588J v=24.25m/s Δx=? PE, KE , and Range • After the string breaks the guy will go flying. How far away from here will he land? v=24.25m/s A (h=50m) t=2.02s PE=980J Δx=? x vt B (h=20m) PE=392J KE=588J v=24.25m/s x (24.25m / s)( 2.02s) x 48.98m Thermo-Spec. Heat • An object’s temperature is determined by how much energy it has. • This is called Specific Heat Formula: Q=mcΔT How can the amount of energy something has be increased? Thermo-Spec. Heat • This is called Specific Heat Formula: Q=mcΔT • Which has more energy, 5kg of water at 25°C or 500kg of water at 25°C. OR ? Thermo-Spec. Heat • This is called Specific Heat Formula: Q=mcΔT More Mass • Which has more energy, 5kg of water at 25°C or 500kg of water at 25°C. OR ? Thermo-Spec. Heat • Which has more energy, a pot of water at 80°C or a swimming pool at 25°C? OR Thermo-Spec. Heat • Which has more energy, a pot of water at 80°C or a swimming pool at 25°C? Because mass is so HUGE OR Q mcT Thermo-Spec. Heat • How much energy would a refrigerator have to take from a 0.5 kg glass of water to lower the temperature 5°C. The specific heat of water is 4180 J/kg°C? Thermo-Spec. Heat • How much energy would a refrigerator have to take from a 0.5kg glass of water to lower the temperature 5°C. The specific heat of water is 4180 J/kg°C? m= 0.5kg c=4180 J/kg°C ΔT=5°C Q mcT Thermo-Spec. Heat • How much energy would a refrigerator have to take from a 0.5kg glass of water to lower the temperature 5°C. The specific heat of water is 4180 J/kg°C? m= 0.5kg c=4180 J/kg°C ΔT=5°C Q mcT Q (0.5kg)( 4180 J / kgC )(5C ) Q 10450 J Unit 4- Electricity and Magnetism • • • • Static Electricity Magnetic Fields Circuits Magnets Electric Field Maps • Remember, electric field lines tell you What Would Protons Do? Electric Field Maps • What is the charge of these two objects? Electric Field Maps • What is the charge of these two objects? + - Electric Field Maps • If they were poles of magnets, which would be North and which would be South? Electric Field Maps • If they were poles of magnets, which would be North and which would be South? N S Electrostatic Force • The force between any two charged particles can be found using this formula: kc Q1Q2 FE 2 d Nm2 kc 8.99 E 9 2 C (constant) Electrostatic Force • What is the force between two electrons separated a distance of .5m? Electrostatic Force • What is the force between two electrons separated a distance of .5m? • Recall from the formula sheet the charge on an electron is -1.60E-19C so… Electrostatic Force • What is the force between two electrons separated a distance of .5m? Q1= -1.60E-19C Q2= -1.60E-19C d=.5m kc Q1Q2 FE d2 Nm2 kc 8.99 E 9 2 C Electrostatic Force • What is the force between two electrons separated a distance of .5m? Q1= -1.60E-19C Q2= -1.60E-19C d=.5m kc Q1Q2 (8.99 E 9)( 1.6 E 19)( 1.6 E 19) FE 2 F E d 0.52 2.3E 28 FE .25 FE 9.21E 28 N Static Electricity • Consider the following: Pith Ball Neutral Conductor Negative Charged Rod Static Electricity • As the rod is brought close, this happens: Pith Ball Neutral Conductor Negative Charged Rod Static Electricity • What is the charge on this part of the ball? Pith Ball Neutral Conductor Negative Charged Rod Static Electricity • The negative rod repels the electrons away in the conductor, leaving the right side + Pith Ball + Neutral Conductor Negative Charged Rod Static Electricity • The negative end of the conductor does the same to the pith ball - + Pith Ball - + Neutral Conductor Negative Charged Rod Static Electricity • So THIS part of the pith ball has a POSITIVE charge. - + Pith Ball - + Neutral Conductor Negative Charged Rod Right Hand Rules • This rule tells you about force on a wire when it’s in a magnetic field Right Hand Rule • The diagram will have current (I), and magnetic field (B), perhaps similar to: I B goes into page Right Hand Rule • Look at the two variables, I and B. Your Index finger goes with I, and your Baby finger goes with B I B goes into page Right Hand Rule • Now stick your thumb out. That is the direction of the force. What is it here? I B goes into page Right Hand Rule • Now stick your thumb out. That is the direction of the force. What is it here? I B goes into page Electromagnetic Induction • If you move a magnet through a coil of wire, you get electricity. Electromagnetic Induction • How can we get more electricity from the same setup? Electromagnetic Induction • Increase the number of coils or… Electromagnetic Induction • Move the magnet faster!! Ohm’s Law V=IR • Ohm’s Law told us about the most basic circuits. A 9V 3Ω Ohm’s Law V=IR • What is the value for current that the Ammeter would measure in this circuit? A 9V 3Ω Ohm’s Law V=IR V=IR 9V=I(3Ω) 3A=I 9V A 3Ω Equivalent Resistance • Equivalent Resistance of Parallel circuits can be calculated with: 1 1 1 RT R1 R2 Equivalent Resistance • What is the equivalent resistance of this circuit? 3Ω 4Ω Equivalent Resistance • What is the equivalent resistance of this circuit? 1 1 1 1 1 1 RT R1 R2 RT 3 1 0.58 RT 3Ω 4 1 (0.58) RT 1.72 RT 4Ω Proving Ohm’s Law • If we wanted to prove Ohm’s Law, we could conduct an experiment to do so… Proving Ohm’s Law • We would need an adjustable power supply, Ammeter, and Resistor. • Would set it up like this: A Proving Ohm’s Law • As the Voltage is slowly increased, we would take several measurements on the Ammeter. Voltage Current 2V 4A 4V 8A 6V 12A A increase observe Proving Ohm’s Law Volts • If we make a graph of the Voltage vs. Current data, and get the equation of the line… y=mx+b Current Proving Ohm’s Law Volts • It should look like Ohm’s Law (V=IR) with the Slope being equal to the resistor, which we can check V=R I +0 Current Proving Ohm’s Law Volts • Yep, slope equals resistance, Ohm’s Law Proved V=R I +0 Current Unit 5- Waves, Optics, Modern • • • • • • Wave properties Lenses Diagrams Snell’s Law Photoelectric Effect Fission & Fusion Velocity and Wavelength • Recall that all electromagnetic waves travel at the speed of light. • v= 3.0E8m/s Velocity and Wavelength • If the frequency of a cell signal is 5.0E15Hz, what is it’s wavelength, λ? Velocity and Wavelength • If the frequency of a cell signal is 5.0E15Hz, what is it’s wavelength, λ? v=3.0E8m/s f=5.05E15Hz λ=? Velocity and Wavelength • If the frequency of a cell signal is 5.0E15Hz, what is it’s wavelength, λ? v=3.0E8m/s f=5.05E15Hz v f 3.0 E8 (5.0 E15) 6 E 8m λ=? Period and Frequency Amplitude • Given the following Amplitude vs. Time graph, find the wave’s frequency. 2 4 Time (s) Period and Frequency • Given the following Amplitude vs. Time graph, find the wave’s frequency. Step 1: Find Period. Amplitude T= 2sec 2 4 Time (s) Period and Frequency Amplitude • Given the following Amplitude vs. Time graph, find the wave’s frequency. Step 2: Find Frequency T= 2sec f=1/T 2 4 Time (s) Period and Frequency Amplitude • Given the following Amplitude vs. Time graph, find the wave’s frequency. Step 2: Find Frequency T= 2sec f=1/T f=1/2sec 2 4 Time (s) f=0.5Hz Snell’s Law • If a beam of light travels from air into water, and the angle of incidence is 25°, what will the angle of refraction be? (nair=1.00 and nwater=1.33) Snell’s Law • If a beam of light travels from air into water, and the angle of incidence is 25°, what will the angle of refraction be? (nair=1.00 and nwater=1.33) n1 sin 1 n2 sin 2 Snell’s Law • If a beam of light travels from air into water, and the angle of incidence is 25°, what will the angle of refraction be? (nair=1.00 and nwater=1.33) n1 sin 1 n2 sin 2 Snell’s Law • If a beam of light travels from air into water, and the angle of incidence is 25°, what will the angle of refraction be? (nair=1.00 and nwater=1.33) n1 sin 1 n2 sin 2 1.00 sin 25 1.33 sin 0.42 1.33 sin 0.31 sin 18.1 Photoelectric Effect • Demonstrated the wave/particle duality of light because it demonstrated one of light’s wave properties: That the wave energy of the photon is dependent on the light’s intensity. Fusion and Fission • Both Fusion and Fission share these characteristics: • 1) In either reaction, Mass is converted into Energy. • 2) In either process, the end Mass is LESS than the beginning mass. The Lens Equation • If an object’s image appears to be 25cm from a curved mirror, and the mirror’s focal length is 10cm, what is the object’s actual distance? The Lens Equation • If an object’s image appears to be 25cm from a curved mirror, and the mirror’s focal length is 10cm, what is the object’s actual distance? f=10cm q=25cm p=? 1 1 1 p q f The Lens Equation • If an object’s image appears to be 25cm from a curved mirror, and the mirror’s focal length is 10cm, what is the object’s actual distance? f=10cm q=25cm p=? 1 1 1 p q f 1 1 1 p 25cm 10cm 1 1 1 p 10 25 1 0.06 p p 16.66cm Plane Mirrors • The reflection of an object in front of a plane mirror is called a REAL image. Survival Skills with Lenses • A fire can be started with just a lens and a little bit of Physics. Survival Skills with Lenses • First you will need a converging lens to focus sunlight. The light (and heat) will be sent to the focal point of the lens Focal Point Survival Skills with Lenses • If you turn the lens around, you can also use it as a magnifying lens… Survival Skills with Lenses • If the focal length of the lens is 8cm and the lady bug is 12cm away, what is the magnification of the lens? Survival Skills with Lenses • Step 1: Find the image distance (q) 1 1 1 p q f Survival Skills with Lenses • Step 1: Find the image distance (q) 1 1 1 p q f 1 1 1 12 q 8 1 .042 q q 24cm Survival Skills with Lenses • Step 2: Find the magnification q M p 24 M 12 M 2 Final Exam All done. GOOD LUCK.
