1. Physics review

2. Motion Equations: V f = V o + at *note V o is initial V V f 2 = V o 2 + 2ax *note x = displacement x = V o t + ½ at 2 V avg = V o + V f 2 when v o = 0 X = ½ at 2 V f 2 = 2ax Look back at your graphing motion notes!!!!!

3. Graphing motion Know how to translate a x –vs- t graph to a v –vs- t graph to an a –vs- t graph Don’t be a dope look at the slope!!!! Recall area under the curve calculations Look at the units of area to determine what that area represents

4. Projectile motion Break V o down into the x and y components and use the motion equations For X DO NOT use x = V o t + ½ at 2 But rather x = V o t (there is no x component a) Do not forget your mad trig skills! No trig on the test? Or so I have been told

5. Newtons laws An object in motion, remains in motion, unless acted on by an external force. The sum of all forces on an object = Mass x acceleration (ΣF = ma) For every force, there is an equal and opposite force

6. Fundamental forces of the Universe!!!!! F g = the force of gravity due to g fields – req. 2 masses for force F ele = the force of electricity (electrostatics) due to E field – req. 2 charges for force F B = the force of magnetism due to B field – 2 magnets – or a magnetic field and a moving charge

7. Free body diagrams 5. The car in #4 is speeding up from 50 to 60 mph. 6. The same car is skidding to a stop. 7. A boy uses a rope to pull a 10 kg box across the floor at constant velocity. 8. In #7, the boy ’ s 20 kg sister is sitting on the box.

8. A boy pulls at a 30˚ angle with a force of 100N on a box of Mass 50kg. Find the resulting acceleration if force of friction = 0. F appl FgFg FNFN F fr + + - - Fx = Fappl cos30 – Ffr = ma 100N cos30 = (50kg)a a = 1.73 m/s 2 NOTE: Fy = F N + Fappl sin30 – Fg = mg F N + 100 sin30 – (50)(9.8) = 0 (there is no motion or acceleration in the y-direction) F N = -50 + 490 = 440N

9. You are given an Atwood’s Machine. There are two masses hanging from a massless pulley. M1 is 15kg and M2 is 28kg. Calculate acceleration of the masses and the force of tension in the rope. m1 F g1 + m2 F g2 - FTFT FTFT F T – Fg 2 = m 2 a -F T + Fg 1 = m 1 a NOTE: a is the same for both masses. a= -2.96 m/s 2 a is to the left mass. + - Fg 1 – Fg 2 = (m1 + m2)a (15kg)(9.8) – (28)(9.8) = (43kg)(a) ΣF = Σm (a)

10. The ladder problem: A ladder leaning on a FRICTIONLESS wall: Ladder mass = 15kg Pig mass= 100kg Pig is ¾ up the ladder Θ = 60 O What coefficient of friction is required on the ground to stop the ladder from slipping?? Θ

11. Ladder Forces Identify all the forces acting on the ladder! F wall F N F fr Fgl Fgp F fr = μ F n so to calculate μ you need F fr and F n !!!!!!!!!

12. Sum the Fx and Fy components Σ Fx =0 =F fr - F wall F fr = F wall Σ Fy = 0 = F N – (F gl + F gp ) F N = F gl + F gp To calculate the needed F fr you need to use τ to find F wall !!!!!

13. Calculate Torque F wall (cos30) F gp (cos 60) F gl (cos60) Σ τ= 0 = F wall (cos30)(L) - F gl (cos60)(1/2 L) – F gp (cos60)(3/4L) F wall (.866) = 36.75 + 367.5 F wall (cos30) = F gl (cos60)(1/2) + F gp (cos60)(3/4) F wall = 467N The red arrows represent the perpendicular forces that are used in the calculation of torque. Recall that τ = F d 60 o

14. Sum the Fx and Fy components F fr = F wall F N = F gl + F gp F fr = F wall = 467N F N = F gl + F gp = (15kg)(9.8) + (100kg)(9.8) = 1127N F fr = µF N so 467 = µ1127, so µ = 0.41

15. Thermodynamics

16. The heat that is transferred From one system to another can be calculated through the equation: Q = m c Δt Where Q = heat energy m = the mass of the material in the system c = heat capacity of the material (material specific amt. of heat required to raise the Temp. of one g. of a sub., one unit of Temp. Δt = the change in temperature (in Kelvin)

17. When you add heat to a system The energy of the system will change (go up) or the system will do work – or both! Q = ΔE + W This is the 1 st law of thermodynamics! Energy can never be created or destroyed

18. PV = nRT The Ideal Gas Law Pressure x volume = #moles x gas constant x the temperature So in the demo when we increase the temp. (a lot) and kept the pressure constant (it was an open system) then the volume of the gas increases rapidly (thermal expansion due to an increase in temperature).

19. The 2 nd law The total entropy of the universe is always increasing Therefore, no engine can be 100% efficient No perpetual motion machines!

20. The 3 rd Law As the Temp. of an object approaches absolute zero – the entropy of the system approaches a constant A perfect crystal at zero K has zero entropy

21. Energy profile A 100g ball is thrown vertically up at 10m/s, rises to a height of 4.8m, then returns to the ground. Consider air friction. Green = Thermal E Black = total E Blue = KE, Red = gPE Note at pt. 2: KE + gPE + TE = Total E E t 2

22. Circular motion Speed = d/t = 2πr / T a c = v 2 /r = 4π 2 r / T 2 F c = ma = mv 2 /r (when we sub in from above) = m4π 2 r / T 2 Recall that F c = F g or F t or F fr

23. Newton’s law of gravitation: And you (really) F g = (Mass Earth)(Mass you) G r 2 (r is distance between center of masses) (G = universal constant = 6.67X10 -11 N * m 2 / kg 2 ) SO: F g = G M E m / r 2 So F g provided F c for objects in space This breaks down for v-small objects (sub atomic)

24. How does this apply to satellites? There is only one speed that a satellite can have in order to maintain an orbit with a constant r. Set the F g equation = to the F c equation GmM E /r 2 = mv 2 / r So: v = GM E / r Notice mass independent!!!! Kick some mass!

25. Kepler r 3 (in a.u.) = T 2 (in earth years) 1 r 3 = GM T 2 4 π 2 r in meters, T in seconds here!!!! The orbit of every planet is an ellipse with the Sun at one of the two foci."orbit of every planet is an ellipse with the Sun at one of the two foci." or

26. Fg = Fc GMm = mv 2 r 2 r Keep in mind that v = 2 π r / T So sub in to get GM = 4 π 2 r 2 r T 2

27. Momentum (m 1 v 1 + m 2 v 2 ) o = (m 1 v 1 + m 2 v 2 ) f ( (50)(10) + (80)(?) = (50)(0) + (80)(0) Therefore V 2 initial = - 6.25 m/s !!! Minus because we are using the coordinate system KE is not conserved here!

28. The Impulse – momentum theorem F Δt = ΔP = mv f - mv o

29. Ballistics pendulum Δh=.25m Mass ball = 0.056kg mass box = 0.196 kg ???? V o of the ball ?????

30. Think about momentum here Set momentum initial = momentum final (m ball v ball + m box v box ) = ((m ball + m box )(v final )) (0.056kg)(?) + 0 = (.056 +.196)( ?) Hey mister – you have 2 unknowns here! But we do know ½ mv f 2 = mgh Solve for v f (= 2.22m/s ) plug into P equation Solve for v o (= 10. m/s)

31. Electrostatics Recall the lab – induction/conduction Recall the grounding w/induction – remove ground – remove induction demo

32. Electric field lines Always move away from + chg. and to – chg. + - Example of a DIPOLE (2 point charges with same magnitude but different signs)

33. Parallel charged plates direction of field direction of e- motion Battery ++++++++++++++ ++++++ + * e- will accelerate through 1.5V hole = particle accelerator - * EPE e- = qV = 2.4 E -19 J e- * ΔEPE = KE = ½ mv 2 - - - - - - - - - - - - - - - - - - - - - * v ~ 7 x 10 5 m/s Recall q e- = 1.6 x 10 -19 J If V = 1, then EPE e- = 1.6 x 10 -19 J = 1 electron Volt (eV) = unit of Energy

34. Flow of charges = current (I) = charge/time = Δq/Δt = columbs/sec = Ampere Resistance (R) = voltage/current = V/I = Ohm’s law As stated earlier: V = energy/charge = ΔEPE/q o = Work/q o = FIId/q o = (q o E) d/q o = E d

35. Power Power = change in energy / time interval P = (Δq)V / Δt Recall that (Δq) / Δt = current (I) Therefore: P = I V (current  Voltage) Units of power are the Watt (W)

36. Resistors in series R total = R 1 + R 2 … 6Ω 3Ω = 9Ω 6V 6V (I) must be the same through both (= V/Σ R) (V) for each individual resistor = IR (for that one resistor)

37. Resistors in Parallel 1/R total = 1/R 1 + 1/R 2… 4Ω = 2.67Ω 8Ω 6V 6V 1/R t = 1/4 + 1/8 = 3/8 so R t = 8/3 = 2.67Ω I 1 = V/R 1 = 6/4 = 1.5 Amps I 2 = 6/8 =.75A I t = I 1 + I 2 = 1.5Amps +.75 Amps = 2.25 Amps V 1 = I 1 R 1 = (1.5) (4) = 6V

38. Kirchoff Rule #1 The Junction rule: The sum of currents coming to a junction = the sum of the currents leaving a junction

39. Rule #2 The Loop Rule: The sum of all potential differences around a loop = 0 That is to say the V battery = V drop total It may be wise to look over your circuits sheets Were you calculate I, R, V

40. Notice How the compass aligns on magnetic field lines The north point of a compass points to the S end of the Earth magnet (and to whatever magnetic field it is in) Like this

41. Right Hand Rule for B Fields Thumb points in the direction of the current Curl fingers in the direction of the B field + OUT IN

42. Just like the force that an electric field exerts on a charge There is a force exerted upon a moving charge by a magnetic field. The magnitude of the magnetic field is defined by the force it exerts on a moving charge B = F B / q o (v ) Units: N/C (m/s) = N(s)/C(m) = N/amps(m) = Tesla

43. Therefore a moving charge in a B field F B = q v B Units: N c m/s T B + Direction of F on charge is out!!!!!

44. A moving charge in a wire F B = I l B Units: N c/s m T B Direction of Force is out once again

45. Well – we will start with a simple example and define a few things I have a B field inside a partial loop of wire. I complete the circuit with a bar, and move the bar. I induce a separation of charges in the bar (a EMF) ! This results in a current in the circuit (conventional current). Apply the 2 nd right hand rule here

46. Wave types Longitudinal wave = motion of particle in same direction as motion of energy ( ) Sound waves – Transverse wave = motion of particle is perpendicular to motion of the energy ( ) Slinky, light -

47. Superposition - ex waves moving toward each other – cancel – then continue… Velocity – of string – v = F t / (m/L) Resonance – E at correct moment - Amp Tacoma bridge collapse Diffraction – waves bend around corners… Interference – constructive/destructive waves ppt. slides 8 and 9

48. λ = 1m λ’ =.5m λ’ (m/wave) = λ (m/wave) - V siren (m/s) x period (s/wave) λ’ = λ - (V siren)(T) Δ λ = λ – λ’ = λ - λ - V siren (λ/V sound) Δ λ = V siren (λ) / V sound DOPPLER

49. Mode (n) = basic unit of oscillation: L of the wave =( n/2)(λ) node anti-node Fundamental (f 0 ) 1 st mode Lowest f of periodic waveform: L =n/2 λ = ½ λ 2 nd mode (3 nodes): L = n/2 λ = 2/2 λ = 1λ 3 rd mode (4 nodes) L = 3/2λ 4 th mode (5 nodes) L = 4/2λ = 2λ

50. Sound waves Longitudinal waves = particle motion in same direction as energy motion Hearing ~ 20 to 20,000 Hz Loudness = amplitude Pitch = frequency Strings: v = Ft m/l length density

51. Refraction (into/out of water) ni sinθi = nr sinθr

52. TIR n i sinθi = n r sinθr 1.33 sinθi = 1 sin 90.

54. Closed tube? Like in lab 1 st mode = ¼ λ L = ¼ λ L = ¾ λ L = 5/4 λ antinode 1 st harmonic (overtone) 2 nd harmonic 3 rd harmonic

55. Electromagnetic waves

56. Then at what angle does the wave intersect the normal on the opposite side of the prism? 32 o 20.4 o ?? O 60 o 90-20.4 = 69.6 o 69.6 o 30 o Unknown angle = 69.6 – 30 = 39.6 o

57. The Signs f is + with converging lens d o is + when obj. and light are on same side of lens (almost always) d i is + when image on opp. Side of lens from real object h o is + always (so h i is negative here) M (magnification) = h i / h o = -d i /d o h o = 1.5cm high d o = 6cm focus = 3cm Focus = 3cm d i = 6cm hihi

58. Overall magnification due to (ma)(mb) Image formed by one lens serves as object for second lens M a = -di a /do a M b = -di b /do b M tot = (-di a /do a )(-di b /do b )