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Physics 7B - AB Lecture 10 June 5 Overview Practice Final Problems

Physics 7B - AB Lecture 10 June 5 Overview Practice Final Problems. Good news everyone! In four days you will be sitting physics exam. Oooh yes. Quiz 4 Re-evaluation Request Due TODAY Quiz 5 & 6 Due June 9 at the time of Final. Quiz 6 Rubrics on the website.

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Physics 7B - AB Lecture 10 June 5 Overview Practice Final Problems

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  1. Physics 7B - ABLecture 10June 5OverviewPractice Final Problems Good news everyone! In four days you will be sitting physics exam. Oooh yes.....

  2. Quiz 4 Re-evaluation Request Due TODAY Quiz 5 & 6 Due June 9 at the time of Final Quiz 6 Rubrics on the website Review session starts TODAY. Schedule on the course website

  3. 7B Final June 9 Mon 1- 3pm Review session starts TODAY.

  4. Final location Haring Hall, Rm. 1227 A through D Haring Hall, Rm. 2205 E through Q Surge 3 Bldg., Rm. 1309 R through Z Separated by familyname:

  5. Final checklist • Pens and pencils • CalculatorWe will not have spare calculators, make sure you bring yours • Photo ID (Student or Government ID) Without it you can’t sit the final, and you will fail the course. Formulas will be provided with the final The pages will be separated -- write your name on every single page when you first get your final

  6. Fluid (PF1&2, Q1, CDQ1) • Circuits (PF4, Q2, CDQ2) Physics breaks into two separate “parts” in 7B: • Momentum (PF6, Q5, CDQ5, CDQ7) • Forces: how to change momentum (PF6&7, Q4) • Angular momentum (Q5,CDQ7) • Torque: how to change ang mom (PF8,Q5,CDQ6) • Simple Harmonic Motion:A specific type of net force Energy density model No net change in energy density/around a circuit Forces and its relation to change in motion • Exponential decay (PF3, Q3,CDQ4) • Osmosis (Q3) Then we have learned four techniques • Vectors (PF5, Q3, Q4) • Components, use of trigonometry (PF5, Q5) • Force diagrams (PF5&7, Q6), Extended force diagrams (PF8, Q6,CDQ6) • Linear/Angular Momentum Charts (Q4, CDQ5)

  7. Today’s lecture Overview of the material, using practice final problems as examples • These review notes are supposed to go over the course, but as you have seen everything before, pieces of the course can be “mixed up” -- this is good practice for the final. • These notes cover a lot of the class, but not all. For example, practice final does not have any problem on diffusion. See CD Quiz 3 for an example. Diffusion occurs when there is concentration gradient of a specie of particles. What flows can be particels or water depending on the membrane property Wait a while… Permeable, semipermeable membrane

  8. Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks  = 25  = 45 Hammmock + Students + Strings = single object What are the contact/non contact forces exerted on the system?

  9. Identifying forces There are contact forces and non contact forces The only non contact force we worry in 7B is gravitational pull of the Earth exerted on all objects, i.e. FEarth on ball Contact force can be exerted by anything that is in contact with your object, i.e. Fstring on ball Fstring on ball Fy, string on ball = 100N Fx, string on ball = m|a| = 10kg(1.5m/s) = 15N FEarth on ball

  10. Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks  = 25  = 45 Hammmock + Students + Strings = single object What are the contact/non contact forces exerted on the system?

  11. Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks FPost on hammock FEarth on hammock = 1000N

  12. Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks FPost on hammock FEarth on hammock = 1000N

  13. Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks  = 25  = 45 A static problem, i.e., torques as well as forces are all balanced (another way of saying this is, net torque is zero & net force is zero) They have to be balanced compnents wise.

  14. Torque, Extended force diagram Practice Final 7 Another static problem This one is harder. Deltoid muscle is attached to the man’s arm and pulls on his arm in this direction Attached at 15cm from the shoulder joint 40N = FEarth on arm Center of Mass of the arm at 30cm from the shoulder joint

  15. Torque, Extended force diagram Practice Final 7 Another static problem Deltoid muscle is attached to the man’s arm and pulls on his arm in this direction Attached at 15cm from the shoulder joint 40N = FEarth on arm Center of Mass of the arm at 30cm from the shoulder joint A static problem, i.e., torques as well as forces are all balanced (another way of saying this is, net torque is zero & net force is zero)

  16. Torque, Extended force diagram Practice Final 7 Deltoid muscle is attached to the man’s arm and pulls on his arm in this direction Attached at 15cm from the shoulder joint 40N = FEarth on arm Center of Mass of the arm at 30cm from the shoulder joint What’s tangential component of the force of Deltoid muscle on the arm? Draw extended force diagram

  17. Torque, Extended force diagram Practice Final 7 Note: An arrow is a vector, a dotted arrow is a component of a vector 80N =Ftangential FMuscle on arm 40N = FEarth on arm

  18. Torque, Extended force diagram Practice Final 7 Note: An arrow is a vector, a dotted arrow is a component of a vector 80N =Ftangential FMuscle on arm Fshoulder joint on arm 40N = FEarth on arm

  19. Torque, Extended force diagram Practice Final 7 Note: An arrow is a vector, a dotted arrow is a component of a vector 80N =Ftangential FMuscle on arm Fshoulder joint on arm 40N = FEarth on arm Now both torque and forces are balanced!

  20. Torque, Extended force diagram Practice Final 7 Muscle goes limp and the arm starts to swing (Now torque is not balanced! ) What is its  after 0.1s? 40N = FEarth on arm

  21. Torque, Extended force diagram Practice Final 7 Muscle goes limp and the arm starts to swing (Now torque is not balanced! ) What is its  after 0.1s? 40N = FEarth on arm || ∆ t = |∆L|= (0.3m)(40N)(0.1sec) = 1.2Nms Lf = I  = 1.2Nms So then if we knew Iarm, we can figure out  !

  22. Recipes for torque • We know how to find each force • Every force in the problem also contributes a torque.(This torque may turn out to be zero) • The magnitude of the torque is on obj= (Ftangential) rwhere r is the distance between where the force is applied and the pivot point Pivot FEarth on pentagon

  23. Pivot r FEarth on pentagon Magnitude of torque is rFEarth on obj, perp Direction is into the screen (RHR) Do this for each force, then add all the torques up to find the net torque.(Some forces are easy: applied either at or through the pivot)

  24. Forces, Force diagram, Acceleratin/Velocity Practice Final 7 Compact shaped person jumping off a window of a burning building He falls freely for 1.5 sec, then the cushion exerts a constant force to bring the person to rest in 0.3sec Hint : Assume |vPerson| = 14.7m/s right before he hits the cushion Cushion Position y (m) Velocity v (m/s) Acceleration a (m/s2)

  25. Forces, Force diagram, Acceleration/Velocity Practice Final 7 Compact shaped person jumping off a window of a burning building (1) Force diagram during the free fall FEarth on person Cushion 2) Force diagram while the cushion is bringing the person to rest FCushion on person FEarth on person

  26. Forces, Force diagram, Acceleration/Velocity Practice Final 7 Compact shaped person jumping off a window of a burning building (1) Force diagram during the free fall FEarth on person Cushion Acceleration a (m/s2) 2) Force diagram while the cushion is bringing the person to rest Check whether the net force, i.e.Fon person, is consistent with a person Remember Fon person = m a person! FCushion on person FEarth on person

  27. Forces, Momentum Practice Final 6 Initial mToyota = 1000kg 20m/s to Left mCadillac = 2000kg 20m/s to Right Final mCadillac+Toyota = 3000kg Stuck together Traveling either to Right or to Left, or remain stationary We don’t know the direction/speed of travel after the collision. What we do know is: pC+Tinitial = pC+Tfinal

  28. Forces, Momentum Practice Final 6 Initial mToyota = 1000kg 20m/s to Left mCadillac = 2000kg 20m/s to Right Total | pC+Tinitial | = 20000kgm/s pC+Tinitial pC |pC | = mC |vC |= 40000kgm/s pT | pT | = mT |vT |= 20000kgm/s Final mCadillac+Toyota = 3000kg Stuck together

  29. Forces, Momentum Practice Final 6 Initial mToyota = 1000kg 20m/s to Left mCadillac = 2000kg 20m/s to Right Total | pC+Tinitial | = 20000kgm/s pC+Tinitial pC 40000kgm/s pT 20000kgm/s Final mCadillac+Toyota = 3000kg Stuck together Travelling to Right at 6.66m/s pC+Tfinal

  30. Forces, Momentum Practice Final 6 Initial mToyota = 1000kg 20m/s to Left mCadillac = 2000kg 20m/s to Right Then find ∆p of each car from: ∆p = m ∆v = m (vfinal–vinitial) pC 40000kgm/s pT 20000kgm/s Pay attention to the direction of vectors when adding/subtracting Final mCadillac+Toyota = 3000kg Stuck together Travelling to Right at 6.66m/s pC+Tfinal

  31. Forces, Momentum Practice Final 6 Initial mToyota = 1000kg 20m/s to Left mCadillac = 2000kg 20m/s to Right Then find ∆p of each car from: ∆p = m ∆v = m (vfinal– vinitial) pC 40000kgm/s pT 20000kgm/s Pay attention to the direction of vectors when adding/subtracting Final mCadillac+Toyota = 3000kg Stuck together Travelling to Right at 6.66m/s ∆pToyota comes out to be equal to ∆pCadillac even with its smaller mass because Toyota changes its direction after the collision pC+Tfinal

  32. Forces, Momentum Practice Final 6 Initial mToyota = 1000kg 20m/s to Left mCadillac = 2000kg 20m/s to Right Then find ∆v of each car: pC 40000kgm/s pT 20000kgm/s Pay attention to the direction of vectors when adding/subtracting Final mCadillac+Toyota = 3000kg Stuck together Travelling to Right at 6.66m/s ∆vToyota comes out to be greater than ∆vCadillac because Toyota changes its direction after the collision pC+Tfinal

  33. Forces, Momentum Practice Final 6 Initial mToyota = 1000kg 20m/s to Left mCadillac = 2000kg 20m/s to Right What about Fave experienced by each car during the collision? pC 40000kgm/s pT SAME ! 20000kgm/s Final One approach (Newton’s 1st law): Fave ∆t = ∆p mCadillac+Toyota = 3000kg Stuck together Travelling to Right at 6.66m/s Alternative approach (Newton’s 3rd law): FC on T = – FT on C !!! pC+Tfinal

  34. Fluids/Circuits Basic rules for looking at fluids/circuits : 1. Energy (density) conservation ∆P + (1/2)∆(v2) +  g∆h = Epump/volume – IR OR (the same as) ∆V =  – IR A1 v1 = A2 v2 2. Current entering = current leaving Junction rule 3. Pressures where two fluids systems touch are equal Voltages that are connected by wire (no circuit element in between) are equal

  35. P1 = 200kPa v1 = 10m/s Fluids, Circuit Practice Final 1 w = 1000kg/m3 A2 = 0.5A1 v1 vs v2 ??

  36. P1 = 200kPa v1 = 10m/s Fluids, Circuit Practice Final 1 w = 1000kg/m3 A2 = 0.5A1 A1 v1 = A2 v2 v2 = (A1 /A2)v1 = 2v1 = 20m/s !

  37. P1 = 200kPa v1 = 10m/s Fluids, Circuit Practice Final 1 w = 1000kg/m3 v2 = 20m/s Keep raising the end 2, ∆P + (1/2)∆(v2) +  g∆h = 0 At what h, P2 is equal to zero?

  38. P1 = 200kPa v1 = 10m/s Fluids, Circuit Practice Final 1 w = 1000kg/m3 v2 = 20m/s ∆P + (1/2)∆(v2) +  g∆h = 0 Substitute P2 = 0, to find h2 (say h1 = 0) h2 = 5m.

  39. Fluids, Circuit Practice Final 2 Why does a hose have a nozzle at the end?

  40. Fluids, Circuit, Forces Practice Final 2 v2 v1 v2 > v1 Why does a hose have a nozzle at the end? A1 v1 = A2 v2 !! In order to keep the flow rate constant throughout the fluid circuit, fluid velocity will increase at a narrowed nozzle. With greater velocity of the fluid coming out of the nozzle, the water will reach farther, allowing the firefighter to fight the fire far from the fire.

  41. Fluids, Circuit, Forces Practice Final 2 v2 v1 v2 > v1 What is the direction of the net force on the small amount of water at three different locations? Example of a problem that combines concepts from different models.

  42. Fluids, Circuit, Forces Practice Final 2 0 These are the direction of the net force on the small amount of water at three different locations. Think about how fluid velocity v is changing at each location, as net force is in the same direction as ∆ v.

  43. Fluids/Circuits 32 20 12 24 Oops… One more technique to remember : Know how to find equivalent resistance, this is an essential technique for analyzing circuits

  44. 32 20 12 24 Circuit Practice Final 3 Req of the whole circuit = 40  ∆V3 = 3V VB ??

  45. 32 20 12 24 Circuit Practice Final 3 Find I3 = 0.25A Find ∆V2 = 5V Find ∆V3 = 8V Find I1 = 0.25A Find I = 0.5A Find ∆V4 = 12V Find ∆VB 20V Finally! Req of the whole circuit = 40  ∆V3 = 3V VB ??

  46. 32 20 12 24 Circuit Practice Final 3 20V Close the switch… The circuit becomes simpler What happens to: ∆V1 I2 ∆V3 ∆V4 ?? Two things that don’t change : VB , resistor values Req decreases, and so Ieq increases

  47. Exponential decay Practice Final 3 Circuit A and B are similar but slightly different RC circuits Circuit A has no battery and the capacitor is charged to a voltage V. Circuit B has a battery having voltage V and the capacitor is uncharged.

  48. Exponential decay Practice Final 3 What was Capacitor C again?? Capacitor: A capacitor stores electrical energy by accumulating charge on two conducting plates. It can also release the stored energy very quickly.

  49. Exponential decay ± (Amount of stuff at t = 0 ) e ±kt Stuff (t) Anytime we see change in something is directly proportional to the amount of that something, we have exponential decay (growth) Ex. Microorganisms in a culture dish, a virus of sufficient infectivity, human population, nuclear chain reaction, charge/discharging capacitor The same statement mathematically: Solution : In each time interval of 1/k (time constant), amount of stuff reduces to 1/e of each previous value/ or grows by a factor of e

  50. Exponential decay Practice Final 3 Cgarged to V Uncharged Vc = Ve ±(1/RC)t C

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