Presentation Transcript

1. Western SOS Physics 1029 Review Session

2. A little about myself…

   ? BMSc
3. Applications of physics… Other classes: Medical Sciences 4900/4930, Physiology 3140, Genetics, Biochemistry MCAT Everyday life
4. Your test 3 Problem Solving Questions 1 from chapters 11,12- fluids 1 from chapters13,14- electricity 1 from chapter 16- elasticity Multiple Choice Section know equations well- mainly for relationships. Math will be on though know conceptual application of concepts know units of variables Cheat Sheet use it to your advantage 8 ½ by 11 sheet – double sided
5. Chapter 11 Pascal’s Law p2- p1 = -ρg (y2-y1) p= patm + ρgd not for fluid air doesn’t matter what the shape of the container is Standard unit for pressure- Pa Question: What the pressure in water 10m below the surface (Density of water= 1X 103 kg/m3, Patm= 1.01 X 105 Pa) P= 1.97 atm or 1.993X105 Pa

6. Archimedes Principle

   Fbuoy Fg Fbuoy= ρfluidVbody g Fnet = Fbuoy – Fg
7. An example A giant toy shark with volume of 1 m3 is dropped into a bucket of water. The toy shark’s density is 800 kg/ m3. What happens to the toy shark? What is the net force? Strategy: Find buoyant and gravitational forces Fbuoy = pfluid V toyshark g= 1000 (1) (10)= 10,000 N Fg = mg= ptoysharkVtoyshark g = (800) (1) (10) = 8000N Fnet = 8,000N – 10,000N = 2,000 N
8. Surface Tension Surface energy σ= ∆E/∆A (J/m2 ) ∆A= lx ∆ly W= ∆E = σ ∆A= F ∆ly Thus σ= F/lx (N/m)
9. Pressure in a bubble- Laplace’s Law The pressure difference between the inside and outside of a fluid with a curved surface is inversely proportional to the radius of curvature of the curved surface ∆p= pi – po Bubble: 4σ/r Droplet: 2σ/r Homog cylinder: σ/r
10. Chapter 12 Ideal dynamic fluid model No turbulences occur during flow (sufficiently slow flow) No sound waves develop in the flowing fluid No friction occurs with the walls of the tube
11. Equation of continuity A1 V1 = A2 V2 = constant ∆V/∆t Questions If volume flow rate is 83 cm3 /s and the cross sectional area of an arteriole is 3.8 cm2 , what is the speed of the blood rushing through the arteriole? AA VA = ∆V/∆t= 83 VA = 83/ AA = 83/ 3.8 = 21.8 cm/ s
12. Bernouillis Principle An increase in the speed of an ideal dynamic fluid in a tube is accompanied by a drop in the pressure during laminar flow p1 + ½ ρ v12 = p2 + ½ ρ v12 = constant
13. Viscous Flow Interactions with wall of container Flow is fastest in middle of tube In a fixed tube, pressure of fluid decreases with distance down the tube Bernoullis law does not apply
14. Viscosity Because there is a resistance (R) to fluid flow, a force (Fext) must be exerted on the fluid Fext= ή A (∆V/∆y), ή [Ns/m2] I- area of submerged plates II- relative speed of plates to each other III- distance between the plates (inversely proportional) Fext decreases with increased temperature (ή decreases) Question A 1.0 mm thick coat of glycerine is placed between two microscopic slides of width 2cm and length 7 cm. Find the force required to move the slides at a constant speed of 20 cm/s relative to each other. (ήglycerine= 1.5Ns/m2)
15. Question cont’d Question F= ή A (∆V/∆y) = 1.5 (1.4X 10-3m2) (0.2m/s) 1 X 10-3 m = 0.42 N
16. Poiseuille’s Law For a viscous fluid (Newtonian Fluid), the volume flow rate is not constant throughout the tube. Two forces act on the liquid: Force in the direction of flow- caused by pressure gradient Resistance force- because of viscosity ∆V/∆t = π (rt4)∆p 8ή (l) Volume flow rate is proportional to the fourth power of the radius of the tube Question: what happens to the volume flow rate for a newtonian fluid when the radius of the tube is doubled? (A= 16X)
17. Ohm’s Law Be careful to not get confused with the electrical one for non- cylindrical tubes Volume flow rate is proportional to the pressure difference for a viscous fluid and inversely proportional to the flow resistance ∆V/∆t= ∆ p/R
18. Chapter 13 Static Electricity Coulomb’s Law To find the force between charges have to find net force on a charge if more than two charges are present F= k q1q2 OR F= _1_q1q2 r2 4πЄ0 r2 Where k (9 X109 Nm2/Cb2), Є0 (8.85X10-12Cb2/N m2) Electron= -1.6 X 10-19 Cb, Proton= +1.6 X 10-19Cb
19. Question What is the magnitude and direction of the electric force between a +5e particle and a +3e particle if they are 7 nm away? F= k q1q2 OR F= _1_q1q2 r2 4πЄ0 r2 q1= +5e= 5 (1.6 X 10-19) = 8 X 10-19Cb q2= +3e= 3 (1.6 X 10-19) = 4.8 X 10-19 Cb r= 7 nm = 7 X 10-9 m F= (9 X 109) (8 X 10-19Cb) (4.8 X 10-19 Cb) 7 X 10-9 m F= 7.1 X 10-11 N away from each other
20. Question 2 Fnet ? + + +
21. Question 2 Fnet ? + + +
22. Electric Field E= _1_qfixed 4πЄ0 r2 Fixed point sends out electric field lines surrounding the charge. As get further away from charge (increase r), magnitude of E decreases Note F= q E
23. Electric field of a dipole Magnitude of Eletric field of a dipole drops in all directions proportional to 1/r3 [compare with E of a single charge- 1/r2] Lim Enet= qd____ X d 2πЄ0 X3 Electrical dipole moment, µ µ= qd
24. Electric Field between Parallel Plates Independent of position Proportional to charge density E= σ/ Є0 σ= Q/A [Cb/m2] charge density Question: A flat surface of a plate has a charge density of +5µCb/m2. What the electric field very close to the surface of the plate? E= σ/ 2Є0 (1/2 because looking at single plate, not both) E= (5 X 10-6Cb)/ ((2) (8.85X 10-12)) E= 2.8 X 105 N/Cb
25. Electrical Energy in Parallel Plates - Eel= qtest(E) (y) Electrical energy in a parallel plate arrangement is a linear function of distance from the plate with the opposite charge as the test charge Note In order to move a test charge closer to the plate with the same charge, Work must be done on the charge. + - + - + + + - - +
26. Electric Potential Energy- single charged particle Eel= _1_qtestQ 4πЄ0 r Therefore Work: W= qtestQ (1 – 1) 4πЄ0rfri Eel + + Q + + + qtest +
27. Electric potential- single point charge Like electric field was for electric force. But now we’re talking about energy V= Eel/ qtest Electric potential is defined as the electric potential energy per unit charge OR V= _1_Q 4πЄ0 r If there are multiple charges have to find the sum of the electric potentials.
28. Electric potential- parallel plates V= E y= (σ/Є0) y
29. Question 2 Vnetat P ? + P + +
30. Conservation of Energy Ekin,init + Eel, init = Ekin, final + Eel, final Kinetic energy= ½ mv2 Eel decrease, Ekin increases + + Q - + + qtest +
31. Chapter 14- Flowing Charge
32. Capacitors- biological membrane Capacitance- ability to store charge (across a membrane, parallel plates) Three quantities characterize a capacitor: areal charge density (σ), the capacitance (C) and the dielectric constant (κ) Units of capacitance Cb/V (Farad, F) C= Q/V= Є0 A b C is proportional to: charge stored across capacitor, area of capacitor Ci is inversely proportional to: distance (b) between the plates
33. Work stored in capacitor W= ½ Q ∆Vfinal ∆Vfinal- potential difference Q- total amount of charge transferred Therefore Eel= ½ QV = ½ (CV) V= ½ CV2
34. Question A parallel plate capacitor with capacitance C=13.5pF is charge to a potential difference of 12.5 V between its plates. The charging batter is then disconnected and a piece of porcelain is placed between the plates What is the potential energy of the device before the porcelain piece was added? What is its potential energy after the porcelain piece has been added? a) Eel= ½ CV2= ½ 13.5X10-12 F (12.5V)2= 1.06X 10-19 J b) Κ of porcelain is 6.5 Eel,fin= (1/κ) (El, init) = 1.06X 10-19 J/ 6.5= 1.6 X 10-10J
35. Current, Resistivity and Resistance I= ∆Q/ ∆t E= ρ J Resistivity (ρ)is the proportionality factor between the magnitude of the electric field (E) and the charge density (J) units Ωm Resistance R= ρ (l/A) - Units Ω
36. Chapter 16- Elastics Young’s Modulus F/A [stress] = Y (∆l/l) [strain] Y = stress/ strain
37. Springs- Force and Energy Hooke’s Law Felast= - k (x-xeq) Eelast= ½ k x2
38. Conservation of energy in springs Combined elastic potential and kinetic energy is a constant if no external forces are acting on the spring Max elastic potential- at extremes of vibration path (amplitude). There is no kinetic energy at this point Etot= Elast= ½ kA2 Max speed is when spring passes through equilibrium position. There is no elastic potential energy at this point Etot= ½ mv2