Upload
lali
View
39
Download
3
Embed Size (px)
DESCRIPTION
Western SOS Physics 1029 Review Session. A little about myself…. ?. BMSc. Applications of physics…. Other classes: Medical Sciences 4900/4930, Physiology 3140, Genetics, Biochemistry MCAT Everyday life . Your test. 3 Problem Solving Questions 1 from chapters 11,12- fluids - PowerPoint PPT Presentation
Citation preview
Western SOS Physics 1029 Review Session
A little about myself…
?BMSc
Applications of physics…
• Other classes: Medical Sciences 4900/4930, Physiology 3140, Genetics, Biochemistry• MCAT• Everyday life
Your test3 Problem Solving Questions
1 from chapters 11,12- fluids1 from chapters13,14- electricity1 from chapter 16- elasticity
Multiple Choice Sectionknow equations well- mainly for relationships. Math will be on thoughknow conceptual application of conceptsknow units of variables
Cheat Sheetuse it to your advantage8 ½ by 11 sheet – double sided
Chapter 11
p2- p1 = -ρg (y2-y1)
p= patm + ρgd
Pascal’s Law
• 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
Archimedes Principle
Fbuoy
FgFbuoy= ρfluid Vbody g
Fnet = Fbuoy – Fg
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 toy shark g= 1000 (1) (10)= 10,000 NFg = mg= ptoyshark Vtoyshark g = (800) (1) (10) = 8000N
Fnet = 8,000N – 10,000N = 2,000 N
Surface Tension
Surface energyσ= ∆E/∆A (J/m2 )∆A= lx ∆lyW= ∆E = σ ∆A= F ∆lyThus σ= F/lx (N/m)
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/rDroplet: 2σ/rHomog cylinder: σ/r
Chapter 12
Ideal dynamic fluid model
1. No turbulences occur during flow (sufficiently slow flow)2. No sound waves develop in the flowing fluid3. No friction occurs with the walls of the tube
Equation of continuityA1 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= 83VA = 83/ AA = 83/ 3.8 = 21.8 cm/ s
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 v1
2 = constant
Viscous Flow
Interactions with wall of containerFlow is fastest in middle of tubeIn a fixed tube, pressure of fluid decreases with distance down the tube
Bernoullis law does not apply
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 platesII- relative speed of plates to each otherIII- 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)
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
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:
1. Force in the direction of flow- caused by pressure gradient2. 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)
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
Chapter 13 Static ElectricityCoulomb’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 10-19 Cb
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 10-19 Cbq2= +3e= 3 (1.6 X 10-19) = 4.8 X 10-19 Cbr= 7 nm = 7 X 10-9 m
F= (9 X 109) (8 X 10-19 Cb) (4.8 X 10-19 Cb) 7 X 10-9 mF= 7.1 X 10-11 N away from each other
Question 2
Fnet ?
+ +
+
Question 2
Fnet ?
+ +
+
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
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
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 10-6 Cb)/ ((2) (8.85X 10-12))E= 2.8 X 105 N/Cb
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.
+
+
+
++
-
-
-
--
+
Electric Potential Energy- single charged particle
Eel= _1_ qtestQ 4πЄ0 r Therefore Work:
W= qtestQ (1 – 1) 4πЄ0 rf ri
qtestQ+
+ +
++
+
Eel
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.
Electric potential- parallel plates
V= E y= (σ/Є0) y
Question 2
Vnet at P ?
+ +
+P
Conservation of Energy
Ekin,init + Eel, init = Ekin, final + Eel, final
Kinetic energy= ½ mv2
qtestQ-
+ +
++
+
Eel decrease, Ekin increases
Chapter 14- Flowing Charge
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 capacitorCi is inversely proportional to: distance (b) between the plates
Work stored in capacitor
W= ½ Q ∆Vfinal
∆Vfinal- potential differenceQ- total amount of charge transferred
Therefore
Eel= ½ QV = ½ (CV) V= ½ CV2
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 platesa) What is the potential energy of the device before the porcelain piece
was added?b) 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 6.5Eel,fin= (1/κ) (El, init) = 1.06X 10-19 J/ 6.5= 1.6 X 10-10J
Current, Resistivity and ResistanceI= ∆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
Chapter 16- Elastics
Young’s Modulus
F/A [stress] = Y (∆l/l) [strain]
Y = stress/ strain
Springs- Force and Energy
Hooke’s Law
Felast= - k (x-xeq)
Eelast= ½ k x2
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