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