Western sos physics 1029 review session
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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

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Western SOS Physics 1029 Review Session

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Western sos physics 1029 review session

Western SOS Physics 1029 Review Session

A little about myself

A little about myself…



Applications of physics

Applications of physics…

  • Other classes: Medical Sciences 4900/4930, Physiology 3140, Genetics, Biochemistry

  • MCAT

  • Everyday life

Your test

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

Chapter 11

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

Archimedes principle

Archimedes Principle



Fbuoy= ρfluidVbody g

Fnet = Fbuoy – Fg

An example

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

Surface tension

Surface Tension

Surface energy

σ= ∆E/∆A (J/m2 )

∆A= lx ∆ly

W= ∆E = σ ∆A= F ∆ly

Thus σ= F/lx (N/m)

Pressure in a bubble laplace s law

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

Chapter 12

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

Equation of continuity

Equation of continuity

A1 V1 = A2 V2 = constant ∆V/∆t


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

Bernouillis principle

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

Viscous flow

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



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)


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 cont’d


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

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)

Ohm s law

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 electricity

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



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

Question 2

Question 2

Fnet ?




Question 21

Question 2

Fnet ?




Electric field

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


F= q E

Electric field of a dipole

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

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

Electrical energy in parallel plates

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


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

Electric Potential Energy- single charged particle

Eel= _1_qtestQ

4πЄ0 r

Therefore Work:

W= qtestQ (1 – 1)











Electric potential single point charge

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

Electric potential- parallel plates

V= E y= (σ/Є0) y

Question 22

Question 2

Vnetat P ?





Conservation of energy

Conservation of Energy

Ekin,init + Eel, init = Ekin, final + Eel, final

Kinetic energy= ½ mv2

Eel decrease, Ekin increases









Chapter 14 flowing charge

Chapter 14- Flowing Charge

Capacitors biological membrane

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


C is proportional to: charge stored across capacitor, area of capacitor

Ci is inversely proportional to: distance (b) between the plates

Work stored in capacitor

Work stored in capacitor

W= ½ Q ∆Vfinal

∆Vfinal- potential difference

Q- total amount of charge transferred


Eel= ½ QV = ½ (CV) V= ½ CV2



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?


Eel= ½ CV2= ½ 13.5X10-12 F (12.5V)2= 1.06X 10-19 J


Κ of porcelain is 6.5

Eel,fin= (1/κ) (El, init) = 1.06X 10-19 J/ 6.5= 1.6 X 10-10J

Current resistivity and resistance

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

Chapter 16 elastics

Chapter 16- Elastics

Young’s Modulus

F/A [stress] = Y (∆l/l) [strain]

Y = stress/ strain

Springs force and energy

Springs- Force and Energy

Hooke’s Law

Felast= - k (x-xeq)

Eelast= ½ k x2

Conservation of energy in springs

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

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