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
A little about myself…
?
BMSc
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
Pascal’s Law
p2 p1 = ρg (y2y1)
p= patm + ρgd
Archimedes Principle
Fbuoy
Fg
Fbuoy= ρfluidVbody g
Fnet = Fbuoy – Fg
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 energy
σ= ∆E/∆A (J/m2 )
∆A= lx ∆ly
W= ∆E = σ ∆A= F ∆ly
Thus σ= F/lx (N/m)
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
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
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
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
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)
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
F= ή A (∆V/∆y)
= 1.5 (1.4X 103m2) (0.2m/s)
1 X 103 m
= 0.42 N
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 1019) = 8 X 1019Cb
q2= +3e= 3 (1.6 X 1019) = 4.8 X 1019 Cb
r= 7 nm = 7 X 109 m
F= (9 X 109) (8 X 1019Cb) (4.8 X 1019 Cb)
7 X 109 m
F= 7.1 X 1011 N away from each other
Fnet ?
+
+
+
Fnet ?
+
+
+
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
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

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.
+

+

+
+
+


+
Eel= _1_qtestQ
4πЄ0 r
Therefore Work:
W= qtestQ (1 – 1)
4πЄ0rfri
Eel
+
+
Q
+
+
+
qtest
+
V= E y= (σ/Є0) y
Vnetat P ?
+
P
+
+
Ekin,init + Eel, init = Ekin, final + Eel, final
Kinetic energy= ½ mv2
Eel decrease, Ekin increases
+
+
Q

+
+
qtest
+
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
W= ½ Q ∆Vfinal
∆Vfinal potential difference
Q total amount of charge transferred
Therefore
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?
a)
Eel= ½ CV2= ½ 13.5X1012 F (12.5V)2= 1.06X 1019 J
b)
Κ of porcelain is 6.5
Eel,fin= (1/κ) (El, init) = 1.06X 1019 J/ 6.5= 1.6 X 1010J
Young’s Modulus
F/A [stress] = Y (∆l/l) [strain]
Y = stress/ strain
Hooke’s Law
Felast=  k (xxeq)
Eelast= ½ k x2
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