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FINAL EXAM
PHYS 3511 (Biological Physics)
DATE/TIME:
April 11, 2015 (1:00 p.m. – 4:00 pm)
PLACE: AT 2019
Only non-programmable calculators are allowed.
Name: ___________________________________
ID: ___________________________________
Please read the following instructions:
• This midterm has 13 pages. Make sure none are missing.
• There are two sections. Section 1 has 10 multiple choice, true/false, or short
answer questions. In section 2 you must all 8 questions.
• You must show all works, in the provided blank space below each question.
• Write your name and student ID in the provided space above.
• There is a two-page equation sheet on page 12 and 13. You may tear the
equation sheet from the midterm.
Read questions carefully before attempting the solutions.
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Useful Equations
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!
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Newton’s Laws F net = ∑ F = 0 (equilibrium); F net = ma (Nonzero net force); 3rd Law if object
A exerts a force on object B, B exerts a force on A of equal magnitude and opposite in direction.
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Newton’s law for rotation τ = Iα . Friction f s ≤ µ s n , f k = µ k n . Fluid friction: 1) turbulent
!
D
!
Fdrag = ρ Av 2 ; 2) laminar flow Fdrag = ς v (v speed), ς is drag coefficient. Momentum: P = mv ;
2
!
in a collision of isolated particles total momentum is Ptotal constant; Newton 2nd law in terms of
!
!
! !
dp
1
momentum Fnet =
. KE = mv 2 ; work: Force*distance, W = F • S = ( F cosθ ) s = F|| s ;
dt
2
1
1
W net = ΔK = mv 22 − mv12 ( W net is net work); gravitational PE, U = mgh ; conservation of
2
2
energy Etotal = U PE + K ≡ constant. Unit of energy Joules ≡ J Power P = work / time = Fv , in
J / s ≡ watt (W).ATP hydrolysis (produces ADP) releases 29kJ/mol of heat. Reverse
ADP → ATP requires 29kJ/mol.
THERMODYNAMICS:1st law ΔU = Q + W ; 2nd law isolated system evolves to state of
Vf
maximum entropy. W = −PΔV or W = − ∫ P dV ; PV = NkBT = nRT , kB = 1.381 × 10 −23 J / K ,
Vi
R = 8.314 J/K*mol, Avogadro number N A = 6.023 × 10 23 particle/mole . Isothermal
(
)
process ΔU = 0,W = NkBT ln Vi / V f ;Isobaric W = −PΔV ; adiabatic PV 5 / 3 ≡ constant,
VT ≡ constant, Q = 0. U = (3 / 2)nRT . Entropy S = Q / T ; Enthalpy H = U + PV; Gibbs free
energy G = H − TS , ΔG = ΔH − T ΔS , stable state has lowest Gibbs free energy; spontaneous
N
if ΔG < 0 . Osmotic pressure Π = posm = ck BT , c = is the number concentration of solutes.
V
kT
kT
Stokes-Einstein Relation: drag coefficient ς = 6πη R ; diffusion coefficient D = B = B ;
ς
6πη R
3/2
Einstein diffusion: x 2 = 2Dt (1D), and r 2 = 6Dt (3D). Diffusion data D = 10 −9 m 2 / s for
small molecules like oxygen; η = 10 −3 Pais for water, and for blood η = 2.5 × 10 −3 Pais .
Δc
Concentration gradient flux: Fick’s law j = −D
in particle / sim 2 , c is in particle / m 3 ;
Δx
continuity flux density j = cv , v is the average speed of the fluid. Total flux is I = jA in unit of
particle / s , where A is the cross-section area of flow medium. Volume flux jv = L p Δp (in m/s),
(
)
where L p is filtration coefficient in mis −1Pa −1 , with total volume flow jv A (in m 3 is −1 ).
Fourier law of heat flow: Q / t = λ AΔT / L , λ is the thermal conductivity coefficient.
Heat
Q
=
Heat Capacity: C =
in J/K or Q = McΔT , c is the specific heat
change in temperature ΔT
capacity in JiK −1 ikg −1 .
Surface equations: Atmospheric pressure patm = 1.01 × 10 5 Pa ; Gauge pressure
pgauge = pabsolute − patm ; Buoyant force Fbuoyant = ρ fluidVobject g ; surface energy (surface tension)
between two phases σ = ΔE / ΔA , unit J / m 2 ≡ N / m ; Pascal Law p = patm + ρ gd , d is the
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depth measured from surface downward; Laplace Formula ΔP = 4σ / R (hollow bubble), and
ΔP = 2σ / R (air bubble in liquid).
Fluid Flow equation: Equation of continuity ΔV / Δt = Av = constant , v is the average fluid
speed; Bernoulli’s law for an ideal fluid p + (1 / 2 ) ρv 2 = constant ; Poiseuille’s law for
ΔV
π 4 Δp
m3
=
rtube
, (unit
)
Δt 8η
l
s
Auditory System: speed of wave c = λ f ; energy density of sound wave ε total = (1 / 2 ) ρ A 2ω 2 ;
Newtonian fluid in a cylindrical tube
Intensity I = cε total ; max pressure amplitude ΔPmax = ( cρω ) A ; SPL 20 log10 ( P / P0 ) , with
P0 = 2 × 10 −5 Pa ; IL 10 log10 ( P / P0 ) , with I 0 = 1× 10 −12 J i m −2 i s −1 ; speed of wave c = T / ρ ℓ for
my version of place theory, with T being the tension in N/m and ρ ℓ = ρ A , ρ ≡ fluid mass
density in kg i m −3 , and A is cross-section area. Resonance: closed tube harmonics λn = 2L / n ,
fn = nc / 2L ; half-open tube harmonics fn = nc / 4L with n = 1, 3, 5, 7…
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!
Electricity: electric force due to electric field F = qE , E in units of Volts/m = V/m; electric
(
)
field of plate in vacuum E = σ / ε 0 , ε 0 = 8.85 × 10 −12 C 2 / Nim 2 , and in material
E = σ / ε , ε = κε 0 , where κ is the dielectric constant; electric potential difference between two
+/- plates ΔV = Ed , d is the plate separation. Capacitance; C = ε 0 A / b in vacuum, unit is Farad
or F; C = ε A / b in material; Q = CΔV ; capacitance per unit area C a = C / A = ε / b , and charge
stored per unit area in a capacitor q a = q / A = CΔV / A = C a ΔV . Also ΔEenergy = qΔV .
q2 1
3 q2 1
E
=
(Born), where ε = 80ε 0 ; ion self-energy el
5 4πε r
8πε r
Electrophoresis: terminal velocity vt = qE / ( 6πη R ) of a macromolecule of charge q.
Electric Current: resistance R = ρ L / A (unit Ω ); resistivity ρ (unit Ωim ); conductance
G = 1 / R (unit Ω−1 ); conductivity σ = 1 / ρ (unit Ω−1m −1 ); conductance per area g = G / A (unit
Electric self-energy: E el =
(
)
Ω−1m −2 ); axoplasm resistivity ρ = ( kBT ) / Dq 2 c .
Membrane rest potential: Nernst potential Vi Nernst = (Vin − Vout ) =
(
)
(
)
kBT ci,out
; Ohmic leakage
ln
ze
ci,in
(
)
Ohmic
= VKNernst
− ΔV / RK + , I Na
= VNaNernst
− ΔV / RNa+ , I ClOhmic
= − VClNernst
− ΔV / RCl −
current I KOhmic
+
+
+
+
−
−
coulomb C
≡ ≡ ampere ≡ A
s
s
Action Potential: Electrotonus decay constant λ = abρmembrane / 2 ρaxoplasm ; time constant
ΔV is the membrane rest potential. Unit of current (I) is
τ = ερmembrane ; Signal potential decay ΔV / ΔVHH = exp− ( x / λ ) , ΔVHH is initial electric potential
stimulation, ΔV potential at distance x from initial stimulation; Signal propagation speed
v = λ /τ .
DATA e = 1.6 × 10 −19 C ; 1M = 1mole / L ; 1L = 10 −3 m 3 ; Avogardo Number
N A = 6.023 × 10 23 particle / mole ; pressure unit 1atm = 1.01325 × 10 5 pa
4
Geometrical Relations: 1) sphere, surface area A = 4π r 2 and volume V = π r 3 ; 2) cylinder
3
2
surface area A = 2π rL and volume V = π r L .
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