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Boise State - BIOL - 191
Pho synthe istos11/07/11Quick Revision :Where does photosynthesis occur Leaf X-section :PalisademesophyllChloroplast under electron microscopeChloroplast 11/07/11Warm-up QuestionsWarm-upThe following microphotograph is atransverse section
Boise State - BIOL - 191
LeafStructureandPhotosynthesisStandardGradeBiologyPhotosynthesis Greenplants(producers)canuselight energytomaketheirownfood Thisprocessiscalledphotosynthesis Greenplantsaregreenbecausetheycontainachemicalcalledchlorophyll.Thischemicalisusedtotrapli
Boise State - BIOL - 191
HydrogeLightenergy nOxygenChlorophyllChemicalATPenergywaterADP + PiHydrogenCarbonDioxideATPADP and PiGlucoseThis powerpoint was kindly donated towww.worldofteaching.comhttp:/www.worldofteaching.com is hometo over a thousand powerpoints
Boise State - BIOL - 191
Day makingPlants length food 1.Joe lives for basketball.He plays basketball every morning.He eats breakfast after playing, then goes to school.His mum says he needs the energy the food provides.At school he wrote down the food he had for breakfast.
Boise State - BIOL - 191
PlantsPlantsStructure LeavesLeavesstemrootsrootsLeavesLeavesFunction of leavesFunction Trap light energy for photosynthesisProducing sugar from photosynthesis Exchange of gases Exchangeoxygen and carbon dioxideoxygenStructureStructureW
Boise State - BIOL - 191
LightDependentReactionObjectivesUnderstandingtheprocessesofphotoionisation,photophosphorylation&photolysis.Knowledgeofwheretheseprocessestakeplace.Knowledgeofthedifferencebetweencyclic&noncyclicphosphorylation.OverviewofphotosynthesisOverviewofpho
Boise State - BIOL - 191
AbsorptionofLightbyChlorophyllHigherBiologyLightabsorptionbychlorophyllMeasuringlightabsorption Aspectrometer isusedtomeasuretheamountofabsorptionateachwavelength Anaction spectrumshowstherateofphotosynthesisatdifferentlightintensitiesChlorophyllA
Boise State - BIOL - 191
Slide 1StructureofPlantsSlide 21.2.3.A. Functions of RootsAnchor & supportplant in the groundAbsorb water &mineralsHold soil in placeRoot HairsFibrous RootsSlide 3B. Root TypesTap Root1. Fibrous Roots:2. Tap Roots larger centralbranch
Boise State - BIOL - 191
PhotomicrographsPlantTissuesBOTANYAlgaeCladophoraspeciesFilamentinformSeveralcellsBOTANYAlgaeAcetabulariaspeciesManycellsBOTANYAlgaeOedogoniumspeciesBOTANYAlgaeSpirogyraspeciesBOTANYSpirogyraspeciesFilamentousalgaeConjugation(sexualr
Boise State - BIOL - 191
PlantsEvolutionaryPastCyanobacteriaBluegreenalgaePhotosyntheticResponsibleforthemarkedincreaseinOxygen2000millionyearsagoCyanobacteriaEvidenceDNA shows the firstgreen plants evolvedfrom prokaryotesbetween 2500 and1000 million yearsago.Same
UNF - PHYSICS - 4360C
(b) T =T=1211 mg 1 m 2 g2mv = mvt = m =222 c2 2 .22 D 2(.145kg )2m 9.8 2 skg( 2 ) (.22 ) ( 2 ) (.0366 )m2= 87 J3 t 23 Fdx = cv dx = c v dt = c vt tanh dt t t t 31= cvt tanh 2 + tanh d 2t t 31= cvt tanh 2 + ln cosh
UNF - PHYSICS - 4360C
F2F1 2F2t1 + t1 ( t t1 ) +( t t1 )m2m2mF 2 F 2 F 2 5F 2At t = 2t1 : x =t1 + t1 + t1 =t1mm2m2mdv dv dxdvc3a== = v = v2dt dx dtdxm1cv 2 dv = dxm1vxmaxcv v 2 dv = 0 mdx1c2v 2 = xmaxmx=2.111xmax2.122mv 2=cGoing up:
UNF - PHYSICS - 4360C
v=vt v(vt2.142+v122), vt =gmg=kc2Going up: Fx = mg c2 v 2dvc22a = v = g kv , k =dxmvxvdvv g kv 2 = 0 dxv1ln ( g kv 2 ) = xv2kg + kv 2= e 2 kx2g + kvgg2v 2 = + v e 2 kx kk2Going down: Fx = mg + c2 vdvv = g + kv
UNF - PHYSICS - 4360C
22c2 v c1 c1 + 4mgc2t1=ln22mc1 + 4mgc2 2c2 v c1 + c1 + 4mgc2(t2c1 + 4mgc2m)12( 2c v + c += ln( 2c v + c 2v0) (c + 4mgc ) ( c +21c1 + 4mgc221c122)+ 4mgc )112c1 + 4mgc2c1222as t , 2c2 vt + c1 c1 + 4mgc2 = 0c1 c1
UNF - PHYSICS - 4360C
2.17dvdv= mv = f ( x )i g ( v )dtdxmvdv= f ( x ) dxg (v)mBy integration, get v = v ( x ) =If F ( x, t ) = f ( x )i g ( t ) :dxdtd 2xd dx = m = f ( x )i g ( t )2dtdt dt This cannot, in general, be solved by integration.If F ( v , t ) =
UNF - PHYSICS - 4360C
Integrating11A = tuu mand substituting e v = uAln 1 + e v t m(b) t = T @ v = 0A v = ln 1 + e v T mA vm1 e v e v = 1 + e TT=AmdvAvdvA= dx(c) v = v = e vvdxmem(a) v = v 1again, let u = e vdu = udvordv =duuv=1ln u1 d
UNF - PHYSICS - 4360C
3r 2= 1gr4Using Mathcad, solve the above non-linear d.e. lettingwhich reduces to: r +1 103 and R 0.01mm (small raindrop). Graphsshow thatv r t and r t 210
UNF - PHYSICS - 4360C
CHAPTER 3OSCILLATIONS3.1x = 0.002 sin 2 ( 512 s 1 ) t [ m ]mmxmax = ( 0.002 ) ( 2 ) ( 512 ) = 6.43 ss22 mmxmax = ( 0.002 ) ( 2 ) ( 512 ) 2 = 2.07 104 2 s s 3.2x = 0.1sin t [m]When t = 0, x = 0and = 5 s 13.3mx = 0.1 cos t smx = 0.5
UNF - PHYSICS - 4360C
3.63.711l=T =s 2.5 s9.82g6For springs tied in parallel:Fs ( x ) = k1 x k2 x = ( k1 + k2 ) x1(k + k ) 2= 1 2 mFor springs tied in series:The upward force m is keq x .Therefore, the downward force on spring k2 is keq x .The upward force
UNF - PHYSICS - 4360C
Fr = mg +mkmkdkcosxm = mg +tM +mM +mM +mFor the block to just begin to leave the bottom of the box atthe top of the vertical oscillations, Fr = 0 at xm = d :mkd0 = mg M +mg ( M + m)d=k3.9x = e t A cos ( d t )dx= e t A d sin ( d t ) e
UNF - PHYSICS - 4360C
3.1117 2 mx = 02317 = and 2 = 2222222 r = 2 = 4 mx + 3 mx +(a)Amax =(b)=2d = 2 2 =25 24 d =522A15 2e Td =3.12F2m d r = 2 =121ln 2 = f d ln 2Td1 d = ( 2 2 ) 2(a)122So, = ( + )2d1111 2 2 2 ln 2 2 2f = fd
UNF - PHYSICS - 4360C
21Td d 1 2=== 1 + 2 2 2 d 4 n TFor large n,Td1 1+ 2 28 nT13.14((a) r = 22122)22 = 2 2 = 0.70721 1 22 1 4(b) Q = d === 0.86622 2 2 2 ( 2 )222(c) tan = 2= 2 2 =2 43122() 2 = tan 1 = 146.331222
UNF - PHYSICS - 4360C
3.16(b) Q =2 2d=222 =1,LC=R2L2 1 R 2 LC 4 L L 1Q== 2 R R C 42 2L L R(c) Q = = C =RR 2 3.17Fext = F sin t = Im F ei t and x ( t ) is the imaginary part of the solution to:mx + cx + kx = F ei ti.e.i t x ( t ) = Im Ae
UNF - PHYSICS - 4360C
= tan 1 ( c 2m )m ( 2 2 ) c + kUsing sin 2 + cos 2 = 1 ,2F22= m ( 2 2 ) c + k + 2 ( c 2m )2AFA=cfw_m ( ) c + k + ( c 2m ) 2222122and x ( t ) = Ae t cos ( t ) + the transient term.3.19l A2 T 21 g8(a)for A =(b)412l1.041g,
UNF - PHYSICS - 4360C
1 Tf ( t ) = c + 2T f ( t ) cos ( n t ) dt cos n tn T2T1+ 2T f ( t ) sin ( n t ) dt sin n t ,n T21Tn = 1, 2, . . ., and c = 2T f ( t ) dtT 2Now, due to the equality of terms in n :2 Tf ( t ) = c + 2T f ( t ) cos ( n t ) dt cos n tn T2T2
UNF - PHYSICS - 4360C
3.22In steady state, x ( t ) = An e (i n t n )nAn =Fnm1( 2 n 2 2 )2 + 4 2 n 2 2 24F,n = 1, 3, 5, . . .andNow Fn =n9 2Q = 100 so 2 40, 00024F1A1 =1m22229 22( 9 ) + 4 40000 FA1 2m 24F1A3 =13m2 2 22 9 ( 9 2 9 2 ) +
UNF - PHYSICS - 4360C
3.24The equation of motion is F ( x ) = x x 3 = mx . For simplicity, let m=1. Then(a)(b)x = x x 3 . This is equivalent to the two first order equations x = y andy = x x3The equilibrium points are defined byx x 3 = x (1 x ) (1 + x ) = 0Thus, the p
UNF - PHYSICS - 4360C
3.25 + sin = 0d 2 cos = 0dt 22Integrating:2= cos 0)0T = 4 2 = 2 ( cos cosord1 2 ( cos cos ) 2Time for pendulum to swing from = 0 to = isNowsubstitute sin =sinsin2so =and after some algebra 2at = 2cos = 1 2 sinand use the iden
UNF - PHYSICS - 4360C
CHAPTER 4GENERAL MOTION OF A PARTICLEIN THREE DIMENSIONS-Note to instructors there is a typo in equation 4.3.14. The range of the projectile is v 2 sin 2v 2 sin 2 2 NOT . 0R=x= 0gg-4.1V V VjkxyzF = c iyz + xz + kxyj(a) F = V = i()(b)
UNF - PHYSICS - 4360C
er1 F = 2r sin r kr n4.3e r0e r sin = 0 conservative0(a)i F =xxyjycx 2k= k ( 2cx x )zz3jycxzy2kzxy2cx x = 01c=2(b)i F =xzya l so4.4(a) x cx = i 2 2 +yyx cx 2 2 =0yycz z+=0y2 y2 1 1 + k cz + z j2y
UNF - PHYSICS - 4360C
v2 = v2 2( + + )m122v = v 2 ( + + ) m(b)v2 2( + + ) = 0m122v = ( + + ) m(c)mx = Fx = Vxmx = Vmy = = 2 yyVmz = = 3 z 2z4.5(a)jF = ix + ydr = idx + dyjon the path x = y :(1,1)(0,0 )11110000F dr = Fx dx + Fy
UNF - PHYSICS - 4360C
(1,1)(and, with x = 1on this path1,0 )(1,1)(0,0 )1F dr = dy = 10F dr = 0 + 1 = 1F is not conservative.4.6From Example 2.3.2, V ( z ) = mgre2( re + z )zV ( z ) = mgre 1 + re From Appendix D,(1 + x )11= 1 x + x2 + z z2V ( z ) = mg
UNF - PHYSICS - 4360C
h=re 1 2 2re v 2zre g22h=re re2v 21 z22greFrom Appendix D,( h, z1(1 + x ) 2 = 1 +h=x x2 +28re re v 2zv4++ z +2 2 2 g 4 grehre )v 2z v 2z 1 +2 g 2 gre 1v2 v2 From Example 2.3.2, h =1 2 g 2 gre v2 v2 1And with (1 x ) 1
UNF - PHYSICS - 4360C
sin = gbv2cos 2 = 1 sin 2 =v 4 g 2b 2v4gb 2 v 4 g 2b 2 gb 2 v 2+= 2+v22 gv 22v2gMeasured from the ground,gb 2 v 2hmax = b + 2 +2v2ghmax = gb The mud leaves the wheel at = sin 1 2 v4.8x = R cos so t =andx = v x t = ( v cos ) tR
UNF - PHYSICS - 4360C
Now sin = cos = cos + 4 2 4 2 2 4 2 2v 2 Rmax =cos 2 + 2g cos 4 2Again using Appendix B, cos 2 = cos 2 sin 2 = 2 cos 2 1v2 2v 2 1 1Rmax =cos + + =cos 2 + + 1222g cos 2 2 g cos Using cos + = sin ,22vRmax =(1 sin )g (1
UNF - PHYSICS - 4360C
2sin 2 =2gh= 1 22csc v01gh 1 + 2 = v0 1 + ghv0 2gh Finally csc 2 = 2 1 + 2 v0 (b)Solving for Rmax Rmax =hhh==2sin 1 2 sin 1 2 csc2 Substituting for csc 2 and solving v2gh Rmax = 0 1 + 2 g v0 4.10 We can again use the results
UNF - PHYSICS - 4360C
zzmaxv0h0h1x0xx1RWe have reversed the trajectory so that h0 (= 9.8 ft), and x0 , the height and range withinwhich Mickey can catch the ball represent the starting point of the trajectory. h1 (=3.28ft) is the height of the ball when Mickey strik
UNF - PHYSICS - 4360C
u zmax = h1oru 2 zmax ( 2 ) = 2 zmax ( 2 .0475 ) = 3.9 zmax . This result is the correct one 3.9 zmax= 0.821 = 39.40x1Now solve for x0 using a relation identical to (4) Thus, tan =h0 = x0( x tan )tan 024 zmaxAgain we obtain a quadratic expres
UNF - PHYSICS - 4360C
(a) The maximum height occurs ator1v cos sin = gT2v21H=cos 2 sin 2 2g2or atdz=0dt1v cos sin 2T=gmaximum at fixed dH=0ddH v 2 11212= 2 cos sin cos cos sin sin = 0d 2 g 222Using the above trigonometric identities, we get1
UNF - PHYSICS - 4360C
projection on the xy-plane make an angle with the x-axis:x = s sin cos ,and Fx = Fr sin cos = mxy = s si n si n ,and Fy = Fr sin sin = myz = s cos ,and Fz = mg + Fr cos = mzSince Fr = c2 s = c2 ( x + y 2 + z 2 ) , the differential equations of moti
UNF - PHYSICS - 4360C
2 x z 8x z 2 +g3g 2z = v sin and 2 x z = v 2 sin 2 :xmax =Forxmax =v 2 sin 2 4v 3 sin 2 sin +g3g 24.16x = A cos ( t + ) ,xx = A sin ( t + )from x = 0 ,from x = A ,y = B cos ( t + ) ,y =0x = A cos ty = B sin ( t + )12121 2kB = ky +
UNF - PHYSICS - 4360C
V= 4 2 mxyy = B cos ( 2 t + )my = V= 9 2 mzzz = C cos ( 3 t + )mz = Since x = y = z = 0 at t = 0 , = = = 2x = A cos t = A sin t2x = A cos tSince v 2 = x 2 + y 2 + z 2 and x = y = z ,vx== A3vA=3vx=sin t3y = B sin 2 t ,y = 2 B c
UNF - PHYSICS - 4360C
tmin =4.182=2Equation 4.4.15 istan 2 =2 AB cos A2 B 2Transforming the coordinate axes xyz to the new axes xyz by a rotation aboutthe z-axis through an angle given, from Section 1.8:x = x cos + y sin ,y = x sin + y cosor,x = x cos y sin , and
UNF - PHYSICS - 4360C
1r1 = ( d x ) + y 2 + z 2 22 2x x2 + y2 + z 2 2+r = ( d 2 dx + x + y + z ) = d 1 dd21nFrom Appendix D, (1 + x ) = 1 + nx + n ( n 1) x 2 + 22 2x x + y2 + z2 1 r1 = d 1 + + 1d2 2 d 2 2 2 12222 22222222 2 x 2x3 2x x + y + z x
UNF - PHYSICS - 4360C
zkB(F = q E+vBEjiv0())jv B = ix + y + kz kB = iyB xBjF = iqyB + q ( E xB )jymx = Fx = qyBqBxx =ymxqBmy = Fy = qE qxB = qE qB x +my22qE qBx qB eE eBx eB y=+ y y=mm mmm meEeBy +2 y = +x ,=mm1 eEy = 2 + x + A co
UNF - PHYSICS - 4360C
hbh mv 2 mgmgR = mg = h ( b 2h ) = b ( 3h b )bbbthe particle leaves the side of the sphere when R = 0bbh = , i.e.,above the central plane33cos =4.2212mv + mgh = 02at the bottom of the loop, h = b12mv = mgb ,2v = 2 gbsohbmv 2
UNF - PHYSICS - 4360C
CHAPTER 5NONINERTIAL REFERENCE SYSTEMS5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net forceacting on him is zero. The scale exerts an upward force, N , whose value is equal to thescale reading - the weight, W, of
UNF - PHYSICS - 4360C
gg = g A = g ij10For small oscillations of a simple pendulum:1T = 2g2gg = g + = 1.005 g 10 2T = 25.5 (a)11= 1.9951.005 ggf = mg is the frictional force acting on theA0box, sof mA0 = ma(b)(a)f( a is the acceleration of the box re
UNF - PHYSICS - 4360C
= i R sin t + R cos t x + i yjjx = y R sin ty = x + R cos there i = 1 !(c) Let u = x + iyu = x + iy = y R sin t i x + iR cos t= yi u =+i x u + i u = R sin t + iR cos t = i ReitTry a solution of the formu = Ae i t + Beitu = i Ae i t + iBeit
UNF - PHYSICS - 4360C
A, A are the accelerations of the asteroid and the Earth in the fixed, inertial frameof reference.1st : examine:A A r= R R ( R R )= ( ) R = ( 2 2 ) Rnote: = k , = kThus:a = ( 2 2 ) R 2 vTherefore: jyix + = ( 2 2 ) iR cos ( ) t R sin ( ) t 2 x +
UNF - PHYSICS - 4360C
= k , with = 0r = ber , so ( r ) = b 2 er v = ve , so v = ver From eqn 5.2.14,a = a + 2 v + ( r ) and putting in terms from abovev 2 2 v b 2ar = bFor no slipping F s mg , so a s gv 2+ 2 v + b 2 s gb2vm + 2 bvm + b 2 2 b s g = 0vm = b 2b 2
UNF - PHYSICS - 4360C
5.10(See Example 5.3.3)m 2 x = mxx ( t ) = Ae t + Be tx ( t ) = Ae t Be tBoundary Conditions:lx2 ( 0) = = A + B2x ( 0 ) = 0 = ( A B )A=l4B=lx ( t ) = cosh t2lllx (T ) = + = cosh T222(a)(b) cosh T = 2lx ( t ) = sinh t2when the bea
UNF - PHYSICS - 4360C
v = zv yi + zvx yvx k jj( v )horiz = zvyi + zvx ( v )horiz112222= ( z2v y + z2vx ) 2 = z ( vx + v y ) 2 = zvFcor = 2m v(F )cor5.13horiz= 2m ( v )horiz = 2m z v , independent of the direction of v .From Example 5.4.1 11 8h 3 2 = xh
UNF - PHYSICS - 4360C
da = r + r + r + 2 r + 2 r dt rot()() a = r + ( r ) + 2 ( r ) + ( r ) + ( r ) + r + r Now( r )is to and r . Let this define a direction n : r = r nSince n , ( r ) is in the plane defined by and r and ( r ) = n r = r .Since ( r ) ( r ) = 2
UNF - PHYSICS - 4360C
1z ( t ) = gt 2 + v t sin + v t 2 cos cos = 022v sin 2v sin t=org 2 v cos cos gWe have ignored the second term in the denominatorsince v would have to beimpossibly large for the value of that term to approach the magnitude gFor example, for = 4
UNF - PHYSICS - 4360C
(jHence: a = a 2 v = 3 2 x i 2 k ix + ySo5.19)ja = ix + y = 3 2 xi + 2 yi 2 xj2x 2 y 3 x = 0y + 2 x = 0(mr = qE + q v BEquation 5.2.14)r = r + r + 2 v + ( r )v = v + r Equation 5.2.13q =B so = 02mqmr q B v B ( r ) = qE + q ( v + r
UNF - PHYSICS - 4360C
Collecting terms and neglecting terms in 2 :gg x + x cos t + y + y sin t = 0llgg x + x sin t y + y cos t = 0ll24hourssin 24T== 73.7 hourssin 195.21T=5.22Choose a coordinate system with the origin at the center of the wheel, the x and
UNF - PHYSICS - 4360C
r = 0Vb VtVt VtVtVt Vt k sin cos + V sin i cos j k sin cosbbbbbb22V Vt Vt V Vt Vt+ k cos 2 j cos ( r ) = k sin 2+ i sinbb bbb22VV ( r ) = k nbSince the origin of the coordinate system is traveling in a circle of r
UNF - PHYSICS - 4360C
CHAPTER 6GRAVITATIONAL AND CENTRAL FORCES6.14m = V = rs331 3m 3rs = 4 F=Gmm( 2rs )222Gm 2 4 3 G 4 3 43== m4 3m 4 3 2FFGm 2 4 3 3==mW mg4g 3 121F 6.672 1011 N m 2 kg 2 4 11.35 g cm 3 1 kg 106 cm3 3=3 (1 kg ) 3W4 9.8
UNF - PHYSICS - 4360C
The gravitational force on the particle is due only to the mass of the earth that isinside the particles instantaneous displacement from the center of the earth, r.The net effect of the mass of the earth outside r is zero (See Problem 6.2).4M = r334
UNF - PHYSICS - 4360C
6.56.6GMm mv 2GMv2 ==so2rrrfor a circular orbit r, v is constant.2 rT=v4 2 r 2 4 2 3T2 =r r3=2vGM2 rvFrom Example 6.5.3, the speed of a satellite in circular orbit is (a) T =1 gR 2 2v= e r3T=2 r 21g 2 Re1 T 2 gRe2 3r =