22 Pages

Plant structure adaptations and responses

Course: BIOL 191, Fall 2010
School: Boise State
Rating:
 
 
 
 
 

Word Count: 1000

Document Preview

1 Structure of Plants Slide Slide 2 1. 2. 3. A. Functions of Roots Anchor & support plant in the ground Absorb water & minerals Hold soil in place Root Hairs Fibrous Roots Slide 3 B. Root Types Tap Root 1. Fibrous Roots: 2. Tap Roots larger central branching roots hold soil in place to prevent soil erosion root reaches deep water sources underground Ex. Grasses Ex. Trees, Carrots,...

Register Now

Unformatted Document Excerpt

Coursehero >> Idaho >> Boise State >> BIOL 191

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
1 Structure of Plants Slide Slide 2 1. 2. 3. A. Functions of Roots Anchor & support plant in the ground Absorb water & minerals Hold soil in place Root Hairs Fibrous Roots Slide 3 B. Root Types Tap Root 1. Fibrous Roots: 2. Tap Roots larger central branching roots hold soil in place to prevent soil erosion root reaches deep water sources underground Ex. Grasses Ex. Trees, Carrots, & Dandelions Slide 4 C. The Structure of a Root Root Hairs Phloem Xylem Meristem Root Cap 1. Root Hairs: increase surface area for water & mineral absorption 2. Meristem: region where new cells are produced 3. Root Cap: protects tip of growing root Slide 5 A. Functions of Stems 1.Support system for plant body 2.Transport system carries water & nutrients 3.Holds leaves & branches upright Looking at the Each light and dark picture to the left: tree ring equals one year of annual growth. Whatrings for fast years had Light the most rain? for spring growth, dark slow summer growth. What years tell of Smaller rings past droughts the experienced that h orst drought? wave occurred. Slide # 6 A. Functions of Leaves 1. Main photosynthetic organ 2. Broad, flat surface increases surface area for light absorption 3. Have systems to prevent water loss Stomata open in day but close at night or when hot to conserve water waxy cuticle on surface 1. System of gas exchange Allow CO2 in and O2 out of leaf Elephant Ear Plant Slide # 7 B. Leaf Structures Leaf Cross-Section 1.Cuticle: waxy layer; covers upper surface Cuticle 1.Veins: transports water, nutrients and food Made of xylem and phloem 1.Mesophyll: contains cells that perform photosynthesis b/c they contain Chloroplasts. Mesophyll Protects leaf against water loss Veins Stoma (Opening) 2 Guard Cells Surround each Stoma Stoma- singular Stomata-plural Slide # 8 More Plant Parts 4. Guard cells: cells that open and close the stoma 4. Stomata: openings in leafs surface; when open: GAS EXCHANGE: Allows CO2 in & O2 out of leaf TRANSPIRATION: Allows excess H2O out of leaf Guard Cells Stoma Slide # 9 What goes O2 out? Function of Stomata What process involves Guard Cells Guard Cells using CO2 and H2O releasing H2O O2 as a waste product? Photosynthesis What goes in? Stoma CO2 What is the plant using this process to make? Stoma Open Stoma Closed Carbohydrates-glucose If the plant needs water for photosynthesis, why is water coming out of the stoma? Slide # 10 Function of Guard Cells These stomata (leaf Guard Cells openings) naturally allow water to evaporate out. Guard Cells Why would the plant close stomata with guard cells? Prevent excess water loss through transpiration. (conserveStoma Open water) So what is the point of having stomata? Allow gas exchange for photosynthesis Stoma Closed Slide # 11 C. Plants find a use for Transpiration 1. Transpiration: loss of excess water from plant leaves 2. Significance: a. Transpiration causes enough pressure to help pull water (& required nutrients) up stem from roots. b. As part of the water cycle, trees transpire water back into the atmosphere. c. Transpiration provides much of the daily rain in rainforest. A B A average size maple tree can transpire liters 200 of water per hour during the summer. Transpiration is the #1 driving force for pulling water up stems from roots. Slide # 12 Structure of a Flower 1.Pistil:female reproductive structure a.Stigma: sticky tip; traps pollen b.Style: slender tube; transports pollen from stigma to ovary c.Ovary: contains ovules; ovary develops into fruit d.Ovule: contains egg cell which develops into a seed when fertilized Stamen Anther Filament Ovule Stigma Pistil Style Ovary Petal Sepal Slide # 13 Structure of a Flower 2.Stamen: male reproductive structure a.Filament: thin stalk; supports anther b.Anther: knob-like structure; produces pollen c.Pollen: contains microscopic cells that become sperm cells Stamen Anther Filament Ovule Stigma Pistil Style Ovary Petal Sepal Slide # 14 Structure of a Flower 3.Sepals: encloses & protects flower before it blooms Stamen Anther Filament Stigma Pistil Style Ovary 4.Petals: usually colorful & scented; attracts pollinators Ovule Petal Sepal Slide # 15 Cross Pollination How does pollination happen? Pollen from an anther is caught by the stigma, travels through style to the ovules in the ovary. What is the result of pollination? A Fruit: An ovary containing seeds. Slide # 16 Chapter 25 Plant Responses and Adaptations Slide #17 Hormone Action on Plants A. Plant cells can produce hormones: which are chemical messengers that travel throughout the plant causing other cells called target cells to respond. B. In plants, hormones control: Movement of hormone Hormoneproducing cells Target cells 1. Plant growth & development 2. Plant responses to environment Cells in one blooming flower signals other blooms using hormones to Slide # 18 C. Plant cells will send signals to one another to tell them: 1. When trees to drop their leaves. 2. When to start new growth. 3. When to cause fruit to ripen. 4. When to cause flowers to bloom. 5. When to cause seeds to sprout. Tree Budding Fruit Ripening Cactus Blooming Leaf Drop Sprouting Corn Seeds Slide # 19 D. Ethylene causes Fruit to Ripen 1.Fruit tissues release a small amount of ethlyene 2.Causes fruits to ripen. 3.As fruit become ripe, they produce more and more ethlyene, accelerating the ripening process. Ethylene released by apples and tomatoes causes fruit to age quickly. Slide # 20 Plant Tropisms 1. Tropism: the way a plant grows in response to stimuli in the environment. a.Phototropism: growth response to light -Plants bend towards light a.Geotrophism: growth response to gravity -plant roots grow down with gravity, shoots (stems) grow up against gravity and out of the soil. a.Thigmotropism: growth response to touch -vines grow up around trees, venus flytrap closes when leaves are touched Slide # 21 What type of tropism is shown in these pictures? tr eo G Ph to o op tr sm i hi T pi o i sm rop tot ho P sm tr o gm ph o sm i eo G op tr ism Th tr o gm i ph o sm i This powerpoint was kindly donated to www.worldofteaching.com http://www.worldofteaching.com is home to over a thousand powerpoints submitted by teachers. This is a completely free site and requires no registration. Please visit and I hope it will help in your teaching.
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

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 =
UNF - PHYSICS - 4360C
This is the same expression as derived in Problem 6.3 for a particle dropped into ahole drilled through the earth. T 1.4 hours.6.8The Earths orbit about the Sun is counter-clockwise as seen from, say, the northstar. Its coordinates on approach at the
UNF - PHYSICS - 4360C
4M d = r33GMm 4F ( r ) = 2 mGr3r6.101 1 k=errduk= e kdru=d 2u k 2 k= e = k 2ud 2 rFrom equation 6.5.10 d 2u1+ u = k 2u + u = 2 2 f ( u 1 )2duml u122f ( u ) = ml ( k + 1) u 3f (r ) = ml 2 ( k 2 + 1)r3The force varies as th
UNF - PHYSICS - 4360C
kIf 1 + 2 < 0 , ml kIf 1 + 2 = 0 , ml du= C1du = c1 + c21r=c1 + c2kIf 1 + 2 > 0 , ml u = A cos(c + d 2u cu = 0 ,d 2d 2u=0d 2)d 2u+ cu = 0 ,d 2kr = A cos 1 + 2 + ml6.12c > 0 , for which u = aeb is a solution.c>0111=r
UNF - PHYSICS - 4360C
For r = a , and the force to be central, try = bt nf ( ) = m 2ab 2 n 2t 2 n 2 + ab 2 n ( n 1) t 2 n 2 For a central field f ( ) = 02n + ( n 1) = 0n=131 = bt 3(a)6.14Calculating the potential energy 4 a2 dv = f ( r ) = k 3 + 5 drrr 2 a2
UNF - PHYSICS - 4360C
1r = a cos ( r = a @ = 0 )33(b) at =r 0 the origin of the force. To find how long it takes 2avvl= 2==11ra 2 cos 2 a cos 2 33a1dt = cos 2 dv3ThusT=320a3a 23 a212cos d = cos d = 4vv3v01 9k 2Since v = 2 2a 113 2
UNF - PHYSICS - 4360C
dV V 1 r (b) dr1 = 2r1 dV = 2 60 1% = 120% !(a)() dr1 2 r1 r1 r V r1 The approximation of a differential has broken down a correct result can be obtained bycalculating finite differences, but the implication is clear a 1% error in boost causes
UNF - PHYSICS - 4360C
6.17From Example 6.10.1 12vr2 2 = 1 + q 2 ( qd sin ) where q =and d =dveaeare dimensionless ratios of the comets speed and distance from the Sun in terms of theEarths orbital speed and radius, respectively (q and d are the same as the factor
UNF - PHYSICS - 4360C
But the total orbital energy iskkE=SoE = 2 a2a2aSince planetary orbits are nearly circularkkV ~andT~a2aaE aand=Thus, E TaTaav=2We obtainav6.20 (a) V =1Vdt0krFrom equation 6.5.4, l = r 2dlr 2 d= 2 or dt =dt rl2 kr