Vibrations and Waves
|
|
|
- Alfred Thomas
- 6 years ago
- Views:
Transcription
1 Chaper 3 3 Vbraons and Waes PROBEM SOUIONS 3. (a) ang o he rgh as pose, he sprng orce acng on he bloc a he nsan o release s F s 30 N N or 7 N o he le A hs nsan, he acceleraon s a F s 7 N 0.60 g 8 s or a 8 s o he le 3. When he objec coes o equlbru (a dsance 0 below he unsreched poson o he end o he sprng), F g and he orce consan s 0 0 g 0.5 g 9.80 s N.59 N 3.3 (a) Snce he collson s perecl elasc, he ball wll rebound o he hegh o.00 beore cong o res oenarl. I wll hen repea hs oon oer and oer agan wh a regular perod. Fro a, wh 0 0, he e requred or he ball o reach he ground s 0.00 a 9.80 s 0.90 s hs s one-hal o he e or a coplee ccle o he oon. hus, he perod s.8 s. (c) No. he ne orce acng on he objec s a consan gen b F = g (ecep when s conac wh he ground). hs s no n he or o Hooe s law. 3. (a) he sprng consan s Page 3.
2 Chaper 3 Fs g 50 N N F Fs.0 03 N N he graph wll be a sragh lne passng hrough he orgn wh a slope equal o.0 03 N. 3.5 When he sse s n equlbru, he enson n he sprng F us equal he wegh o he objec. hus, g gng g 7.5 N s 0. g 3.6 (a) he ree-bod dagra o he pon n he cener o he srng s gen a he rgh. Fro hs, we see ha F 0 F sn or F 375 N sn 35.0 sn N Snce he bow requres an appled horzonal orce o 375 N o hold he srng a 35.0 ro he ercal, he enson n he sprng us be 375 N when he sprng s sreched 30.0 c. hus, he sprng consan s F 375 N N 3.7 (a) When he bloc coes o equlbru, F 0 g 0 gng 0 g 0.0 g 9.80 s 75 N 0.06 or he equlbru poson s 0.06 below he unsreched poson o he lower end o he sprng. When he eleaor (and eerhng n ) has an upward acceleraon o a.00 s second law o he bloc ges, applng Newon s F 0 g a or 0 F g a Page 3.
3 Chaper 3 where = 0 a he equlbru poson o he bloc. Snce 0 g 0 [see par (a)], hs becoes a and he new poson o he bloc s a 0.0 g.00 s 75 N. 0. c or. c below he equlbru poson. (c) When he cable breas, he eleaor and s conens wll be n ree-all wh a = g. he new equlbru poson o he bloc s ound ro F 0 g g, whch elds 0 0 snapped, he bloc was a res relae o he eleaor a dsance. When he cable below he new equlbru poson. hus, whle he eleaor s n ree-all, he bloc wll oscllae wh aplude 0.8 abou he new equlbru poson, whch s he unsreched poson o he sprng s lower end. 3.8 (a) When he gun s red, he elasc poenal energ nall sored n he sprng s ransored no nec energ o he projecle. hus, s necessar o hae 0 0 or g 5.0 s N he agnude o he orce requred o copress he sprng 8.00 c and load he gun s Fs 99 N N 3.9 (a) Assue he rubber bands obe Hooe s law. hen, he orce consan o each band s F s 5 N N hus, when boh bands are sreched 0.0, he oal elasc poenal energ s PEs N J Conseraon o echancal energ ges KE PE s KE PE s, or Page 3.3
4 Chaper J, so 60 J g 9 s 3.0 (a) F a a 30 N N wor done PEs 575 N J 3. Fro conseraon o echancal energ, KE PE g PE s KE PE g PE s or 0 gh gng gh 0.00 g 9.80 s N 3. Conseraon o echancal energ, ( KE PEg PEs ) ( KE PEg PE s), ges , or N s 000 g 3.3 An unnown quan o echancal energ s conered no nernal energ durng he collson. hus, we appl conseraon o oenu ro jus beore o jus aer he collson and oban + M(0) = (M + )V, or he speed o he bloc and ebedded bulle jus aer collson s V M g.0 g 300 s.9 s Now, we use conseraon o echancal energ ro jus aer collson unl he bloc coes o res. hs ges 0 M V, or V M.0 g.9 s N Page 3.
5 Chaper 3 3. (a) A eher o he urnng pons, = A, he consan oal energ o he sse s oenarl sored as elasc poenal energ n he sprng. hus, E A. When he objec s dsance ro he equlbru poson, he elasc poenal energ s PEs and he nec energ s KE. A he poson where KE = PE s, s necessar ha or (c) When KE = PE, conseraon o energ ges E KE PEs PEs PEs 3PE s, or A 3 A 3 or A (a) A au dsplaceen ro equlbru, all o he energ s n he or o elasc poenal energ, gng E, and a E a 7.0 J N A he equlbru poson ( = 0), he sprng s oenarl n s relaed sae and PE s = 0, so all o he energ s n he or o nec energ. hs ges KE 0 a E 7.0 J (c) I, a he equlbru poson, = a = 3.5 /s, he ass o he bloc s E a 7.0 J 3.5 s 7.90 g (d) A an poson, he consan oal energ s, E KE PE, so a = 0.60 s E 7.0 J N g.57 s (e) A = 0.60, where =.57 /s, he nec energ s Page 3.5
6 Chaper g.57 s 6. J KE () A = 0.60, where KE = 6. J, he elasc poenal energ s PEs E KE 7.0 J 6. J 0.9 J or alernael, N J PEs (g) A he rs urnng pon (or whch < 0 snce he bloc sared ro res a = +0.0 and has passed hrough he equlbru a = 0) all o he reanng energ s n he or o elasc poenal energ, so E E loss 7.0 J.0 J 33.0 J and 33.0 J 33.0 J N (a) F 83.8 N N E PEs 83.8 N J (c) Whle he bloc was held saonar a = 5.6 c, F F F 0, or he sprng orce was equal n agnude and opposel dreced o he appled orce. When he appled orce s suddenl reoed, here s a ne orce F s =.58 N dreced oward he equlbru poson acng on he bloc. hs ges he bloc an acceleraon hang agnude s a F s.58 N 0.50 g 8.3 s (d) A he equlbru poson, PE s = 0, so he bloc has nec energ KE = E = 0.5 J and speed E 0.5 J 0.50 g.00 s Page 3.6
7 Chaper 3 (e) I he surace was rough, he bloc would spend energ oercong a reardng rcon orce as oed oward he equlbru poson, causng o arre a ha poson wh a lower speed han ha copued aboe. Copung a nuber alue or hs lower speed requres nowledge o he coecen o rcon beween he bloc and surace. 3.7 Fro conseraon o echancal energ, KE PE PE KE PE PE g s g s we hae A, or A (a) he speed s a au a he equlbru poson, = 0. a A 9.6 N s 0.0 g When = 0.05, 9.6 N s 0.0 g (c) When = +0.05, 9.6 N s 0.0 g (d) I a, hen A A hs ges A = A /, or A 3 /.0 c 3 / 3.5 c. 3.8 (a) KE = 0 a = A, so E KE PE 0 A, or he oal energ s s E 50 N A 0.5 J Page 3.7
8 Chaper 3 he au speed occurs a he equlbru poson where PE s = 0. hus, E a, or a E 50 N A s 0.50 g (c) he acceleraon s a F/ /. hus, a = a a a = a = A. a a A A 50 N s 0.50 g 3.9 he au speed occurs a he equlbru poson and s a A hus, A a 6.0 N s.00 g, and Fg g.00 g 9.80 s 39. N 3.0 A 0.0 N g s 3. (a) he oon s sple haronc because he re s roang wh consan eloc and ou are loong a he unor crcular oon o he bup projeced on a plane perpendcular o he re. Noe ha he angenal speed o a pon on he r o a rollng re s he sae as he ranslaonal speed o he ale. hus, 3.00 s and he angular eloc o he re s car r 3.00 s rad s hereore, he perod o he oon s 0.0 rad s 0.68 s Page 3.8
9 Chaper 3 3. (a) r s 0.68 s.00 s Hz (c).00 s 3. rad s 3.3 he angle o he cran pn s. Is -coordnae s Acos Acos where A s he dsance ro he cener o he wheel o he cran pn. hs s o he correc or o descrbe sple haronc oon. Hence, one us conclude ha he oon s ndeed sple haronc. 3. he perod o braon or an objec-sprng sse s. hus, = 0.3 s and = 35. g = g, he orce consan o he sprng s g 0.3 s 8. N 3.5 he sprng consan s ound ro Fs g 0.00 g 9.80 s N When he objec aached o he sprng has ass = 5 g, he perod o oscllaon s 0.05 g 0.63 s.5 N 3.6 he sprngs copress 0.80 c when supporng an addonal load o = 30 g. hus, he sprng consan s g 30 g 9.80 s N When he ep car, M = g, oscllaes on he sprngs, he requenc wll be Page 3.9
10 Chaper N M.0 03 g. Hz 3.7 (a) he perod o oscllaon s where s he sprng consan and s he ass o he objec aached o he end o he sprng. Hence, 0.50 g.0 s 9.5 N I he car s released ro res when s.5 c ro he equlbru poson, he aplude o oscllaon wll be A =.5 c =.5 0. he au speed s hen gen b a A A 9.5 N 0.50 g s (c) When he car s c ro he le end o he rac, has a dsplaceen o = c c =.0 c ro he equlbru poson. he speed o he car a hs dsance ro equlbru s 9.5 N A s 0.50 g 3.8 he general epresson or he poson as a uncon o e or an objec undergong sple haronc oon wh = 0 a = 0 s Asn. hus, 5. c sn 8.0, we hae ha he aplude s A = 5. c and he angular requenc s 8.0 rad s. (a) he perod s 8.0 s s he requenc o oon s 0.5 s.0 s-.0 Hz (c) As dscussed aboe, he aplude o he oon s A 5. c (d) Noe: For hs par, our calculaor should be se o operae n radans ode. Page 3.0
11 Chaper 3 I =.6 c, hen A.6 c 5. c sn sn sn radans and 0.5 rad 0.5 rad 8.0 rad s. 0 s 0 3 s s 3.9 (a) A he equlbru poson, he oal energ o he sse s n he or o nec energ and a E, so he au speed s a E 5.83 J 0.36 g 5.98 s he perod o an objec-sprng sse s, so he orce consan o he sprng s 0.36 g 0.50 s 06 N (c) A he urnng pons, = A, he oal energ o he sse s n he or o elasc poenal energ, or E A, gng he aplude as A E 5.83 J 06 N For a sse eecung sple haronc oon, he oal energ a be wren as s a E KE PE A, where A s he aplude and a s he speed a he equlbru poson. Obsere ro hs epresson, ha we a wre a A. (a) I, hen a E a becoes a a and ges Page 3.
12 Chaper A a A or A 3 I he elasc poenal energ s PE s = E/, we hae A or A and A 3.3 (a) A = 3.50 s, F N rad 5.00 (3.00 ) cos.58 (3.50 s).0 N s, or F.0 N dreced o he le he angular requenc s 5.00 N.00 g.58 rad s and he perod o oscllaon s 3.97 s..58 rad s Hence he nuber o oscllaons ade n 3.50 s s N 3.50 s 3.97 s (a) F 7.50 N N 50 N g. rad s Page 3.
13 Chaper 3. rad s 3.56 Hz 3.56 Hz 0.8 s (c) A = 0, = 0 and = , so he oal energ o he oscllaor s E KE PEs 0 50 N J (d) When = A, 0 so E KE PE 0 s A. hus, A E 0.33 J 50 N/ c (e) A = 0, KE a E, or a E 0.33 J g. s a a F a A 50 N g 5.0 s Noe: o sole pars () and (g), our calculaor should be se n radans ode. () A = s, Equaon 3.a ges he dsplaceen as 50 N A cos A cos 5.00 c cos s 0.99 c g (g) Fro Equaon 3.b, he eloc a = s s A sn A sn 50 N 50 N g g sn s.0 Page 3.3
14 Chaper 3 and ro Equaon 3.c, he acceleraon a hs e s a A cos A cos 50 N 50 N cos s.59 s g g Fro Equaon 3.6, A A Hence, A A cos A cos A sn Fro Equaon 3., a A cos A cos 3.3 (a) he hegh o he ower s alos he sae as he lengh o he pendulu. Fro g, we oban 9.80 s 5.5 s g 59.6 On he Moon, where g =.67 /s, he perod wll be s g.67 s 3.35 he perod o a pendulu s he e or one coplee oscllaon and s gen b g, where l s he lengh o he pendulu. (a) 3.00 n 60 s 0 oscllaons n.50 s he lengh o he pendulu s g.50 s 9.80 s Page 3.
15 Chaper he perod n oo s / g and he perod n Cabrdge s C C / g C. We now ha = C =.000 s, ro whch, we see ha g C g g, or C C 0.99 C g (a) he perod o he pendulu s g. hus, on he Moon where he ree-all acceleraon s saller, he perod wll be longer and he cloc wll run slow. he rao o he pendulu s perod on he Moon o ha on Earh s Moon Earh gmoon gearh 9.80 g g.63 Earh Moon.5 Hence, he pendulu o he cloc on Earh aes.5 cs whle he cloc on he Moon s ang.00 c. Aer he Earh cloc has ced o.0 h and agan reads :00 dngh, he Moon cloc wll hae ced o.0 h/.5 = 9.80 h and wll read 9 : 8 AM (a) he lower eperaure wll cause he pendulu o conrac. he shorer lengh wll produce a saller perod, so he cloc wll run aser or gan e. he perod o he pendulu s 0 / g a 0 C, and a 5.0 C s / g. he rao o hese perods s 0 / 0 /. Fro Chaper 0, he lengh a 5.0 C s 0 Al0, so 0 6 Al( ) C 5.0 C 0 C hs ges hus n one hour (3 600 s), he chlled pendulu wll gan s. s (a) Fro g, he lengh o a pendulu wh perod s = g /. Page 3.5
16 Chaper 3 On Earh, wh =.0 s, 9.80 s.0 s c I =.0 s on Mars, 3.7 s.0 s c he perod o an objec on a sprng s, whch s ndependen o he local ree-all acceleraon. hus, he sae ass wll wor on Earh and on Mars. hs ass s 0 N.0 s 0.5 g 3.0 he apparen ree-all acceleraon s he ecor su o he acual ree-all acceleraon and he negae o he eleaor s acceleraon. o see hs, consder an objec ha s hangng ro a ercal srng n he eleaor and appears o be a res o he eleaor passengers. hese passengers belee he enson n he srng s he negae o he objec s wegh, or g app where app g s he apparen ree-all acceleraon n he eleaor. An obserer locaed ousde he eleaor apples Newon s second law o hs objec b wrng F g a e where e ae g g app, whch ges app e a s he acceleraon o he eleaor and all s conens. hus, g g a. (a) I we choose downward as he pose drecon, hen a 5.00 s n hs case and g s.8 s app (downward). he perod o he pendulu s e s g.8 s app Agan choosng downward as pose, a 5.00 s and g s.80 s (downward) n hs case. he perod s now gen b e app s g.80 s app Page 3.6
17 Chaper 3 (c) I a 5.00 s horzonall, he ecor su gapp g a e s e as shown n he sech a he rgh. he agnude s g app 5.00 s 9.80 s.0 s and he perod o he pendulu s s g.0 s app 3. (a) he dsance ro he boo o a rough o he op o a cres s wce he aplude o he wae. hus, A = 8.6 c and A.3 c he horzonal dsance ro a cres o a rough s a hal waelengh. Hence, 5.0 c and 0. c Fgure P3. (c) he perod s 8.0 s s (d) he wae speed s 0. c 8.0 s 87 c s.87 s 3. (a) he aplude s he agnude o he au dsplaceen ro equlbru (a = 0). hus, A.00 c he perod s he e or one ull ccle o he oon. hereore,.00 s (c) he perod a be wren as requenc s, so he angular Fgure P3..00 s rad s Page 3.7
18 Chaper 3 (d) he oal energ a be epressed as E A. hus, a A /, and snce, hs becoes a A and elds a a A rad s.00 c c s (e) he sprng eers au orce, F, when he objec s a au dsance ro equlbru,.e., a A.00 c. hus, he au acceleraon o he objec s a A F aa A rad s.00 c.93 c s () he general equaon or poson as a uncon o e or an objec undergong sple haronc oon wh = 0 when = 0 s = A sn ( ). For hs oscllaor, hs becoes.00 c sn 3.3 (a) he perod and he requenc are recprocals o each oher. hereore, 0.9 MHz s s 9.8 ns s s (a) he requenc o a ranserse wae s he nuber o cress ha pass a gen pon each second. hus, 5.00 cress pass n.0 seconds, he requenc s s s Hz he waelengh o he wae s he dsance beween successe aa or successe na. hus,.76 and he wae speed s s s 3.5 he speed o he wae s Page 3.8
19 Chaper 3 5 c.5 c s 0.0 s and he requenc s 0.0 b 30.0 s.33 Hz hus,.5 c s.33 Hz 3.9 c 3.6 Fro =, he waelengh (and sze o salles deecable nsec) s 30 s Hz he requenc o he wae (ha s, he nuber o cress passng he cor each second) s.00 s- and he waelengh (dsance beween successe cress) s 8.50 c. hus, he wae speed s 8.50 c.00 s- 7.0 c s 0.70 s and he e requred or he rpples o rael 0.0 oer he surace o he waer s s 58.8 s 3.8 (a) When he boa s a res n he waer, he speed o he wae relae o he boa s he sae as he speed o he wae relae o he waer,.0 s. he requenc deeced n hs case s.0 s Hz ang easward as pose, wae, boa wae, waer boa, waer (see he dscusson o relae eloc n Chaper 3 o he eboo) ges wae, boa.0 s.0 s 5.0 s and boa, wae wae, boa 5.0 s hus, Page 3.9
20 Chaper 3 boa wae, 5.0 s Hz 3.9 he down and bac dsance s = he speed s hen d oal s 0.0 s F Now, 0.00 g g so F g 0.0 s 80.0 N 3.50 he speed o he wae s s 5.0 s and he ass per un lengh o he rope s / g. hus, ro F, we oban 5.0 s g 9 N F 3.5 (a) he speed o ranserse waes n he cord s F, where s he ass per un lengh. Wh he enson n he cord beng F =.0 N, he wae speed s F F F.0 N g. s he e o rael he lengh o he cord s s 0.8 s 3.5 (a) In ang 6 round rps, he pulse raels he lengh o he lne es or a oal dsance o. he speed o he pulse s hen Page 3.0
21 Chaper s 8.6 s he speed o ranserse waes n he lne s F, so he enson n he lne s g F s 73.8 N 3.53 (a) he ass per un lengh s g g Fro F, he requred enson n he srng s F 50.0 s 0.00 g 30.0 N, F 8.00 N 0.00 g 5.8 s 3.5 he ass per un lengh o he wre s g g and he speed o he pulse s s.3 s hus, he enson n he wre s F -3.3 s.50 0 g.9 N Bu, he enson n he wre s he wegh o a 3.00-g objec on he Moon. Hence, he local ree-all acceleraon s g F.9 N 3.00 g.6 s 3.55 he perod o he pendulu s / g, so he lengh o he srng s Page 3.
22 Chaper s.00 s g hen ass per un lengh o he srng s hen g g When he pendulu s ercal and saonar, he enson n he srng s F Mball g 5.00 g 9.80 s 9.0 N and he speed o ranserse waes n s F 9.0 N g 8.5 s 3.56 I s he ass per un lengh or he rs srng, hen s ha o he second srng. hus, wh F F F, he speed o waes n he second srng s F F F 5.00 s 7.07 s 3.57 (a) he enson n he srng s F g 3.0 g 9.80 s 9 N. hen, ro F, he ass per un lengh s F 9 N s 0.05 g When =.00 g, he enson s F g.0 g 9.80 s 0 N and he speed o ranserse waes n he srng s F 0 N 0.05 g 0 s 3.58 I he enson n he wre s F, he ensle sress s Sress = F/A, so he speed o ranserse waes n he wre a be wren as Page 3.
23 Chaper 3 F A Sress Sress / /( A ) Bu, Sress A V =olue, so A dens. hus,. When he sress s a s au, he speed o waes n he wre s ( Sress) 9 a.70 0 Pa a g/ 586 s 3.59 (a) he speed o ranserse waes n he lne s F, wh beng he ass per un lengh. hereore, F F F.5 N g 3. s he worer could hrow an objec, such as a snowball, a one end o he lne o se up a pulse, and use a sopwach o easure he e aes a pulse o rael he lengh o he lne. Fro hs easureen, he wo rer would hae an esae o he wae speed, whch n urn can be used o esae he enson (a) In ang n round rps along he lengh o he lne, he oal dsance raeled b he pulse s n n. he wae speed s hen n Fro F as he speed o ranserse waes n he lne, he enson s F M n M n n M 3.6 (a) Consruce nererence produces he au aplude Aa A A 0.50 Desruce nererence produces he nu aplude Page 3.3
24 Chaper 3 An A A We are gen ha Acos( ) (0.5 )cos(0. ). (a) B nspecon, he aplude s seen o be A 0.5 he angular requenc s 0. rad s. Bu, so he sprng consan s (0.30 g)(0. rad s) 0.7 N (c) Noe: Your calculaor us be n radans ode or par (c). A = 0.30 s, 0.5 cos 0. rad s 0.30 s 0.3 (d) Fro conseraon o echancal energ, he speed a dsplaceen s gen b A. hus, a = 0.30 s, when = 0.3, he speed s 0. rad s (0.5 ) (0.3 ) 0. s 3.63 (a) he perod o a brang objec-sprng sse s, so he sprng consan s.00 g s 9 N I =.05 s or ass, hs ass s (9 N/)(.05 s) 6. g 3.6 (a) he perod s he recprocal o he requenc, or 96 s s 5.0 s sound 33 s 96 s (a) he perod o a sple pendulu s g, so he perod o he rs sse s Page 3.
25 Chaper s g 9.80 s he perod o ass-sprng sse s, so he perod o he second sse s =, hen l g and he sprng consan s g.0 g 9.80 s N 3.66 Snce he sprng s lgh, we neglec an sall aoun o energ los n he collson wh he sprng, and appl conseraon o echancal energ ro when he bloc rs sars unl coes o res agan. hs ges KE PE g PE s KE PE g PE s, or a gh hus, a gh g 9.80 s N Choosng PE g = 0 a he nal hegh o he bloc, conseraon o echancal energ ges KE PE PE KE PE PE, or g s g s g 0, where s he speed o he bloc aer allng dsance. (a) When = 0, he non-zero soluon o he energ equaon ro aboe ges 3.00 g 9.80 s a a g or g a N When = 5.00 c = , he energ equaon ges g, or Page 3.5
26 Chaper N s s 3.00 g 3.68 (a) We appl conseraon o echancal energ ro jus aer he collson unl he bloc coes o res. KE PE s KE PE s ges 0 0 MV or he speed o he bloc jus aer he collson s V M 900 N g.50 s Now, we appl conseraon o oenu ro jus beore pac o edael aer he collson. hs ges 0 MV bulle bulle or M bulle bulle V.00 g 00 s.5 s 00 s g he echancal energ conered no nernal energ durng he collson s bulle bulle E KE KE MV or E g 00 s 00 s.00 g.50 s E 37 J 3.69 Choose PE g = 0 when he blocs sar ro res. hen, usng conseraon o echancal energ ro when he blocs are released unl he sprng reurns o s unsreched lengh ges KE PE PE KE PE PE, or g s g s Page 3.6
27 Chaper 3 g sn 0 g g 5 g 9.80 s 0.00 sn 0 30 g 9.80 s N 0.00 eldng. s 3.70 (a) When he gun s red, he energ nall sored as elasc poenal energ n he sprng s ransored no nec energ o he bulle. Assung no loss o energ, we hae, or 9.80 N s g Fro a, he e requred or he pelle o drop.00 o he loor, sarng wh 0 0, s 0.00 a 9.80 s 0.5 s he range (horzonal dsance raeled durng he lgh) s hen s 0.5 s 8.9 Page 3.7
28 Chaper he ree-bod dagra a he rgh shows he orces acng on he balloon when s dsplaced dsance s = along he crcular arc ollows. he ne orce angenal o hs pah s F F Bsn g sn B g sn ne For sall angles, sn s/ Also, g He V g and he buoan orce s acng on he balloon s B ar V g. hus, he ne resorng orce F ne ar He Vg s Obsere ha hs s n he or o Hooe s law, F s, wh ar He Vg hus, he oon wll be sple haronc and he perod s gen b HeV He Vg g ar He ar He hs elds s s 3.7 (a) When he bloc s gen soe sall upward dsplaceen, he ne resorng orce eered on b he rubber bands s F F F sn, where an ne For sall dsplaceens, he angle wll be er sall. hen sn an s, and he ne resorng orce Page 3.8
29 Chaper 3 F Fne F he ne resorng orce ound n par (a) s n he or o Hooe s law F =, wh = F/. hus, he oon wll be sple haronc, and he angular requenc s F 3.73 Newon s law o graaon s GM F, where M r r 3 3 hus, F= 3 G r whch s o Hooe s law or, F r, wh = 3 G 3.7 he nner p o he wng s aached o he end o he sprng and alwas oes wh he sae speed as he end o he brang sprng. hus, s au speed s.7 0 N nner, a sprng, a 3 A 0.0 c 0.5 c s g reang he wng as a rgd bar, all pons n he wng hae he sae angular eloc a an nsan n e. As he wng rocs on he ulcru, he nner p and ouer ps ollow crcular pahs o deren rad. Snce he angular eloces o he ps are alwas equal, we a wre r ouer ouer r nner nner he au speed o he ouer p s hen Page 3.9
30 Chaper 3 r ouer ouer, a nner, a rnner c s.3 c s (a) 500 N.00 g 5.8 rad s Appl Newon s second law o he bloc whle he eleaor s accelerang: F Fs g a Wh F and a g 3, hs ges g g 3, or s g.00 g 9.80 s N c 3.76 (a) Noe ha as he sprng passes hrough he ercal poson, he objec s ong n a crcular arc o radus. Also, obsere ha he -coordnae o he objec a hs pon us be negae 0 so he sprng s sreched and eerng an upward enson orce o agnude greaer han he objec s wegh. hs s necessar so he objec eperences a ne orce oward he po o suppl he needed cenrpeal acceleraon n hs poson. hs s suarzed b Newon s second law appled o he objec a hs pon, sang F g Conseraon o energ requres ha E KE PEg, PEs, KE PEg, PE s,, or E 0 g 0 g, reducng o g (c) Fro he resul o par (a), obsere ha ( )( g ). Subsung hs no he resul ro par ges g( ) ( )( g). Aer epandng and regroupng ers, hs becoes ( ) (3 g ) ( 3 g ) 0, whch s a quadrac equaon a b c 0, wh a 50 N N b 3g g 9.80 s 50 N N and Page 3.30
31 Chaper 3 c 3g g 9.80 s.50 N Applng he quadrac orula, eepng onl he negae soluon [see he dscusson n par (a)] ges b b ac a or 0.0 (d) Because he lengh o hs pendulu ares and s longer hroughou s oon han a sple pendulu o lengh, s perod wll be longer han ha o a sple pendulu he au acceleraon o he oscllang sse s a A A a he rcon orce eered beween he wo blocs us be capable o accelerang bloc B a hs rae. When bloc B s on he erge o slppng, n g a and we us hae s s a s s a aa A sg hus, A s g s.50 Hz c Page 3.3
Chapters 2 Kinematics. Position, Distance, Displacement
Chapers Knemacs Poson, Dsance, Dsplacemen Mechancs: Knemacs and Dynamcs. Knemacs deals wh moon, bu s no concerned wh he cause o moon. Dynamcs deals wh he relaonshp beween orce and moon. The word dsplacemen
PHYS 1443 Section 001 Lecture #4
PHYS 1443 Secon 001 Lecure #4 Monda, June 5, 006 Moon n Two Dmensons Moon under consan acceleraon Projecle Moon Mamum ranges and heghs Reerence Frames and relae moon Newon s Laws o Moon Force Newon s Law
Motion in Two Dimensions
Phys 1 Chaper 4 Moon n Two Dmensons adzyubenko@csub.edu hp://www.csub.edu/~adzyubenko 005, 014 A. Dzyubenko 004 Brooks/Cole 1 Dsplacemen as a Vecor The poson of an objec s descrbed by s poson ecor, r The
Today s topic: IMPULSE AND MOMENTUM CONSERVATION
Today s opc: MPULSE ND MOMENTUM CONSERVTON Reew of Las Week s Lecure Elasc Poenal Energy: x: dsplaceen fro equlbru x = : equlbru poson Work-Energy Theore: W o W W W g noncons W non el W noncons K K K (
Response of MDOF systems
Response of MDOF syses Degree of freedo DOF: he nu nuber of ndependen coordnaes requred o deerne copleely he posons of all pars of a syse a any nsan of e. wo DOF syses hree DOF syses he noral ode analyss
Homework 8: Rigid Body Dynamics Due Friday April 21, 2017
EN40: Dynacs and Vbraons Hoework 8: gd Body Dynacs Due Frday Aprl 1, 017 School of Engneerng Brown Unversy 1. The earh s roaon rae has been esaed o decrease so as o ncrease he lengh of a day a a rae of
CHAPTER 2 Quick Quizzes
CHAPTER Quck Quzzes (a) 00 yd (b) 0 (c) 0 (a) False The car may be slowng down, so ha he drecon o s acceleraon s oppose he drecon o s elocy (b) True I he elocy s n he drecon chosen as negae, a pose acceleraon
WebAssign HW Due 11:59PM Tuesday Clicker Information
WebAssgn HW Due 11:59PM Tuesday Clcker Inormaon Remnder: 90% aemp, 10% correc answer Clcker answers wll be a end o class sldes (onlne). Some days we wll do a lo o quesons, and ew ohers Each day o clcker
UNIT 1 ONE-DIMENSIONAL MOTION GRAPHING AND MATHEMATICAL MODELING. Objectives
UNIT 1 ONE-DIMENSIONAL MOTION GRAPHING AND MATHEMATICAL MODELING Objeces To learn abou hree ways ha a physcs can descrbe moon along a sragh lne words, graphs, and mahemacal modelng. To acqure an nue undersandng
2015 Sectional Physics Exam Solution Set
. Crrec answer: D Ne: [quan] denes: uns quan WYSE cadec Challenge 05 Secnal Phscs Ea SOLUTION SET / / / / rce lengh lengh rce enu ass lengh e a) / ass ass b) energ c) wrk lengh e pwer energ e d) (crrec
Water Hammer in Pipes
Waer Haer Hydraulcs and Hydraulc Machnes Waer Haer n Pes H Pressure wave A B If waer s flowng along a long e and s suddenly brough o res by he closng of a valve, or by any slar cause, here wll be a sudden
Displacement, Velocity, and Acceleration. (WHERE and WHEN?)
Dsplacemen, Velocy, and Acceleraon (WHERE and WHEN?) Mah resources Append A n your book! Symbols and meanng Algebra Geomery (olumes, ec.) Trgonomery Append A Logarhms Remnder You wll do well n hs class
First-order piecewise-linear dynamic circuits
Frs-order pecewse-lnear dynamc crcus. Fndng he soluon We wll sudy rs-order dynamc crcus composed o a nonlnear resse one-por, ermnaed eher by a lnear capacor or a lnear nducor (see Fg.. Nonlnear resse one-por
x y θ = 31.8 = 48.0 N. a 3.00 m/s
4.5.IDENTIY: Vecor addiion. SET UP: Use a coordinae sse where he dog A. The forces are skeched in igure 4.5. EXECUTE: + -ais is in he direcion of, A he force applied b =+ 70 N, = 0 A B B A = cos60.0 =
5-1. We apply Newton s second law (specifically, Eq. 5-2). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( )
5-1. We apply Newon s second law (specfcally, Eq. 5-). (a) We fnd he componen of he foce s ( ) ( ) F = ma = ma cos 0.0 = 1.00kg.00m/s cos 0.0 = 1.88N. (b) The y componen of he foce s ( ) ( ) F = ma = ma
Solution in semi infinite diffusion couples (error function analysis)
Soluon n sem nfne dffuson couples (error funcon analyss) Le us consder now he sem nfne dffuson couple of wo blocks wh concenraon of and I means ha, n a A- bnary sysem, s bondng beween wo blocks made of
Thus the force is proportional but opposite to the displacement away from equilibrium.
Chaper 3 : Siple Haronic Moion Hooe s law saes ha he force (F) eered by an ideal spring is proporional o is elongaion l F= l where is he spring consan. Consider a ass hanging on a he spring. In equilibriu
Chapter 3: Vectors and Two-Dimensional Motion
Chape 3: Vecos and Two-Dmensonal Moon Vecos: magnude and decon Negae o a eco: eese s decon Mulplng o ddng a eco b a scala Vecos n he same decon (eaed lke numbes) Geneal Veco Addon: Tangle mehod o addon
Chapter Lagrangian Interpolation
Chaper 5.4 agrangan Inerpolaon Afer readng hs chaper you should be able o:. dere agrangan mehod of nerpolaon. sole problems usng agrangan mehod of nerpolaon and. use agrangan nerpolans o fnd deraes and
Normal Random Variable and its discriminant functions
Noral Rando Varable and s dscrnan funcons Oulne Noral Rando Varable Properes Dscrnan funcons Why Noral Rando Varables? Analycally racable Works well when observaon coes for a corruped snle prooype 3 The
Section 14 Forces in Circular Motion
Secion 14 orces in Circular Moion Ouline 1 Unifor Circular Moion Non-unifor Circular Moion Phsics 04A Class Noes Wh do objecs do wha he do? The answer we have been invesigaing is forces If forces can eplain
THEORETICAL AUTOCORRELATIONS. ) if often denoted by γ. Note that
THEORETICAL AUTOCORRELATIONS Cov( y, y ) E( y E( y))( y E( y)) ρ = = Var( y) E( y E( y)) =,, L ρ = and Cov( y, y ) s ofen denoed by whle Var( y ) f ofen denoed by γ. Noe ha γ = γ and ρ = ρ and because
Physics Notes - Ch. 2 Motion in One Dimension
Physics Noes - Ch. Moion in One Dimension I. The naure o physical quaniies: scalars and ecors A. Scalar quaniy ha describes only magniude (how much), NOT including direcion; e. mass, emperaure, ime, olume,
CHAPTER II AC POWER CALCULATIONS
CHAE AC OWE CACUAON Conens nroducon nsananeous and Aerage ower Effece or M alue Apparen ower Coplex ower Conseraon of AC ower ower Facor and ower Facor Correcon Maxu Aerage ower ransfer Applcaons 3 nroducon
Physics 3 (PHYF144) Chap 3: The Kinetic Theory of Gases - 1
Physcs (PYF44) ha : he nec heory of Gases -. Molecular Moel of an Ieal Gas he goal of he olecular oel of an eal gas s o unersan he acroscoc roeres (such as ressure an eeraure ) of gas n e of s croscoc
Notes on the stability of dynamic systems and the use of Eigen Values.
Noes on he sabl of dnamc ssems and he use of Egen Values. Source: Macro II course noes, Dr. Davd Bessler s Tme Seres course noes, zarads (999) Ineremporal Macroeconomcs chaper 4 & Techncal ppend, and Hamlon
Page 1 o 13 1. The brighes sar in he nigh sky is α Canis Majoris, also known as Sirius. I lies 8.8 ligh-years away. Express his disance in meers. ( ligh-year is he disance coered by ligh in one year. Ligh
How does the momentum before an elastic and an inelastic collision compare to the momentum after the collision?
Experent 9 Conseraton o Lnear Moentu - Collsons In ths experent you wll be ntroduced to the denton o lnear oentu. You wll learn the derence between an elastc and an nelastc collson. You wll explore how
t A. 3. Which vector has the largest component in the y-direction, as defined by the axes to the right?
Ke Name Insrucor Phsics 1210 Exam 1 Sepember 26, 2013 Please wrie direcl on he exam and aach oher shees of work if necessar. Calculaors are allowed. No noes or books ma be used. Muliple-choice problems
1. The graph below shows the variation with time t of the acceleration a of an object from t = 0 to t = T. a
Kinemaics Paper 1 1. The graph below shows he ariaion wih ime of he acceleraion a of an objec from = o = T. a T The shaded area under he graph represens change in A. displacemen. B. elociy. C. momenum.
s in boxe wers ans Put
Pu answers in boxes Main Ideas in Class Toda Inroducion o Falling Appl Old Equaions Graphing Free Fall Sole Free Fall Problems Pracice:.45,.47,.53,.59,.61,.63,.69, Muliple Choice.1 Freel Falling Objecs
I. OBJECTIVE OF THE EXPERIMENT.
I. OBJECTIVE OF THE EXPERIMENT. Swissmero raels a high speeds hrough a unnel a low pressure. I will hereore undergo ricion ha can be due o: ) Viscosiy o gas (c. "Viscosiy o gas" eperimen) ) The air in
Midterm Exam. Thursday, April hour, 15 minutes
Economcs of Grow, ECO560 San Francsco Sae Unvers Mcael Bar Sprng 04 Mderm Exam Tursda, prl 0 our, 5 mnues ame: Insrucons. Ts s closed boo, closed noes exam.. o calculaors of an nd are allowed. 3. Sow all
The Single Particle Path Integral and Its Calculations. Lai Zhong Yuan
Te Sngle Parcle Pa Inegral and Is Calculaons La Zong Yuan Suary O Conens Inroducon and Movaon Soe Eaples n Calculang Pa Inegrals Te Free Parcle Te Haronc Oscllaor Perurbaon Epansons Inroducon and Movaon
Let s treat the problem of the response of a system to an applied external force. Again,
Page 33 QUANTUM LNEAR RESPONSE FUNCTON Le s rea he problem of he response of a sysem o an appled exernal force. Agan, H() H f () A H + V () Exernal agen acng on nernal varable Hamlonan for equlbrum sysem
Chapter 3 Kinematics in Two Dimensions
Chaper 3 KINEMATICS IN TWO DIMENSIONS PREVIEW Two-dimensional moion includes objecs which are moing in wo direcions a he same ime, such as a projecile, which has boh horizonal and erical moion. These wo
Energy Problems 9/3/2009. W F d mgh m s 196J 200J. Understanding. Understanding. Understanding. W F d. sin 30
9/3/009 nderanding Energy Proble Copare he work done on an objec o a.0 kg a) In liing an objec 0.0 b) Puhing i up a rap inclined a 30 0 o he ae inal heigh 30 0 puhing 0.0 liing nderanding Copare he work
The ray paths and travel times for multiple layers can be computed using ray-tracing, as demonstrated in Lab 3.
C. Trael me cures for mulple reflecors The ray pahs ad rael mes for mulple layers ca be compued usg ray-racg, as demosraed Lab. MATLAB scrp reflec_layers_.m performs smple ray racg. (m) ref(ms) ref(ms)
One-Dimensional Kinematics
One-Dimensional Kinemaics One dimensional kinemaics refers o moion along a sraigh line. Een hough we lie in a 3-dimension world, moion can ofen be absraced o a single dimension. We can also describe moion
Including the ordinary differential of distance with time as velocity makes a system of ordinary differential equations.
Soluons o Ordnary Derenal Equaons An ordnary derenal equaon has only one ndependen varable. A sysem o ordnary derenal equaons consss o several derenal equaons each wh he same ndependen varable. An eample
10. A.C CIRCUITS. Theoretically current grows to maximum value after infinite time. But practically it grows to maximum after 5τ. Decay of current :
. A. IUITS Synopss : GOWTH OF UNT IN IUIT : d. When swch S s closed a =; = d. A me, curren = e 3. The consan / has dmensons of me and s called he nducve me consan ( τ ) of he crcu. 4. = τ; =.63, n one
[ ] 2. [ ]3 + (Δx i + Δx i 1 ) / 2. Δx i-1 Δx i Δx i+1. TPG4160 Reservoir Simulation 2018 Lecture note 3. page 1 of 5
![[ ] 2. [ ]3 + (Δx i + Δx i 1 ) / 2. Δx i-1 Δx i Δx i+1. TPG4160 Reservoir Simulation 2018 Lecture note 3. page 1 of 5](/thumbs/85/92299540.jpg "[ ] 2. [ ]3 + (Δx i + Δx i 1 ) / 2. Δx i-1 Δx i Δx i+1. TPG4160 Reservoir Simulation 2018 Lecture note 3. page 1 of 5")
TPG460 Reservor Smulaon 08 page of 5 DISCRETIZATIO OF THE FOW EQUATIOS As we already have seen, fne dfference appromaons of he paral dervaves appearng n he flow equaons may be obaned from Taylor seres
Main Ideas in Class Today
Main Ideas in Class Toda Inroducion o Falling Appl Consan a Equaions Graphing Free Fall Sole Free Fall Problems Pracice:.45,.47,.53,.59,.61,.63,.69, Muliple Choice.1 Freel Falling Objecs Refers o objecs
J i-1 i. J i i+1. Numerical integration of the diffusion equation (I) Finite difference method. Spatial Discretization. Internal nodes.
umercal negraon of he dffuson equaon (I) Fne dfference mehod. Spaal screaon. Inernal nodes. R L V For hermal conducon le s dscree he spaal doman no small fne spans, =,,: Balance of parcles for an nernal
0 time. 2 Which graph represents the motion of a car that is travelling along a straight road with a uniformly increasing speed?
1 1 The graph relaes o he moion of a falling body. y Which is a correc descripion of he graph? y is disance and air resisance is negligible y is disance and air resisance is no negligible y is speed and
IB Physics Kinematics Worksheet
IB Physics Kinemaics Workshee Wrie full soluions and noes for muliple choice answers. Do no use a calculaor for muliple choice answers. 1. Which of he following is a correc definiion of average acceleraion?
Elastic and Inelastic Collisions
laic and Inelaic Colliion In an LASTIC colliion, energy i conered (Kbefore = Kafer or Ki = Kf. In an INLASTIC colliion, energy i NOT conered. (Ki > Kf. aple: A kg block which i liding a 0 / acro a fricionle
Best test practice: Take the past test on the class website
Bes es pracice: Take he pas es on he class websie hp://communiy.wvu.edu/~miholcomb/phys11.hml I have posed he key o he WebAssign pracice es. Newon Previous Tes is Online. Forma will be idenical. You migh
NEWTON S SECOND LAW OF MOTION
Course and Secion Dae Names NEWTON S SECOND LAW OF MOTION The acceleraion of an objec is defined as he rae of change of elociy. If he elociy changes by an amoun in a ime, hen he aerage acceleraion during
In the complete model, these slopes are ANALYSIS OF VARIANCE FOR THE COMPLETE TWO-WAY MODEL. (! i+1 -! i ) + [(!") i+1,q - [(!
ANALYSIS OF VARIANCE FOR THE COMPLETE TWO-WAY MODEL The frs hng o es n wo-way ANOVA: Is here neracon? "No neracon" means: The man effecs model would f. Ths n urn means: In he neracon plo (wh A on he horzonal
Two Dimensional Dynamics
Physics 11: Lecure 6 Two Dimensional Dynamics Today s lecure will coer Chaper 4 Saring Wed Sep 15, W-F oice hours will be in 3 Loomis. Exam I M oice hours will coninue in 36 Loomis Physics 11: Lecure 6,
Physics 201 Lecture 15
Phscs 0 Lecue 5 l Goals Lecue 5 v Elo consevaon of oenu n D & D v Inouce oenu an Iulse Coens on oenu Consevaon l oe geneal han consevaon of echancal eneg l oenu Consevaon occus n sses wh no ne eenal foces
Work Power Energy. For conservaive orce ) Work done is independen o he pah ) Work done in a closed loop is zero ) Work done agains conservaive orce is sored is he orm o poenial energy 4) All he above.
( ) () we define the interaction representation by the unitary transformation () = ()
Hgher Order Perurbaon Theory Mchael Fowler 3/7/6 The neracon Represenaon Recall ha n he frs par of hs course sequence, we dscussed he chrödnger and Hesenberg represenaons of quanum mechancs here n he chrödnger
Two Dimensional Dynamics
Physics 11: Lecure 6 Two Dimensional Dynamics Today s lecure will coer Chaper 4 Exam I Physics 11: Lecure 6, Pg 1 Brie Reiew Thus Far Newon s Laws o moion: SF=ma Kinemaics: x = x + + ½ a Dynamics Today
Unit 1 Test Review Physics Basics, Movement, and Vectors Chapters 1-3
A.P. Physics B Uni 1 Tes Reiew Physics Basics, Moemen, and Vecors Chapers 1-3 * In sudying for your es, make sure o sudy his reiew shee along wih your quizzes and homework assignmens. Muliple Choice Reiew:
Q2.4 Average velocity equals instantaneous velocity when the speed is constant and motion is in a straight line.
CHAPTER MOTION ALONG A STRAIGHT LINE Discussion Quesions Q. The speedomeer measures he magniude of he insananeous eloci, he speed. I does no measure eloci because i does no measure direcion. Q. Graph (d).
1. VELOCITY AND ACCELERATION
1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.3 Velociy-Time Graph Gradien = acceleraion Area under
Physics 101 Fall 2006: Exam #1- PROBLEM #1
Physics 101 Fall 2006: Exam #1- PROBLEM #1 1. Problem 1. (+20 ps) (a) (+10 ps) i. +5 ps graph for x of he rain vs. ime. The graph needs o be parabolic and concave upward. ii. +3 ps graph for x of he person
ESS 265 Spring Quarter 2005 Kinetic Simulations
SS 65 Spng Quae 5 Knec Sulaon Lecue une 9 5 An aple of an lecoagnec Pacle Code A an eaple of a knec ulaon we wll ue a one denonal elecoagnec ulaon code called KMPO deeloped b Yohhau Oua and Hoh Mauoo.
1. Kinematics I: Position and Velocity
1. Kinemaics I: Posiion and Velociy Inroducion The purpose of his eperimen is o undersand and describe moion. We describe he moion of an objec by specifying is posiion, velociy, and acceleraion. In his
total If no external forces act, the total linear momentum of the system is conserved. This occurs in collisions and explosions.
Lesson 0: Collsons, Rotatonal netc Energy, Torque, Center o Graty (Sectons 7.8 Last te we used ewton s second law to deelop the pulse-oentu theore. In words, the theore states that the change n lnear oentu
Physics 201 Lecture 2
Physcs 1 Lecure Lecure Chper.1-. Dene Poson, Dsplcemen & Dsnce Dsngush Tme nd Tme Inerl Dene Velocy (Aerge nd Insnneous), Speed Dene Acceleron Undersnd lgebrclly, hrough ecors, nd grphclly he relonshps
Displacement ( x) x x x
Kinemaics Kinemaics is he branch of mechanics ha describes he moion of objecs wihou necessarily discussing wha causes he moion. 1-Dimensional Kinemaics (or 1- Dimensional moion) refers o moion in a sraigh
( )a = "t = 1 E =" B E = 5016 V. E = BHv # 3. 2 %r. c.) direction of induced current in the loop for : i.) "t < 1
99 3 c dr b a µ r.? d b µ d d cdr a r & b d & µ c µ c b dr µ c µ c b & ' ln' a +*+* b ln r ln a a r a ' µ c b 'b* µ c ln' * & ln, &a a+ ncreang no he page o nduced curren wll creae a - feldou of he page
Sklar: Sections (4.4.2 is not covered).
COSC 44: Dgal Councaons Insrucor: Dr. Ar Asf Deparen of Copuer Scence and Engneerng York Unversy Handou # 6: Bandpass Modulaon opcs:. Phasor Represenaon. Dgal Modulaon Schees: PSK FSK ASK APK ASK/FSK)
University Physics with Modern Physics 14th Edition Young TEST BANK
Universi Phsics wih Modern Phsics 14h Ediion Young SOLUTIONS MANUAL Full clear download (no formaing errors) a: hps://esbankreal.com/download/universi-phsics-modern-phsics- 14h-ediion-oung-soluions-manual-/
TP A.14 The effects of cut angle, speed, and spin on object ball throw
echnical proof echnical proof TP A.14 The effecs of cu angle, speed, and spin on objec ball hrow supporing: The Illusraed Principles of Pool and illiards hp://billiards.colosae.edu by Daid G. Alciaore,
Phys 221 Fall Chapter 2. Motion in One Dimension. 2014, 2005 A. Dzyubenko Brooks/Cole
Phys 221 Fall 2014 Chaper 2 Moion in One Dimension 2014, 2005 A. Dzyubenko 2004 Brooks/Cole 1 Kinemaics Kinemaics, a par of classical mechanics: Describes moion in erms of space and ime Ignores he agen
UNIVERSITAT AUTÒNOMA DE BARCELONA MARCH 2017 EXAMINATION
INTERNATIONAL TRADE T. J. KEHOE UNIVERSITAT AUTÒNOMA DE BARCELONA MARCH 27 EXAMINATION Please answer wo of he hree quesons. You can consul class noes, workng papers, and arcles whle you are workng on he
Learning Objectives. Self Organization Map. Hamming Distance(1/5) Introduction. Hamming Distance(3/5) Hamming Distance(2/5) 15/04/2015
/4/ Learnng Objecves Self Organzaon Map Learnng whou Exaples. Inroducon. MAXNET 3. Cluserng 4. Feaure Map. Self-organzng Feaure Map 6. Concluson 38 Inroducon. Learnng whou exaples. Daa are npu o he syse
Lecture 16 (Momentum and Impulse, Collisions and Conservation of Momentum) Physics Spring 2017 Douglas Fields
Lecure 16 (Momenum and Impulse, Collisions and Conservaion o Momenum) Physics 160-02 Spring 2017 Douglas Fields Newon s Laws & Energy The work-energy heorem is relaed o Newon s 2 nd Law W KE 1 2 1 2 F
Lecture 18: The Laplace Transform (See Sections and 14.7 in Boas)
Lecure 8: The Lalace Transform (See Secons 88- and 47 n Boas) Recall ha our bg-cure goal s he analyss of he dfferenal equaon, ax bx cx F, where we emloy varous exansons for he drvng funcon F deendng on
CS434a/541a: Pattern Recognition Prof. Olga Veksler. Lecture 4
CS434a/54a: Paern Recognon Prof. Olga Veksler Lecure 4 Oulne Normal Random Varable Properes Dscrmnan funcons Why Normal Random Varables? Analycally racable Works well when observaon comes form a corruped
Bandlimited channel. Intersymbol interference (ISI) This non-ideal communication channel is also called dispersive channel
Inersymol nererence ISI ISI s a sgnal-dependen orm o nererence ha arses ecause o devaons n he requency response o a channel rom he deal channel. Example: Bandlmed channel Tme Doman Bandlmed channel Frequency
3.6 Derivatives as Rates of Change
3.6 Derivaives as Raes of Change Problem 1 John is walking along a sraigh pah. His posiion a he ime >0 is given by s = f(). He sars a =0from his house (f(0) = 0) and he graph of f is given below. (a) Describe
Conservation of Momentum. The purpose of this experiment is to verify the conservation of momentum in two dimensions.
Conseraion of Moenu Purose The urose of his exerien is o erify he conseraion of oenu in wo diensions. Inroducion and Theory The oenu of a body ( ) is defined as he roduc of is ass () and elociy ( ): When
Mechanics Physics 151
Mechancs Physcs 5 Lecure 9 Hamlonan Equaons of Moon (Chaper 8) Wha We Dd Las Tme Consruced Hamlonan formalsm H ( q, p, ) = q p L( q, q, ) H p = q H q = p H = L Equvalen o Lagrangan formalsm Smpler, bu
Variants of Pegasos. December 11, 2009
Inroducon Varans of Pegasos SooWoong Ryu bshboy@sanford.edu December, 009 Youngsoo Cho yc344@sanford.edu Developng a new SVM algorhm s ongong research opc. Among many exng SVM algorhms, we wll focus on
A B C D September 25 Exam I Physics 105. Circle the letter of the single best answer. Each question is worth 1 point
2012 Sepember 25 Eam I Physics 105 Circle he leer of he single bes answer. Each uesion is worh 1 poin Physical Consans: Earh s free-fall acceleraion = g = 9.80 m/s 2 3. (Mark wo leers!) The below graph
II. Light is a Ray (Geometrical Optics)
II Lgh s a Ray (Geomercal Opcs) IIB Reflecon and Refracon Hero s Prncple of Leas Dsance Law of Reflecon Hero of Aleandra, who lved n he 2 nd cenury BC, posulaed he followng prncple: Prncple of Leas Dsance:
Mechanics Physics 151
Mechancs Physcs 5 Lecure 9 Hamlonan Equaons of Moon (Chaper 8) Wha We Dd Las Tme Consruced Hamlonan formalsm Hqp (,,) = qp Lqq (,,) H p = q H q = p H L = Equvalen o Lagrangan formalsm Smpler, bu wce as
KINEMATICS IN ONE DIMENSION
KINEMATICS IN ONE DIMENSION PREVIEW Kinemaics is he sudy of how hings move how far (disance and displacemen), how fas (speed and velociy), and how fas ha how fas changes (acceleraion). We say ha an objec
PHYSICS 149: Lecture 9
PHYSICS 149: Lecure 9 Chaper 3 3.2 Velociy and Acceleraion 3.3 Newon s Second Law of Moion 3.4 Applying Newon s Second Law 3.5 Relaive Velociy Lecure 9 Purdue Universiy, Physics 149 1 Velociy (m/s) The
2/20/2013. EE 101 Midterm 2 Review
//3 EE Mderm eew //3 Volage-mplfer Model The npu ressance s he equalen ressance see when lookng no he npu ermnals of he amplfer. o s he oupu ressance. I causes he oupu olage o decrease as he load ressance
MECHANICAL PROPERTIES OF FLUIDS NCERT
Chaper Ten MECHANICAL PROPERTIES OF FLUIDS MCQ I 10.1 A all cylinder is filled wih iscous oil. A round pebble is dropped from he op wih zero iniial elociy. From he plo shown in Fig. 10.1, indicae he one
Introduction to Numerical Analysis. In this lesson you will be taken through a pair of techniques that will be used to solve the equations of.
Inroducion o Nuerical Analysis oion In his lesson you will be aen hrough a pair of echniques ha will be used o solve he equaions of and v dx d a F d for siuaions in which F is well nown, and he iniial
Practice Problems - Week #4 Higher-Order DEs, Applications Solutions
Pracice Probles - Wee #4 Higher-Orer DEs, Applicaions Soluions 1. Solve he iniial value proble where y y = 0, y0 = 0, y 0 = 1, an y 0 =. r r = rr 1 = rr 1r + 1, so he general soluion is C 1 + C e x + C
s = rθ Chapter 10: Rotation 10.1: What is physics?
Chape : oaon Angula poson, velocy, acceleaon Consan angula acceleaon Angula and lnea quanes oaonal knec enegy oaonal nea Toque Newon s nd law o oaon Wok and oaonal knec enegy.: Wha s physcs? In pevous
THE PREDICTION OF COMPETITIVE ENVIRONMENT IN BUSINESS
THE PREICTION OF COMPETITIVE ENVIRONMENT IN BUSINESS INTROUCTION The wo dmensonal paral dfferenal equaons of second order can be used for he smulaon of compeve envronmen n busness The arcle presens he
~v = x. ^x + ^y + ^x + ~a = vx. v = v 0 + at. ~v P=A = ~v P=B + ~v B=A. f k = k. W tot =KE. P av =W=t. W grav = mgy 1, mgy 2 = mgh =,U grav
PHYSICS 5A FALL 2001 FINAL EXAM v = x a = v x = 1 2 a2 + v 0 + x 0 v 2 = v 2 0 +2a(x, x 0) a = v2 r ~v = x ~a = vx v = v 0 + a y z ^x + ^y + ^z ^x + vy x, x 0 = 1 2 (v 0 + v) ~v P=A = ~v P=B + ~v B=A ^y
Bag for Sophia by Leonie Bateman and Deirdre Bond-Abel
Bag for Sopha 2012 by Leone Baeman and Derdre Bond-Abel Ths bag was desgned o go wh he beauful feled wool scarf of our book Elegan Quls, Counry Charm. Make boh and you ll have he perfec ensemble o wear
Transcription: Messenger RNA, mrna, is produced and transported to Ribosomes
Quanave Cenral Dogma I Reference hp//book.bonumbers.org Inaon ranscrpon RNA polymerase and ranscrpon Facor (F) s bnds o promoer regon of DNA ranscrpon Meenger RNA, mrna, s produced and ranspored o Rbosomes
Different kind of oscillation
PhO 98 Theorecal Qeson.Elecrcy Problem (8 pons) Deren knd o oscllaon e s consder he elecrc crc n he gre, or whch mh, mh, nf, nf and kω. The swch K beng closed he crc s copled wh a sorce o alernang crren.
CHAPTER 10 ROTATIONAL MOTION
CHAPTER 0 ROTATONAL MOTON 0. ANGULAR VELOCTY Consder argd body rotates about a fxed axs through pont O n x-y plane as shown. Any partcle at pont P n ths rgd body rotates n a crcle of radus r about O. The
Opening Shock and Shape of the Drag-vs-Time Curve
Openng Shock and Shape o he Drag-vs-Te Curve Jean Povn Physcs Deparen, San Lous Unversy, S. Lous MO Conac: povnj@slu.edu 314-977-8424 Talk presened a he 19 h AIAA Aerodynac Deceleraor Syses Conerence Wllasburg,
Homework 2 Solutions
Mah 308 Differenial Equaions Fall 2002 & 2. See he las page. Hoework 2 Soluions 3a). Newon s secon law of oion says ha a = F, an we know a =, so we have = F. One par of he force is graviy, g. However,
PHYS 1443 Section 003 Lecture #22
PHYS 443 Section 003 Lecture # Monda, Nov. 4, 003. Siple Bloc-Spring Sste. Energ of the Siple Haronic Oscillator 3. Pendulu Siple Pendulu Phsical Pendulu orsion Pendulu 4. Siple Haronic Motion and Unifor
PHY2053 Summer C 2013 Exam 1 Solutions
PHY053 Sue C 03 E Soluon. The foce G on o G G The onl cobnon h e '/ = doubln.. The peed of lh le 8fulon c 86,8 le 60 n 60n h 4h d 4d fonh.80 fulon/ fonh 3. The dnce eled fo he ene p,, 36 (75n h 45 The
DEEP UNFOLDING FOR MULTICHANNEL SOURCE SEPARATION SUPPLEMENTARY MATERIAL
DEEP UNFOLDING FOR MULTICHANNEL SOURCE SEPARATION SUPPLEMENTARY MATERIAL Sco Wsdom, John Hershey 2, Jonahan Le Roux 2, and Shnj Waanabe 2 Deparmen o Elecrcal Engneerng, Unversy o Washngon, Seale, WA, USA