Bài tập vật lý 10 nâng cao

LE TRONG TUONG (Chu bien) Ll/ONG TAT OAT - LE C H A N HUNG PHAM OiNH THIET - BUI TRONG TUAN i^elp f m^^ fkm m r m m m ng cao NHA XUAT BAN GIAO DUC VIET NAM Lfi TRONG TUONG (Chu bien) LUONG TAT DAT - Lfi C H A N HUNG PHAM DINH THI^T - BUI TRONG T U A N Bai tap VAT U10 Ndng cao (Tdi bdn ldn thit tu) NHA XUAT BAN GIAO DUG VIET NAM Ban quyen thuoc Nha xuat ban due Giao Viet Nam 01-2010/CXB/639-1485/GD Ma s6': NB006T0 Ldi MOI BAIJ= Cac em hoc sinh than Imen Cuon Bai tap Vat10ll nang cao la mgt bo phanhihi cO cOa sach giao khoa Vatli10 nang cao,no se giup cac hoc tot han mon Vat ll Sach duac chia lam phan, mdi hai umg vdi cac cht/ang cCia sach phan g6m cac chuang tuang giao khoaVatli10 nang cao Phan motgom cac bai tapvi du va debai Cac baitapvidu la nhumg baitap.di^n hinh cac chuang nen trinh bay tiet chi Cac em co the tim phuang thay phap giai chung cho cac bai tap chuang Cac em hay co Phan hai la cac hudng dan giai va ldi gang tim each giai cac bai tap, dUng voi doc ph Hay xem lai bai hoc mdi chua ducfctim each giai mot Ch!khi nao thuc su khdng dUdc timeach bai tap nao dd giaicac em mdi xem hudng din hoac ldi giai Cudi mol chuang cd cac bai tap ve thi nghiem chung ta Hay thu gan gui vdl ddi song hangcQa hien cac thi nghiem trinh bay d day Cac so l cOa chinh cac em cung se nhumg la bai tap hay Chuc cac em hoc gioi va cang yeu thich Vatli Cac tac gia DEBAI Phan mot DONG HOC CHAT DI M I - BAI TAP VI DU Ba! M6t cha^t dilm chuyin d6ng’tr6n m6t dvrofag thang D6 thi chuyin d6ngcua no duac ve tr6n Hinh LL Hay m6 ta chuyin ddng chSft dilm cua Tinh van tdc trungbinhva tdc d6 trung binh chft cuadilm c : 0sau s-^ls;0s-^4s;ls-^5s;0s-^5s khoang thoi gian x(cm)M Hinh 1.1 Bdi gidi Trong khoang thod gian tur dintt == s, 0d6s thi chuyin ddng m6t l dudng thang I6n diva lam mdt aj gocv6itrueOt Nhu vay ch^t dilmchuyin ddng thing diutheo chilu duong true cua toa dd, tut vitri cd toa dd bang dinvi tried toa dd bang cm Van ch&tdc dilm cua bang : V = tanttj = = cm/s TiJfluc t = din s t = 2,5 s, dd thi duorng lathang mdt di xudng va la Ot Nhu vay chatdilmchuyin ddngdiutheo ehilu nguoelai, tctc goc a2 vdi true la theo chieu am cuatrue toa dd, tur vi tri x = cm den vi tri : la cua chat dilm -2-4 , , V = tana-= = - cm/s - 1,5 ’ ’ liic t = 4d6 s, thi la mdt dudng nam ngang song Txt luc =t2,5 s den true thdi gian, chat diem diing d vi tri yen cd toa x = - cm Tu: liic t= sdint= s, dd thi la mdt dudng thang di ldn a3 v vdi true Ot Nhu vay chat dilm chuyin ddng thangduofng deu eua theotru c toa tit vi tri X = -2 em den vi tri x = em eh^t Vandilm tde la eua: - (-2)^ , = cm/s V = tana ,= ’ , ’ Van tde trung binh tinh duoc theo edng thiic : Vtb = dd ddi X, khoang thdi gian X9 t2 tj ’ ^ ^ , Tdc dd trung binhtinh dugc theo cdng thtic : T^-’ -»-> u-u quang dudng di duoc Toe tmng binh= S; khoang thdi gian thi = om ; t2 luc = s thi X2 = cm, hay la a) Luctj= s Xj At =t2- tl= s -= 10 s Dd ddi khoang thdi :gian dd la Ax = X2 - X]= - = cm Vay : ^tb= ^ = Y ^ "^ ’^"^Z^Quang dudng dugfc di khoang thdi gian dd la : As = |x2- XJI = - = cm vay : As Tdc dd trung binh = = cm/s b) Liic =tj0 s Xj thi = em ; t2 luc = s X2 thi = - cm At = t - t4i - =0= s Ax ==X2 vay: Vtb Xj= - 2-0 - =-2 cm _ _ ^^ -2 _ -0,5cm/s ~ At ~ Do chuyin ddng khdng theo mdt ehilu eho nen ta tmh dugc quang nhu sau: s tj = s, quang dudng di Asi dugc = |x2 la- x11 = |4 - 0| = Tilf = tl0 din Tuf t’l = s den tj= s, quang dudng diAs2 dugc =|0 la - 4| = em t’{ = Tii s din t’l" = 2,5 s, quang dudng di AS3duge = 1-2la - 0| = cm tj"Ttr - 2,5din s t2 = s, chatdilm diing lai d =dilm - cm, x quang dudng di AS4 dugc - la vay quang dudng di dugc khoang ti thdi 0= sgian den t2=tuf s la : As = Asi+ As2+ AS3 =4 + 2+= 10 cm Tdc dd trung binh khoang thdi gian dd la : Tdc dd trung binh = = =2,5 cm/s At Ta nhanthaygia tri van eua tdc trung binh va tdc dd trung binh tr khoang thdi gian dd la khae c) Tuong tu, khoang thdi t2-gian ti= At - =1 = s, ta ed Ax = X2 - Xi =0 - =4- cm ’ , Ax ^^^ = AT = -4 = "^’^"^/^ As = As2+ AS3 + AS4 + AS5 = + + + = cm As Tdc dd trung binh -: ==-r= cm/s At d) Trong khoang thdi gian At t2- ti = 5= - = s, ta cd : Ax = X2 - Xl = - = cm Vtb = T = c"^/s As = Asi+ As2+ AS3 + AS4 + Asg= + + + + = A 10 Tdc dd trung binh = - = - =2,4 em/s At Bai Mdt xe nho trugt trdn mang nghidng Oign ddm truekhf toa dd Ox tri mang va cd ehilu duong hudng xudng phia dudi Biet rang, gia tdc cua la cm/s ,vaWe xe di ngang qua gdcvan toa tdcdd, cua nd VQ = la- cm/Si Vilt phuong tnnh chuyin ddng l^ygdc cua thdi xe, gian liic xe la di ngan qua gdc toa dd Hdi xe chuyin ddng theo hudng nao, sau bao lau thi xe nam d vi tri nao ? Sau dd xe chuyin ddng nhu thetinh nao van ?tdc Haycua xe sau s luc diing Liic lai dd xe nam d vi tri nao ? Bdi gidi Phuong trinh chuyin evia ddng xe la phuong tnnh chuyin ddng ddi diuvdi van tdc ban VQ =ddu - em/s, gia tdc bang em/s va XQ = em Phuang trinh dd la : X = - 6.t 8.t^ + (1) 2 a) Xe chuyin ddng phia di trdn ldn theo ehilu am cua true Ox va dii van tde bang khdng Ta cd : v = Vo+ at = - + 8.t = (2) Tut ddsuyra: 8.t = , hay t = - = 0,75 s o b) Vi tri cua xe luc : la X = - 0,75 8.(0,75)^ + - = - 2,25 em > a) Sau dat van tde bang thi xe chuyin ddn diu theo ddng chil nha ngugc lai xudng phiadudi (chilu duong eiia true Ox) xe diing b) van tde eua xe dugc tmh theo edng thiieWc(2) Sau 3l tlie la d thdi dilm t = 0,75 + = 3,75 s, bang van tdc cua xe liic dd : v =VQ + at = - 8.3,75 + = 24 cm/s c) Liic dd vi tri ciia : xe la X = - 3,75 ^+.8.(3,75)2 = 33,75 cm 7.23.Nhung chd’t long khdng udtcdc dfto 7.24* Giam di 1,2 lto, ^^^^^ nghia la : =1,2 a(d 80*’C) 7.25.0,075 N/m 7.26 0,0365 N rugu Luc bl etog 7.27* Luc etog bl mat cua nude lto hon luc etog ma mat eua nude thuc hidn edng duong Ai =0,0728.0,04.0,02 =582,4.10"^ J A2 = cdng - 0,0241.0,04 Lue etog bl mat cfia rugu thucam hidn = - 192,8.10""^ J vay edng tdng cdng 582,4.10"’’ la : A = 192,8.1 = 389,6.10"^ J 7.28.816kg/m^ 7.29 0,022 N/m 7.30 Trutog hgp A, mue ehdt long dtog cao 14,7 mm 7.31.Cdu A 7.32 Nudc sdi, vi nhidt hod hod lto hon toilu so vdd to 7.33*.- Cdu A khdng dung vi cd cung nhidt cho thi mdt tfch kh cua nd Midng totog khdng ttog ma cdn giam di Chtog lam ndng chay mdt cue nudc thi da tfchViriing cfia da lto nudchon thi tfch ridng cua nudc ndn cue nude thitfch dd tan euarandthi giam thi (ndngchdt cha - Cdu B khdng dfing vi ta cung nhidt dl khdi chtog ban) thi toidt cfia khdi ehdt dang ndng chay - Cdu C dung vi cung nhiet la su truyin ntog lugng d - Cdu D khdng dfing vi frong ba cdu frdn chi cd mdt c 7.34 Cd tbl, chi cto hfit khf dl giam dp sudt tac dung 7.35.cauB Hudng ddn : Dung Bdng dp sudt vd khdi lugng rieng ct a nhiet khdc (xem bai 56 SGK) va ndi suy tt 25 C-30 C 7.36.6 25 C thi dfige, 35 cdnCthi d khdng, vi toidt 35*’Ccao dd hon nhid tdd ban cua CO2 156 7.37.1 atm ; dp bed sudt bao hoa ehi phu thudc nhidt dd, khdng thitfch phu 7.38* a) b) e) Cd Lam Nen Lam toflng each sau ddy : lato dtog tfch din 284 K dang nhidt dd’n 23,3 lft ddng thdd ca hai each trdn, nghia la vfia nen v 7.39 4,8 tdn kJ/kg 7.40 60 7.41.120,62 kJ 7.42 64,4 kg ; dung phuang trito Cla-pd-rdn -Men-dd-ld= 6,77 p = : ^ va tfto duge p kg/m^ RT -2 7.43.a = 10 g/m o ; A = 12,8 g/m (tra Bdng dp sudt vd khdi r% nudc bdo hod d nhiet khdc = =78% SGK); f 7.44 Tang thdm 23% 7.45.11J[ C Nhiet dd ma d dd d kfnh cfia sd md di la diem dm m^ 7.46 0,4 7.47 64% 7.48 72% va 76% 7.49*.17 C Hu&ng ddn : Hai nhidt kl ehi nhidt dd gidng toau 100% thi Luc dd, Im^ khdng khf dm cd 12,8 g hod 2bnudc C thi lugng h lugfng rieng cu nude bao hoa la : 17,3 g/m (xem Bdng dp sudt vd khd hai nudc hdo 6hod nhiet khdc SGK) Vay dd f = i | 4X 100= 74%.Nhidtk lkhd chi 20 Cma dd dm ti ddi 74% tiii la th 17,3 Bdng tra dd dm ti ddi (xem bai 56 SGK) su chdto Idchtoidt dd gifla nhidt k l3 la C.Vay ddp sd17laC 7.50* Trutog hgp C, dd dm ri dd’i la 92,4% 157 7.51 - Kep hai tdm thuy tinh song song each btog (trudc’dd phai lau sach hai mat ddi didn dl dam bao dft - Nhung toe mdt canh cua hai tdm cham mat nudc cu dang ldn d gifla hai tdm thuy tinh - Do cao cfia phdn nudc dtog ldn va tfto hd sd m^t cua nud 4a pghd , - , = h , suy ara = ^-^ theo cong thuc pgd 7.52 - Nd’i hai d’ng thuy tinh thdng vdd btog d’ng - Dd nudc vaodng,chdctobtog, doddcao himat tfi thotog ddii midng dng to - Dung nut bit kfn ddu dng tod, ntog dng cao ld khf ddudng tod bi giam va bi nen dtog nhi + BChd’i + Mat thoang hai benIdch chdto Ah + Do cao cdt khf bdn dng h2.tod la - Tfto Pa(dp sudt khf quyln) theo phuong ttinh dang nh ^pghjAh ^’ hi-h2 ,^ - Dfing thudc hi, Ah, h2se tfto dugc dp sudt khf quyln 7.53.- Nhung d’ng thing dtog vao chdt ldng - Do dd cao hitfi mat thoang dd’n midng dng kjif ban (cfia ddu) cdt - Gifl nguydn vi tri d’ng, bit ddu trdn cfia d’ng (gi phuong dinkhi ddu dudi - Tfi tfi tod’c dng ldn caothtog theo dtog h2 Cua cao cdt khf bi giam d ddu gdn ldndinmat thodng Do dd d’ng, va h la dd cao cdt chdt ldng ttong dng so vdd ma - Day la qua trito dan ding ed tbltoidt, tfnh duge khdi lugng cdng thfic : - ^ Pa(h2 - hi) ^ ghh2 - Dung thudc dugc cdc h^ h2, gidh.tri 158 CO s d cun NHICTDONG LlfCHOC 8.1.Trutog hgp D 8.2.Vidn thfi todt, vi toan bd ddng nang chuyin sang ndi 8.3.Hdn hgp khf d cudi ki nen Nlu coi cac Iftutog thi khf ndi khf nhu cua khf chi phu thudc toidt dd Nhidt dd cfia cac khf nhidt dd cua cdc khf d cud’i ki thoat, 8.4.800 W 9v’ 8.5*At = ; dd c la toidt dung ridng cua chi 8c 8.6.0,11K 8.7.8,15 K 8.8.357 m/s 8.9* Theo cdch a thi qua trito AiBi la qua trito dtog nhidt Theo each b thi nto nhanh khf ndng ldn (vi toilt toa khdng kip),dutog bilu didn laAiCBi dutog bdn trdn dutog AiBi(Hinh 8.lG) Theo ninlto hon vi dutog bilu each b cdng diln qud trinh ndm ttdn dutog ding toidtAiBi 8.10 Day C 8.11*.Cach A Hudng ddn : Dfing dd thi p - V dl bilu thi cdng cua khf t ndu rdi so sanh cdng d mdi cdch sosato btog cdc cdchdidn tfch na cac dutog bilu didn 159 8.12 Nhidt lugng cung cdp cho khf ttong qud trinh dtog lam ndng khf cdn dung dl khf sinh edng dan 8.13.396 kJ Tfto theo cdng thfic : A’ = thi pAVtfch ; AVttog la phdn thdm khdng khf phdng duge sudd ndng dtog dp Kh khdi phdng qua cac khe hd 8.14*.3,3.10^ J;6,1.10^ J Ifthfi todt ddi vdd qud tr Hudng ddn : Phuong trinh cua nguydn la Q = AU - A Trude hit, ta phai tfto cdng A’ cfia khf dt pVi = vRTi pV2 = VRT2 Trfi vdd trang thai ddu va cudi cua khf, tavavilt hai phfiong trito cho toau, p(V2- Vi) ta =ed vR(T2- Ti) V I phai c phep ta tfto cdng cua khf dto dtog dp vi v l ttdi la = 3324000J vay : A’800.8,31.500 = A’ « 3,3.10 J Cdng khf sinh 8.15* 49 J Hudng ddn Trudehitdp dung dito Sde-lo ludtdd’i vdd hai qua tr = T2T4,ttong dd T la toidt dd cua qua tri tfch, ta timT^dugc Mud’n tfnh cdng ma khf thuc hidn chu trito, ta nhdt i234 Bilu thfie tfto edng cua chu tiito T2+T4 -la 2(T2T4)^^^] A’ = R[ Xem thdm|iutog dto cua bai 8.14 8.16 760 W 8.17 cau C 8.18 24% ; 42% kJ;28,4kJ 8.19.113,5 8.20 84 kJ ; 14,1 kJ 8.21.Cau C:vat bang chi, vi chi ed nhidt dung ridng be to 8.22 Cdu C : vdt bang niken, vi niken cd khdi lugng ridn 8.23*.- Cdu A sai vi ndi nang cua Iftudng hd khf khdng bao thi gdmntog tuong tde giua cdc hat cdu tao ndn he 160 - Cdu B sai vi toidt lugng truyin cho hd cdn cd tbl l cdc hat edu tao ndn hd Vf du ta cdp toidt dl lam cho Lfic dd toidt dd khdng tang ma tbl ndng tang vi cd s bdn bd - Cau C dung Vf du ta nto khf thi ta vfia lam khf n tfch khf - cauD sai vi nlu cho Iftutog khf dan vao chan khdng thi ndi khdng ddi ma tbl tfch eua hd tang ldn 8.24* - Kf hidu binh cd nudc ban ddu la X, bito la Y - Dat X vao nudc ndng, Y vao nudc lanh, nudc se chay khf bi nd ndng va eo lai lato) - Chd dd’n ngtog chay, ddi chd X vao nudc lato, Y tihhd Y ngdn ndn nude khdng chay ngugc ldn dug dng thuy khdng khf tran tfi binh Y sang binh X - Laiddi chd hai binh, nudc lai chay thdm tfi X sang Y - Cd tbl lap lai nhilu lto dl ttog hidu qua MOT SO nni TnpTONG HOP V CO NHICT Ap dung dinh luat bao toto co ntog : mv = mgh dd v la vto tdc cua qua cdu d dilm thdp nhdtB H h = HB = / - /cosa V -* M I B Tfidd:v^= g / ( l - c o s a ) (1) pt Tai vi tri thdp todt cfia qua IG):edu (Hito Hinh I G T - P = Fh^ f v^ T = mg + - (2) Thay (1) vao (2) ta dugc : T = mg(3 - 2cosa) = 2,49 N Do lto cfia T khdng phu thudc gia tri cfia chilu dai Vi m^ = mg va lue ma sdt khdng dang kl, ndn chdc chto keo vdt A di ldn Ap dung dinh ludt II vdt: Niu-ton cho md Vdt B : mg - T = ma Vdt A : T - mgsina = ma Giai he ta dugc : a= -g(l-sina) (1) T = mg(l + sina) (2) lite etog Vdy cdng sudt tfic thdd cua T tai thdd dilm t la : g’=Tv= Tat Thay (1) va (2) vao (3) ta dugc : g’ = -^mg^tcos^a « W 162 ’ (3) a) Giai btog phuong phap ddng luc hgc Dua vao phep phto tfch luc va dp dung dito ludt II gia tdc cua vat lfic di ldn (ehilu duong ldy theo chi = -g(sina + pjcosa) Qutog dutog vdt di ldn : s= 0-v? ^ ’ 2ai v2 V? (1) 2g(sina + pj cosa) Gia tde cua vat luc di : L xudng a2 = g(sina - Ptcosa) Quangdutog vat di xud’ng : v?-0 vi S = -4 2a2 (2) : 2g(sina - PfCosa) Giai hd phuang trinh (1), (2) pjvavdd s,cac ta dugc to : _ Pt = ^ ^tana = 0,28 ^ (3) vf+v^ 2 vf +V^ 4gsma 2 vf + VT Do dd h= s.sina = -^ «0,64 m (4) 4g b) Giai bang phuong phdp ntog lugng Chgn md’c tfto tbl nang d A Khi vat di ldn, cdng cua luc ma sdt btog dd bid’n thid ^ms = Wg- W^, hay -sptmgeosa : = mgh mv^ ^^ (5) Khi vat di xudng, edng cua luc ma sat cung btog dd bi Ams = WA - WB ,hay -sptingeosa : = - ^ - mgh (6) Giai hd (5), (6), ta cung kit dat qua dugc toucdc (3), (4) 163 4* Vdt chiu tac dung cua frgng lue P va tuylh phto N cua luc ban phdpcd P + N = ma Khi vdt rdi bl mat ban cdu thiddN = ,do P = ma Chid’u xud’ng phuong cfia kuihBO ban : mgcosa mv Hinh ehilu cfia a trdn phfiOng ban kfnh la gia tdc hutog tdm Dp dd : V = grcosa = gh (1) Ap dung dinh ludt bao toto eo ntog Hinh 2G mgr = mgh mv + (2) Thay (1) vao (2), ta thu duge : h r Luc ddu cac khf bong bdng xa phdng cd nhidt dd C)va lue (hod thd cfia ngudd cd37nhidt dd luc ddy Ac-si khdng khf lto hon frgng lugng cfia bong bdng xa ph bdng bay ldn Sau dd, bong bdng xa phdng giam nhidt dd toa nhidt va thu tod tbl tfch bong bdng lai ndn luc ddy Ae-si trgng lugng cfia bong bdng xa phdng Din thimdt khdng lficddi nao d frgng lugng cfia bong bdng xa phdng lto hon luc ddyla Ac-si-mit, kitqua vdn tdc di ldn eua bong bdng giam dto rdi tfi tfi rod O frang thdi khdng trgng lugng, khdi thuy ngto (khdn CO lai thato dang hinh cdu ; cdn khdi nudc (dfto u frdn toto bd mat frong ehfia eua binh Mdi giay blp didn cung cdp cho nudc ttong dm mdt nhid Q = 0,8.1000 = 800J Nhidt lugng lam hod hod mdt lugng nudc btog : 800 m = - = « 0,354 g L 2,26.10^ 164 I Thi tfch eua lugng bednudc dugc tao ra1 gidy ttongla : ^8,31.373 , - = ^^^^^o< 0,000595 m »3^nnn^ 0,000 m3 V = mRT = 0,354 1^ P 18 1,013.10^ Tdc dd hod phut raid : V 0,000595 , ^ = ^= -o;ooor^’^^"^/^ = Hod nudc sdi thoat ngoai khdng qua trito tfto cdng cua hod nudc Ig (tao chid’m tfithitfch V theo cdng th nudc) A = pAV = pV = - R T ^ = : ^ ^ - 172,2 J p 18 8* F^ ldn khf cdu Luc bing trgng Ta tfnh luc ddy Ac-si-met thitfch khdng khf dm bi chid’m chd Bdy gid ta tfto khdi lugng ridng cua khdng khf dm p’ -3 ridng cua khdng khf khd p cdng bedvdi nudclugng ed ttong m khdngkhi Theo cdng thfic f = -^ = ^’^’ "^^y ^ = ^’^ ^ Theo Bdng dp sudt vdlugfng khdi rieng cua hai nudc bdo hod d khdc (xem SGK, bai 56) thi A = 17,3 g/m vay : a = 0,8.17,3 gim = 13,84 Bay gid ta tfnh khd’i lugng ridng eua khdng khf khd Cla-pd-rdn - Men-dd-ld-ep, ta rfit : m P=V pp 10^29 , , =1^=831293=1’’’^/"^ Tfi dd p’ = p 1191 + a += 13,84 1=204,84 givn FA = (p’.64)g 77,110.9,8 = = 755,678 N « 216,6 N Lfic ntog khf cdu (77,110 la : F- =55).9,8 Dd dto cua ddy thep dugc tfnh theo edng thfie cfia din F i.^^ A/ / I F cn 216,6 ^ ^ ^ ^ ^ _ = E-r-, suy = /g: A/= 50 z 0,025 « m S /g "ES 2,1.10" 2.10-^ vay A/«2,5 cm Ggimi va m2 la khdi lugng cue nudc dd va lugng hod nu toidt lugng kl Ta cd mi : + m2 = 500 g Suy : m2 = 500 mi 165 Ta thilt ldp phuong trito c t o bang toidt lugng : - Nhidt lugng ma cue nude da va nude ttong nhidt lugng Qi = mic’iLO - (-5)]miX + + m^ c’l (25 - 0) +C2300 [25 -(-5)] - Nhidt lugng ma lugng hoi nudc toa la : Q2 = (500 - m i ) L+ (500 - m ^ c ’ i ’ ( 1-0205 ) Vi cac dd l t oQieua va Q2 phai btog toau, ta cd : 5mici + m^X + 25miei+ 30.300e2 (500 = mi)L + 75(500 mi)ci dd c\va Cj la nhidt dung ridng cua nudc da 02 valanude, nhidtc dung ridng cua todm Thay cdc gia tri da bilt vao phuong trito frdn va giai Khdi lugng cue nude mi =dd423,1 g Khd’i lugng hod m2 nude = 500 - 423,1 = 76,9 g 10 Nguydn tdc sudd dm ddng lue duge bilu thi d so dd dudd A^ H T - Tl , T-Ti I ->A’ = Qi h Qi Q^ = Qi - A’ Qi = A Q2=A vTi T1-T2 ^ Opng CO T - T i ^_ Q i | /" T nhift H# Mng sudi , suy : M&y lenh (T-Ti)T2 = Q T(Ti - T2) ’2 J T-T, Ta vilt dugc tou trdn lAI= la IA’Ivi = Qi = - ^ ( v dTi > Ti) Hinh 3G He thd’ng sudi toto toidt lugng Q2 + Qi: Q = O ’ - A + O - Q ^-^^ Q1-A + Q2-Q1 J ^^-^i^^2_Q Ti(T-Ti) +^iT(Ti-T2)~^iT(Ti-T2) Tdm lai: Q = Q^ + Q’l= Qi^ [ j ^ ^ Ap d , n g s d : Q=i 6000kcal thi Q= Q i i | [ | | ^ ) Q « 2,98.6000 « 18000 kcal Nhu vay sudi dm ddng luc lgi ban toilu 166 (gdp g t o ba ldn MUC LUC Trang Ldd ndi ddu PH^N MOT : DE BAI ChuangI Ddng hge ehdt dilm Chuang II Ddng luc hgc chdt dilm -ChuangIII Titohge vat rdn Chuang IVCac dinh luat bao toan Co hge ehdt luu ChuangV Chuang VI Chd’t khf Chuang VII Chdt rdn va chdt long Su chuyin tbl ’ChuangVIIICo sd cfia nhidt ddng luc hgc Mdt sdbdi tap tdng hgfpve ca vd nhiet PHXN HAI :Hir6NG D A N GIAI VA DAP SO ChuangI Ddng hge chdt dilm Chuang II Ddng lue hge ehdt dilm Chuang III Tinh hge vdt rdn Chuang IV Cae dinh luat bao toan Co hge cha’t luu ChuangV Chuang VI Chdt khf long tbl Chuang VII Chat rto va cha’t Sirchuyin Chuang VIIICo sd cua nhidt ddng luc hge Mdt sdbdi tap tdng hgp vd venhiet ca 18 31 40 59 65 71 82 88 90 90 106 122 129 146 150 154 159 162 167 Chiu trdch nhiem xudt : Chu bdn tich HDQT kiem Tdng Giim d6cNGO TRAN AI d6ckiemT6ngbien tap N G U Y I I N QUt THAO Pho T6ng Giam Bien tap ldn ddu :NGUYfeN TIEN BINH - VU THANH MAI Bien tap bdn: tdi D THI BICH L I £ N Bien tap thudt: Id CAO LAN PHlTONG - DINH XUAN DUNG Trinh bdy bia : TA THANH TtJNG in : D THI BICH LifiN Sica bdn Che bdn: CONG TY CPTHI^T K£ VA PHAT HANH SACH GIAO DUC BAI TAP VAT Ll 10 NANG CAO Ma s d : NB006T0 In 10.000 ban, (Qe44BT) khd 17 x 24 cm In tai Cong ty.co phan in Son La ban Sd in:10SL sd xuat 01-2010/CXB/639-1485/GD In xong va nop luu chilu thang nam 2010; 168 Xl-B QUALITY CROWN HUAN VaONG MIEN KIM CUONG C H A T LI/0NGQUOC TE CHl/CJNG HO CHI MINH SACH BAI TAP LCfP 10 BAI TAP OAI SO 10 B A I T A P T I N H O C I O BAI TAP HiNH HOC10 B A I T A P T I ^ N G A N H I O B A I T A P V A T L I I O BAI TAP TI^NG PHAP 10 BAI TAP HOA HOC 10 BAI TAP TI^NG NGA 10 BAI TAP NGffVAN 10 (tapmot,taphai) SACH BAI T A P L O P 10 - NANG CAO 10 BAI TAP OAI S BAI TAP HINH HOC BAI TAP 10 BAI TAP VAT LI 10 BAI TAP HOA HOC 10 NGUrVAN 10 (tap mot,tap hai) BAI TAP TI^NG ANH 10 Ban dpc co the mua sach tai: Cac Cong ty Sach - tnroTig Thiethoc bia cac dia phuofng Cong ty CP Diu tu triSn va Phat GiaodueHa Noi, 187BGiang Vo, TR Ha Noi CP.Dautu va Phat trien Giao due Phuomg 231Nguyin Nam, Van Cu, Quan 5, TR Cong ty GiaodueDa Nlng, 15NguySn Chi Thanh, TP Da Nlng Cong ty CP Dau tu tri8n va Phat hoac cac cua hang sach cua Nha xuat ban Giao Nam : due Viet 187 Giang Vo ; 232 Tay Son ; Tign 23 ;Trang 25 Han Thuyen ; 32E Kim Ma ; 14/3 Nguyin Khan Tai TR Da Nang : 78 Pasteur ; 247 Hai Phong - Tai TRH6 Chi Minh 104 Mai Thi Luu 2A ;Dinh Tien Hoang, Quan ; 240 TrknBinh Trong231 ; Nguyen Van Cu, Quan ; Binh Thai, Quan 11 - Tai TR Can Tho 5/5 Duong 30/4 - Tai TR Ha Noi : Website:www.nxbgd.com.vn 9 " 00 9 Gid: 7.600 [...]... ddy cao su - Ldn lugt treo thdm eac qua nang 1, 2, 3 8 lam dd ddy cao su6 co lai - Lam ngugc lai, bdt ddn cae qua 1nang 8, 7, Sau khi ldy sd lieu nhilu ldn trong qud trinh dan ra vd eo lai cua Ff ddy, ban da ve dugc dd thi F(x) (Hinh 2.22), trong dd dudng 1 fing vdi qua trinh ddy cao su dan ra, dudng 2 fing vdi qua trinh ddy cao su eo lai Hay phdn tfch dd thi dl rut ra tfnh chdt dan hdi cua ddy cao. .. ddng dge theo tr toa dd Bilt gia tdc ciia nd khdng ddi la 8 cm/s nd dgd’c a) Vi tri ciia nd sau 2 s b) Vdn tdc ciia nd sau 3 s 11 1.16 Mdt electron ed vdn tdc3 .10^ ban m/s ddu Ndu la nd chiu mdt gia 8 .10^ ’^m/s2thi: a) Sau bao ldu nd dat duge 5,4 .10^ vdnm/s tdc? b) Quang dudng nd di dugc la bao nhidu trong khoang th 1.17 Mdt may bay phan luc khi ha canh cd vdn tdc tilp d datmay dugc dl giam td’e dd, gia... nhidu ? 1.19 Mdt ngudi nem mdt qua bdng tuf mat ddt ldn cao theo van tdc 4 m/s a) Hdi khoang thdi gian giiJa hai thdi dilm ma cua’qua van tdc bdngcdcung dd ldn bang 2,5 m/s la ?bao nhidu b) Dd cao liic dd bang bao nhidu ? 1.20* Mdt vat rod tu do, trong giay cudi cung rod duge 3 liic bat ddu reddd’n luc cham ddt 1.21 Ngudi ta tha mdt hdncita das6 d dd cao 8vdd turmdt m mat so ddt (vdn mdt hdn bi thep rod... 0,3 10, 2 30 0,4 15,8 25 0,5 23,0 20 0,6 31,4 15 0,7 40,8 0,8 5,6 45 40 35 * 10 M 5 51,5 0,1 0,2 0,3 0,4 0,5 0,6 t(s) 0,7 0,8 Hinh 1.4 1.44 Mdt ban da lam thf X (cm) nghidm chuyin ddng cua bgt khf trong d’ng thing va ghi duge sd lidu rdi ve dd thi nhu Hinh 1.5 Mdi dudng thing trdn dd thi ung vdd mdt gdc nghidng ciia dng Hay phdn tfch va xdc dinh tinhchdt cua chuyin ddng cua bot khf Hinh 1.5 2-BTVL 10( NC)-A... liic dugc vat di duge theo phuang nam ngang ketha tiicho tdi khi ch b) Khi h1000 = m, hay tfnh VQ dl /1500 = m Bd qua anh hudng cua khdng khf 2.25*.Titmdt dilm d dd cao h = 18 m so vddVomat ddt va each tudng nha mdt/=khoang 3 m, ngudi ta nem mdt hdn sdi theo phuang nam ngang vdi vdn td’e ban ddu Vg Trdn tudng ed mdt ciia sd ehilu cao a = 1 m, mep dudd eua cita each mat ddt mdt khoang b = 2 m (Hinh 2.7)... eua dinh ngudi edng idn ghingdi tai dilm cao nhdt va dilm thdp nhdt eu cao nhdt, ddu eua ngudi phi cdng hudng ddt, ghidxudng bdn trdn) 2.38.Mdt vdt duge dat d mep mdt chile ban xoay Hdi sd vd ban bang bao nhidu thi vat se vang ra khdi banhinh ban ?trdn Cho ban kfnh r = 0,4 m, hd sd ma sat nghi bdng m/s^.0,4 vd g = 26 2.39.Trongthilt bi b Hinh binh 2.13, Mnh = 10 rem Ta dl mdt tru cd bdn kfnh binh cfia... khoang nd thdi tronggian dd bang bao nhie b) Ta edthi tinh dugcvantd’e trung binh ciia nd trong khoang dugc khdng ? Giai thich nhd eae sd tr6n lidu 1 .10 Mdt electron chuyin ddng trong dng den hinh eua mdt may tang td’e diudan tur vantdc3.10m/sdinvan tdc 5 .10 m/s trtn md tinh dudng thang bang 2 cm Hay: a) Gia tdc dlectron cua trong chuyin ddng dd Electron dihitquang dudng dd b) Thdi gian ~1.11.Mdt d td... VQ cd hudng nhuhinh ve Hay chgn cau Hinh 2.3 md ta dung vl chuyin ddng 1 :eiia vat diuldn A vat 1 chuyin ddng chdm ddntrdn, dd’n mdt dd cao nhd xud’ng dudd chuyin ddng nhanhdiu ddn B Vdt1 chuyin ddng diuldn phfa trdn C Vdt 1 chuyin ddng cham ddn trdn diuldn dinmdt dd cao nhdt din dumg lai D Cd thixay ra mdt trong 3 kha nang ndi trdn,ldn tuy cuaVg thudc v Ggi y: Khi lam loai bai trac nghidm, cdn vdn dung... tde cua vdt cudi tai th 2 .10 Mdt lue F truyin eho vdt mi ed mdt khdigia lugng tdc bang 8 m/s cho mdt vdt khac ed khdi m2 mdt lugng gia tde bang Nlu 4 m/s dem ghep hai vdt dd lai thanh mdt vdt thi lue dd truyin cho v bao nhidu ? 2.11.Mdt vdt cd khdi lugng 3 kg dang chuyin diuvdi ddng vdnthing VQ tdc = 2 m/ thi chiu tde dung eua mdt luc 9 N VQ ciing Hdi chilu vdt se vdi chuyin 10 m tilp theo trong thdi... 2.15.Hai ngudi keo mdt sgi ddyf Hinh 2.6 theo hai hudng ngugc nhau, mdi ngudi keo mdt lue 50 N Hdi sgi diit ddy cdkhdng nlu nd chi chiu duge lu hay la80N? 22 2.16 Tfnh gia tde rod tu do d dd cao 5 km nita va band kfnh dd cao Trai bang D Cho gia tdc rod tu do d mat ddtm/s^, la ban g = kfnh 9,80 Trdi Ddt R = 6400km 2.17 Hay tra cliu cdc bang dii lidu d phu luc 2 cua SGK d ddn giiia Mat Trdi va Trdi Ddt 2.18 ... NB006T0 Ldi MOI BAIJ= Cac em hoc sinh than Imen Cuon Bai tap Vat10ll nang cao la mgt bo phanhihi cO cOa sach giao khoa Vatli10 nang cao, no se giup cac hoc tot han mon Vat ll Sach duac chia lam phan,... 1,67 .10 kg chuyin ddng vdi van Vp = 1 .10 m/s tdd va cham vao hat nhan heh (thudng gg ndm ydn Sau va cham, prdtdn Ifii vddgiat van tdc v’p= 6 .10 m/s cdn hat bay vlphfa trudcvan vdd tdcVa =4 .10 m/s... dd cao m s, Trong dd, mdt thang 000 N ldn may dfia mdt khd’i lugng nang dd cao 10 m s.Hay so sanh cdng, cdng sudt cua ngudi va may hidn da thuc 48 4.17 Ngudi ta mudn ndng mdt hdm 200 kg ldn cao