Izibonelo ezingu-24 zemibuzo ye-Fluid Dynamics
Ithiyori kaTorriceli
1. Ibhavu elikhulu eligcwele amanzi kanye nepayipi njengoba kuboniswe esithombeni. Uma u-g = 10 ms -2 , khona-ke ijubane lokufafaza kwamanzi avela epayipini...
A. 3 ms-1 
B. 8 ms -1
C. 9 ms -1
D. 30 ms -1
E. 900 ms -1
Ingxoxo
Kuyaziwa ukuthi :
Ukuphakama (h) = 85 cm – 40 cm = 45 cm = 0,45 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Umbuzo : Isivinini sokufafaza kwamanzi kusuka empompini (v)
Impendulo :
Ithiyori kaTorricelli ithi ijubane lokufafaza kwamanzi emgodini osebangeni elingu-h ukusuka ebusweni bamanzi lifana nejubane lamanzi awela ngokukhululeka esuka ekuphakameni okungu-h.
Ijubane lesifutho samanzi libalwa kusetshenziswa ifomula yokunyakaza kokuwa kwamahhala v t 2 = 2 gh
v t 2 = 2 gh = 2(10)(0,45) = 9
v t = √9 = 3 m/s
Impendulo efanele ngu-A.
2. Ithangi lamanzi liqukethe imbobo odongeni (bheka isithombe). Isivinini samanzi njengoba ephuma emgodini… (g = 10 ms -2 )
A. 12 ms-1
B. 10 ms -1
C. 6 ms -1
D. 5 ms -1
E. 2 ms -1
Ingxoxo
Kuyaziwa ukuthi :
Ukuphakama (h) = 1,5 m – 0,25 m = 1,25 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Umbuzo : Isivinini samanzi uma ephuma emgodini (v)
Impendulo :
v t 2 = 2 gh = 2(10)(1,25) = 25
v t = √25 = 5 m/s
Impendulo efanele ngu-D.
3. Idamu linamanzi angu-1 meter ukuphakama (g = 10 ms -2 ) futhi kukhona imbobo evuzayo odongeni (bheka isithombe). Isivinini samanzi aphuma emgodini...
A. 1 ms-1
B. 2 ms -1
C. 4 ms -1
D. 8 ms -1
E. 10 ms -1
Ingxoxo
Kuyaziwa ukuthi :
Ukuphakama (h) = 1 m – 0,20 m = 0,8 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Umbuzo : Isivinini samanzi uma ephuma emgodini (v)
Impendulo :
v t 2 = 2 gh = 2(10)(0,8) = 16
v t = √16 = 4 m/s
Impendulo efanele ngu-D.
4. Isilinda sigcwele uketshezi olufanele. Umgodi omncane (oncane kakhulu kunomthamo wesilinda) utholakala odongeni lwawo, ngamasentimitha angama-20 ukusuka phezulu, okuvumela uketshezi ukuthi luphume (njengoba kuboniswe esithombeni). Liyini ijubane lejethi yamanzi evela emgodini omncane?
A. 1,0 ms-1 
B. 2,0 ms –1
C. 3,0 ms –1
D. 5,0 ms –1
E. 5,5 ms –1
Ingxoxo
Kuyaziwa ukuthi:
Ibanga lomgodi ukusuka ebusweni (h) = 20 cm = 0,2 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Umbuzo: Ijubane lamanzi elivela emgodini omncane lihamba ngokushesha kangakanani?
Impendulo:
Ithiyori kaTorriceli ithi ijubane lokuphuma kwamanzi emgodini lifana nejubane lamanzi ehla ngokukhululeka esuka endaweni ephakeme h. Ngokwezibalo:
Impendulo efanele ingu-B.
5. Ithangi lamanzi linengxenye enkulu evundlile, enomgodi omncane ku-A. Isivinini samanzi aphuma emgodini A...
A. ngokulingana ngqo no-h
I-B. ilingana ngqo ne-h 1
I-C. ilingana ngqo ne-√h
I-D. ilingana ngqo ne- h2
I-E. ilingana ngqo ne-(h 1 – h 2 )
Ingxoxo
Isibalo sikaBernoulli:
P 1 + 1/2 ρ v 1 2 + ρ gh 1 = P 2 + 1/2 ρ v 2 2 + ρ gh 2
Incazelo: P 1 = ingcindezi 1, v 1 = isivinini 1, h 1 = ukuphakama 1, ρ = ukuminyana, g = ukusheshisa kwamandla adonsela phansi, P 2 = ingcindezi 2, v 2 = isivinini 2, h 2 = ukuphakama 2
Lesi sibalo singasetshenziswa ukuthola isibalo sokubala ijubane lomgodi ophuma emanzini u-A. Uma ingxenye ephezulu yethangi ivulekile, khona-ke u-P 1 = ingcindezi yomoya ngaphezu kobuso bamanzi. Uma kukhona imbobo encane endaweni A, khona-ke u-P 2 = ingcindezi yomoya ngaphandle kwembobo encane. Ingcindezi yomoya noma kuphi eduze kwethangi kumele ifane, ngakho-ke u-P 1 = P 2 . Ngakho-ke, u-P 1 no-P 2 bangasuswa kusibalo sikaBernoulli.
1/2 ρ v 1 2 + ρ gh 1 = 1/2 ρ v 2 2 + ρ gh 2
Imbobo ku-A incane ngakho amanzi aphuma ngembobo anejubane elikhulu, kanti indawo engaphezulu yebhavu inkulu ngakho ijubane lokwehla kwezinga lamanzi lincane. Ngakho-ke inani lika-v 2 lisondela ku-zero ukuze i-1/2 ρ v 2 2 isuswe kulesi sibalo.
Incazelo: v 1 = isivinini samanzi aphuma emgodini A, g = ukusheshisa ngenxa yamandla adonsela phansi = 9,8 m/s 2 , h = ibanga phakathi komgodi A kanye nobuso bamanzi.
Impendulo efanele ngu-C.
6. Ithubhu liqukethe uketshezi olufanele. Odongeni kunezimbobo ezimbili ezincane (ezincane kakhulu kunengxenye evundlile yethubhu) ukuze uketshezi luphume (njengoba kuboniswe esithombeni). Isilinganiso esiphakathi kuka-x 1 no-x 2 …..
A. 2: 3
B. 3: 5
C. 2: 5
D. 4: 5
Isahluko 3: 4
Ingxoxo
Okokuqala, bala ijubane lamanzi aphuma epayipini. Ithiyori kaTorriceli ithi ijubane lamanzi aphuma epayipini lifana nejubane lamanzi lapho ewa ngokukhululeka esuka endaweni ephakeme lapho u-h 1 = 20 cm = 0,2 amamitha kanye no-h 2 = 50 cm = 0,5 amamitha.
Impendulo efanele ngu-D.
Isibalo Sokuqhubeka
7. Amanzi ageleza esuka epayipini A aye epayipini B futhi aqhubekela epayipini C. Isilinganiso sendawo enqamulayo ka-A nendawo enqamulayo ka-C singu-8:3. Uma izinga lokugeleza epayipini A lilingana no-v, khona-ke izinga lokugeleza epayipini C...
A. 3/8 v
B. V
C. 8/3 v
D. 3 v
E. 8 v
Ingxoxo
Kuyaziwa ukuthi:
Indawo enqamulayo A (A A ) = 8
Indawo enqamulayo C (A C ) = 3
Izinga lokugeleza epayipini A (v A ) = v
Umbuzo: Izinga lokugeleza epayipini C (v C )
Impendulo:
Ifomula yesibalo sokuqhubeka:
A A v A = A C v C
8 v = 3 v C
v C = 8/3 v
Impendulo efanele ngu-C.
8. Ipayipi elinobubanzi obungu-12 cm lincipha kolunye uhlangothi libe ububanzi obungu-8 cm. Uma ijubane lokugeleza engxenyeni enkulu yobubanzi bepayipi lingu-10 cm/s, khona-ke ijubane lokugeleza ekugcineni okuncane lingu…..
A. 22,5 cm/s
B. 4,4 cm/s
C. 2,25 cm/s
D. 0,44 cm/s
E. 0,225 cm/s
Ingxoxo
Kuyaziwa ukuthi:
Ububanzi 1 (d 1 ) = 12 cm, irediyasi 1 (r 1 ) = 6 cm
Ububanzi 2 (d 2 ) = 8 cm, irediyasi 2 (r 2 ) = 4 cm
Izinga lokugeleza 1 (v 1 ) = 10 cm/s
Okufunwayo: izinga lokugeleza 2 (v 2 )
Impendulo:
Indawo ephambanayo 1 (A 1 ) = π r 2 = π 6 2 = 36π cm 2
Indawo ephambanayo 2 (A 2 ) = π r 2 = π 4 2 = 16π cm 2
Isibalo sokuqhubeka koketshezi:
A 1 v 1 = A 2 v 2
(36π)(10) = (16π) v 2
(36)(10) = (16) v 2
360 = (16) v 2
v 2 = 360/16
v 2 = 22,5 cm/s
Impendulo efanele ngu-A.
9. Uketshezi lugeleza ngepayipi njengoba kuboniswe esithombeni esilandelayo. Uma indawo enqamulayo A1 = 8 cm2 , A2 = 2 cm2 kanye nesivinini soketshezi v2 = 2 m/s khona-ke ubukhulu be-v1 bungu …
A. 0,5 ms-1
B. 1,0 ms -1
C. 1,5 ms -1
D. 2,0 ms -1
E. 2,5 ms -1
Ingxoxo
Kuyaziwa ukuthi :
Indawo enqamulayo 1 (A 1 ) = 8 cm 2
Indawo enqamulayo 2 (A 2 ) = 2 cm 2
Isivinini soketshezi esigabeni sesi-2 (v 2 ) = 2 m/s
Okufunwayo : isivinini soketshezi esigabeni 1 (v 1 )
Impendulo :
Isibalo sokuqhubeka koketshezi :
A 1 v 1 = A 2 v 2
8 v 1 = (2)(2)
8 v 1 = 4
v 1 = 4 / 8 = 0,5 m/s
Impendulo efanele ngu-A.
10. Bheka isithombe! Uma ububanzi besigaba esikhulu buphindwe kabili kunobubanzi besigaba esincane, ijubane lokugeleza koketshezi epayipini elincane...
A. 01 ms-1
B. 04 ms −1
C. 08 ms −1
D. 16 ms −1
E. 20 ms −1
Ingxoxo
Kuyaziwa ukuthi :
Ububanzi obukhulu obunqamulayo (d 1 ) = 2
Irediyasi yesigaba esikhulu esiphambeneyo (r 1 ) = ½ d 1 = ½ (2) = 1
Indawo enkulu ephambanayo ( A 1 ) = π r 1 2 = π (1) 2 = π (1) = π
Ububanzi obuncane obunqamulayo (d 2 ) = 1
Irediyasi encane enqamula isigaba (r 2 ) = ½ d 2 = ½ (1) = ½
Indawo encane ephambanayo ( A 2 ) = π r 2 2 = π (1/2) 2 = π (1/4) = ¼ π
Ijubane lokugeleza koketshezi esigabeni esikhulu ( v 1 ) = 4 m/s
Umbuzo : ijubane lokugeleza koketshezi engxenyeni encane ( v 2 )
Impendulo :
Isibalo sokuqhubeka koketshezi :
A 1 v 1 = A 2 v 2
π 4 = ¼ π (v 2 )
4 = ¼ (v 2 )
v 2 = 8 m/s
Impendulo efanele ngu-C.
Isimiso kanye neSilinganiso sikaBernoulli
11. Isitatimende esingezansi sihlobene namandla okuphakamisa endizeni, okungukuthi...
A. Umfutho womoya ongaphezu kwephiko mkhulu kunomfutho womoya ongaphansi kwephiko.
B. Umfutho womoya ngaphansi kwamaphiko awuthinti ukuphakama kwendiza.
C. Ijubane lokugeleza komoya ngaphezu kwephiko likhulu kunejubane lokugeleza komoya ngaphansi kwephiko.
D. Isivinini sokugeleza komoya ngaphezu kwephiko sincane kunesivinini sokugeleza komoya ngaphansi kwephiko.
E. Isivinini sokugeleza komoya asithinti ukuphakama kwendiza.
Ingxoxo
Ifomula yokucindezela: P = F / A, lapho P = ingcindezi, F = amandla, A = indawo engaphezulu.
Ngokusekelwe kule fomula, ingcindezi ilingana ngqo namandla. Ngakho-ke uma ingcindezi yomoya inkulu, amandla omoya asunduzayo nawo azoba makhulu. Amandla yiwo abangela ukuba izinto zinyakaze, hhayi ingcindezi.
Isimiso sikaBernoulli sithi uma isivinini somoya siphezulu khona-ke ingcindezi iphansi kanti uma isivinini somoya siphansi khona-ke ingcindezi iphezulu.
Ukuze indiza iphakamise, ukusunduza komoya ngaphansi kwephiko kumele kube kukhulu. Ukusunduza kukhulu uma ingcindezi yomoya ngaphansi kwephiko inkulu. Ukucindezela komoya ngaphansi kwephiko kukhulu uma ijubane lomoya ngaphansi kwephiko liphansi.
Impendulo efanele ngu-C.
12. Naka izitatimende ezilandelayo!
(1) amandla okuphakamisa endizeni
(2) iphampu ye-hydraulic
(3) isifutho somiyane
(4) amabhaluni omoya oshisayo angasuka
Isitatimende esingenhla esisekelwe emthethweni kaBernoulli yilesi...
A. (1) kanye no-(2)
B. (1) kanye no-(3)
C. (2) kanye no-(4)
D. (1), (2), kanye (4)
E. (1), (3), kanye no-(4)
Ingxoxo
Izitatimende ezinezimiso zokusebenza ezisekelwe emthethweni kaBernoulli yilezi: (1) ukuphakamisa izindiza kanye (3) izifutho zomiyane. (2) Amaphampu e-hydraulic asebenza ngesimiso somthetho kaPascal. (4) Amabhaluni omoya oshisayo angahamba ngesimiso somthetho ka-Archimedes (ukuhamba).
Impendulo efanele ingu-B.
13. Esithombeni, amanzi ampontshwa nge-compressor enomfutho ongu-120 kPa angene epayipini elingezansi (1) bese egeleza phezulu ngesivinini esingu-1 ms -1 (g = 10 ms -2 kanti ubuningi bamanzi bungu-1000 kg.m -3 ). Umfutho wamanzi epayipini eliphezulu (II)….
A. 52,5 kPa
B. 67,5 kPa
C. 80,0 kPa
D. 92,5 kPa
E. 107,5 kPa
Ingxoxo
Kuyaziwa ukuthi:
Irediyasi yepayipi elikhulu (r 1 ) = 12 cm
Irediyasi yepayipi elincane (r 2 ) = 6 cm
Umfutho wamanzi epayipini elikhulu (ikhasi 1 ) = 120 kPa = 120.000 Pascal
Ijubane lamanzi epayipini elikhulu (v 1 ) = 1 ms -1
Ukuphakama kwepayipi elikhulu (h 1 ) = 0 m
Ukuphakama kwepayipi elincane (h 2 ) = 2 m
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms -2
Ubuningi bamanzi = 1000 kg.m -3
Umbuzo: Umfutho wamanzi epayipini 2 (ikhasi 2 )
Impendulo:
Ijubane lamanzi epayipini 2 libalwa kusetshenziswa i-Continuity Equation:
A 1 v 1 = A 2 v 2
(π r 1 2 )(v 1 ) = (π r 2 2 )(v 2 )
(r 1 2 )(v 1 ) = (r 2 2 )(v 2 )
(r 1 2 )(v 1 ) = (r 2 2 )(v 2 )
(12 2 )(1 m/s) = (6 2 )(v 2 )
144 = 36 v 2
v 2 = 144 / 36
v 2 = 4 m/s
Ingcindezi yamanzi epayipini 2 ibalwa kusetshenziswa i-equation kaBernoulli:
14. Isitatimende esingezansi esiphathelene namandla okuphakamisa endizeni silungile...
A. Umfutho womoya ongaphezu kwephiko mkhulu kunomfutho womoya ongaphansi kwephiko.
B. Umfutho womoya ngaphansi kwamaphiko awuthinti ukuphakama kwendiza.
C. Ijubane lokugeleza komoya ngaphezu kwephiko likhulu kunejubane lokugeleza komoya ngaphansi kwephiko.
D. Isivinini sokugeleza komoya ngaphezu kwephiko sincane kunesivinini sokugeleza komoya ngaphansi kwephiko.
E. Isivinini sokugeleza komoya asithinti ukuphakama kwendiza.
Ingxoxo
Qhathanisa ingxoxo yombuzo odlule.
Impendulo efanele ngu-C.
15. Amaphiko endiza aklanyelwe ukuhlinzeka ngokukhushulwa okuphezulu kakhulu, njengoba kuboniswe esithombeni. Uma u-v kuyijubane lokugeleza komoya kanti u-P uyingcindezi yomoya, khona-ke ngokwesimiso sikaBernoulli, umklamo wenziwe ngendlela yokuthi...
A. vA > vB ngakho-ke u-PA > PB
B. v A > v B ukuze P A < P B
C. v A < v B ukuze P A < P B
D. v A < v B ukuze P A > P B
E. v A > v B ukuze P A = P B
Ingxoxo
Ukuze iphiko lendiza liphakanyiswe phezulu, ukusunduzwa komoya phansi kwephiko kumele kube kukhulu kunokusunduzwa okuphezulu kwephiko.
Amandla okusunduza ngaphansi kwephiko makhulu uma ingcindezi yomoya ngaphansi kwephiko inkulu.
Isimiso sikaBernoulli sithi ingcindezi yomoya ngaphansi kwephiko inkulu uma isivinini somoya ngaphansi kwephiko sincane.
Ngakho-ke v A > v B ukuze P A < P B
Impendulo efanele ingu-B.
16. Naka lesi sitatimende esilandelayo!
(1) i-venturimeter
(2) isifutho somiyane
(3) i-barometer
(4) i-thermometer
Isitatimende esihlobene nokusetshenziswa komthetho kaBernoulli sithi...
A. (1) kanye no-(2)
B. (1) kanye no-(3)
C. (1) kanye no-(4)
D. (2) kanye (3)
E. (2) kanye no-(4)
Ingxoxo
Izitatimende ezinezimiso zokusebenza ezisekelwe emthethweni kaBernoulli yizi (1) i-venturimeter kanye (2) i-mosquito sprayer. (3) i-barometer enesimiso sokusebenza esisekelwe kungcindezi ye-hydrostatic (4) i-mercury noma i-alcohol enesimiso sokusebenza esisekelwe ekukhuleni.
Impendulo efanele ngu-A.
17. Naka lesi sitatimende esilandelayo!
(1) iphampu ye-hydraulic
(2) i-carburetor
(3) i-venturimeter
(4) i-thermometer
Amathuluzi asebenzisa isimiso sawo sokusebenza esisekelwe emthethweni kaBernoulli...
A. (1) kanye no-(2)
B. (1) kanye no-(3)
C. (1) kanye no-(4)
D. (2) kanye (3)
E. (2) kanye no-(4)
Ingxoxo
Izitatimende ezinezimiso zokusebenza ezisekelwe emthethweni kaBernoulli yizi (2) ama-carburetor kanye (3) ama-venturimeter. (1) amaphampu e-hydraulic anezimiso zokusebenza ezisekelwe ekucindezelweni kwe-hydrostatic, umthetho kaPascal. (4) ama-thermometer e-mercury noma e-alcohol anezimiso zokusebenza ezisekelwe ekunwetshweni.
Impendulo efanele ngu-D.
18. Bheka isithombe esilandelayo!
Indawo yepayipi elikhulu ingu-5 m ngaphezu komhlabathi kanti ipayipi elincane lingu-1 m ngaphezu komhlabathi. Ijubane lokugeleza kwamanzi epayipini elikhulu lingu-36 km.h -1 ngokucindezela okungu-9,1 x 10 5 Pa, kuyilapho ingcindezi epayipini elincane ingu-2.10 5 Pa, bese kuthi ijubane lamanzi epayipini elincane libe…. (ubuningi bamanzi = 10 3 kg.m -3 )
A. 10 ms-1
B. 20 ms -1
C. 30 ms -1
D. 40 ms -1
E. 50 ms -1
Ingxoxo
Kuyaziwa ukuthi:
Umfutho wamanzi epayipini elikhulu (ikhasi 1 ) = 9,1 x 10 5 I-Pascal = 910.000 I-Pascal
Umfutho wamanzi epayipini elincane (ikhasi 2 ) = 2 x 10 5 Pascal = 200.000 Pascal
Ijubane lamanzi epayipini elikhulu (v 1 ) = 36 km/hora = 36(1000)/(3600) = 36000/3600 =10 m/s
Ukuphakama kwepayipi elikhulu (h 1 ) = -4 amamitha
Ukuphakama kwepayipi elincane (h 2 ) = 0 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms -2
Ubuningi bamanzi = 1000 kg.m -3
Umbuzo: Isivinini samanzi epayipini elincane (v 2 )
Impendulo:
Ijubane lamanzi epayipini elincane (v 2 ) libalwa kusetshenziswa i-equation kaBernoulli:
Impendulo efanele ngu-D.
19. Ipayipi elinobubanzi obungu-15 cm lixhunywe kwelinye ipayipi elinobubanzi obungu-5 cm. Zombili zisesimweni esivundlile. Uma izinga lokugeleza kwamanzi epayipini elikhulu lingu-1 ms -1 ekucindezelweni okungu-10 5 N m -2 , khona-ke ingcindezi epayipini elincane (ubuningi bamanzi 1 g cm -3 ) ingu ...
A. 10.000 N m -2
B. 15.000 N m -2
C. 30.000 N m -2
D. 60.000 N m -2
E. 90.000 N m -2
Ingxoxo
Kuyaziwa ukuthi:
Irediyasi yepayipi elikhulu (r 1 ) = 15 cm = 0,15 m
Irediyasi yepayipi elincane (r 2 ) = 5 cm = 0,05 m
Umfutho wamanzi epayipini elikhulu (ikhasi 1 ) = 10 5 N m -2 = 100.000 N m -2
Ijubane lamanzi epayipini elikhulu (v 1 ) = 1 ms -1
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms -2
Ubuningi bamanzi = 1 gr cm -3 = 1000 kg m -3
Womabili amapayipi asesimweni esivundlile ukuze umehluko ekuphakameni kwamapayipi (Δh) = 0.
Umbuzo: Ukucindezela epayipini elincane (ikhasi 2 )
Impendulo:
Ijubane lamanzi epayipini 2 libalwa kusetshenziswa i-Continuity Equation:
A 1 v 1 = A 2 v 2
(π r 1 2 )(v 1 ) = (π r 2 2 )(v 2 )
(r 1 2 )(v 1 ) = (r 2 2 )(v 2 )
(0,15 2 )(1 m/s) = (0,05 2 )(v 2 )
0,0225 = 0,0025 v 2
v 2 = 0,0225 / 0,0025
v 2 = 9 m/s
Ingcindezi epayipini elincane (ikhasi 2 ) ibalwa kusetshenziswa i-equation kaBernoulli:
Impendulo efanele ngu-D.
Isimiso sikaBernoulli
20. Bheka isithombe sesifutho sokufutha omiyane esithombeni esilandelayo! U-P umele ingcindezi kanti u-v umele izinga lokugeleza koketshezi oluxosha omiyane. Ukuze ukhiphe uketshezi oluxosha omiyane, udinga...
A. vA =vB kanye no-PA =PB
B. v A > v B kanye ne-P A > P B
C. v A > v B kanye ne-P A < P B
D. v A < v B kanye ne-P A > P B
E. v A < v B kanye ne-P A < P B
Ingxoxo
v A > v B
p A < p B
Impendulo efanele ngu-C.
21. Amaphiko endiza aklanyelwe ukuba abe nokuphakama okuphezulu, njengoba kuboniswe esithombeni. Uma u-v kuyijubane lokugeleza komoya kanti u-P uyingcindezi yomoya, khona-ke ngokomthetho kaBernoulli umklamo wenziwe ngendlela yokuthi...
A. vA > vB ngakho-ke u-PA > PB
B. v A > v B ukuze P A < P B
C. v A < v B ukuze P A < P B
D. v A < v B ukuze P A > P B
E. v A > v B ukuze P A = P B
Ingxoxo
Ukuze iphiko lendiza liphakanyiswe phezulu, ukusunduzwa komoya phansi kwephiko kumele kube kukhulu kunokusunduzwa okuphezulu kwephiko.
Ukucindezelwa ngaphansi kwephiko kukhulu uma ingcindezi yomoya ngaphansi kwephiko ikhulu. P = F/A, lapho ingcindezi P ilingana ngqo no-F. Uma ingcindezi inkulu, kulapho amandla eba maningi khona.
Isimiso sikaBernoulli sithi ingcindezi yomoya ngaphansi kwephiko inkulu uma isivinini somoya ngaphansi kwephiko sincane.
Ngakho-ke v A > v B ukuze P A < P B
Impendulo efanele ingu-B.
I-Torriceli
22. Ithangi lamanzi elingamasentimitha ayi-100 ukuphakama ligcwele amanzi ngokuphelele. Umgodi u-Q utholakala ngamasentimitha ayi-10 ngaphezu komhlabathi. Ibanga eliqondile lendiza yamanzi (x) li…
A. 0,2 m
B. 0,3 m
C. 0,6 m
D. 0,9 m
E. 1,0 m
Ingxoxo
Kuyaziwa ukuthi:
Ibanga lomgodi ukusuka ebusweni bamanzi (h) = 100 cm – 10 cm = 90 cm = 0,9 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Umbuzo: Ibanga lendiza yamanzi elivundlile (x)
Impendulo:
Isivinini samanzi aphuma emgodini
Ithiyori kaTorriceli ithi ijubane lejethi yamanzi eliphuma emgodini lifana nejubane lamanzi ehla ngokukhululeka esuka endaweni ephakeme h. Ngokwezibalo:
Incazelo: v = isivinini, g = ukusheshisa kwamandla adonsela phansi, h = ibanga lomgodi ukusuka ebusweni bamanzi
Isivinini sejethi yamanzi ephuma emgodini:
Isikhathi sokufinyelela kwamanzi ebusweni bomhlaba
Ukuhamba kwamanzi esuka emgodini aze afike phansi ukunyakaza kwe-parabolic, lapho indlela injengoba kuboniswe esithombeni ohlangothini. Ukunyakaza kwe-parabolic kubhekwa njengokuqukethe izinhlobo ezimbili zokunyakaza, okungukuthi ukunyakaza okuqondile nokuvundlile. Ukunyakaza okuvundlile kuhlaziywa njengo ukunyakaza okuqondile okufanayo futhi ukunyakaza kwesiqondiso esiqondile kulokhu kuhlaziywa njengoba ukunyakaza kokuwa kwamahhala.
Okokuqala, bala isikhathi lapho amanzi esemoyeni usebenzisa ifomula yokuwa yamahhala.
Kuyaziwa ukuthi:
Ukuphakama kwembobo (y) = 10 cm = 0,1 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Kubuziwe: isikhawu sesikhathi (t)
Impendulo:
y = 1/2 gt 2
0,1 = 1/2 (10) t 2
0,1 = 5 t 2
t 2 = 0,1 / 5
t 2 = 0,02
t = √0,02 imizuzwana
Ibanga elivundlile (x)
Kuyaziwa ukuthi:
Ijubane lokuqala (v o = v ox ) = 3 √2 m/s
Isikhathi esiphakathi (t) samanzi emoyeni = √0,02 imizuzwana
Umbuzo: Ibanga elivundlile (x)
Impendulo:
v = x / t
x = vt = (3 √2) ( √0,02) = (3)(1,41)(0,14) = 0,59 = 0,6 amamitha
Impendulo efanele ngu-C.
23. Bheka isithombe esilandelayo!
Ithangi liqukethe amanzi afinyelela ukuphakama okungu-1 m. Ephuzwini P odongeni lwethangi kukhona umgodi omncane kakhulu, ngakho-ke isivinini samanzi aphumayo… (g = 10 ms -2 )
A. 5 ms-1
B. 4 ms -1
C. 2,5 ms -1
D. 2 ms -1
E. 1,5 ms -1
Ingxoxo
Kuyaziwa ukuthi:
Ibanga lomgodi ukusuka ebusweni bamanzi (h) = 100 cm – 80 cm = 20 cm = 0,2 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Umbuzo: Isivinini sokuphuma kwamanzi (v)
Impendulo:
Isivinini sejethi yamanzi ephuma emgodini:
Impendulo efanele ngu-D.
I-Venturimeter
24. Ingxenye enkulu yepayipi le-venturimeter inendawo enqamulayo A1 = 6cm2 futhi ingxenye encane yamapayipi inendawo enqamulayo u-A2 = 5cm2Ijubane lamanzi angena epayipini le-venturimeter lingu... h = 20 cm, g = 10 m/s2.
A. 2 m/s
B. 3 m/s
C. 4 m/s
D. 5 m/s
E. 6 m/s
Ingxoxo:
Impendulo efanele ingu-B.
Umthombo wombuzo:
Imibuzo Yefiziksi Yezivivinyo Zikazwelonke Zesikole Samabanga Aphezulu/Isikole Samabanga Aphezulu Sokufundela Umsebenzi
