Izibalo ze-Fluid - izinkinga nezixazululo

Izibalo ze-Fluid - izinkinga nezixazululo

Umfutho woketshezi

1. Uyini umehluko phakathi komfutho wegazi ongaphansi kwamanzi phakathi kobuchopho kanye nezinyawo zomuntu okuphakama kwakhe kungu-165 cm (ake sithi ukuminyana kwegazi = 1.0 × 10 ³ kg/m3 ^, ukusheshisa ngenxa yamandla adonsela phansi = 10 m/ ^s2 )

Kwaziwa:

Ukuphakama (h) = 165 cm = 165/100 m = 1.65 amamitha

Ubuningi begazi (ρ) = 1.0 × 10 ³ kg/ ^m3

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Okufunayo: ukucindezela koketshezi

Isixazululo:

P = ρ gh

P = (1.0 × 10 ³ )(10)(1.65)

P = (1.0 × 10 ⁴ )(1.65)

P = 1.65 x 10 ⁴ N/m ²

Ipayipi Eliphezulu

2. Ipayipi le-AU liqala ligcwaliswe ngamanzi kunepayipi elilodwa eligcwele uwoyela, njengoba kuboniswe esithombeni esingezansi. Ubuningi bamanzi bungu-1000 kg/m3 ^. Uma ukuphakama kwamafutha kungu-8 cm kanti ukuphakama kwamanzi kungu-5 cm, kungakanani ubuningi bamafutha?

Kwaziwa:

Ubuningi bamanzi = 1000 kg.m ^-3

Ukuphakama kwamanzi (h ₂ ) = 5 cm

Ukuphakama kwamafutha (h ₁ ) = 8 cm

Okufunwayo: ukuminyana kwamafutha

Isixazululo:

ρ ₁ gh ₁ =ρ ₂ gh ₂

ρ ₁ ihora ₁ =ρ ₂ ihora ₂

(1000)(5) = (ρ ₂ )(8)

5000 = (ρ ₂ )(8)

ρ ₂ = 625 kg.m ^-3

3. Ipayipi le-AU laqala lagcwaliswa ngophalafini labe selifaka amanzi. Uma isisindo sikaphalafini singamagremu angu-0.8/cm3 ^kanti ubuningi bamanzi buyigremu eli-1/cm3 ^kanti indawo enqamulayo ingu-1.25 cm2 ^. Thola ukuthi kufanele kufakwe amanzi angakanani ukuze umehluko wokuphakama kobuso bukaphalafini ube ngu-15 cm.

A. 9 ml

B. 12 ml

C. 15 ml

D. 18 ml

Kwaziwa:

Ubuningi be-parafini (ρ ₁ ) = 0.8 gram/ ^cm3

Ubuningi bamanzi (ρ ₂ ) = 1 gram/ ^cm3

Indawo yesigaba sepayipi e = 1.25 cm ²

Umehluko wokuphakama kobuso be-kerosene (h ₁ ) = 15 cm

Okufunwayo: Umthamo wamanzi

Isixazululo:

Ukuphakama kwamanzi (h2 ₎ :

ρ ₁ gh ₁ = ρ ₂ gh ₂

(0,8)(15)(1)(h ₂ )

h ₂ = 12 cm

Umthamo wamanzi:

V = ( Indawo yesigaba sepayipi e )(ukuphakama kwamanzi)

V = (1.25 cm ² )(12 cm)

V = 15 cm ³

Ilitha eli-1 = 1 dm ³ = 10 ³ cm ³

I-millilitha eli-1 = 10 ^-3 amalitha s = (10 ^-3 )(10 ³ ) cm ³ = 1 cm ³

Umthamo wamanzi ungama-15 cm ³ = amamililitha ayi-15

Impendulo efanele ithi C.

4. Ipayipi U eligcwele amanzi anobukhulu obungu-1000 kg/m3 ^. Ikholomu eyodwa yepayipi U eligcwele i-glycerin enobukhulu obungu-1200 kg/m3 ^. Uma ukuphakama kwe-glycerin kungu-4 cm, nquma umehluko wokuphakama kwamakholomu womabili epayipi.

A. 0.8 cm

B. 4 cm

C. 8 cm

D. 12 cm

Kwaziwa:

Ubuningi bamanzi (ρ ₁ ) = 1000 kg/ ^m3

Ubuningi be-glycerin (ρ ₂ ) = 1200 kg/ ^m3

Ukuphakama kwe-glycerin (h ₂ ) = 4 cm

Okufunwayo: Umehluko wokuphakama kwamakholomu womabili epayipi.

Isixazululo:

Ukuphakama kwekholomu yepayipi (h ₁ ):

ρ ₁ ihora ₁ = ρ ₂ ihora ₂

(1000)(h ₁ ) = (1200)(4)

(1000)(h ₁ ) = 4800

h ₁ = 4.8 cm

Umehluko wokuphakama kwamakholomu womabili epayipi U = h ₁ – h ₂ = 4.8 cm – 4 cm = 0.8 cm

Impendulo eyiyo ngu-A.

5. Ipayipi i -U inamaphethelo amabili avulekile agcwele amanzi anesisindo esingu - 1 g/cm3 ^. Indawo yesigaba eceleni kwepayipi iyafana, okungu-1 cm2 ^. Umuntu ushaya kolunye uhlangothi lwepayipi ukuze ubuso bamanzi kolunye unyawo bukhuphuke ngo-10 cm ukusuka endaweni yalo yokuqala. Uma ukusheshisa okubangelwa amandla adonsela phansi kungu-10 m/s2 khona-ke nquma amandla asetshenziswa yilowo muntu.

A. 20 kilodynes

B. 10 kilodynes

C. 2 kilodynes

D. 1 kilodyne

Kwaziwa:

Shintsha wonke amayunithi abe uhlelo lwamazwe ngamazwe.

Ubuningi bamanzi (ρ ₁ ) = 1 gr/cm ³ = 10 ^-3 kg / 10 ^-6 m ³ = 10 ³ kg/m ³

Indawo yesiphambano sepayipi (A) = 1 cm ² = 10 ^-4 m ²

Ushintsho lwekholomu yepayipi (h) = 10 cm = 1 dm = 10 ^-1 m

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms ^-2 = 10 ¹ ms ^-2

Umthamo wamanzi ahambayo (V) = (A)(h) = (1 cm ² )(10 cm) = 10 cm ³ = (10 ¹ )(10 ^-6 m ³ ) = 10 ^-5 m ³

Okufunwayo: Amandla (F) asetshenziswa ngumuntu.

Isixazululo:

Amandla asebenze yilowo muntu = isisindo samanzi anokuphakama okungu-10 cm

F = w

F = mg —–> Isibalo sobuningi : m = ρ V

F = ρ V g

F = (10 ³ )(10 ^-5 )(10 ¹ )

F = (10 ⁴ )(10 ^-5 )

F = 10 ^-1 Newton —–> 1 Newton = 10 ⁵ amadyne

F = (10 ^-1 )(10 ⁵ dyne)

F = 10 ⁴ amadayini

F = amakhilodine ayi-10

Impendulo efanele ngu-B.

6. Ipayipi elinomumo ongu-Y lifakwa libheke phansi ukuze unyawo lwesobunxele nonyawo lwesokudla kucwiliswe ezinhlotsheni ezimbili zoketshezi. Ngemva kokuba izinyawo zombili zicwiliswe oketshezini, khona-ke ingxenye ephezulu yepayipi lika-Y ivalwa ngomunwe bese idonswa phezulu, ukuze imilenze emibili yepayipi lika-Y igcwaliswe ngekholomu yoketshezi oluhlukahlukene olunobukhulu obuphezulu. Uma ubukhulu boketshezi lokuqala bungu-0.80 gram.cm ^-3 kanti ubukhulu besibili bungu- 0.75 gram.cm ^-3 , kanti ikholomu engezansi yoketshezi ingu-8 cm, khona-ke nquma umehluko wokuphakama phakathi kwamakholomu amabili oketshezi ku-U pip e.

A. 1.0666 cm

B. 0.9375 cm

C. 0.3533 cm

D. 0.5333 cm

Kwaziwa:

Ubuningi boketshezi lokuqala (ρ ₁ ) = 0,80 gram.cm ^-3

Ubuningi boketshezi lwesibili (ρ ₂ ) = 0,75 gram.cm ^-3

Ukuphakama koketshezi oluphansi (h ₁ ) = 8 cm

Okufunwayo: Umehluko wokuphakama phakathi kwamakholomu amabili oketshezi ku-U pip e

Isixazululo:

Ukuphakama koketshezi oluphezulu (h2 ₎ :

ρ ₁ ihora ₁ = ρ ₂ ihora ₂

(0.80)(8) = (0.75)(h ₂ )

6.4 = 0.75 (h ₂ )

h ₂ = 6.4 / 0.75

h ₂ = 8.5 cm

Umehluko wokuphakama koketshezi = h ₂ – h ₁ = 8.5333 cm – 8 cm = 0.5333 cm

Impendulo efanele ngu-D.

Amandla amakhulu

7. Itshe elinomthamo ongu-0.5 m ³ elibekwe oketshezini olunobukhulu obungu-1.5 gr cm ^–3 . Ukusheshisa ngenxa yamandla adonsela phansi kungu-10 ms ^-2 . Iyini amandla okuphapha?

Kwaziwa:

Umthamo wetshe (V) = 0.5 m ³

Ubuningi bamanzi (ρ) = 1.5 gr cm ^–3 = 1500 kg m ^-3

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms ^-2

Kufunwa: amandla anamandla ( _FA )

Isixazululo:

Isibalo samandla agelezayo:

F _A = ​​​​ρ g V = (1500 kg m ^-3 )(10 ms ^-2 )(0.5 m ³ ) = 7500 kg m/s ² = 7500 AmaNewton

I-Float

8. Ibhulokhi leqhwa liyantanta olwandle njengoba kuboniswe esithombeni esingezansi. Ubuningi bolwandle bungu-1.2 gr cm ^–3 kanti ubuningi beqhwa bungu-0.9 gr c ^–3 . Ubuningi beqhwa emanzini olwandle = ……. x ubuningi beqhwa emoyeni.

Kwaziwa:

Ubuningi bolwandle (ρ _ulwandle ) = 1.2 gr cm ^–3

Ubuningi beqhwa (ρ _ice ) = 0.9 gr c ^–3

Okufunwayo: Umthamo weqhwa emanzini olwandle = ……. x umthamo weqhwa emoyeni.

Isixazululo:

Umthamo weqhwa olwandle = 0.75

Umthamo weqhwa emoyeni = 0.25

Umthamo weqhwa emanzini olwandle = 3 x umthamo weqhwa emoyeni (3 x 0.25 = 0.75).

9. Into intanta oketshezini lapho u-2/3 wento useketshezini. Uma ubuningi bento bungu-0.6 gr cm3 ^, pho ubuningi bamanzi buyini.

Kwaziwa:

Ingxenye yento ewuketshezi = 2/3

Ubuningi bento = 0.6 gr cm ³ = 600 kg m ³

Okufunwayo: ubuningi boketshezi (x)

Isixazululo:

Ubuningi boketshezi bungu-900 kg ^m3

10. Izinkuni zintanta emanzini, lapho ingxenye engu-3/5 yezinkuni isemanzini. Uma ubuningi bamanzi bungu-1 × 10 ³ kg/m3 ^, buyini ubuningi bokhuni?

Kwaziwa:

Ingxenye yento emanzini = 3/5

Ubuningi bamanzi = 1×10 ³ kg/m ³ = 1000 kg/m ³

Okufunwayo: Ubuningi bokhuni (x)

Isixazululo:

Ubuningi bokhuni bungu-600 kg/m ³ = 6 x 10 ² kg/m ³

1. Kuyini uketshezi oluqhubekayo?
   • Impendulo: I-fluid statics, eyaziwa nangokuthi i-hydrostatics, iyigatsha le-fluid mechanics elifunda uketshezi oluphumuzayo kanye namandla asetshenziswa uketshezi olunganyakazi ezintweni ezicwilisiwe kanye nezindonga zesitsha.
2. Ingcindezi kuketshezi ihluka kanjani ngokujula?
   • Impendulo: Kuketshezi olunganyakazi, ingcindezi iyanda ngokulandelana ngokujula ngenxa yesisindo sekholomu yoketshezi ngaphezu kwanoma yikuphi ukujula okunikeziwe. Ushintsho ekucindezelweni ngokujula lunikezwa yi $\Delta P=\rho gh$, lapho $\rho$ ukuminyana koketshezi, $g$ ukusheshisa kwamandla adonsela phansi, futhi $h$ ukujula.
3. Iyini isimiso sikaPascal?
   • Impendulo: Isimiso sikaPascal sithi ushintsho ekucindezelweni okusetshenziswa kuketshezi oluvalekile ludluliselwa lungapheli kuzo zonke izingxenye zoketshezi nasezindongeni zesitsha salo.
4. Isebenza kanjani i-hydraulic lift ngokusekelwe ezimisweni ze-fluid statics?
   • Impendulo: I-hydraulic lift isebenzisa isimiso sikaPascal. Uma amandla amancane esetshenziswa episton encane, idala ingcindezi kuketshezi. Lokhu kucindezela kudluliselwa kunganciphi kulo lonke uketshezi, kusebenzisa amandla amakhulu kakhulu episton enkulu, okwenza i-lift ikwazi ukuphakamisa izinto ezisindayo ngomzamo omncane kakhulu.
5. Iyini amandla okuphapha futhi ahlobene kanjani ne-fluid statics?
   • Impendulo: Amandla okuphapha ngamandla aphezulu asetshenziswa uketshezi kunoma iyiphi into ecwilisiwe. Ngokomthetho ka-Archimedes, amandla okuphapha entweni alingana nesisindo soketshezi olususwe yinto.
6. Kungani izinto zintanta noma zicwila oketshezini?
   • Impendulo: Ukuthi into iyantanta noma iyacwila kuncike ebuhlotsheni obuphakathi kwamandla okuntanta kanye nesisindo sento. Uma amandla okuntanta (ngenxa yoketshezi olususiwe) makhulu kunesisindo sento, izontanta. Uma isisindo sento sikhulu, izocwila.
7. Uyini umqondo wokucindezela kwe-hydrostatic?
   • Impendulo: Ukucindezela kwe-hydrostatic yingcindezi ekhishwa uketshezi oluphumulile ngenxa yamandla adonsela phansi. Iyanda ngokulandelana ngokujula koketshezi, futhi ibalwa ngokuthi P=P₀​+ρgh, lapho P₀​ ukucindezela okungaphezulu, $\rho$ ukuminyana koketshezi, $g$ ukusheshisa kwamandla adonsela phansi, futhi $h$ ukujula.
8. Ingcindezi yomoya ihlobene kanjani ne-fluid statics?
   • Impendulo: Umkhathi ungacatshangwa njengoketshezi. Umfutho womoya umfutho okhishwa yisisindo somoya ngaphezu kwephuzu elithile. Uyancipha ngokuphakama, ngendlela efanayo nendlela umfutho oketshezini oncipha ngayo njengoba umuntu ekhuphukela phezulu kukholamu yoketshezi.
9. Iyiphi indima edlalwa ukuma kwesitsha ekusatshalalisweni kwengcindezi yoketshezi olungaguquki ngaphakathi kwaso?
   • Impendulo: Ku-fluid statics, ingcindezi ekujuleni okunikeziwe incike kuphela ekuphakameni kwekholomu yoketshezi ngaphezu kwalokho kujula, hhayi ekumeni kwesitsha. Ngakho-ke, ingcindezi ekujuleni okuthile iyafana kungakhathaliseki ukuma kwesitsha.
10. Iyini ukubaluleka kwe-hydrostatic paradox?

• Impendulo: I-hydrostatic paradox iqokomisa ukuthi kuma-static e-fluid, amandla asetshenziswa uketshezi oluqhubekayo phansi kwesitsha ancike kuphela ekuphakameni kwekholomu ye-fluid, hhayi kumthamo wayo noma ukuma kwesitsha. Ngakho-ke, izitsha ezihlukene kakhulu ezinokuphakama okufanayo koketshezi zinomfutho ofanayo esisekelweni sazo, noma ngabe ziphethe amanani ahlukene oketshezi.