Izinkinga kanye nezixazululo zoketshezi
1. Isitsha esigcwele amanzi futhi kukhona umgodi, njengoba kuboniswe esithombeni esingezansi. Uma ukusheshisa ngenxa yamandla adonsela phansi kungu-10 ms-2, lingakanani ijubane lamanzi adlula kulowo mgodi?
Kwaziwa:
Ukuphakama (h) = 85 cm – 40 cm = 45 cm = 0.45 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Kufunwa: Isivinini samanzi (v)
Isixazululo:
Ithiyori kaTorricelli ithi amanzi ashiya umgodi ngesivinini esifanayo nento ewa ngokukhululeka isuka ekuphakameni okufanayo. Ukuphakama (h) = 85 cm – 40 cm = 45 cm = 0.45 amamitha
Ijubane lamanzi libalwa kusetshenziswa i-equation ye- ukunyakaza kokuwa kwamahhala :
vt2 = 2 gh
v t 2 = 2 gh = 2(10)(0.45) = 9
v t = √9 = 3 m/s
2. Isitsha esigcwele amanzi futhi kukhona umgodi, njengoba kuboniswe esithombeni esingezansi. Uma ukusheshisa ngenxa yamandla adonsela phansi kungu-10 ms-2, lingakanani ijubane lamanzi adlula kulowo mgodi?
Kwaziwa:
Ukuphakama (h) = 1.5 m – 0.25 m = 1.25 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Kufunwa: Isivinini samanzi (v)
Isixazululo:
v t 2 = 2 gh = 2(10)(1.25) = 25
v t = √25 = 5 m/s
3. Isitsha esigcwele amanzi futhi kukhona umgodi, njengoba kuboniswe esithombeni esingezansi. Uma ukusheshisa ngenxa yamandla adonsela phansi kungu-10 ms-2, lingakanani ijubane lamanzi adlula kulowo mgodi?
Kwaziwa:
Ukuphakama (h) = 1 m – 0.20 m = 0.8 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Kufunwa: Isivinini samanzi (v)
Isixazululo:
v t 2 = 2 gh = 2(10)(0.8) = 16
v t = √16 = 4 m/s
4. Isitsha esigcwele amanzi futhi kukhona umgodi, njengoba kuboniswe esithombeni esingezansi. Uma ukusheshisa ngenxa yamandla adonsela phansi kungu-10 ms-2, lingakanani ijubane lamanzi adlula kulowo mgodi?
Kwaziwa:
Ukuphakama (h) = 20 cm = 0.2 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Okufunayo: Isivinini samanzi (v)
Isixazululo:
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5. Isitsha esigcwele amanzi bese kuba khona kukhona amabili holes, njengoba kuboniswe emfanekisweni ongezansi. Iyini isilinganiso sika-x1 kuya ku-x2?
I-Solution

Isikhathi sokuwa kwamanzi ngaphandle kokuphuma emgodini 1:
h = 1/2 ku2
0.8 = 1/2 (10) t 2
0.8 = 5 t 2
t2 = 0.8/5 = 0.16
t = imizuzwana emi-0.4
Isikhathi sokuwa kwamanzi ngaphandle kokuphuma emgodini 2:
h = 1/2 ku2
0.5 = 1/2 (10) t 2
0.5 = 5 t 2
t2 = 0.5/5 = 0.1
t = √0.1 isekhondi
Ibanga elivundlile (x):
x1 =v1 t1 = (2)(0.4) = amamitha angu-0.8
x2 =v2 t2 = (√10)(√0.1) = (10)(0.1) = 1 imitha
Isilinganiso sika-x1 kuya ku-x2 :
x1 : x2 = 0.8 : 1 = 8 : 10 = 4 : 5
6. Amanzi ageleza ngepayipi elinobubanzi obuhlukahlukene, u-A kuya ku-B bese kuba u-C. Isilinganiso sika-A kuya ku-C singu-8: 3. Uma ijubane lamanzi epayipini u-A lingu-v, liyini ijubane lamanzi epayipini u-C.
Kwaziwa:
Indawo ye-A (A)A= = 8
Indawo ye-C (A)C= = 3
Isivinini samanzi epayipini A (v)A) = v
Okufunayo: Isivinini samanzi epayipini C (v)C)
Isixazululo:
Isibalo sokuqhubeka:
A A v A = A C v C
8 v = 3 vC
vC = 8/3 v
7. Uma ijubane lamanzi epayipini elinobubanzi obungu-12 cm lingu-10 cm/s, lingakanani ijubane lamanzi epayipini elinobubanzi obungu-8 cm?
Kwaziwa:
Ububanzi 1 (d)1) = 12 cm, irediyasi 1 (r1= 6cm
Ububanzi 2 (d)2) = 8 cm, irediyasi 2 (r2= 4cm
Isivinini samanzi 1 (v1) = 10 cm/s
Kufunwa: Isivinini samanzi 2 (v2)
Isixazululo:
Indawo 1 (A)1) = π r2 = π 62 = 36π cm2
Indawo 2 (A)2) = π r2 = π 42 = 16π cm2
Isibalo sokuqhubeka:
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
8. Amanzi ageleza ngepayipi elinobubanzi obuhlukahlukene, njengoba kuboniswe esithombeni esingezansi. Uma indawo 1 (A)1= 8cm2, A2 = 2cm2 kanye nesivinini samanzi epayipini 2 = v2 = 2 m/s pho lingakanani ijubane lamanzi epayipini 1 = v1.
Kwaziwa:
Indawo 1 (A)1= 8cm2
Indawo 2 (A)2= 2cm2
Isivinini samanzi epayipini 2 (v)2) = 2 m/s
Kufunwa: isivinini samanzi epayipini 1 (v1)
Isixazululo:
Isibalo sokuqhubeka:
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
9. Uma ububanzi bepayipi elikhulu buphindwe kabili kunobubanzi bepayipi elincane, lingakanani ijubane loketshezi epayipini elincane?
Kwaziwa:
Ububanzi bepayipi elikhulu (d)1= = 2
Irediyasi yepayipi elikhulu (r)1) = ½ d1 = ½ (2) = 1
Indawo yepayipi elikhulu (A1) = π r12 = π (1)2 = π (1) = π
Ububanzi bepayipi elincane (d)2= = 1
Irediyasi yepayipi elincane (r)2) = ½ d2 = ½ (1) = ½
Indawo yepayipi elincane (A2) = π r22 = π (1/2)2 = π (1/4) = ¼ π
Isivinini soketshezi epayipini elikhulu (v1) = 4 m/s
Kufunwa: Isivinini soketshezi epayipini elincane (v2)
Isixazululo:
Isibalo sokuqhubeka:
A 1 v 1 = A 2 v 2
π 4 = ¼ π (v 2 )
4 = ¼ (v 2 )
v 2 = 8 m/s
Isimiso sikaBernoulli kanye nezibalo
I-10. WAmanzi ampontshwa nge-120 kPa i-compressor ingena epayipini elingezansi (1) bese igeleza iye phezulu ngesivinini esingu-1 m/s. Ukusheshisa ngenxa yamandla adonsela phansi 10 m/s kanye nokuminyana kwamanzi is 1000 kg / m-3. Kuyini tumfutho wamanzi epayipini eliphezulu (II)).
Kwaziwa:
Irediyasi yepayipi elingezansi (r)1= 12cm
Irediyasi yepayipi elingezansi (r)2= 6cm
Umfutho wamanzi epayipini elingezansi (p1) = 120 kPa = 120,000 amaPascal
Isivinini samanzi epayipini elingezansi (v1) = 1 ms-1
Ukuphakama kwepayipi elingezansi (h1= 0m
Ukuphakama kwepayipi eliphezulu (h2= 2m
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms -2
Ubuningi bamanzi = 1000 kg.m -3
Okufunayo: Umfutho wamanzi epayipini 2 (p)2)
Isixazululo:
Ijubane lamanzi epayipini 2 libalwa ngesibalo sokuqhubeka:

Umfutho wamanzi epayipini 2 ubalwa kusetshenziswa i-equation kaBernoulli:

11. A lipayipi elikhulu elingamamitha ama-5 ngaphezu komhlabathi kanye nepayipi elincane elingamamitha ama-1 ngaphezu komhlabathi. Ijubane lamanzi epayipini elikhulu lingu-36 km / h ngokucindezela okungu-9.1 x 105 Pa, kuyilapho ingcindezi epayipini elincane ingu-2.105 I-Pa. Kuyini ijubane lamanzi iipayipi elincane? Water ubuningi = 103 kg / m3
Kwaziwa:
Umfutho wamanzi epayipini elikhulu (p1) = 9.1 x 105 Ama-Pascal = ama-Pascal angu-910,000
Umfutho wamanzi epayipini elincane (p2) = 2 x 105 Ama-Pascal = ama-Pascal angu-200,000
Isivinini samanzi epayipini elikhulu (v)1) = 36 km/h = 36(1000)/(3600) = 36000/3600 =10 m/s
Ukuphakama kwepayipi elikhulu (h1) = -4 amamitha
Ukuphakama kwepayipi elincane (h2) = 0 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms -2
Ubuningi bamanzi = 1000 kg/m3
Okufunayo: Isivinini samanzi epayipini elincane (v2)
Isixazululo:
Isivinini samanzi epayipini elincane (v2) kubalwa kusetshenziswa i-equation kaBernoulli:

12. I-pshayela nge irediyasi of I-15 cm ixhunywe nelinye ipayipi ngerediyasi ye 5 cm. Zombili zisesimweni esivundlile. TIjubane lokugeleza kwamanzi epayipini elikhulu lingu-1 m/s ngokucindezela okungu-105 N/m2. Kuyini the amanzi ingcindezi epayipini elincane (1 g cm-3)
Kwaziwa:
Irediyasi yepayipi elikhulu (r)1) = 15 cm = 0.15 m
Irediyasi yepayipi elincane (r)2) = 5 cm = 0.05 m
Umfutho wamanzi epayipini elikhulu (p1= = 105 N m-2 = 100.000 N m-2
Isivinini samanzi epayipini elikhulu (v1) = 1 ms-1
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms -2
Water Ubuningi = 1 igremu cm-3 = 1000 kg m-3
Umehluko wokuphakama (Δh) = 0.
Okufunayo: Ukucindezela epayipini elincane (p2)
Isixazululo:
Ijubane lamanzi epayipini 2 libalwa kusetshenziswa i-equation of continuity:

Umfutho wamanzi epayipini elincane (p2) kubalwa kusetshenziswa i-equation kaBernoulli:

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