Misalai 5 na Dokar Bernoulli
1 P1 dan v1 shine matsin lamba da saurin iskar da ke sama da fikafikin, P2 dan v2 shine matsin lamba da saurin iskar da ke ƙarƙashin fikafikan. Domin jirgin sama ya tashi, yanayin shine...
A. P1 =P2 dan v1 = v2
B. P1 <P2 dan v1 > v2
C.P1 <P2 dan v1 < v2
D.P1 > P2 dan v1 > v2
E.P1 > P2 dan v1 < v2
Tattaunawa:
An san cewa:
P1 = Matsin iska a kan fikafikai
P2 = Matsin iska a ƙarƙashin fikafikai
v1 = saurin iska a kan fikafikai
v2 = saurin iska a ƙarƙashin fikafikai
An tambaya :
Wadanne sharuɗɗa ne jirgin zai ɗaga ko jirgin zai yi sama?
Jawab :
Dokar Bernoulli ko kuma a kira shi daidai Ka'idar Bernoulli ya bayyana cewa idan saurin ruwa ya yi yawa, matsin ruwan ya yi ƙasa, akasin haka, idan saurin ruwa ya yi ƙasa, matsin ruwan ya yi yawa. Ruwa abubuwa ne da za su iya gudana. Misalan ruwa sune iska da ruwa.
Idan gudun iska ya yi yawa, matsin iska ya yi ƙasa, akasin haka, idan gudun iska ya yi ƙasa, matsin iska ya yi yawa.
Idan gudun iska da ke sama da fikafikin ya fi saurin iska da ke ƙasa da fikafikin, matsin iska da ke sama da fikafikin ya yi ƙasa da matsin iska da ke ƙasa da fikafikin. Saboda matsin iska da ke ƙasa da fikafikin ya fi girma, matsin iska da ke ƙasa da fikafikin ma ya fi girma. Akasin haka, matsin iska da ke sama da fikafikin ya yi ƙasa, don haka matsin iska da ke sama da fikafikin ma ya yi ƙasa (P = F/A - mafi girman matsin lamba, P, mafi girman ƙarfin, F).
Tura iska a ƙarƙashin fikafikan ya fi girma, tursasawar iska a sama da fikafikan ya fi ƙanƙanta, don haka akwai ƙarfin da aka samu ko kuma ƙarfin da aka mayar zuwa sama. Wannan ƙarfin gaba ɗaya yana kan gaba wanda ke sa jirgin ya ɗaga..
Jadi sharuɗɗan da jirgin zai ɗauka don tashi shine P1 <P2 kuma v1 > v2
Amsar da ta dace ita ce B.
2.
A cikin hoton, ana tura ruwa ta hanyar na'urar matsa lamba ta 120 kPa zuwa cikin bututun ƙasa (1) kuma yana gudana sama da gudun 1 m/s.-1 (g = 10 ms-2 kuma yawan ruwan shine 1000 kg.m-3). Matsin ruwa a cikin bututun sama (II) shine….
A. 52,5 kPa
B. 67,5 kPa
C. 80,0 kPa
D. 92,5 kPa
E. 107,5 kPa
Tattaunawa
An san cewa:
Babban radius na bututu (r)1= 12 cm
Ƙaramin radius na bututu (r)2= 6 cm
Matsin ruwa a cikin manyan bututu (p1) = 120 kPa = 120.000 Pascals
Gudun ruwa a cikin manyan bututu (v)1) = 1 ms-1
Babban tsayin bututu (h)1= 0m
Ƙaramin tsayin bututu (h)2= 2m
Saurin gudu saboda nauyi (g) = 10 ms-2
Yawan ruwa (ρ) = 1000 kg.m-3
An tambaya: Matsin ruwa a cikin bututu 2 (p2)
Amsa:
Ana ƙididdige saurin ruwa a cikin bututun 2 ta amfani da Lissafin Ci gaba:
A1 v1 = A ba2 v2
(π r12)(v1) = (π r22)(v2)
(r12)(v1) = (r22)(v2)
(122)(1 m/s) = (62)(v2)
144 = 36 v2
v2 = 144/36
v2 = 4m/s
Ana ƙididdige matsin lamba na ruwa a cikin bututun 2 ta amfani da lissafin Bernoulli:
120.000 + ½ (1000)(12) + (1000)(10)(0) = p2 + ½ (1000)(42) + (1000)(10)(2)
120.000 + ½ (1000)(1) + (1000)(10)(0) = p2 + ½ (1000)(16) + (1000)(10)(2)
120.000 + 500 + 0 = p2 + (500)(16) + 20.000
120.000 + 500 = p2 +8000+20.000
120.500 = p2 + 28.000
p2 = 120.500-28.000
p2 = Pascals 92.500
p2 = 92,5 kPa
Amsar da ta dace ita ce D.
3. Kalli hoton da ke ƙasa!
Babban bututun yana da nisan mita 5 daga ƙasa, ƙaramin bututun kuma yana da nisan mita 1 daga ƙasa. Yawan kwararar ruwa a cikin babban bututun shine kilomita 36/h.-1 tare da matsin lamba na 9,1 x 105 Pa, yayin da matsin lamba a cikin ƙaramin bututu shine 2.105 Pa, to saurin ruwa a cikin ƙaramin bututu shine…. (yawan ruwa = 103 kg.m-3)
A. 10 ms-1
B. 20 ms-1
C. 30 ms-1
D. 40 ms-1
E. 50 ms-1
Tattaunawa
An san cewa:
Matsin ruwa a cikin manyan bututu (p1= 9,1 x 105 Pascal = Pascals 910.000
Matsin ruwa a cikin ƙananan bututu (p2= 2 x 105 Pascal = Pascals 200.000
Gudun ruwa a cikin manyan bututu (v)1) = 36 km/h = 36(1000)/(3600) = 36000/3600 = 10 m/s
Babban tsayin bututu (h)1) = -4 mita
Ƙaramin tsayin bututu (h)2) = mita 0
Saurin gudu saboda nauyi (g) = 10 ms-2
Yawan ruwa (ρ) = 1000 kg.m-3
An tambaya: Gudun ruwa a cikin ƙaramin bututu (v)2)
Amsa:
Gudun ruwa a cikin ƙaramin bututu (v)2) ana ƙididdige shi ta amfani da lissafin Bernoulli:
910.000 + ½ (1000)(10)2) + (1000) (10) (0) = 200.000 + ½ (1000) (v)22) + (1000)(10)(-4)
910.000 + 50.000 + 0 = 200.000 + 500 v22 - 40.000
960.000 = 160.000 + 500 v22
800.000 = 500 v22
800.000 / 500 = v22
1600 = ku22
v2 = √1600
v2 = 40m/s
Amsar da ta dace ita ce D.
4. A cikin hoton, ana tura ruwa ta hanyar na'urar matsa lamba ta 120 kPa zuwa cikin bututun ƙasa (1) kuma yana gudana sama da gudun 1 m/s.-1 (g = 10 ms-2 kuma yawan ruwan shine 1000 kg.m-3). Matsin ruwa a cikin bututun sama (II) shine….
A. 52,5 kPa
B. 67,5 kPa
C. 80,0 kPa
D. 92,5 kPa
E. 107,5 kPa
Tattaunawa
An san cewa:
Babban radius na bututu (r)1= 12 cm
Ƙaramin radius na bututu (r)2= 6 cm
Matsin ruwa a cikin manyan bututu (p1) = 120 kPa = 120.000 Pascals
Gudun ruwa a cikin manyan bututu (v)1) = 1 ms-1
Babban tsayin bututu (h)1= 0m
Ƙaramin tsayin bututu (h)2= 2m
Saurin gudu saboda nauyi (g) = 10 ms-2
Yawan ruwa (ρ) = 1000 kg.m-3
An tambaya: Matsin ruwa a cikin bututu 2 (p2)
Amsa:
Ana ƙididdige saurin ruwa a cikin bututun 2 ta amfani da Lissafin Ci gaba:
A1 v1 = A ba2 v2
(π r12)(v1) = (π r22)(v2)
(r12)(v1) = (r22)(v2)
(122)(1 m/s) = (62)(v2)
144 = 36 v2
v2 = 144/36
v2 = 4m/s
Ana ƙididdige matsin lamba na ruwa a cikin bututun 2 ta amfani da lissafin Bernoulli:
120.000 + ½ (1000)(12) + (1000)(10)(0) = p2 + ½ (1000)(42) + (1000)(10)(2)
120.000 + ½ (1000)(1) + (1000)(10)(0) = p2 + ½ (1000)(16) + (1000)(10)(2)
120.000 + 500 + 0 = p2 + (500)(16) + 20.000
120.000 + 500 = p2 +8000+20.000
120.500 = p2 + 28.000
p2 = 120.500-28.000
p2 = Pascals 92.500
p2 = 92,5 kPa
Amsar da ta dace ita ce D.
5. Kalli hoton da ke ƙasa! Matsayin babban bututun yana da mita 5 sama da ƙasa, ƙaramin bututun kuma yana da mita 1 sama da ƙasa. Saurin kwararar ruwa a cikin babban bututun shine kilomita 36/h.-1 tare da matsin lamba na 9,1 x 105 Pa, yayin da matsin lamba a cikin ƙaramin bututu shine 2.105 Pa, to saurin ruwa a cikin ƙaramin bututu shine…. (yawan ruwa = 103 kg.m-3)
A. 10 ms-1
B. 20 ms-1
C. 30 ms-1
D. 40 ms-1
E. 50 ms-1
Tattaunawa
An san cewa:
Matsin ruwa a cikin manyan bututu (p1= 9,1 x 105 Pascal = Pascals 910.000
Matsin ruwa a cikin ƙananan bututu (p2= 2 x 105 Pascal = Pascals 200.000
Gudun ruwa a cikin manyan bututu (v)1) = 36 km/h = 36(1000)/(3600) = 36000/3600 = 10 m/s
Babban tsayin bututu (h)1) = -4 mita
Ƙaramin tsayin bututu (h)2) = mita 0
Saurin gudu saboda nauyi (g) = 10 ms-2
Yawan ruwa (ρ) = 1000 kg.m-3
An tambaya: Gudun ruwa a cikin ƙaramin bututu (v)2)
Amsa:
Gudun ruwa a cikin ƙaramin bututu (v)2) ana ƙididdige shi ta amfani da lissafin Bernoulli:
910.000 + ½ (1000)(10)2) + (1000) (10) (0) = 200.000 + ½ (1000) (v)22) + (1000)(10)(-4)
910.000 + 50.000 + 0 = 200.000 + 500 v22 - 40.000
960.000 = 160.000 + 500 v22
800.000 = 500 v22
800.000 / 500 = v22
1600 = ku22
v2 = √1600
v2 = 40m/s
Amsar da ta dace ita ce D.
Tushen tambaya:
Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a
