Misalai 9 na Dokar Pascal
1. Yankin giciye na kowane jack na hydraulic shine 0,04 m2 da kuma 0,10 m2Idan ƙarfin shigarwar shine Newtons 5, menene matsakaicin ƙarfin fitarwa?
Tattaunawa:
An san cewa:
A1 = 0,04 m2
A2 = 0,10 m2
F1 = 5N
An tambaya: F2 ?
Amsa:
F1/A1 = F2/A2
5/0,04 = F2/ 0,10
125=F2/ 0,10
F 2 = (125)(0,10) = 12,5 N
2. Radius ɗin ƙaramin ɓangaren giciye na jack ɗin hydraulic shine 2 cm kuma radius na babban ɓangaren giciye shine 25 cm. Nawa ƙarfin da aka yi amfani da shi ga ƙaramin ɓangaren giciye don ɗaga mota mai nauyin kilogiram 2000?
Tattaunawa:
An san cewa:
r1 = 2 cm = 0,02 m
r2 = 25 cm = 0,25 m
A1 = (3,14)(0,02)2 = 0,001256 m2
A2 = (3,14)(0,25)2 = 0,19625 m2
F2 = w = mg = (2000)(9,8 m/s2) = 19600 N
An tambaya: F1 ?
Amsa:
F1/A1 = F2/A2
F1/0,001256 = 19600/0,19625
F1/ 0,001256 = 99.872,6
F 1 = 125,44 N
3. Matsakaicin matsin lamba akan jack ɗin hydraulic shine atm 10. Menene matsakaicin nauyi (kg) da za a iya ɗagawa idan diamita na fitarwa shine cm 20?
Tattaunawa:
An san cewa:
1 atm = 1,013 x 105 N / m2
p aunawa = 10 atm = (10)(1,013 x 105 N / m2= 10,13 x 105 N / m2
F1 = 10,13 x105 N / m2
A1 = 1 m2
A2 = (3,14)(0,10 m)2 = 0,0314 m2
An tambaya: F2 ?
Amsa:
p = F/A ko F1/A1 = F2/A2
1013000/1 = F2/ 0,0314
1.013.000=F2/ 0,0314
F 2 = N 31.808
Matsakaicin nauyi:
w = mg
31.808 = m (9,8)
m = 3.245,7 kg
4. Kalli hoton da ke ƙasa!
Yankin giciye A 1 = 10 cm 2
Yankin giciye A 2 = 100 cm 2
Ƙarfin (F 1 ) da dole ne a yi amfani da shi don riƙe F 2 = 100 N don tsarin ya kasance cikin daidaito shine ...
A. 1.000 Newtons
B. 100 Newtons
C. 10 Newtons
D. 1 Newtons
Tattaunawa
An san cewa:
Yankin giciye 1 (A 1 ) = 10 cm 2
Yankin giciye 2 (A 2 ) = 100 cm 2
Ƙarfi 2 (F 2 ) = 100 Newtons
An tambaya: Ƙarfi 1 (F 1 )
Amsa:
Kayan aikin da ke cikin hoton da ke sama jack ne na hydraulic. Yadda jack na hydraulic ke aiki ya dogara ne akan dokar Pascal. Dokar Pascal ta bayyana cewa matsin lamba da aka sanya wa ruwa yana yaɗuwa ta hanyar ruwan ta kowace hanya, tare da dukkan sassan ruwan suna fuskantar matsin lamba iri ɗaya.
Tsarin matsi:
P = F / A
Bayanin dabara: P = matsin lamba, F = ƙarfi, A = yankin saman.
Matsin da ruwa ke fuskanta a cikin ƙaramin bututu (P)1) = F1 / ZUWA1.
Matsin da ruwa ke fuskanta a cikin babban bututu (P)2) = F2 / ZUWA2
Matsin ruwan a cikin ƙaramin bututu (P)1) yayi daidai da matsin lambar ruwa a cikin babban bututun (P2).
P1 =P2
F1 / ZUWA1 = F2 / ZUWA2
F 1 / 10 cm 2 = 100 N / 100 cm 2
F 1 / 10 = 1 N
F 1 = (10)(1 N)
F 1 = 10 Newton
Amsar da ta dace ita ce C.
5. Kalli hoton da ke ƙasa!
Idan yankin giciye na A 1 = 0,001 m2 da A 2 = 0,1 m2 , to girman ƙarfin matsin lamba a A 2 shine...
A. 100 N
B. 1.000 N
C. 10.000 N
D. 100.000 N
Tattaunawa
An san cewa:
Yankin giciye 1 (A 1 ) = 0,001 m 2
Yankin giciye 2 (A 2 ) = 0,1 m 2
Ƙarfin da ke kan yankin giciye 1 (F 1 ) = 100 Newton
Tambaya: Ƙarfi a yankin giciye 2 (F 2 )
Amsa:
Dokar Pascal ta bayyana cewa matsin lamba da aka sanya wa ruwa yana yaɗuwa ta hanyar ruwan ta kowace hanya, inda dukkan sassan ruwan ke fuskantar irin wannan matsin lamba.
P1 =P2
F1 / ZUWA1 = F2 / ZUWA2
100 N / 0,001 m 2 = F 2 / 0,1 m 2
100 N / 0,001 = F 2 / 0,1
100.000 N = F 2 / 0,1
F 2 = (0,1)(100.000 N)
F 2 = N 10.000
Amsar da ta dace ita ce C.
6. Kalli hoton!
An sanya motar da nauyinta ya kai N 16.000 a kan piston B kamar yadda aka nuna a hoton. Don ɗaga motar, ana buƙatar ƙarfin F na….
A. 50 N
B. 80 N
C. 200 N
D. 400 N
Tattaunawa
Bayyana dukkan na'urori zuwa na'urorin ƙasashen waje
An san cewa:
Nauyin mota (w) = 16.000 N
Faɗin saman B (A B ) = 4000 cm 2 = 4000 / 10.000 m 2 = 4 / 10 m 2 = 0,4 m 2
Faɗin saman A (A A ) = 50 cm 2 = 50 / 10.000 m 2 = 0,005 m 2
An tambaya: Force F
Amsa:
Ana ƙididdige ƙarfin F ta amfani da dabarar dokar Pascal.
F / A A = w / A B
F / 0,005 m 2 = 16.000 N / 0,4 m 2
F / 0,005 = 16.000 N / 0,4
F / 0,005 = N40.000
F = (40.000 N)(0,005)
F = 200 Newton
Amsar da ta dace ita ce C.
7. Kalli hoton kayan aikin hydraulic mai siffar U a ƙasa!
Nauyin toshe X akan ƙaramin bututun don daidaita kayan aikin hydraulic shine….
A. 5 N
B. 25 N
C. 10.000 N
D. 25.000 N
Tattaunawa
An san cewa:
Faɗin saman ƙaramin silinda (A)1= 5 cm2
Faɗin saman babban silinda (A)2= 100 cm2
Ƙarfin nauyi na tubalin da ke kan babban silinda (F)2) = 500 Newton
Tambaya: Ƙarfin nauyi na tubalin da ke kan ƙaramin bututu (F)1)
Amsa:
Kayan aikin da ke cikin hoton da ke sama jack ne na hydraulic. Yadda jack na hydraulic ke aiki ya dogara ne akan dokar Pascal. Dokar Pascal ta bayyana cewa matsin lamba da aka sanya wa ruwa yana yaɗuwa ta hanyar ruwan ta kowace hanya, tare da dukkan sassan ruwan suna fuskantar matsin lamba iri ɗaya.
Idan aka danna saman ƙaramin bututun, ana matse ruwan da ke cikin ƙaramin bututun. Matsin da ruwan da ke cikin ƙaramin bututun ke fuskanta yana wucewa zuwa ruwan da ke cikin babban bututun. Ruwan da ke cikin babban bututun yana fuskantar matsi iri ɗaya da ruwan da ke cikin ƙaramin bututun. Akasin haka, matsin da ruwan da ke cikin babban bututun ke fuskanta yana wucewa zuwa ruwan da ke cikin ƙaramin bututun. Ruwan da ke cikin ƙaramin bututun yana fuskantar matsi iri ɗaya da wanda ruwan da ke cikin babban bututun ke fuskanta.
Tsarin matsi:
P = F / A
Bayanin dabara: P = matsin lamba, F = ƙarfi, A = yankin saman.
Matsin da ruwa ke fuskanta a cikin ƙaramin bututu (P)1) = F1 / ZUWA1.
Matsin da ruwa ke fuskanta a cikin babban bututu (P)2) = F2 / ZUWA2
Matsin ruwan a cikin ƙaramin bututu (P)1) yayi daidai da matsin lambar ruwan a cikin babban silinda P2).
P1 =P2
F1 / ZUWA1 = F2 / ZUWA2
F1 / 5cm ku2 = 500 N / 100 cm2
F1 / 5 = 500 N / 100
F1 / 5 = 5 N
F1 = (5)(5) N
F1 = 25N
Amsar da ta dace ita ce B.
8.
Kula da tsarin bututun Pascal mai zuwa!
Idan ƙarfin matsi F1 = 300 N. Girman ƙarfin F2 domin daidaito ya faru shine….
A. 72 N
B. 1250 N
C. 1500 N
D. 3600 N
Tattaunawa
An san cewa:
Faɗin saman ƙaramin silinda (A)1= 12 cm2
Faɗin saman babban silinda (A)2= 50 cm2
Girman ƙarfin matsin lamba akan ƙaramin bututu (F)1) = 300 Newton
Tambaya: Girman ƙarfin matsin lamba akan babban bututun (F)2)
Amsa:
Matsin ruwan a cikin ƙaramin bututu (P)1) yayi daidai da matsin lambar ruwa a cikin babban bututun (P2).
P1 =P2
F1 / ZUWA1 = F2 / ZUWA2
300 N / 12 cm2 = F2 / 50cm ku2
300 N / 12 = F2 / 50
25 N = F2 / 50
F2 = (50)(25) N
F2 = 1250N
Amsar da ta dace ita ce B.
9.
Kalli hoton kayan aikin Pascal mai zuwa!
Girman ƙarfin (F) da dole ne a yi amfani da shi don tabbatar da daidaiton matsayin kayan aikin shine….
A. 100 Newtons
B. 250 Newtons
C. 600 Newtons
D. 4800 Newtons
Tattaunawa
An san cewa:
Faɗin saman ƙaramin silinda (A)1= 4 cm2
Faɗin saman babban silinda (A)2= 16 cm2
Girman nauyi a kan babban bututu (F)2) = 1000 Newton
Tambaya: Girman ƙarfin da ke kan ƙaramin bututu (F)1)
Amsa:
Matsin ruwan a cikin ƙaramin bututu (P)1) yayi daidai da matsin lambar ruwa a cikin babban bututun (P2).
P1 =P2
F1 / ZUWA1 = F2 / ZUWA2
F1 / 4cm ku2 = 1000 N / 16 cm2
F1 / 4 = 1000 N / 16
F1 / 1 = 1000 N / 4
F1 = 1000 N / 4
F1 = 250N
Amsar da ta dace ita ce B.
Tushen tambaya:
Tambayoyin Jarrabawar Ƙasa ta Kimiyya ta Makarantar Sakandare/Makarantar Sakandare ta Musulunci
Tambayoyin dokar Pascal
1. Yankin giciye na kowane jack na hydraulic shine 0,03 m2 da kuma 0,12 m2Idan ƙarfin shigarwar shine Newtons 6, menene matsakaicin ƙarfin fitarwa?
(24 N)
2. Radius ɗin ƙaramin ɓangaren giciye na jack ɗin hydraulic shine 1,5 cm kuma radius na babban ɓangaren giciye shine 20 cm. Nawa ƙarfin da aka yi amfani da shi ga ƙaramin ɓangaren giciye don ɗaga mota mai nauyin kilogiram 2200?
(121,3 N)
3. Matsakaicin matsin lamba akan jack ɗin hydraulic shine atm 14. Menene matsakaicin nauyi (kg) da za a iya ɗagawa idan diamita na fitarwa shine cm 30?
(10.224 kg)
