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Sejumlah darah yang massa jenisnya 1,05 g / cm 3 dipompakan dari jantung menuju otak dalam debit20cm/s. Selisih ketinggian otak dan jantung adalah 49 cm. Diameter pembuluh darah di jantung0,25mm sedangkan di otak 0,01 mm. Jika tekanan darah saat di jantung 2,0 atm danpercepatangravitasi 10 m / s 2 , maka berapa tekanan darah saat di otak? ( 1 atm = 1 , 01 × 1 0 5 Pa ; 1 Pa = 1 N / m 2 ) .

Sejumlah darah yang massa jenisnya 1,05  dipompakan dari jantung menuju otak dalam debit 20 cm/s. Selisih ketinggian otak dan jantung adalah 49 cm. Diameter pembuluh darah di jantung 0,25 mm sedangkan di otak 0,01 mm. Jika tekanan darah saat di jantung 2,0 atm dan percepatan gravitasi 10 , maka berapa tekanan darah saat di otak? 

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Y. Maghfirah

Master Teacher

Jawaban terverifikasi

Jawaban

besar tekanan ke dua nya adalah 3,28124157 x 10 10 Pa

besar tekanan ke dua nya adalah 3,28124157 x 1010 Pa

Pembahasan

Diketahui: Untuk menjawab soal ini gunakan persamaan kontinuitas untuk menentukan v 1 dan v 2 Substitusikan v 1 dan v 2 ke persamaan bernouli: Sehingga besar tekanan ke dua nya adalah 3,28124157 x 10 10 Pa

Diketahui: 

table attributes columnalign right center left columnspacing 0px end attributes row rho equals cell 1 comma 05 space straight g divided by cm cubed equals 1050 space kg divided by straight m cubed end cell row Q equals cell 20 space cm cubed divided by straight s equals 2 cross times 10 to the power of negative 5 space end exponent straight m cubed divided by straight s end cell row cell increment h end cell equals cell 49 space cm equals 49 cross times 10 to the power of negative 2 end exponent space straight m end cell row cell d subscript 1 end cell equals cell 25 cross times 10 to the power of negative 5 space end exponent straight m end cell row cell d subscript 2 end cell equals cell 1 cross times 10 to the power of negative 5 end exponent space straight m end cell row cell P subscript 1 end cell equals cell 2 space atm equals 2 comma 02 cross times 10 to the power of 5 space Pa end cell end table

Untuk menjawab soal ini gunakan persamaan kontinuitas untuk menentukan v1 dan v2

table attributes columnalign right center left columnspacing 0px end attributes row cell A subscript 1 space v subscript 1 end cell equals Q row cell 1 fourth pi d subscript 1 squared space v subscript 1 end cell equals cell 2 cross times 10 to the power of negative 5 space end exponent end cell row cell v subscript 1 end cell equals cell fraction numerator 8 cross times 10 to the power of negative 5 space end exponent over denominator 3 comma 14 cross times open parentheses 25 cross times 10 to the power of negative 5 space end exponent close parentheses squared end fraction end cell row cell space space space space end cell equals cell fraction numerator 8 cross times 10 to the power of negative 5 space end exponent over denominator 1.962 comma 5 cross times 10 to the power of negative 10 end exponent end fraction end cell row cell space space space space end cell equals cell 4 cross times 10 to the power of negative 3 end exponent cross times 10 to the power of 5 end cell row cell space space space space end cell equals cell 400 space straight m divided by straight s end cell row blank blank blank row cell A subscript 1 v subscript 1 end cell equals cell A subscript 2 v subscript 2 end cell row cell 1 fourth pi d subscript 1 squared cross times 400 end cell equals cell 1 fourth pi d subscript 2 squared cross times v subscript 2 end cell row cell open parentheses 25 cross times 10 to the power of negative 5 space end exponent close parentheses squared cross times 400 end cell equals cell open parentheses 1 cross times 10 to the power of negative 5 end exponent close parentheses squared cross times v subscript 2 end cell row cell 625 cross times 400 end cell equals cell v subscript 2 end cell row cell v subscript 2 end cell equals cell 25 cross times 10 to the power of 4 space straight m divided by straight s end cell end table

Substitusikan v1 dan v2 ke persamaan bernouli:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P subscript 1 plus 1 half ρv subscript 1 squared plus ρgh subscript 1 end cell equals cell straight P subscript 2 plus 1 half ρv subscript 2 plus ρgh subscript 2 end cell row blank blank blank row cell straight P subscript 1 end cell equals cell straight P subscript 2 plus 1 half ρv subscript 2 minus 1 half ρv subscript 1 squared plus ρgh subscript 2 minus ρgh subscript 1 end cell row cell straight P subscript 1 end cell equals cell straight P subscript 2 plus 1 half straight rho open parentheses straight v subscript 2 squared minus straight v subscript 1 squared close parentheses plus ρg increment straight h end cell row cell 2 comma 02 cross times 10 to the power of 5 end cell equals cell straight P subscript 2 plus 1 half cross times 1050 open parentheses open parentheses 25 cross times 10 to the power of 4 close parentheses squared minus 400 squared close parentheses plus open parentheses 1050 cross times 10 cross times 49 cross times 10 to the power of negative 2 end exponent close parentheses end cell row cell 2 comma 02 cross times 10 to the power of 5 end cell equals cell straight P subscript 2 plus 525 open parentheses 624 comma 9984 cross times 10 to the power of 8 close parentheses plus 5145 end cell row cell 2 comma 02 cross times 10 to the power of 5 end cell equals cell straight P subscript 2 plus 328124 comma 16 cross times 10 to the power of 8 plus 5145 end cell row cell 2 comma 02 cross times 10 to the power of 5 end cell equals cell straight P subscript 2 plus 328124 comma 160 cross times 10 to the power of 5 plus 5145 end cell row cell straight P subscript 2 end cell equals cell plus-or-minus open parentheses negative 328124 comma 157 cross times 10 to the power of 5 close parentheses end cell end table undefined 

Sehingga besar tekanan ke dua nya adalah 3,28124157 x 1010 Pa

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