{"id":1558,"date":"2015-11-28T03:21:58","date_gmt":"2015-11-28T03:21:58","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1558"},"modified":"2024-08-23T14:31:40","modified_gmt":"2024-08-23T14:31:40","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-experiencia-de-torricelli-e-vasos-comunicantes-com-liquidos-imisciveis-em-equilibrio","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/hidrostatica\/experiencia-de-torricelli-e-vasos-comunicantes-com-liquidos-imisciveis-em-equilibrio\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-experiencia-de-torricelli-e-vasos-comunicantes-com-liquidos-imisciveis-em-equilibrio\/","title":{"rendered":"Experi\u00eancia de Torricelli e Vasos comunicantes com l\u00edquidos imisc\u00edveis em equil\u00edbrio – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/span><\/span><\/span><\/p>\n

Experi\u00eancia de Torricelli e Vasos comunicantes com l\u00edquidos imisc\u00edveis em equil\u00edbrio<\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) S\u00e3o bolhas de ar que \u00e9 expulso do interior do vidro do conta-gotas.<\/b><\/span><\/span><\/p>\n

b) Com o ar expulso, a press\u00e3o interna diminui e a maior press\u00e3o externa empurra o l\u00edquido para dentro.<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O ar no interior da sanfona e no interior do bulbo de vidro e em contato com a superf\u00edcie do l\u00edquido est\u00e1 igualmente concentrado (igualmente denso) e suportam a mesma press\u00e3o (P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0\u00a0 como 2 (no bulbo de vidro) e 3 (no tubo) est\u00e3o no mesmo n\u00edvel, suportam tamb\u00e9m a mesma press\u00e3o (P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 no tubo , P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>>P<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>, pois 3 suporta maior altura de coluna l\u00edquida que 4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I- Falsa, ser\u00e1 menor, pois no alto da montanha a press\u00e3o \u00e9 menor.<\/b><\/span><\/span><\/p>\n

II- Correta \u2013 veja teoria<\/b><\/span><\/span><\/p>\n

III- Correta \u2013 veja teoria<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A alternativa incorreta \u00e9 a\u00a0<\/b><\/span><\/span>D<\/b><\/span><\/span>, pois com a boca da garrafa totalmente tampada a \u00e1gua n\u00e3o jorrar\u00e1 por nenhum dos dois orif\u00edcios, pois a press\u00e3o externa \u00e9 maior que a interna\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00c0 medida que os sacos sobem sobe seu volume vai aumentando porque a diferen\u00e7a entre a press\u00e3o do ar dentro dos sacos e a press\u00e3o cada vez mais reduzida fora dela gera uma for\u00e7a resultante que empurra o pl\u00e1stico da bexiga para fora\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B.<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo do valor da press\u00e3o atmosf\u00e9rica em Bras\u00edlia\u00a0 —\u00a0 P=d<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>.g.h=13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.9,8.0,67\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 P=8,9.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(Pa)\u00a0 —\u00a0Essa press\u00e3o, em Bras\u00edlia (P=8,9.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>), quando medida com mercurocromo de densidade d<\/b><\/span><\/span>mc<\/b><\/span><\/span><\/sub>=0,99.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>) equivaler\u00e1 \u00e0 uma altura de\u00a0 —\u00a0 P=d<\/b><\/span><\/span>mc<\/b><\/span><\/span><\/sub>.g.h\u2019\u00a0 —\u00a0 8,9.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=0,99.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.9,8.h\u2019\u00a0 —\u00a0 h\u2019=8,9.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\/9,702.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h\u2019=0,92.10<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>h\u2019=9,2m=92dm<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quanto maior a altitude,\u00a0<\/b><\/span><\/span>menor<\/b><\/span><\/span>\u00a0\u00e9 a press\u00e3o e quanto menor a press\u00e3o<\/b><\/span><\/span>\u00a0menor<\/b><\/span><\/span>\u00a0\u00e9 a temperatura de ebuli\u00e7\u00e3o da \u00e1gua\u00a0 —<\/b><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A press\u00e3o ao n\u00edvel do mar corresponde \u00e0 press\u00e3o devida a altura da coluna de ar\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>ar<\/b><\/span><\/span><\/sub>.g.h\u00a0 —\u00a0 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=1,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h\u00a0 —\u00a0 h=78.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>m=7,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>m=7,8km\u00a0 —\u00a0 R<\/b><\/span><\/span>– C<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Ao n\u00edvel do mar o bar\u00f4metro de merc\u00fario equilibra uma coluna de 76cm de Hg\u00a0 —\u00a0 como a press\u00e3o diminui de 1cm de Hg a cada 100m, na altura de 900m a coluna de merc\u00fario diminuir\u00e1 de 9cm e, ter\u00e1 uma altura de P=76-9=67cmHg\u00a0 —\u00a0 sendo a densidade do l\u00edquido 4 vezes menor que a do merc\u00fario, ela equilibrar\u00e1 uma altura 4 vezes maior de l\u00edquido\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P=67 x 4=268cm de l\u00edquido\u00a0 —\u00a0 P=2,68m de l\u00edquido\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>V=4m.5m.3m=60m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 1,2=m\/60\u00a0 —\u00a0 m=72kg\u00a0 —\u00a0 P=mg=72.10\u00a0 —\u00a0 P=720N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo que a press\u00e3o na superf\u00edcie da \u00e1gua (ponto Q) \u00e9 a mesma que aquela devido \u00e0 altura da coluna<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0de \u00e1gua (ponto p)\u00a0 —\u00a0 P<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 densidade d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0e g=9,8m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 — P<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>==d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=<\/b><\/span><\/span>10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.9,8.h<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 h<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=10,3m\u00a0 —\u00a0 portanto, a press\u00e3o atmosf\u00e9rica equilibra uma coluna de \u00e1gua de altura aproximadamente 10m e, essa press\u00e3o n\u00e3o consegue empurrar qualquer coluna de \u00e1gua a uma altura superior a 10m<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Na Lua n\u00e3o existiria atmosfera que exerceria press\u00e3o sobre a coluna de merc\u00fario, de modo que a coluna n\u00e3o subiria.<\/b><\/span><\/span><\/p>\n

b) P<\/b><\/span><\/span>planeta<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>.1g\/2,5.h\u00a0 —\u00a0 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/10=13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10\/2,5.h\u00a0 —\u00a0 h=1,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\/54,4.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=0,0184m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0h=184mm\u00a0<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.10=d<\/b><\/span><\/span>vinho<\/b><\/span><\/span><\/sub>.g.15 —\u00a0 d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>vinho<\/b><\/span><\/span><\/sub>=1,5<\/b><\/span><\/span><\/p>\n

b) menor que 15m\u00a0 —\u00a0 valor compreendido entre 10m e 15m, dependendo da quantidade de \u00e1gua acrescentada ao vinho<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Todos os pontos da superf\u00edcie livre do caf\u00e9 est\u00e3o submetidos \u00e0 mesma press\u00e3o e consequentemente possuem o mesmo n\u00edvel horizontal em rela\u00e7\u00e3o \u00e0 linha de refer\u00eancia do sistema\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Todos os pontos que se encontram na mesma linha horizontal (altura, profundidade)est\u00e3o submetidos \u00e0 mesma press\u00e3o\u00a0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=p<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0(mesmo n\u00edvel)\u00a0 —\u00a0 o g\u00e1s pressiona mais o l\u00edquido que a press\u00e3o atmosf\u00e9rica p\u00f3\u00a0 —\u00a0 p<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>\u00a0> p<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O l\u00edquido 2 \u00e9 mais denso e deve ficar na parte inferior dos vasos\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u00e9 maior que P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0(pressiona mais a coluna de \u00e1gua) e P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u00e9 menor que P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>(pressiona menos a coluna de \u00e1gua)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A press\u00e3o nos pontos M e N \u00e9 a mesma\u00a0 —\u00a0 P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>atm\u00a0<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>.g<\/b><\/span><\/span>..<\/b><\/span><\/span><\/sub>18=P<\/b><\/span><\/span>atm\u00a0<\/b><\/span><\/span><\/sub>+ d<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>.g.10 + d<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>.g.10\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 18.d<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=10.d<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>\u00a0+10.d<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 8d<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=10d<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=8\/10=0,8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A press\u00e3o do g\u00e1s no botij\u00e3o \u00e9 a mesma que no ponto A (P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>), que por sua vez \u00e9 a mesma<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

que no ponto M\u00a0 —P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>M\u00ad<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>.g.h=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.1,04\u00a0 —\u00a0 P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 14,44.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 1,44.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>=2,44.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0Pa<\/b><\/span><\/span><\/p>\n

b) Em a\u00a0 —\u00a0 P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>=F\/S\u00a0 —\u00a0 2,44.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=F\/2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F=48,8N<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A press\u00e3o nos pontos M e N da figura \u00e9 a mesma\u00a0 —\u00a0 P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g.0,8 = P<\/b><\/span><\/span>atm\u00a0<\/b><\/span><\/span><\/sub>\u00a0+<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g.0,5\u00a0 —\u00a0 0,8.d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0=0,5<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sub>d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,8.d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0,5.2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/0,8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1,25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kig\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

b) P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g.0,8=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 1,25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,8\u00a0 —\u00a0 P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 10.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>+ 0,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=1,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Press\u00e3o causada pelo petr\u00f3leo, P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0n\u00e3o participa\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g.h=\u00a0 1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>= 40.10\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=400Pa\u00a0 —<\/b><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/p>\n

23<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ \u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g.h = P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+\u00a0<\/b><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g.0,75h\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0,75\u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/\u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0,75\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=500mL=500.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>L=500.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>dm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=500.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=S.h\u00a0 —\u00a0 5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=S.h\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0—\u00a0 h=5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\/S\u00a0 — d<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=0,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=8.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 — \u00a0P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>.g.h= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 8.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10.5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\/S\u00a0 —\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 4\/S\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>\u00eambolo<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ F\/S=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ p\/S\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ p\/S<\/b><\/span><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 4\/S=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ p\/S\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0p=4N<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe na \u00faltima coluna que o produto\u00a0<\/b><\/span><\/span>\u00a0PxV \u00e9 praticamente constante\u00a0<\/b><\/span><\/span>(Lei de Boyle para transforma\u00e7\u00f5es isot\u00e9rmicas)<\/b><\/span><\/span><\/p>\n

b) Supondo que cada 8 unidades tenha 0,5cm\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 8u – 0,5m\u00a0 —\u00a0 24u \u2013 h\u00a0 —\u00a0 h=1,5m\u00a0 —\u00a0 P=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+\u00a0<\/b><\/span><\/span>dgh=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 14.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.1,5\u00a0 —\u00a0 P=3,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(Pa)<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo a se\u00e7\u00e3o reta circular (S=1,0.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>) no diafragma a mesma, e como a densidade do l\u00edquido A \u00e9 maior que a do l\u00edquido B, a intensidade da for\u00e7a resultante sobre o diafragma vale\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/S\u00a0 —\u00a0 F<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.S\u00a0 —\u00a0 F<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.S\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.S<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>– P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.S=S(P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=S.(d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g.h \u2013 d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g.h)=1,0.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.(5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,1 \u2013 0,5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,1)=<\/b><\/span><\/span><\/p>\n

1,0.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.4,5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=4.500.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0N<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo, onde os pontos R e S est\u00e3o no mesmo n\u00edvel e suportam a mesma<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

press\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>S\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0\u00a0<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.(21 + d)\u00a0 —\u00a0 P<\/b><\/span><\/span>S\u00a0<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>.g.h\u00a0 —\u00a0 P<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=OS\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.(21 + d)= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>.g.h\u00a0 —\u00a0 d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.(21 + d)= \u00a0d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>.h\u00a0 —\u00a0\u00a0 observe que h + d=42\u00a0 —\u00a0 d=42 – h\u00a0 —\u00a0 d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.(21 + 42 – h)=d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>.h\u00a0 —\u00a0 1.63 \u2013 1.h=0,8h\u00a0 —\u00a0 1,8h=63\u00a0 —\u00a0 h=63\/1,8\u00a0 —\u00a0 h=35cm\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) As press\u00f5es exercidas pelos l\u00edquidos I e II no fundo do tubo (pontos M e N) s\u00e3o iguais —\u00a0 P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N\u00a0\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>\u00a0. g . h<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>\u00a0= P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\u00a0. g . h<\/b><\/span><\/span>II\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 1800 . g . 0,20 = d<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\u00a0. g . 0,60\u00a0 —\u00a0 \u00a0\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\u00a0= 600 kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

b) Nos pontos compreendidos entre -80cm at\u00e9 -40cm a press\u00e3o \u00e9 constante e \u00e9 a atmosf\u00e9rica de valor P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 np ponto 80cm a press\u00e3o tamb\u00e9m \u00e9 a atmosf\u00e9rica de valor P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 nos pontos de abscissas -20cm at\u00e9 + 20cm a press\u00e3o \u00e9 constante e vale\u00a0 —\u00a0 P<\/b><\/span><\/span>-20<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm\u00a0<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>= 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 1800.10.0,20=1,036.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0 —\u00a0 de -20cm at\u00e9 + 20cm \u2013 P=1,036.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 entre esses valores a varia\u00e7\u00e3o \u00e9 linear (reta inclinada)\u00a0 —\u00a0 veja gr\u00e1fico abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) Para que os n\u00edveis dos l\u00edquidos se igualem, a superf\u00edcie livre do l\u00edquido I deve subir 20cm, estacionando na abscissa<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

-60cm e a do l\u00edquido II descer 20cm, estacionando na abscissa +60cm\u00a0 — nos pontos M e N a press\u00e3o \u00e9 a mesma e, no ponto M vale\u00a0 —\u00a0 P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>.g.h=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 1.800.10.0,40=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 0,072.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>= 1,072.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 press\u00e3o no ponto de abscissa +60cm\u00a0 —\u00a0 P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>+60cm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>.g.h\u00a0 —\u00a0 1,072.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>+60cm<\/b><\/span><\/span><\/sub>\u00a0+ 600.10.0,40\u00a0 —\u00a0 P<\/b><\/span><\/span>+60cm<\/b><\/span><\/span><\/sub>=1,072.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u2013 0,024.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>+60cm<\/b><\/span><\/span><\/sub>=1,048.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 gr\u00e1fico abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Na figura II, na extremidade do orif\u00edcio de altura h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2,0m a press\u00e3o que atua \u00e9 a press\u00e3o atmosf\u00e9rica P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=dgh<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.2,0 + P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u2013 2,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=8,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

b) Chamando de S a \u00e1rea da base do tanque\u00a0 —\u00a0 lei geral dos gases perfeitos\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.S(6,0 \u2013\u00a0<\/b><\/span><\/span>h<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/300 = 8,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.S.(6,0 \u2013 2,0)\/360\u00a0 —\u00a0 h<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10\/3 m=3,3m<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=<\/b><\/span><\/span>P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ dgh=76cmHg + 30cmHg\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub>=106cmHg\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ \u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g.h\u00a0 —\u00a0 P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ \u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g (h \u2013 0,25h)\u00a0 —\u00a0 P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ \u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g.h=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>+ \u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g.(h-0,25)\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>A.<\/b><\/span><\/span><\/sub>h= \u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.(h-0,25h)\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.h= \u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.0,75h\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/\u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>0,75\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

Para que o \u00f3leo escoe com facilidade da lata, voc\u00ea deve fazer dois furos na mesma, um para que a press\u00e3o atmosf\u00e9rica empurre o \u00f3leo para fora da lata e o outro para que o \u00f3leo saia da mesma.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

32-\u00a0<\/b><\/span><\/span><\/span>a) O escoamento do l\u00edquido se interrompe quando a press\u00e3o no fundo do recipiente se iguala \u00e0 press\u00e3o no bico da torneira, que \u00e9 a press\u00e3o atmosf\u00e9rica p<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 chamando de p a press\u00e3o exercida sobre a superf\u00edcie livre do l\u00edquido nesta situa\u00e7\u00e3o\u00a0 —\u00a0 pelo teorema de Stevin\u00a0 —\u00a0 p<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= p + \u03c1<\/b><\/span><\/span>\uf020<\/b><\/span><\/span>g H\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 P=\u03c1gH\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u0394P=\u03c1gH<\/b><\/span><\/span><\/p>\n

c) Como a press\u00e3o atmosf\u00e9rica diminui com a altitude, a cidade localizada \u00e0 maior altitude \u00e9 aquela na qual \u00e9 menor a press\u00e3o atmosf\u00e9rica\u00a0 — \u00a0<\/b><\/span><\/span>a cidade B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

33-<\/b><\/span><\/span><\/span> <\/b>a) Falsa\u00a0 —\u00a0 uma bomba de suc\u00e7\u00e3o \u00e9 uma bomba que retira o ar de sua extremidade superior, produzindo o v\u00e1cuo\u00a0 —\u00a0 com ar\u00a0 expulso,a press\u00e3o interna diminui e a maior press\u00e3o externa empurra o l\u00edquido para cima\u00a0 —\u00a0 isso ocorre independente da espessura dos canos e da pot\u00eancia do motor que produz a suc\u00e7\u00e3o (v\u00e1cuo).<\/b><\/span><\/span><\/p>\n

b) Correta\u00a0 — Se a experi\u00eancia de Torricelli for realizada ao n\u00edvel do mar, mas com \u00e1gua, cuja densidade \u00e9 13,6 vezes menor que a do merc\u00fario, a altura da coluna de \u00e1gua ser\u00e1 aproximadamente igual a 10,3m.<\/b><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 veja justificativa a.<\/b><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 veja justificativa b.<\/b><\/span><\/span><\/p>\n

e) Falsa.<\/b><\/span><\/span><\/p>\n

R- B\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) O l\u00edquido s\u00f3 vai parar de escoar pela torneira quando a press\u00e3o (P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) em todos os pontos do fundo do recipiente for igual \u00e0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

press\u00e3o na ponta da torneira que est\u00e1 aberta e essa press\u00e3o \u00e9 a press\u00e3o atmosf\u00e9rica (P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 isso ocorre porque todos os pontos de mesmo n\u00edvel de altura (mesma horizontal) suportam a mesma press\u00e3o (teorema de Stevin)\u00a0 —\u00a0 chamando de P a press\u00e3o na interface ar-l\u00edquido e de P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>(press\u00e3o atmosf\u00e9rica) a press\u00e3o na parte inferior do l\u00edquido, pelo teorema de Stevin\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= P + \u03c1.g.H\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P= P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0– \u03c1.g.H.<\/b><\/span><\/span><\/p>\n

b) Veja a figura abaixo que mostra a press\u00e3o e o volume do ar contido no recipiente antes e depois do volume \u2206V de \u00e1gua ser<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

transferido\u00a0 — \u00a0numa evolu\u00e7\u00e3o isot\u00e9rmica a temperatura \u00e9 constante (T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= T)\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= P.V\/T\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V = P.(V + \u2206V)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V\/(V + \u2206V).<\/b><\/span><\/span><\/p>\n

c) Observe no gr\u00e1fico que a menor press\u00e3o atmosf\u00e9rica (P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=\u03c1.g.H) \u00e9 a da cidade B\u00a0 —\u00a0 como a press\u00e3o atmosf\u00e9rica diminui com a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

altitude,quanto menor a press\u00e3o atmosf\u00e9rica maior \u00e9 a altitude da cidade\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a cidade B est\u00e1 numa altitude maior.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/a><\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Experi\u00eancia de Torricelli e Vasos comunicantes com l\u00edquidos imisc\u00edveis em equil\u00edbrio 01-\u00a0a) S\u00e3o bolhas de ar que \u00e9 expulso do interior do vidro do conta-gotas. b) Com o ar expulso, a press\u00e3o interna diminui e a maior press\u00e3o externa empurra o l\u00edquido para dentro. 02-\u00a0O ar no interior da sanfona e no interior<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1554,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1558","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1558","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1558"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1558\/revisions"}],"predecessor-version":[{"id":10890,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1558\/revisions\/10890"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1554"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1558"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}