{"id":1589,"date":"2015-11-29T12:40:37","date_gmt":"2015-11-29T12:40:37","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1589"},"modified":"2024-08-23T14:27:02","modified_gmt":"2024-08-23T14:27:02","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-equacao-da-continuidade-teorema-de-bernoulli","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/hidrodinamica\/equacao-da-continuidade-equacao-de-bernoulli\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-equacao-da-continuidade-teorema-de-bernoulli\/","title":{"rendered":"Equa\u00e7\u00e3o da Continuidade \u2013 Teorema de Bernoulli – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Equa\u00e7\u00e3o da Continuidade \u2013 Teorema de Bernoulli<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0A vaz\u00e3o \u00e9 a mesma \u2013 regime estacion\u00e1rio\u00a0 —<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0S=100cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Z=7.200L\/h=7.200\/3.600L\/s=2.L\/s\u00a0 —\u00a0 Z=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s\u00a0 —\u00a0 Z=S.v\u00a0 —\u00a0 2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>= 10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>v\u00a0 —\u00a0 v=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 v=0,2m\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0\u0394V=18.10.2=360m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394t=10h\u00a0 —\u00a0 S=25.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Z=\u0394V\/\u0394t=360\/10\u00a0 —\u00a0 Z=36m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/h\u00a0 —\u00a0 Z=S.v\u00a0 —\u00a0 36=25.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.v\u00a0 —\u00a0 v=36\/25.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 v=14.400m\/h\u00a0 —\u00a0 v=14.400\/3.600=4m\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0O jato de ar que se move com velocidade v, paralelamente ao extremo (A) de um tubo que est\u00e1 imerso em um l\u00edquido, faz com que a press\u00e3o a\u00ed diminua em rela\u00e7\u00e3o ao extremo inferior (ponto B) do tubo.<\/b><\/span><\/span><\/p>\n

A diferen\u00e7a de press\u00e3o entre os pontos A e B empurra o fluido para cima. O ar r\u00e1pido tamb\u00e9m divide o fluido em pequenas gotas, que s\u00e3o empurradas e se espalham para a frente\u00a0 —\u00a0 R- E<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0R- C\u00a0 —\u00a0 veja exerc\u00edcio anterior<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0Vaz\u00e3o\u00a0 —\u00a0 Z=0,01m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s\u00a0 —\u00a0 volume total\u00a0 —\u00a0 \u0394V=10×1.500=15.000L=15.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394V=15m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Z= \u0394V\/ \u0394t\u00a0 —\u00a0 10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=15\/ \u0394t\u00a0 —\u00a0 \u0394t=1.500s=25min\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- B\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Veja a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2d<\/b><\/span><\/span>1\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2r<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c0(r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03c0(r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 — (r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=(2r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 (r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4(r<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u0394h\/\u0394t=9\/3\u00a0 —\u00a0 v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=3cm\/s\u00a0 —\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4.3=v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=12cm\/s<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Maior \u00e1rea de se\u00e7\u00e3o transversal (1), menor velocidade, maior press\u00e3o\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0(01) correta \u2013 \u201cos acr\u00e9scimos de press\u00e3o sofridos por um ponto de um l\u00edquido em equil\u00edbrio s\u00e3o transmitidos integralmente a todos os pontos do l\u00edquido e das paredes do recipiente que o cont\u00e9m\u201d\u00a0 —\u00a0 Princ\u00edpio de Pascal.<\/b><\/span><\/span><\/p>\n

(02) Falsa, a vaz\u00e3o \u00e9 a mesma, quem aumenta \u00e9 a velocidade de sa\u00edda da \u00e1gua.<\/b><\/span><\/span><\/p>\n

(04)\u00a0 Falsa\u00a0 — Observe na equa\u00e7\u00e3o P + d.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=constante que a press\u00e3o P \u00e9 inversamente proporcional \u00e0 velocidade v, ou seja, quanto menor a \u00e1rea, maior a velocidade e menor a press\u00e3o.<\/b><\/span><\/span><\/p>\n

(08) Correta \u2013 \u00e9 a press\u00e3o atmosf\u00e9rica \u2013 Veja (01)<\/b><\/span><\/span><\/p>\n

(16) Correta \u2013 a partir da sa\u00edda, as part\u00edculas de \u00e1gua ficam sujeitas \u00e0 for\u00e7a peso, desprezando-se a resist\u00eancia do ar, e descrevem um arco de par\u00e1bola.<\/b><\/span><\/span><\/p>\n

( 01 + 08 + 16) = 25<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>(01) Correta \u2013 P=dgh (teorema de Stevin)\u00a0 —\u00a0 como h diminui, com d e g constantes, a press\u00e3o P tamb\u00e9m diminui.<\/b><\/span><\/span><\/p>\n

(02) Falsa, \u00e9 a press\u00e3o atmosf\u00e9rica.<\/b><\/span><\/span><\/p>\n

(04) Correta \u2013 a press\u00e3o no ponto B (press\u00e3o atmosf\u00e9rica) \u00e9 maior que a press\u00e3o no ponto A.<\/b><\/span><\/span><\/p>\n

(08) Correta \u2013 veja (04)<\/b><\/span><\/span><\/p>\n

(16) Falsa\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0e P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mesmo n\u00edvel horizontal \u2013 teorema de Stevin<\/b><\/span><\/span><\/p>\n

(01 + 04 + 08) = 13<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Z<\/b><\/span><\/span>constante<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0= S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.v=S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.2v\u00a0 —\u00a0 S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=2S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A vaz\u00e3o total das duas torneiras \u00e9 Z=5L\/min + 3L\/min=8L\/min\u00a0 —\u00a0 Z=8.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/min\u00a0 —\u00a0 Z=S.v\u00a0 —\u00a0 8.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=0,8.v\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

v=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\/min\u00a0 —\u00a0 v=1cm\/min\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0A vaz\u00e3o \u00e9 sempre a mesma independente da espessura da mangueira\u00a0 —\u00a0 no lan\u00e7amento horizontal a velocidade v \u00e9 a mesma e trata-se de um movimento uniforme de equa\u00e7\u00e3o\u00a0 —\u00a0 S=S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ vt<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 S=v.t\u00a0 —\u00a0 se o alcance S \u00e9 quadruplicado, a velocidade v tamb\u00e9m \u00e9 quadruplicada\u00a0\u00a0 —\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c0.(R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v=\u03c0.(R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.4v\u00a0 —\u00a0 (R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/(R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1\/4\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Z=S.v=\u03c0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v=3,14.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.33\u00a0 —\u00a0 Z=104cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Z=0,104L\/s (transporta 0,104 litros em cada 1 segundo)<\/b><\/span><\/span><\/p>\n

b) Z=<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>V\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>t\u00a0 —\u00a0 0,104=5\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>t=48s<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Comprimento da canal\u00a0 —\u00a0 s=4m\u00a0 —\u00a0 tempo de percurso\u00a0 —\u00a0 t=1min=60s\u00a0 —\u00a0 velocidade da \u00e1gua\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 v=s\/t=4\/60\u00a0 —V=1\/15m\/s\u00a0 —\u00a0 \u00e1rea de se\u00e7\u00e3o transversal do canal\u00a0 —\u00a0 R=1m\u00a0 — \u00a0S=\u03c0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=\u03c0.(1\/2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 S=\u03c0\/8\u00a0 —\u00a0 vaz\u00e3o\u00a0 —\u00a0 Z=S.v=(\u03c0\/8).(1\/15)\u00a0 —\u00a0 Z=\u03c0\/120m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s (m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.s<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0I- Correta \u2013 Teorema de Stevin (P=d.g.h) \u2013 todos os pontos de um mesmo l\u00edquido (mesma densidade) localizados num mesmo n\u00edvel horizontal (no caso,mesma altura), suportam a mesma press\u00e3o.<\/b><\/span><\/span><\/p>\n

II- Falso \u2013 est\u00e3o em alturas diferentes\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub><P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/p>\n

III- Correta –\u00a0<\/b><\/span><\/span>vi<\/b><\/span><\/span>scosidade \u00e9 definida como a resist\u00eancia que um fluido oferece ao seu pr\u00f3prio movimento. Quanto maior for a sua viscosidade, menor ser\u00e1 a sua capacidade de escoar (fluir) e maior ser\u00e1 a for\u00e7a de atrito entre o fluido e as paredes do recipiente onde ele est\u00e1 escoando \u00a0<\/b><\/span><\/span><\/p>\n

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

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

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Z=S.v=200.1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0Z=200m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s<\/b><\/span><\/span><\/p>\n

b) Z=S.v\u00a0 —\u00a0 200=40.v\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0v=5,0m\/s<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0a) 10L -1h\u00a0 —\u00a0 150L \u2013 t h\u00a0 —\u00a0 t=15h<\/b><\/span><\/span><\/p>\n

b) Z=\u0394V\/\u0394t=10.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/1h\u00a0 —\u00a0 Z=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/h\u00a0 —\u00a0 Z=S.v\u00a0 —\u00a0 10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=5.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>.v\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>v=2,0.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\/h
\n<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Vaz\u00e3o\u00a0 —\u00a0 Z=S.v=2,5.30\u00a0 —\u00a0 Z=75cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s\u00a0 —\u00a0 Z=\u0394V\/\u0394t\u00a0 —\u00a0 75=5,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/\u0394t\u00a0 —\u00a0 \u0394t=5,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/75=0,072.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394t=72s<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0Z=v.S=\u221a(2gh).S=\u221a(2.10.5).3.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=10.3.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=3.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>s\u00a0 —\u00a0 Z=3L\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0Equa\u00e7\u00e3o de Bernoulli\u00a0 —\u00a0 tubula\u00e7\u00e3o horizontal \u2013 h=0\u00a0 —\u00a0 s\u00f3 tem energia cin\u00e9tica\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>+ dv<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ dv<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

1,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.(2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.(8)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 — 152.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ 32.10<\/b><\/span><\/span>3\u00a0\u00a0<\/b><\/span><\/span><\/sup>—\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=152.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u2013 32.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=120.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=1,2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span><\/p>\n

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

24-<\/b><\/span><\/span><\/span>\u00a0S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=2S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.5=S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 — \u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=10m\/s \u00a0—\u00a0\u00a0<\/b><\/span><\/span>teorema de Bernoulli\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ d.g.h<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ d.(V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 =<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ d.g.h<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ d.(V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 7.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0+ 8.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10.10 + 8.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.(5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ 8.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10.1 + 8.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.(10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 7.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0+ 80.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0+ 10.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ 8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0+ 40.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 97.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ 48.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=49.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=4,9.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0a) v=180km\/h\/3,6=50\u00a0 —\u00a0 v=50m\/s\u00a0 —\u00a0 redu\u00e7\u00e3o da press\u00e3o\u00a0 —\u00a0 \u0394P=\u03c1v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=1,2.(50)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u0394P=1,5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

b) varia\u00e7\u00e3o de press\u00e3o=for\u00e7a\/\u00e1rea\u00a0 —\u00a0 \u0394P=peso\/S\u00a0 —\u00a0 1,5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=m.10\/5.400\u00a0 —\u00a0 m=81.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>kg\u00a0 —<\/b><\/span><\/span>\u00a0 m=8,1.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>t<\/b><\/span><\/span><\/p>\n

c) \u03c1v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=\u0394P=peso\/\u00e1rea\u00a0 —\u00a0 1,2.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=m.g\/5.400\u00a0 —\u00a0\u00a0 1,2.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=250.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10\/5.400\u00a0 —\u00a0 v=\u221a771,6\u00a0 —\u00a0 v=27,77m\/sx3,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>v=100km\/h<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Z=v.S=\u221a(2gh).S=\u221a(2.10.5).2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Z=\u221a100.2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0Z=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) entrada\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=2\/1=1cm=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 h<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 v<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=1,5m\/s\u00a0 — \u00a0segundo andar\u00a0 —\u00a0 P<\/b><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1\/2=0,5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 h<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=5m\u00a0 —\u00a0 S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 \u03c0.(R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=\u03c0.(R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 (10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.1,5=(5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —<\/b><\/span><\/span><\/p>\n

1,5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=25.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1,5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\/25.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0,06.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=6m\/s<\/b><\/span><\/span><\/p>\n

b) Bernoulli\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ d(v<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>+ d.g.h<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0= P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ d(v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>+ d.g.h<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.(1,5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>+ 10.10.0 = P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>(6)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.5\u00a0 —\u00a0 40.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0+ 0,1125.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0+ 0 = P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ 1,8.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0+ 5.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=40,1125.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0\u2013 6,8.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=33,3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=3,3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa<\/b><\/span><\/span><\/p>\n

c) vaz\u00e3o\u00a0 —\u00a0 Z=S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u03c0.(R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.6=3,14.(5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.6\u00a0 —\u00a0 Z=471.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=4,71.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span>\u00a0Z=4,71.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s\u00a0 ou\u00a0 Z=0,471L\/s<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0a)<\/b><\/span><\/span>\u00a0Falsa\u00a0<\/b><\/span><\/span>\u00a0– se as alturas estivessem no mesmo n\u00edvel, as velocidades do vento em cada uma delas seriam iguais e, assim n\u00e3o haveria diferen\u00e7a de press\u00e3o para empurrar o ar, n\u00e3o havendo ventila\u00e7\u00e3o dentro da toca.<\/b><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span>\u00a0Correta<\/b><\/span><\/span>\u00a0\u2013 o arbusto diminui a velocidade do vento na abertura 1 aumentando, nela, a press\u00e3o. Assim, a diferen\u00e7a de press\u00e3o entre as aberturas ser\u00e1 aumentada, favorecendo a ventila\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

c) Como as alturas s\u00e3o constantes, a diferen\u00e7a de energia potencial gravitacional tamb\u00e9m \u00e9 constante\u00a0\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ dv<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2= P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ dv<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2\u00a0 — \u0394P= P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=d\/2(v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>)\u00a0 — \u0394P \u00e9 diretamente proporcional \u00e0 diferen\u00e7a do m\u00f3dulo do quadrado das velocidades\u00a0 —<\/b><\/span><\/span>\u00a0 Falsa<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

d) Correta \u2013 ocorre da abertura de menor velocidade do vento, maior press\u00e3o (abertura 1) para a abertura de maior velocidade do vento, menor press\u00e3o (abertura 2)\u00a0<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0a) \u0394t=1min e 40s=60 + 40\u00a0 —\u00a0 \u0394t=100s\u00a0 —\u00a0 \u0394S=20m\u00a0 —\u00a0 v=\u0394S\/\u0394t=100\/20\u00a0 —\u00a0 v=5m\/s\u00a0 —\u00a0 vaz\u00e3o\u00a0 —\u00a0 Z=\u0394V\/\u0394t<\/b><\/span><\/span><\/p>\n

=500.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/100\u00a0 —\u00a0 Z=5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/s\u00a0 —\u00a0 energia utilizada para elevar a \u00e1gua a uma altura h=20m num local onde g=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394W=dgh=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.20\u00a0 —\u00a0 \u0394W=2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0 P<\/b><\/span><\/span>o\u00fatil<\/b><\/span><\/span><\/sub>=\u0394W.Z=2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>o\u00fatil<\/b><\/span><\/span><\/sub>=1.000W (J\/s)\u00a0 —\u00a0 rendimento (\u03b7=P<\/b><\/span><\/span>o\u00fatil<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>ototal<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 0.5=1.000\/P<\/b><\/span><\/span>ototal\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>ototal<\/b><\/span><\/span><\/sub>=2.000W<\/b><\/span><\/span><\/p>\n

b) V=S.h\u00a0 —\u00a0 5.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>=2.h\u00a0 —\u00a0 h=0,25m\u00a0 —\u00a0 P=dgh=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,25\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=2,5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0Z=400cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>s=4.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Z=4.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>s\u00a0 —\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Z\u00a0 —\u00a0 constante\u00a0 —\u00a0 Z=S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

4.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2m\/s\u00a0 —\u00a0 Z=S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4m\/s\u00a0 —\u00a0 Stevin em 2\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=d.g.h=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,5\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=o\u00a0 —\u00a0 dgh=0\u00a0 —\u00a0 Bernoulli\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ d(v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>+ d(v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.4\/2=5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.16\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=13.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=11.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(N.m<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0Veja a figura abaixo\u00a0 —\u00a0 B<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 bomba no len\u00e7ol\u00a0 —\u00a0 B<\/b><\/span><\/span>S<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 bomba no solo\u00a0 —\u00a0 trabalho (energia)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0para elevar a \u00e1gua a uma altura h\u00a0 —\u00a0 W=d.g.h\u00a0 —\u00a0 pot\u00eancia \u00fatil\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>=W.Z\u00a0 —\u00a0 P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>=d.g.h.Z \u00a0—\u00a0 BS \u2013 P<\/b><\/span><\/span>uS<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.100.0,03 =3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>= =3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.1\/750\u00a0 —\u00a0 P<\/b><\/span><\/span>uS<\/b><\/span><\/span><\/sub>=40 hp\u00a0 —\u00a0 rendimento de 80%\u00a0 —\u00a0 \u03b7=P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,8=40\/P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>tS<\/b><\/span><\/span><\/sub>=50 hp\u00a0 —\u00a0 B<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 bomba no len\u00e7ol\u00a0 —\u00a0 P<\/b><\/span><\/span>uL<\/b><\/span><\/span><\/sub>=d.g.h.Z=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.250.0,03\u00a0 —\u00a0\u00a0\u00a0 P<\/b><\/span><\/span>uL<\/b><\/span><\/span><\/sub>=100 hp\u00a0 —\u00a0 \u03b7=\u00a0 P<\/b><\/span><\/span>uL<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>tL<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,8=100\/P<\/b><\/span><\/span>tL<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0\u00a0 P<\/b><\/span><\/span>tL<\/b><\/span><\/span><\/sub>=125 hp\u00a0 —\u00a0 utilizando o compressor\u00a0 —\u00a0 h<\/b><\/span><\/span>compressor<\/b><\/span><\/span><\/sub>=1,5h<\/b><\/span><\/span>BS<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>ucompressor<\/b><\/span><\/span><\/sub>=1,5.P<\/b><\/span><\/span>uBS<\/b><\/span><\/span><\/sub>\u00a0 — P<\/b><\/span><\/span>ucompressor<\/b><\/span><\/span><\/sub>=1,5.40\u00a0 —\u00a0 P<\/b><\/span><\/span>ucompressor<\/b><\/span><\/span><\/sub>=65 hp\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03b7=P<\/b><\/span><\/span>ucompressor<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>tcompressor<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>tcompressor<\/b><\/span><\/span><\/sub>=75 hp\u00a0 —\u00a0 0) Falsa\u00a0 —\u00a0 dever\u00e1 ser de 125 hp\u00a0 —\u00a0 1) verdadeira\u00a0 —\u00a0 2) verdadeira\u00a0 —\u00a0 3) verdadeira\u00a0 —\u00a0 4) verdadeira<\/b><\/span><\/span><\/p>\n

32<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> P=d.g.h.Z=d.g.h.\u0394V\/\u0394t\u00a0 — P=0,5cv=0,5.750\u00a0 —\u00a0 P=375W\u00a0 —\u00a0\u00a0 375=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.15.500.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/\u0394t\u00a0 —\u00a0 \u0394t=75.000\/375=200s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Apenas a afirma\u00e7\u00e3o II \u00e9 incorreta, pois os pontos A e B est\u00e3o no mesmo l\u00edquido, mas em alturas distintas e logo pela Lei de Stevin est\u00e3o sob press\u00f5es diferentes.<\/b><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria\u00a0<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Voc\u00ea pode chegar \u00e0 op\u00e7\u00e3o correta sem apelar para equa\u00e7\u00f5es ou leis da F\u00edsica, mas apenas se baseando em fatos do cotidiano\u00a0 —\u00a0 quando voc\u00ea joga \u00e1gua no jardim ou lava o carro com uma mangueira convencional, voc\u00ea coloca o polegar na extremidade de sa\u00edda da \u00e1gua para diminuirmos a \u00e1rea de fluxo\u00a0 —\u00a0 isso, consequentemente, provoca um aumento de press\u00e3o e um aumento na velocidade, lan\u00e7ando a \u00e1gua \u00e0 maior dist\u00e2ncia\u00a0 —\u00a0 fisicamente voc\u00ea pode usar a equa\u00e7\u00e3o da continuidade e a equa\u00e7\u00e3o de Bernoulli\u00a0 —\u00a0 sendo Q a vaz\u00e3o, v a velocidade do fluxo e A a \u00e1rea da sec\u00e7\u00e3o transversal, a equa\u00e7\u00e3o da continuidade\u00a0 —\u00a0 Q = v A. (I)\u00a0 —\u00a0 a equa\u00e7\u00e3o de Bernoulli relaciona o acr\u00e9scimo de press\u00e3o (p) com a altura de bombeamento (h) e com a velocidade de fluxo (v)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

considerando o sangue um fluido incompress\u00edvel de densidade d e que seja bombeado a partir do repouso, desprezando perdas nas paredes\u00a0 —\u00a0 p=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + mgh (II)\u00a0 —\u00a0 observando a equa\u00e7\u00e3o (I) voc\u00ea conclui que, se a vaz\u00e3o \u00e9 constante, diminuindo-se a \u00e1rea de fluxo a velocidade aumenta\u00a0 —\u00a0 na equa\u00e7\u00e3o (II), se a velocidade aumenta, a press\u00e3o tamb\u00e9m aumenta.<\/b><\/span><\/span><\/p>\n

Analisando as op\u00e7\u00f5es:<\/b><\/span><\/span><\/p>\n

a) Falsa\u00a0 —\u00a0 pelo exposto acima.<\/b><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 as for\u00e7as mencionadas formam um par a\u00e7\u00e3o-rea\u00e7\u00e3o.Essas for\u00e7as nunca se anulam, pois agem em corpos diferentes.<\/b><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 a press\u00e3o de 2,5 Pa significa que \u00e9 exercida uma for\u00e7a de 2,5 N em 1 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

d) Correta.<\/b><\/span><\/span><\/p>\n

e) Falsa\u00a0 —\u00a0 considerando a densidade do merc\u00fario, d<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>\u00a0= 13,6 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0kg\/m<\/b><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sup>(n\u00e3o fornecida no enunciado), a press\u00e3o sist\u00f3lica de 120 mmHg = 0,12 mHg, pode ser calculada no Sistema Internacional pelo teorema de Stevin:<\/b><\/span><\/span><\/p>\n

p = d<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>\u00a0g h = 13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,12 \u00a0—\u00a0 p = 1,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0Pa\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados: V = 1.200 L; h = 30 m; L = 200 m\u00a0 —\u00a0 seguindo as instru\u00e7\u00f5es do fabricante, entremos com os dados na tabela para obtermos o valor de H.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como mostrado, obtemos H = 45 m\u00a0 —\u00a0 analisando o gr\u00e1fico dado, temos os valores mostrados: H\u00a0 = 45 m \u00a0—\u00a0 \u00a0Q = 900 L\/h.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Calculando o tempo para encher o reservat\u00f3rio\u00a0 —\u00a0 Q=V\/t\u00a0 —\u00a0 900=1.200\/t\u00a0 —\u00a0 t=1.200\/900\u00a0 —\u00a0 t=4\/3h\u00a0 —\u00a0 t=80min\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

t=1h e 20min\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Tempo total do banho\u00a0 —\u00a0 \u0394t<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0= 6 min e 54 s = 414 s = 6,9 min\u00a0 —\u00a0 tempo com um quarto de volta\u00a0 —\u00a0 \u0394t<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 1 min e 18 s = 78 s = 1,3 min\u00a0 —\u00a0 tempo com o registro fechado\u00a0 —\u00a0 \u0394t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 3 min e 36 s = 216 s = 3,6 min\u00a0 —\u00a0 tempo com vaz\u00e3o total\u00a0 —\u00a0 \u0394t<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0= ?\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

soma dos tempos\u00a0 —\u00a0 \u0394t<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0= \u0394t<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0+ \u0394t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ \u0394t<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 6,9 = 1,3 + 3,6 + \u0394t<\/b><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 \u0394t<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=2 min\u00a0 —\u00a0 c\u00e1lculo do consumo de \u00e1gua, usando os dados da tabela\u00a0 —\u00a0 C<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= 1,3×1,5 + 2×10,8 = 1,95 + 21,6 \u00a0—\u00a0 C<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= 23,55 L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>C\u00e1lculos feitos na quest\u00e3o anterior\u00a0 —\u00a0 o chuveiro ficou ligado durante um curto intervalo de 78 s, despejando 1,95 L\u00a0 —\u00a0 \u00a0a seguir, ficou fechado durante 216 s e, finalmente, com vaz\u00e3o total durante 120 s, despejando 21,6 L\u00a0 —\u00a0 fazendo essas compara\u00e7\u00f5es\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0Art\u00e9rias: s\u00e3o vasos de maior calibre que os demais, de parede espessa que\u00a0saem\u00a0do cora\u00e7\u00e3o levando sangue para os \u00f3rg\u00e3os e<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

tecidos do corpo\u00a0 —\u00a0 capilares sang\u00fc\u00edneos: s\u00e3o vasos de pequeno calibre que ligam as extremidades das \u00a0art\u00e9rias \u00e0s veias\u00a0 —\u00a0 as veias levam o sangue vindo do corpo, ao cora\u00e7\u00e3o e suas paredes s\u00e3o mais finas que as das art\u00e9rias\u00a0 —\u00a0 a art\u00e9ria Aorta \u00e9 a maior do corpo humano, pois al\u00e9m de ser a maior em extens\u00e3o, ela \u00e9 a de maior (espessura, di\u00e2metro) calibre\u00a0 —\u00a0 observe que o vaso I possui maior \u00e1rea (espessura, di\u00e2metro) que o de cada vaso II, ent\u00e3o ele s\u00f3 pode ser a art\u00e9ria aorta\u00a0 —\u00a0 o fluxo de sangue no corpo humano \u00e9 constante, ou seja, em cada vaso, o volume que circula no mesmo intervalo de tempo \u00e9 o mesmo\u00a0 —\u00a0\u00a0 \u03c6<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>\u00a0= \u03c6<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span>olI<\/b><\/span><\/span><\/sub>\/\u2206t = V<\/b><\/span><\/span>olII<\/b><\/span><\/span><\/sub>\/\u2206t\u00a0 — (S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.l<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>)\/ \u2206t = S<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>.l<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\/\u2206t \u00a0(1)<\/b><\/span><\/span>\u00ad<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a velocidade do sangue no interior de cada vaso \u00e9 diferente e vale\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=l<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>\/\u2206t (2)\u00a0 —\u00a0 V<\/b><\/span><\/span>1I<\/b><\/span><\/span><\/sub>=l<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\/\u2206t (3)\u00a0 —\u00a0 comparando (1) com (2) e com (3)\u00a0 —\u00a0 S<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>\u00a0= S<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 240.30 = 240000V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2700\/240000\u00a0 —<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,03cm\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/a><\/span><\/h3>\n

\u00a0<\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Equa\u00e7\u00e3o da Continuidade \u2013 Teorema de Bernoulli 01-\u00a0A vaz\u00e3o \u00e9 a mesma \u2013 regime estacion\u00e1rio\u00a0 —\u00a0 R- B 02-\u00a0S=100cm2=102.10-4\u00a0 —\u00a0 S=10-2m3\u00a0 —\u00a0 Z=7.200L\/h=7.200\/3.600L\/s=2.L\/s\u00a0 —\u00a0 Z=2.10-3m3\/s\u00a0 —\u00a0 Z=S.v\u00a0 —\u00a0 2.10-3= 10-2v\u00a0 —\u00a0 v=2.10-3\/10-2\u00a0 —\u00a0 v=0,2m\/s\u00a0 —\u00a0\u00a0R- C 03-\u00a0\u0394V=18.10.2=360m3\u00a0 —\u00a0 \u0394t=10h\u00a0 —\u00a0 S=25.10-4m2\u00a0 —\u00a0 Z=\u0394V\/\u0394t=360\/10\u00a0 —\u00a0 Z=36m3\/h\u00a0 —\u00a0 Z=S.v\u00a0 —\u00a0 36=25.10-4.v\u00a0 —\u00a0 v=36\/25.104\u00a0 —\u00a0 v=14.400m\/h\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1585,"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-1589","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1589","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=1589"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1589\/revisions"}],"predecessor-version":[{"id":10886,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1589\/revisions\/10886"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1585"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1589"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}