{"id":3280,"date":"2017-09-30T23:58:32","date_gmt":"2017-09-30T23:58:32","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3280"},"modified":"2024-09-04T14:46:35","modified_gmt":"2024-09-04T14:46:35","slug":"resolucao-comentada-hidrostatica-2014-2013","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/hidrostatica-2013-2014\/resolucao-comentada-hidrostatica-2014-2013\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada HIDROST\u00c1TICA \u2013 2014 \u2013 2013"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada <\/span><\/span><\/span><\/p>\n

HIDROST\u00c1TICA \u2013 2014 \u2013 2013.<\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O Empuxo (for\u00e7a vertical e para cima) corresponde ao\u00a0<\/b><\/span><\/span>peso do volume de l\u00edquido (no caso, <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e1gua)deslocad<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span>\u00a0pela parte imersa do bloco e, pelo enunciado esse\u00a0<\/b><\/span><\/span>volume tem massa\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>\u00a0com peso fornecido por\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=m.g=E<\/b><\/span><\/span><\/span>— por outro lado, o bloco encontra-se em\u00a0<\/b><\/span><\/span>equil\u00edbrio (for\u00e7a resultante nula) flutuando na \u00e1gua<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

e assim, seu peso P<\/b><\/span><\/span>bloco<\/b><\/span><\/span><\/sub>\u00a0(vertical e para baixo) deve anular o empuxo E (vertical e para cima) —\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>bloco<\/b><\/span><\/span><\/sub><\/span>=M.g=E=P<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>— m,g = M.g —\u00a0<\/b><\/span><\/span>m=M<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span>imagem<\/b><\/span><\/span><\/span>\u00a0vista pelo observador \u00e9\u00a0<\/b><\/span><\/span>virtual\u00a0<\/b><\/span><\/span><\/span>e est\u00e1\u00a0<\/b><\/span><\/span>sempre acima do objeto\u00a0<\/b><\/span><\/span><\/span>e, ambos na\u00a0<\/b><\/span><\/span>mesma <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

reta vertical,<\/b><\/span><\/span><\/span>\u00a0onde:<\/b><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>—- dist\u00e2ncia vertical da imagem \u00e0 superf\u00edcie de separa\u00e7\u00e3o (interface) \u00e1gua-ar (dist\u00e2ncia aparente, em que o observador enxerga a imagem)<\/b><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>—- dist\u00e2ncia vertical do objeto, at\u00e9 a superf\u00edcie de separa\u00e7\u00e3o (interface) \u00e1gua-ar.<\/b><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>—- \u00edndice de refra\u00e7\u00e3o do meio onde est\u00e1 o observador (no caso, ar)<\/b><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>—- \u00edndice de refra\u00e7\u00e3o do meio onde est\u00e1 o objeto a imagem (no caso, \u00e1gua)<\/b><\/span><\/span><\/p>\n

Pelas informa\u00e7\u00f5es do exerc\u00edcio:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ent\u00e3o, o observador\u00a0<\/b><\/span><\/span>ver\u00e1 a imagem do fundo da piscina a uma dist\u00e2ncia d<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=2,25m abaixo da superf\u00edcie da<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e1gua e, ter\u00e1 a impress\u00e3o de v\u00ea-lo a uma altura\u00a0<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/span>=3,00 \u2013 2,25\u00a0<\/b><\/span><\/span><\/span>= 0,75 m\u00a0<\/b><\/span><\/span><\/span>mais pr\u00f3ximo em rela\u00e7\u00e3o \u00e0 profundidade real —\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

03-\"\"<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>press\u00e3o hidrost\u00e1tica\u00a0<\/b><\/span><\/span><\/span>no fundo de cada recipiente \u00e9 fornecida pelo teorema de Stevin\u00a0<\/b><\/span><\/span>— P= d.g. h\u00a0<\/b><\/span><\/span><\/span>. No volume\u00a0<\/b><\/span><\/span>1 de altura 2m<\/b><\/span><\/span><\/span>— P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>\u00e1gua.<\/b><\/span><\/span><\/sub>g.h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= d<\/b><\/span><\/span>\u00e1gua.<\/b><\/span><\/span><\/sub>g.2 —\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2d<\/b><\/span><\/span><\/span>\u00e1gua.<\/b><\/span><\/span><\/sub><\/span>g.<\/b><\/span><\/span><\/span><\/p>\n

No volume\u00a0<\/b><\/span><\/span>2 de altura 1m\u00a0<\/b><\/span><\/span><\/span>— P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>\u00e1gua.<\/b><\/span><\/span><\/sub>g.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= d<\/b><\/span><\/span>\u00e1gua.<\/b><\/span><\/span><\/sub>g.1 —\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>\u00e1gua.<\/b><\/span><\/span><\/sub><\/span>g.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0Peso\u00a0<\/b><\/span><\/span><\/span>— sendo a\u00a0<\/b><\/span><\/span>densidade<\/b><\/span><\/span><\/span>\u00a0do cilindro do lado\u00a0<\/b><\/span><\/span>direito<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>maior que a do cilindro do lado esquerdo<\/b><\/span><\/span><\/span>, o cilindro do\u00a0<\/b><\/span><\/span>lado direito ter\u00e1 maior massa,\u00a0<\/b><\/span><\/span><\/span>pois d=m\/v (nessa express\u00e3o, como o volume V \u00e9 o mesmo a massa m \u00e9 diretamente proporcional \u00e0 densidade d) — pela express\u00e3o P=m.g, quanto maior a massa maior o peso e\u00a0<\/b><\/span><\/span>o peso do cilindro direito \u00e9 maior que o do cilindro esquerdo\u00a0<\/b><\/span><\/span><\/span>(ambos aplicados no centro geom\u00e9trico) veja<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

primeira figura — assim, o\u00a0<\/b><\/span><\/span>peso da boia ser\u00e1 deslocado para o lado direito\u00a0<\/b><\/span><\/span><\/span>(ponto B)\u00a0<\/b><\/span><\/span><\/span>figura da direita.<\/b><\/span><\/span><\/p>\n

Empuxo<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span>Enunciado do princ\u00edpio de Arquimedes:<\/b><\/span><\/span><\/span><\/p>\n

\u201cTodo corpo\u00a0<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/span><\/span>\u00a0ou parcialmente mergulhado num l\u00edquido em equil\u00edbrio, recebe uma for\u00e7a de\u00a0<\/b><\/span><\/span><\/span>dire\u00e7\u00e3o vertical e sentido\u00a0 para cima denominada de Empuxo<\/b><\/span><\/span><\/span><\/span>, cuja intensidade\u00a0 \u00e9\u00a0<\/b><\/span><\/span><\/span>igual ao peso do volume de l\u00edquido deslocado\u201c<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Considere um l\u00edquido de densidade d<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>\u00a0contido num recipiente R\u2019 no n\u00edvel indicado na figura I, onde est\u00e1 um corpo s\u00f3lido de densidade d<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>tal que d<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>\u00a0> d<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Em seguida o corpo \u00e9 colocado em R\u2019 e\u00a0 fica totalmente imerso (d<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>\u00a0> d<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>), expulsando um volume de l\u00edquido que \u00e9 igual ao pr\u00f3prio volume do corpo. O volume total (V) do l\u00edquido deslocado \u00e9 recolhido no recipiente R (figura II).<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span><\/span>pesar o volume V do l\u00edquido deslocado voc\u00ea obter\u00e1 o valor do empuxo<\/b><\/span><\/span><\/span><\/span>, que \u00e9 uma\u00a0<\/b><\/span><\/span><\/span>for\u00e7a vertical e para cima\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>—\u00a0 empuxo=peso do volume de l\u00edquido deslocado (que \u00e9 o mesmo volume que o do corpo, se totalmente imerso)\u00a0 — E= P<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>=m<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>g\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>=m<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>\/V\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>=d<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 E=d<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/span><\/sub>.g<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio, com a boia totalmente imersa ela deslocar\u00e1 um volume de \u00e1gua igual ao seu pr\u00f3prio volume — o\u00a0<\/b><\/span><\/span>empuxo<\/b><\/span><\/span><\/span>, vertical e para cima, corresponde ao\u00a0<\/b><\/span><\/span>peso do volume de \u00e1gua deslocada\u00a0<\/b><\/span><\/span><\/span>e, como a \u00e1gua \u00e9 um l\u00edquido homog\u00eaneo (mesma densidade em todos seus pontos) o\u00a0<\/b><\/span><\/span>empuxo ser\u00e1 aplicado em seu<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

centro de massa que \u00e9 onde \u00e9 aplicado seu peso e que corresponde ao centro geom\u00e9trico da boia<\/b><\/span><\/span><\/span><\/p>\n

(ponto B)\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sobre a\u00a0<\/b><\/span><\/span>bexiga imersa\u00a0<\/b><\/span><\/span><\/span>agem as for\u00e7as (<\/b><\/span><\/span>peso, vertical e para baixo<\/b><\/span><\/span><\/span>) e o (<\/b><\/span><\/span>empuxo, vertical e para cima<\/b><\/span><\/span><\/span>), tal que:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

06- <\/b><\/span><\/span><\/span>Como, em ambos os casos a\u00a0<\/b><\/span><\/span>esfera est\u00e1 flutuando o empuxo em cada um deles \u00e9 igual ao peso da esfera que \u00e9 o mesmo,\u00a0<\/b><\/span><\/span><\/span>pois\u00a0<\/b><\/span><\/span>P=mg e a massa \u00e9 a mesma para cada esfera\u00a0<\/b><\/span><\/span><\/span>— j\u00e1 o\u00a0<\/b><\/span><\/span>l\u00edquido\u00a0<\/b><\/span><\/span><\/span>2 \u00e9 mais denso que o l\u00edquido 1<\/b><\/span><\/span><\/span>, sendo\u00a0<\/b><\/span><\/span><\/span>mais pesado ficando mais embaixo<\/b><\/span><\/span><\/span>, fazendo com que a esfera fique mais acima.<\/b><\/span><\/span><\/p>\n

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

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o objeto de volume V flutua no l\u00edquido ele est\u00e1 em equil\u00edbrio e, nesse caso, o empuxo E<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(for\u00e7a vertical e para cima) deve anular o peso P )for\u00e7a vertical e para baixo_.<\/b><\/span><\/span><\/p>\n

Empuxo —\u00a0<\/b><\/span><\/span>E =<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span>.g= d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.(3\/4)V.g.<\/b><\/span><\/span><\/span><\/p>\n

Peso do objeto de volume V —\u00a0<\/b><\/span><\/span>P=d<\/b><\/span><\/span><\/span>objeto.<\/b><\/span><\/span><\/sub><\/span>V.g<\/b><\/span><\/span><\/span>=600.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.V.g\u00a0<\/b><\/span><\/span>— P=0,6Vg.<\/b><\/span><\/span><\/span><\/p>\n

E = P\u00a0<\/b><\/span><\/span><\/span>— d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.(3\/4)V.g = 0,6Vg —\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>2,4\/3=<\/b><\/span><\/span>0,8g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O objeto\u00a0<\/b><\/span><\/span>flutua (equil\u00edbrio) com 2\/3 de seu volume imerso\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=V\/3<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span>E = P<\/b><\/span><\/span><\/span>\u00a0— \u03c1.g.V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0=P — \u03c1.g.2V\/3 = \u03c1<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.g.V —\u00a0<\/b><\/span><\/span>\u03c1.2\/3 = \u03c1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

09-<\/b><\/span><\/span><\/span>\u00a0I. Correta\u00a0<\/b><\/span><\/span><\/span>— a atmosfera terrestre \u00e9 uma imensa camada de ar e outros gases, com dezenas de quil\u00f4metros de altura, que s\u00e3o atra\u00eddos pela gravidade da Terra e por isso, sua densidade \u00e9 maior em pontos mais pr\u00f3ximos da superf\u00edcie. Ela atua como uma zona intermedi\u00e1ria entre o espa\u00e7o sideral e a Terra e acompanha todos os movimentos da Terra.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0\u00a0\u00a0\"\"\u00a0\u00a0\u00a0\"\"<\/p>\n

Essa camada gasosa exerce uma press\u00e3o sobre os corpos nela mergulhados, devida ao peso da coluna de ar que se encontra sobre esses corpos, e que \u00e9 chamada de press\u00e3o atmosf\u00e9rica, sendo tanto maior quanto mais o corpo estiver mais perto da superf\u00edcie da Terra.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0<\/b><\/span><\/span><\/span>— princ\u00edpio enunciado por Pascal, f\u00edsico e matem\u00e1tico franc\u00eas (1623 \u2013 1663), conhecido como\u00a0princ\u00edpio de Pascal:<\/b><\/span><\/span><\/p>\n

\u201c\u00a0O acr\u00e9scimo de press\u00e3o exercido num ponto de um l\u00edquido ideal em equil\u00edbrio \u00e9 transmitida integralmente a todos os pontos desse l\u00edquido e tamb\u00e9m \u00e0s paredes do recipiente onde est\u00e1 contido\u201d<\/b><\/span><\/span><\/p>\n

A maioria dos sistemas multiplicadores de for\u00e7as \u00e9 baseado no princ\u00edpio de Pascal e, para explic\u00e1-lo considere um l\u00edquido ideal no interior de dois cilindros verticais de se\u00e7\u00f5es diferentes e interligados. Esses cilindros, em contato com a parte superior do l\u00edquido, possuem dois \u00eambolos de \u00e1reas S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Uma for\u00e7a de intensidade F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0aplicada ao \u00eambolo de menor \u00e1rea (S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>), provocar\u00e1 um aumento de press\u00e3o dado por\u00a0\u0394P=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e, pelo princ\u00edpio de Pascal esse acr\u00e9scimo de press\u00e3o se transmitir\u00e1 integralmente a todos os pontos do l\u00edquido e das paredes, inclusive para o \u00eambolo de maior \u00e1rea (S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>). Ent\u00e3o, o \u00eambolo maior fica sujeito a uma for\u00e7a F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>, tal que\u00a0\u0394P=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>. Pelo princ\u00edpio de Pascal essa varia\u00e7\u00e3o de press\u00e3o se transmite integralmente\u00a0 do \u00eambolo menor ao \u00eambolo maior e s\u00e3o iguais\u00a0 — \u00a0F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

Observe na express\u00e3o acima que, como S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0> S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, tem-se que F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0> F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e, assim, a intensidade da for\u00e7a \u00e9 proporcional \u00e0 \u00e1rea de cada \u00eambolo, ou seja, esse sistema \u00e9 capaz de multiplicar for\u00e7as.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0Exemplos de sistemas multiplicadores de for\u00e7as baseados no Princ\u00edpio de Pascal:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

III. Correta\u00a0<\/b><\/span><\/span><\/span>— os mais densos, mais pesados, ficam em baixo e os menos densos, mais leves, ficam em cima<\/b><\/span><\/span><\/p>\n

IV. Falsa\u00a0<\/b><\/span><\/span><\/span>— o empuxo \u00e9 uma for\u00e7a exercida de\u00a0<\/b><\/span><\/span>baixo para cima\u00a0<\/b><\/span><\/span><\/span>sobre qualquer corpo imerso num fluido e \u00e9 igual ao peso do volume de fluido deslocado \u201cPrincipio de Arquimedes\u201d<\/b><\/span><\/span><\/p>\n

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

10-<\/b><\/span><\/span><\/span>\u00a0Peso de cada corpo\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.v<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>g=<\/b><\/span><\/span>m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>g\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>Empuxo de cada corpo\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>E=d<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.g\u00a0<\/b><\/span><\/span><\/span>— como est\u00e3o flutuando est\u00e3o em equil\u00edbrio e P = E — m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.g = d<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>.g —\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>como a massa do corpo \u00e9 a mesma e o l\u00edquido \u00e9 o mesmo (mesma d<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>) o volume imerso \u00e9 o mesmo para os tr\u00eas s\u00f3lidos.<\/b><\/span><\/span><\/span><\/p>\n

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

11-<\/b><\/span><\/span><\/span>\u00a0Press\u00e3o (P)\u00a0<\/b><\/span><\/span><\/span>\u00e9 uma grandeza f\u00edsica obtida pelo\u00a0<\/b><\/span><\/span>quociente entre a intensidade da for\u00e7a (\"\") e a \u00e1rea\u00a0 (S)\u00a0<\/b><\/span><\/span><\/span>em que a for\u00e7a se distribui<\/b><\/span><\/span><\/span>.\u00a0Observe na\u00a0<\/b><\/span><\/span>figura abaixo que a for\u00e7a (\"\") que produz a press\u00e3o sobre a \u00e1rea S \u00e9 perpendicular \u00e0 mesma.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

No caso do exerc\u00edcio a for\u00e7a \u00e9 a hidrost\u00e1tica, devida \u00e0 press\u00e3o hidrost\u00e1tica e tamb\u00e9m \u00e9<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

perpendicular \u00e0 \u00e1rea da superf\u00edcie em cada ponto.<\/b><\/span><\/span><\/p>\n

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

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) A press\u00e3o na interface \u00e1gua-merc\u00fario no ramo esquerdo do tubo (ponto k), pelo teorema de Stevin \u00e9 dada por P<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Stevin \u00e9 dada por P<\/b><\/span><\/span>K<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm +\u00a0\u00a0<\/b><\/span><\/span><\/sub>\u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.(h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ x + h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>), sendo \u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0a densidade da \u00e1gua.<\/b><\/span><\/span><\/p>\n

b) Como conseq\u00fc\u00eancia do teorema de Stevin, todos os pontos no mesmo n\u00edvel horizontal suportam a <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

mesma press\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span>K<\/b><\/span><\/span><\/sub>\u00a0= P<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>atm +\u00a0\u00a0<\/b><\/span><\/span><\/sub>\u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.(h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ x + h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= \u03c1<\/b><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ \u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.x + P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ \u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= \u03c1<\/b><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub>.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 — h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=(\u03c1<\/b><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub>\u00a0– \u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>)\/ \u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O\u00a0<\/b><\/span><\/span>empuxo\u00a0<\/b><\/span><\/span><\/span>(for\u00e7a vertical e para cima) \u00e9 igual ao\u00a0<\/b><\/span><\/span>peso do volume de l\u00edquido deslocado\u00a0<\/b><\/span><\/span><\/span>e assim, quanto\u00a0<\/b><\/span><\/span>maior o peso\u00a0<\/b><\/span><\/span><\/span>maior ser\u00e1 o volume de l\u00edquido deslocado\u00a0<\/b><\/span><\/span><\/span>e consequentemente,\u00a0<\/b><\/span><\/span>maior ser\u00e1 a altura da superf\u00edcie da \u00e1gua do balde.<\/b><\/span><\/span><\/span><\/p>\n

Maior peso B\u00a0<\/b><\/span><\/span><\/span>(barco + \u00e2ncora dentro, deslocam maior volume de \u00e1gua)\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>menor peso A\u00a0<\/b><\/span><\/span><\/span>(barco sem<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e2ncora, desloca menor volume de \u00e1gua)\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>intermedi\u00e1rio C<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>(volume deslocado somente pelo barco ) + volume deslocado somente pela \u00e2ncora).<\/b><\/span><\/span><\/span><\/p>\n

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

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Press\u00e3o<\/b><\/span><\/span><\/span>\u00a0a que o submarino\u00a0<\/b><\/span><\/span>est\u00e1 sujeito no fundo do plat\u00f4 a uma profundidade de\u00a0<\/b><\/span><\/span><\/span>h=3600m\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>teorema de Stevin — P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>= P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+ d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.g,h<\/b><\/span><\/span><\/span>= 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.10.3600 — P<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>= 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 36.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0—\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=361.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/p>\n

\u00c9 pedida a rela\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>=361.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>= 361.<\/b><\/span><\/span><\/span><\/p>\n

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

15-<\/b><\/span><\/span><\/span>\u00a0I. Correta.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>E=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>\u00a0— observe nessa express\u00e3o que a\u00a0<\/b><\/span><\/span>rela\u00e7\u00e3o entre E e h \u00e9 uma fun\u00e7\u00e3o do primeiro grau.<\/b><\/span><\/span><\/span><\/p>\n

III, Falsa\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>Teorema de Stevin\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span>\u0394P=d.g.h\u00a0<\/b><\/span><\/span><\/span>(a diferen\u00e7a de press\u00e3o \u00e9\u00a0<\/b><\/span><\/span>diretamente proporcional \u00e0 altura h).<\/b><\/span><\/span><\/span><\/p>\n

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

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>05-\u00a0<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span>figura II\u00a0<\/b><\/span><\/span><\/span>indica o valor da\u00a0<\/b><\/span><\/span>press\u00e3o atmosf\u00e9rica P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>hg<\/b><\/span><\/span><\/sub><\/span>.g.h=<\/b><\/span><\/span><\/span>13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,75 —\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>=1,02.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0Pa<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

A press\u00e3o P<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>\u00a0exercida pelo g\u00e1s \u00e9 equilibrada pela press\u00e3o atmosf\u00e9rica (P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa) somada \u00e0 press\u00e3o devida ao merc\u00fario de densidade d<\/b><\/span><\/span>hg<\/b><\/span><\/span><\/sub>=13,6g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0—\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>g<\/b><\/span><\/span><\/sub><\/span>= P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+ d<\/b><\/span><\/span><\/span>hg<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>= 1,02.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,25=1,02.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 34.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=1,02.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 0,34.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0—\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>g<\/b><\/span><\/span><\/sub><\/span>=1,36.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0Pa.<\/b><\/span><\/span><\/span><\/p>\n

R- 01<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Densidade e press\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Correta\u00a0 — A densidade de um corpo (no caso um l\u00edquido) pode ser definida como sendo a grandeza f\u00edsica que fornece a quantidade de massa (mat\u00e9ria)\u00a0 de que est\u00e1 concentrada num determinado volume. Chamando de\u00a0m\u00a0a quantidade de massa contida em certo volume V, a express\u00e3o matem\u00e1tica da densidade \u00e9:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No Sistema Internacional de Unidades (SI) a unidade de densidade \u00e9 o quilograma por metro c\u00fabico (kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>), mas s\u00e3o usados tamb\u00e9m o grama por cent\u00edmetro c\u00fabico (g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>) e o quilograma por litro (kg\/L). Rela\u00e7\u00f5es:<\/b><\/span><\/span><\/p>\n

1g=10-<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>kg\u00a0 —\u00a0 1cm3=10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>m3\u00a0\u00a0\u00a0—\u00a0\u00a01g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0—\u00a0 1g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>kg\/10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>L\u00a0 —\u00a01gcm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=1 kg\/L<\/b><\/span><\/span><\/p>\n

Regra pr\u00e1tica:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

02. Correta\u00a0 —\u00a0 a press\u00e3o sempre vai estar relacionada ao\u00a0m\u00f3dulo\u00a0da for\u00e7a que age\u00a0perpendicularmente \u00e0 superf\u00edcie\u00a0 —\u00a0 portanto, n\u00e3o importa em qual dire\u00e7\u00e3o a for\u00e7a \u00e9 exercida, ela na afeta a dire\u00e7\u00e3o da press\u00e3o, que \u00e9 sempre perpendicular a superf\u00edcie\u00a0 —\u00a0 assim, o que interessa \u00e9 s\u00f3 a intensidade da press\u00e3o.<\/b><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 A press\u00e3o exercida por uma coluna l\u00edquida n\u00e3o depende das dimens\u00f5es do recipiente que a cont\u00e9m, mas apenas da natureza do l\u00edquido, fornecida pela sua densidade (d), do local (g) e da altura da coluna (h)\u00a0 —\u00a0 entre dois pontos A e B de alturas h<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e h<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>, a diferen\u00e7a de press\u00e3o \u00e9 dada por (lei de Stevin)\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0\u2206P=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub>– d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>B\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 \u2206P=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g.(h<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 h<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>).<\/b><\/span><\/span><\/p>\n

08- Falsa — \u00a0s\u00e3o inversamente proporcionais\u00a0 — \u00a0P=F\/A\u00a0 —\u00a0 observe nessa f\u00f3rmula \u00a0que para uma mesma for\u00e7a, quanto menor a \u00e1rea maior ser\u00e1 a press\u00e3o exercida pela for\u00e7a\u00a0 —\u00a0 como exemplos pr\u00e1ticos, observe as figuras abaixo:<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\"\"\u00a0\"\"<\/p>\n

 <\/p>\n

\"\"\"\"<\/p>\n

16. Correta\u00a0 — enunciado do princ\u00edpio de Arquimedes:<\/b><\/span><\/span><\/p>\n

\u201cTodo corpo total ou parcialmente mergulhado num l\u00edquido em equil\u00edbrio, recebe uma for\u00e7a de dire\u00e7\u00e3o vertical e sentido\u00a0 para cima denominada de Empuxo, cuja intensidade\u00a0 \u00e9 igual ao peso do volume de l\u00edquido deslocado\u201c\u00a0 —\u00a0 trata-se do empuxo\u00a0 —\u00a0 E=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>l\u00edquido deslocado<\/b><\/span><\/span><\/sub>.g.<\/b><\/span><\/span><\/p>\n

R- 01,02,04,16)<\/b><\/span><\/span><\/p>\n

 <\/p>\n

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00c1gua\u00a0 —\u00a0 Va=0,01m3=0,01.103dm3\u00a0 —\u00a0 Va=10L\u00a0 —\u00a0 da=1g\/cm3=103kg\/m3=103kg\/103L\u00a0 —\u00a0 da=1kg\/L\u00a0 —\u00a0 da=ma\/Va\u00a0 —\u00a0 1=ma\/10\u00a0 —\u00a0 ma=10L\u00a0 —\u00a0 \u00f3leo\u00a0 —\u00a0 Vo=2000cm3=2000.10-3dm3\u00a0 —\u00a0 Vo=2L\u00a0 —\u00a0 do=0,9g\/cm3=0,9.103kg\/m3=0,9.103kg\/103L\u00a0 —\u00a0 do=0,9kg\/L\u00a0 —\u00a0 do=mo\/Vo\u00a0 —\u00a0 0,9=mo\/2\u00a0 —\u00a0 mo=1,8kg\u00a0 —\u00a0 a massa da mistura vale\u00a0 m=10 + 1,8=11,8kg\u00a0 —\u00a0\u00a0R- A.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Teorema de Stevin \u2013 Press\u00e3o hidrost\u00e1tica – Vasos comunicantes<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Peixe 1\u00a0 —\u00a0 passou de 120m para 90m (desceu)\u00a0 —\u00a0 \u2206P=d.g.\u2206h=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.30=3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa=3atm\u00a0 —\u00a0 a <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

press\u00e3o aumentou\u00a0 —\u00a0 peixe 2\u00a0 —\u00a0 passou de 30m para 90m (subiu)\u00a0 —\u00a0 \u2206P=d.g.\u2206h=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.60=6.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa=6atm\u00a0 —\u00a0 a press\u00e3o diminuiu\u00a0 —<\/b><\/span><\/span><\/p>\n

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

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38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Teorema Fundamental da Hidrost\u00e1tica ou de Teorema de Stevin\u00a0 —\u00a0 \u2206P=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.h, onde P \u00b4a press\u00e3o hidrost\u00e1tica no ponto de sa\u00edda da \u00e1gua (onde est\u00e1 a ducha), g \u00e9 a acelera\u00e7\u00e3o da gravidade local e h \u00e9 altura medida desde qualquer<\/b><\/span><\/span><\/p>\n

Ponto da superf\u00edcie livre da \u00e1gua no interior da caixa d\u2019\u00e1gua e o ponto de sa\u00edda da mesma, onde est\u00e1 a ducha.<\/b><\/span><\/span><\/p>\n

R- E.<\/b><\/span><\/span><\/p>\n

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Experi\u00eancia de Torricelli<\/b><\/span><\/span><\/span><\/p>\n

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35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os 5 primeiros cilindros est\u00e3o ligados e, ap\u00f3s abertas as v\u00e1lvulas eles atingir\u00e3o o equil\u00edbrio hidrost\u00e1tico e a altura final de cada um que deve ser a mesma (h) e dada por\u00a0 —\u00a0 h=(8 + 7 + 6 + 5 + 4)\/5=6dm\u00a0 —\u00a0 observe na figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

que a v\u00e1lvula que une E a F est\u00e1 na altura de 6dm e n\u00e3o escoar\u00e1 \u00e1gua de E para F que permanecer\u00e1 na altura de 3dm\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

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Princ\u00edpio de Pascal<\/b><\/span><\/span><\/span><\/p>\n

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25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se voc\u00ea chamar V1=V\u00a0 —\u00a0 V2=4V\u00a0 —\u00a0 se voc\u00ea chamar h2=h\u00a0 —\u00a0 h1=3h\u00a0 —\u00a0 V2=S2.h2\u00a0 —\u00a0 V1=S1.h1\u00a0 —\u00a0 4V = S2.h (I)\u00a0 —\u00a0 V=S1.3h (II)\u00a0 —\u00a0 (II) em (I)\u00a0 —\u00a0 4,3,S1.h = S2.h\u00a0 —\u00a0 S2=12S1\u00a0 —\u00a0 estando o sistema em equil\u00edbrio a press\u00e3o em cada recipiente \u00e9 a mesma\u00a0 —\u00a0 P1\u00a0= P2\u00a0 —\u00a0 F1\/S1\u00a0= F2\/S2\u00a0 —\u00a0 F1\/S1\u00a0= F2\/12S1\u00a0 —\u00a0 F1\/F2=12\u00a0 —\u00a0\u00a0R- A.<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>A maioria dos sistemas multiplicadores de for\u00e7as \u00e9 baseado no princ\u00edpio de Pascal e, para explic\u00e1-lo considere um l\u00edquido ideal no interior de dois cilindros verticais de se\u00e7\u00f5es diferentes e interligados. Esses cilindros, em contato com a parte superior do l\u00edquido, possuem dois \u00eambolos de \u00e1reas S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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Uma for\u00e7a de intensidade F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0aplicada ao \u00eambolo de menor \u00e1rea (S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>), provocar\u00e1 um aumento de press\u00e3o dado por\u00a0\u0394P=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e, pelo princ\u00edpio de Pascal esse acr\u00e9scimo de press\u00e3o se transmitir\u00e1 integralmente a todos os pontos do l\u00edquido e das paredes, inclusive para o \u00eambolo de maior \u00e1rea (S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>). Ent\u00e3o, o \u00eambolo maior fica sujeito a uma for\u00e7a F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>, tal que\u00a0\u0394P=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>. Pelo princ\u00edpio de Pascal essa varia\u00e7\u00e3o de press\u00e3o se transmite integralmente\u00a0 do \u00eambolo menor ao \u00eambolo maior e s\u00e3o iguais\u00a0 — \u00a0F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

Observe na express\u00e3o acima que, como S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0> S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, tem-se que F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0> F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e, assim, a intensidade da for\u00e7a \u00e9 proporcional \u00e0 \u00e1rea de cada \u00eambolo, ou seja, esse sistema \u00e9 capaz de multiplicar for\u00e7as.<\/b><\/span><\/span><\/p>\n

Substituindo os valores fornecidos\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/4.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=2.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\/0,16\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=50N\u00a0\u00a0—\u00a0\u00a0R- C.<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura que a press\u00e3o na superf\u00edcie de cada recipiente \u00e9 a mesma e \u00e9 a press\u00e3o atmosf\u00e9rica\u00a0 —\u00a0 no<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

recipiente da esquerda\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ dgh\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 dgh=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,4\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 no recipiente da direita\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ dgh\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 dgh=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.1,2\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 1,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u2013 (P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 12.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u2013 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 12.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=8000Pa\u00a0\u00a0—\u00a0\u00a0R- D.<\/b><\/span><\/span><\/p>\n

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Teorema de Arquimedes \u2013 Empuxo<\/b><\/span><\/span><\/span><\/p>\n

67-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) \u00c1rea da base de cada cilindro A<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=A=\u03c0.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=3.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 A=3cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 volume do cilindro 1\u00a0 —\u00a0 V=A.h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=3.5\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=15cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 volume do cilindro 2\u00a0 —\u00a0 V=A.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3.0,5\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1,5cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 massa do <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

cilindro 1\u00a0 —\u00a0 d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —0,2=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/15\u00a0 —\u00a0 m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=3g\u00a0 —\u00a0 massa do cilindro 2\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 8=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/1,5\u00a0 —\u00a0 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=12g\u00a0 —\u00a0\u00a0 massa total\u00a0 —\u00a0 m=3 + 12=15g\u00a0 —\u00a0\u00a0m=15g ou m=15.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>kg.<\/b><\/span><\/span><\/p>\n

b) Peso do cilindro\u00a0 —\u00a0 P=m.g=15.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 P=15.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 c\u00e1lculo do empuxo sobre o cilindro com volume imerso V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>.g=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>.10\u00a0 —\u00a0 E=10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>V<\/b><\/span><\/span>i\u00a0<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 como o est\u00e1 em equil\u00edbrio o peso deve anular o empuxo\u00a0 —\u00a0 P=E\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

15.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0= 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=15.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=15cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 altura (h<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>) da parte imersa\u00a0 —\u00a0 V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=S.h<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 15=3.h<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0\u00a0\u00a0\u00a0—\u00a0\u00a0h<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=5cm.<\/b><\/span><\/span><\/p>\n

68-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Peso do submarino sem lastro\u00a0 —\u00a0 P<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>.g=13.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.10=13.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 empuxo com o submarino totalmente imerso\u00a0 —\u00a0 E=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>.g=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.15.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10=15.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 peso do volume V\u2019 de \u00e1gua (lastro) que o submarino deve receber para que fique totalmente imerso e em equil\u00edbrio\u00a0 —\u00a0 P\u2019=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V\u2019.g=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.V\u2019.10=10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>V\u2019\u00a0 —\u00a0 como\u00a0 ele est\u00e1 em equil\u00edbrio a intensidade da for\u00e7a resultante <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sobre ele deve ser nula\u00a0 —\u00a0 E = P<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0+ P\u2019\u00a0 —\u00a0 15.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0= 13.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>V\u2019\u00a0 —\u00a0 V\u2019=2.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=<\/b><\/span><\/span><\/p>\n

2.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=200m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u00a0—\u00a0\u00a0R- B.<\/b><\/span><\/span><\/p>\n

69-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Peso do transatl\u00e2ntico\u00a0 —\u00a0 P=mg=1,2.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>.10=12.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 para que ele flutue sem tocar o fundo sua altura h deve ser tal que\u00a0 —\u00a0 V=b,L,h=300.40.h=12,10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>h\u00a0 —\u00a0 empuxo\u00a0 —\u00a0 E=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V.g. = <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.12.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>h.10 = 12.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>h\u00a0 —\u00a0 como ele deve estar flutuando (em equil\u00edbrio)\u00a0 —\u00a0 E = P\u00a0 —\u00a0 12.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>h=12.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=12.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\/12.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=10m\u00a0 —\u00a0\u00a0R- A.<\/b><\/span><\/span><\/p>\n

70-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando o ar do interior do bal\u00e3o \u00e9 aquecido, ele se dilata, tornando sua densidade e, consequentemente seu peso menor\u00a0 —\u00a0 quando o peso do bal\u00e3o (vertical e para baixo) fica menor que o empuxo sobre ele (vertical e para cima), o bal\u00e3o sobe\u00a0\u00a0—\u00a0\u00a0R- A.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada HIDROST\u00c1TICA \u2013 2014 \u2013 2013. 01-\u00a0O Empuxo (for\u00e7a vertical e para cima) corresponde ao\u00a0peso do volume de l\u00edquido (no caso, \u00e1gua)deslocado\u00a0pela parte imersa do bloco e, pelo enunciado esse\u00a0volume tem massa\u00a0m\u00a0com peso fornecido por\u00a0P\u00e1gua=m.g=E— por outro lado, o bloco encontra-se em\u00a0equil\u00edbrio (for\u00e7a resultante nula) flutuando na \u00e1gua e assim, seu peso Pbloco\u00a0(vertical e para baixo) deve anular o<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3278,"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-3280","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3280","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=3280"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3280\/revisions"}],"predecessor-version":[{"id":3281,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3280\/revisions\/3281"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3278"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3280"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}