{"id":3702,"date":"2018-02-13T16:54:44","date_gmt":"2018-02-13T16:54:44","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3702"},"modified":"2024-08-20T23:30:14","modified_gmt":"2024-08-20T23:30:14","slug":"resolucao-comentada-hidrostatica-pressao-hidrostatica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/hidrostatica-pressao-hidrostatica\/resolucao-comentada-hidrostatica-pressao-hidrostatica\/","title":{"rendered":"Hidrost\u00e1tica – Resolu\u00e7\u00e3o"},"content":{"rendered":"

\u00a0<\/span>Resolu\u00e7\u00e3o Comentada<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Hidrost\u00e1tica<\/b><\/span><\/span><\/span><\/p>\n

Press\u00e3o Hidrost\u00e1tica<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

 <\/p>\n

\u00a0<\/span>01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A press\u00e3o no fundo do reservat\u00f3rio devida apenas \u00e0 coluna l\u00edquida P<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.g.h \u00e9 chamada depress\u00e3o hidrost\u00e1tica\u00a0e P=P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+ d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.g.h \u00e9 chamada de press\u00e3o total, press\u00e3o absoluta ou simplesmente press\u00e3o\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 o gr\u00e1fico da press\u00e3o total P em fun\u00e7\u00e3o da altura \u201cprofundidade\u201d h, (P=P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>+ d.g.h, \u201c\u00a0esta express\u00e3o \u00e9 chamada de Teorema Fundamental da Hidrost\u00e1tica ou de Teorema de Stevin), que \u00e9 uma fun\u00e7\u00e3o do primeiro \u00a0grau), \u00e9 uma reta inclinada\u00a0 —\u00a0 portanto, a press\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

no fundo do recipiente varia linearmente com a altura e independe da \u00e1rea de se\u00e7\u00e3o reta de cada cilindro\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando voc\u00ea abre a v\u00e1lvula, a \u00e1gua flui da caixa da esquerda para a da direita at\u00e9 que elas tenham a mesma altura\u00a0 —\u00a0 a altura da \u00e1gua na caixa da esquerda diminui e consequentemente a press\u00e3o na base da mesma tamb\u00e9m diminui\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Teorema de Stevin\u00a0 —\u00a0 P= P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+ dgh\u00a0 —\u00a0 4.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.h\u00a0 —\u00a0 h=3.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>h=30m\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 para sofrer na subida vertical uma varia\u00e7\u00e3o de press\u00e3o \u0394P=10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, em \u0394t=1s, ele deve se deslocar\u00a0 —\u00a0 \u0394P=d.g.h\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.h\u00a0 —\u00a0 h=1m\u00a0 —\u00a0 V= \u2206S\/\u2206t\u00a0 —\u00a0 V=\u0394h\/\u0394t\u00a0 —\u00a0 V=1\/1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=1m\/s<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A\u00a0\u00a0press\u00e3o do ar aprisionado no barco \u00e0 profundidade h=(2,20 \u2013 1,70)=0,5m \u00e9 dada por P= P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+ d.g.h\u00a0 —\u00a0 P \u2013 P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0= d.g.h = 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.0,5\u00a0 —\u00a0\u00a0P \u2013 P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>=5,0.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a \u00a0do jato de \u00e1gua da descarga faz com que a \u00e1gua da mesma flua rapidamente pelo sif\u00e3o, levando os detritos embora\u00a0 —\u00a0 depois, a \u00e1gua volta a ficar na mesma altura no sif\u00e3o e no vaso, pois os dois ficam sujeitos \u00e0 mesma press\u00e3o (mesmo princ\u00edpio dos vasos comunicantes), mantendo o n\u00edvel da \u00e1gua constante e impedindo que o cheiro do esgoto invada o\u00a0banheiro\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span> <\/b>De acordo com Teorema de Stevin, pontos de mesmo l\u00edquido em repouso que est\u00e3o na mesma horizontal est\u00e3o sob mesma press\u00e3o — ent\u00e3o, no ponto\u00a0A\u00a0da horizontal que passa pela interface entre a \u00e1gua e o l\u00edquido mais denso, a press\u00e3o deve ser a mesma que no ponto\u00a0B<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

situado na \u00e1gua, na mesma horizontal\u00a0 —\u00a0 p<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= p<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.g<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/sub><\/span>.h<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 como d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>\u00a0> d<\/b><\/span><\/span><\/span>\u00e1gua\u00a0<\/b><\/span><\/span><\/sub><\/span>(dado do exerc\u00edcio)\u00a0 —h<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>\u00a0< h<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>;\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>glic<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1,3 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0= 1,3.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0 h<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10 cm = 10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0m; h<\/b><\/span><\/span><\/span>glic<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1 m\u00a0 —\u00a0\u00a0 g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 p<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1,01.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0Pa\u00a0 —\u00a0 press\u00e3o da coluna de \u00e1gua\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=(10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>).10.(10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=1,0.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>Pa=0,01.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>Pa\u00a0 —\u00a0 press\u00e3o da coluna de glicerina\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>glic<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>glic<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>glic<\/b><\/span><\/span><\/sub><\/span>=(1,3.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>).(10).(1)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>glic<\/b><\/span><\/span><\/sub><\/span>=1,3.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>Pa=0,13.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>Pa\u00a0 —\u00a0 na superf\u00edcie que separa \u00e1gua-glicerina a press\u00e3o a press\u00e3o \u00e9 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>+ P<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=1,01.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 0,01.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1,02.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>Pa\u00a0 —\u00a0 no fundo do recipiente, a press\u00e3o \u00e9 (p<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>) e vale\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P<\/b><\/span><\/span><\/span>glic<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1,01.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 0,01.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>+ 0,13.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1,15.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0Pa.\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

Obs: segundo o Sistema Internacional de Unidades o plural das unidades \u00e9 feito apenas acrescentando a letra s no final, quando n\u00e3o terminada em s. Caso seja terminada em s, n\u00e3o sofre flex\u00e3o, e quando grafadas por extenso, deve ser em letras min\u00fasculas. Exemplos: pascal – pascals; decibel – decibels; newton – newtons.<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: m = 900 kg; A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2.500 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 25.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0; A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 25 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 25.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>; d<\/b><\/span><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub><\/span>\u00a0= 900 kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>; h = 4m\u00a0 —\u00a0 de acordo com o teorema de Stevin, pontos de um mesmo l\u00edquido em repouso que est\u00e3o na <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

mesma horizontal suportam a mesma press\u00e3o\u00a0 — a press\u00e3o no ponto (1) provocada pelo peso do carro \u00e9 igual \u00e0 press\u00e3o no ponto (2) provocada pela for\u00e7a\u00a0\"\"\u00a0, somada \u00e0 press\u00e3o da coluna l\u00edquida\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>r1<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>r2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>esocarro<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F\/A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ d<\/b><\/span><\/span><\/span>\u00b4oleo<\/b><\/span><\/span><\/sub><\/span>.g.h\u00a0 —\u00a0 9000\/25.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=F\/25.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 900.10.4\u00a0 —\u00a0 36.000 \u2013 36.000=F\/25.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F=0\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A diferen\u00e7a de press\u00e3o hidrost\u00e1tica (\u0394P) entre dois pontos de desn\u00edvel h, para um l\u00edquido de densidade d<\/b><\/span><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub><\/span>, \u00e9 dada pelo teorema de Stevin\u00a0 —\u00a0 \u0394P = d<\/b><\/span><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub><\/span>.g.h\u00a0 —\u00a0 assim,, essa diferen\u00e7a s\u00f3 depende da densidade do l\u00edquido, do desn\u00edvel h e da gravidade local g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 h\u00a0= 2.000 m\u00a0 —\u00a0\u00a0g\u00a0= 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03c1\u00a0= 0,9 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0= 0,9.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=9.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 teorema de Stevin\u00a0 —\u00a0 \u2206P = \u03c1\u00a0g\u00a0h =9.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.10.2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u2206P\u00a0= 180.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u00a0\u2206P = 1,8.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>\u00a0Pa\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Voltar para os exerc\u00edcios<\/span><\/a>
\n<\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

\u00a0Resolu\u00e7\u00e3o Comentada Hidrost\u00e1tica Press\u00e3o Hidrost\u00e1tica \u00a0   \u00a001-\u00a0A press\u00e3o no fundo do reservat\u00f3rio devida apenas \u00e0 coluna l\u00edquida Pl\u00edquido=dl\u00edquido.g.h \u00e9 chamada depress\u00e3o hidrost\u00e1tica\u00a0e P=Patm\u00a0+ dl\u00edquido.g.h \u00e9 chamada de press\u00e3o total, press\u00e3o absoluta ou simplesmente press\u00e3o\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<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3700,"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-3702","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3702","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=3702"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3702\/revisions"}],"predecessor-version":[{"id":10676,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3702\/revisions\/10676"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3700"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3702"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}