{"id":1187,"date":"2015-08-25T12:05:06","date_gmt":"2015-08-25T12:05:06","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1187"},"modified":"2024-08-23T13:43:19","modified_gmt":"2024-08-23T13:43:19","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-polias-ou-roldanas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/polias-e-roldanas\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-polias-ou-roldanas\/","title":{"rendered":"Polias ou Roldanas – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Polias ou Roldanas<\/b><\/span><\/span><\/span><\/p>\n

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

01<\/b><\/span><\/span>– Observe que temos duas polias m\u00f3veis\u00a0 —\u00a0 n=2\u00a0 —\u00a0 F =25\/2<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 25 = Q\/2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q = 100N<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

02<\/b><\/span><\/span>– a) Uma roldana m\u00f3vel, o homem deve aplicar uma for\u00e7a 2<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sup>=2<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sup>=2 vezes menor, ou seja, de 550\/2=275N.<\/b><\/span><\/span><\/p>\n

b) Duas vezes menor, ou seja, de 0,5m.<\/b><\/span><\/span><\/p>\n

03<\/b><\/span><\/span>– O dinam\u00f4metro indica a intensidade da for\u00e7a que traciona o fio que est\u00e1 preso ao solo, ou seja, 400N. Observe na figura, que a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

distribui\u00e7\u00e3o de for\u00e7as nos fornece o peso do bloco, 50N. Como P=m.g\u00a0 —\u00a0 50=m.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=5kg<\/b><\/span><\/span><\/p>\n

04<\/b><\/span><\/span>– Peso do carro\u00a0 —\u00a0 P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=m.g\u00a0 —\u00a0 P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=700.10\u00a0 —\u00a0 P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=7000N\u00a0<\/b><\/span><\/span><\/p>\n

Chamando de P o peso do contrapeso, de m sua massa, e colocando todas as for\u00e7as, observamos<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

que sobre o carro agem as for\u00e7as 7P (para cima) o peso do carro P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=7000N (para baixo). Como ele est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 7P=P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 7P=7000\u00a0 —\u00a0 P=1000N e m=100kg.\u00a0<\/b><\/span><\/span><\/p>\n

O cabo central exerce uma for\u00e7a de 2P (veja figura)\u00a0 —\u00a0 F=2.1000\u00a0 —\u00a0 F=2000N<\/b><\/span><\/span><\/p>\n

05<\/b><\/span><\/span>– Observe a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

06<\/b><\/span><\/span>– Tr\u00eas polias m\u00f3veis n=3\u00a0 —\u00a0 2<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sup>=2<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=8 vezes menor<\/b><\/span><\/span><\/p>\n

07<\/b><\/span><\/span>– P=m.g\u00a0 —\u00a0 P=8.10\u00a0 —\u00a0 P=80N\u00a0 —\u00a0 Observe nas figuras abaixo as distribui\u00e7\u00f5es de for\u00e7as.<\/b><\/span><\/span><\/p>\n

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

Observe tamb\u00e9m que no esquema A todas as polias s\u00e3o fixas e o peso do bloco \u00e9 transmitido integralmente (F<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=80N).<\/b><\/span><\/span><\/p>\n

No esquema B temos uma polia m\u00f3vel e o peso do bloco cai pela metade.<\/b><\/span><\/span><\/p>\n

08<\/b><\/span><\/span>– For\u00e7as que agem sobre voc\u00ea\u00a0 —\u00a0 seu peso de intensidade\u00a0 —\u00a0 P=m.g\u00a0 —\u00a0 P=60.10\u00a0 —\u00a0 P=600N\u00a0\u00a0 e a indica\u00e7\u00e3o da balan\u00e7a N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Desce acelerando\u00a0 —\u00a0 P \u2013 N = m.a\u00a0 —\u00a0 600 \u2013 N=60.3\u00a0 — \u00a0-N=180 \u2013 600\u00a0 —\u00a0 N=420N\u00a0 —\u00a0 voc\u00ea se sente \u201cmais leve\u201d (peso aparente).<\/b><\/span><\/span><\/p>\n

09<\/b><\/span><\/span>– a) Quem est\u00e1 sobre a balan\u00e7a \u00e9 a pessoa de massa 70kg e sobre ela agem duas for\u00e7as. Seu peso P para baixo, que ela troca com a Terra, de intensidade P=m.g=70.10\u00a0 —\u00a0 P=700N e a for\u00e7a normal N para cima, que ela troca com a balan\u00e7a.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como ele sobe acelerando, a balan\u00e7a indicar\u00e1 um valor maior que o peso (N>P)\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 N \u2013 P = m.a\u00a0 —\u00a0 N \u2013 700=70.3<\/b><\/span><\/span><\/p>\n

—\u00a0 N=700 + 210\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=910N<\/b><\/span><\/span><\/p>\n

b) A for\u00e7a de tra\u00e7\u00e3o para cima nos cabos est\u00e1 puxando o elevador mais a pessoa (sistema) de peso\u00a0 —\u00a0 P=(1000 + 70).10\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P=10.700N.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0Como ele sobe acelerando T>P\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>m.a\u00a0 —\u00a0 T \u2013 10.700 = 1.070.3\u00a0 —T=10.700 + 3.210\u00a0 —\u00a0 T= 13.910N<\/b><\/span><\/span><\/p>\n

Como temos tr\u00eas cabos, a for\u00e7a de tra\u00e7\u00e3o em cada cabo ser\u00e1 13.910\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=4.636,7N<\/b><\/span><\/span><\/p>\n

c) A pessoa bateria com a cabe\u00e7a no teto e depois cairia com a mesma acelera\u00e7\u00e3o e velocidades que o elevador.<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>– O peso aparente corresponde \u00e0 for\u00e7a normal N\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P \u2013 N=m.a\u00a0 —\u00a0 100 \u2013 N =10.5\u00a0 —\u00a0 N=50N.<\/b><\/span><\/span><\/p>\n

11<\/b><\/span><\/span>– Sobre o corpo agem duas for\u00e7as. A indica\u00e7\u00e3o do dinam\u00f4metro para cima (T=2,5N) e o peso para baixo (P=2,0N). Sendo T>P, ele est\u00e1 subindo e acelerando ou descendo e freando. R- D<\/b><\/span><\/span><\/p>\n

12<\/b><\/span><\/span>– A tra\u00e7\u00e3o m\u00e1xima suportada pelo cabo vale T=m.g=1200.10\u00a0 —\u00a0 T=12.000N.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O peso do elevador \u00e9 de P=1000.10\u00a0 —\u00a0 P=10.000N.\u00a0 —\u00a0 sobe acelerando – T>P\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 T \u2013 P =m.a\u00a0 —\u00a0 12000 \u2013 10.000=<\/b><\/span><\/span><\/p>\n

1.000.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

13<\/b><\/span><\/span>– O peso da pessoa \u00e9 constante e vale\u00a0 —\u00a0 P=m.g\u00a0 —\u00a0 P=80g (g \u00e9 pedido)\u00a0 —\u00a0 Quando o elevador sobe com acelera\u00e7\u00e3o a, N=960N e<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 N \u2013 P=ma\u00a0 —<\/b><\/span><\/span>\u00a0960 \u2013 80g = 80a\u00a0 I<\/b><\/span><\/span><\/p>\n

Quando o elevador desce com acelera\u00e7\u00e3o a, N=640N e F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P \u2013 N=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>80g \u2013 640=80a<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>II<\/b><\/span><\/span><\/p>\n

Somando I com II obtemos\u00a0<\/b><\/span><\/span>a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, que, substitu\u00eddo em I ou em II nos fornece<\/b><\/span><\/span>g=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

Ent\u00e3o o peso da pessoa \u00e9 P=mg\u00a0 —\u00a0 P=80.10\u00a0 —\u00a0 P=800N. Quando a balan\u00e7a registrar 800N est\u00e1 registrando o peso da pessoa e nesse caso o elevador estar\u00e1 em repouso ou subindo ou descendo em MRU.<\/b><\/span><\/span><\/p>\n

Quando a balan\u00e7a registrar zero o elevador e a pessoa est\u00e3o em queda livre.<\/b><\/span><\/span><\/p>\n

14<\/b><\/span><\/span>– Os \u00f3rg\u00e3os internos s\u00f3 se movem ou tendem a se mover, por in\u00e9rcia, quando houver varia\u00e7\u00e3o de velocidade, ou seja, surgir acelera\u00e7\u00e3o. R- D<\/b><\/span><\/span><\/p>\n

15<\/b><\/span><\/span>– Se indica um peso maior, N>P e o elevador pode estar subindo e acelerando ou descendo e freando. R- C<\/b><\/span><\/span><\/p>\n

16<\/b><\/span><\/span>– P=m.g\u00a0 —\u00a0 P=0,8.10\u00a0 —\u00a0 P=8,0N\u00a0 —\u00a0 tra\u00e7\u00e3o m\u00e1xima que o fio suporta\u00a0 —\u00a0 T=9,6N\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 T \u2013 P=ma\u00a0 —\u00a0 9,6 \u2013 8,0=0,8.a\u00a0 —\u00a0 a=1,6\/0,8\u00a0 —\u00a0 a=2,0m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>. Como T>P, ele sobe acelerando ou desce freando. R- C<\/b><\/span><\/span><\/p>\n

17<\/b><\/span><\/span>– Vamos colocar todas as for\u00e7as que agem sobre o homem e sobre o elevador<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 peso do elevador\u00a0 —\u00a0 P<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=Mg\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 P<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>\u00a0\u2013 peso do homem\u00a0 —\u00a0 P<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>=mg\u00a0 —\u00a0 N \u2013 for\u00e7a trocada entre o homem e o piso do elevador\u00a0 —\u00a0 T \u2013 for\u00e7a de tra\u00e7\u00e3o na corda que puxa o homem e o elevador para cima.<\/b><\/span><\/span><\/p>\n

Homem\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 T + N – P<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T + N \u2013 mg = ma\u00a0 I<\/b><\/span><\/span><\/p>\n

Elevador\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=Ma\u00a0 —\u00a0 T \u2013 P<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 N = Ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T \u2013 Mg \u2013 N = Ma\u00a0 II<\/b><\/span><\/span><\/p>\n

Somando I com II\u00a0 —\u00a0 2T \u2013 Mg \u2013 mg = Ma + ma\u00a0 —\u00a0 2T = (M + m)g + (M + m)a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=(M + m).(a + g)\/2<\/b><\/span><\/span><\/p>\n

18<\/b><\/span><\/span>– a=g\/5\u00a0 —\u00a0 Desce acelerando\u00a0 —\u00a0 N<P\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 P \u2013 N=m.a\u00a0 —\u00a0\u00a0\u00a0 m.g \u2013 N=m.g\/5\u00a0 —\u00a0 N=(5mg \u2013 mg)\/5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=4mg\/5\u00a0 I<\/b><\/span><\/span><\/p>\n

P=mg\u00a0 II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Dividindo I por II\u00a0 —\u00a0 N\/P=4mg\/5X1\/mg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N\/P=4\/5\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

19<\/b><\/span><\/span>– Sobre o elevador temos T=4f=4.1350\u00a0 —\u00a0 T=5400N (para cima) e P=mg=600.10\u00a0\u00a0 —\u00a0 P=6000N (para baixo).\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P \u2013 T=m.a\u00a0 —\u00a0 6000 \u2013 5400=600.a\u00a0 —\u00a0 a=600\/600\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=1m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

20<\/b><\/span><\/span>– Considerando o sistema, elevador mais pessoa de massas (M + m)\u00a0 —\u00a0 desce freando\u00a0 —\u00a0 T>P\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 T \u2013 P=(M + m).a\u00a0\u00a0 —\u00a0 T \u2013 (M + m).g= (M + m).a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=(M + m).(a + g)\u00a0 R- B<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

21<\/b><\/span><\/span>– Colocando a for\u00e7a pedida e o peso que interessa:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o bloco de massa 2m sobe freando com acelera\u00e7\u00e3o a, N<P\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0=m.a\u00a0 —\u00a0 P \u2013 N=2m.a\u00a0 —\u00a0 2mg \u2013 N=2m.a\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

N=2m(g \u2013 a).<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span>O astronauta tem que descer com acelera\u00e7\u00e3o igual \u00e0 da gravidade da Lua que \u00e9 20% da terrestre\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 a<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=0,2.10=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 — \u00a0\u00a0observe a figura 1 onde est\u00e3o colocadas as for\u00e7as que agem sobre o astronauta —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0\u2013 T = m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 220.10 – T = 220.2\u00a0 —<\/b><\/span><\/span><\/p>\n

T=2200 \u2013 440\u00a0 —\u00a0 T= 1760N\u00a0 —\u00a0 essa tra\u00e7\u00e3o \u00e9 transmitida at\u00e9 o contrapeso\u00a0 —\u00a0 figura 2\u00a0 —\u00a0 o contrapeso deve subir com a mesma acelera\u00e7\u00e3o que o astronauta est\u00e1 descendo, ou seja, a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2T \u2013\u00a0 P<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2.1760 = m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.10 =<\/b><\/span><\/span><\/p>\n

M<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.2\u00a0 —\u00a0 m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u2248293kg\u00a0 —\u00a0 como, segundo o enunciado existem for\u00e7as de atrito, este resultado \u00e9 aproximado eest\u00e1 mais pr\u00f3xima da alternativa A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span>Para que a dist\u00e2ncia vertical entre os centros de massa de cada um passe de 9m para 4,5m, o contrapeso deve estar verticalmente mais abaixo que o astronauta e cada um deles deve se deslocar de S=4,5\/2=2,25m (veja figura)\u00a0 —\u00a0 escolhendo um<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

deles, por exemplo, o astronauta\u00a0 —\u00a0 ele deve subir S<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=2,25m enquanto que no mesmo tempo o contrapeso desce de 2,25m\u00a0 — \u00a0equa\u00e7\u00e3o do astronauta\u00a0 —\u00a0 S<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 2,25 = 0 + 0.t + 2.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=2,25\u00a0 —\u00a0 t=1,5s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

24<\/b><\/span><\/span>–<\/b><\/span><\/span>Pelo enunciado eles se movem com acelera\u00e7\u00f5es de 0,2gTerra, mas como existe atrito essa acelera\u00e7\u00e3o vai diminuir com a dissipa\u00e7\u00e3o de energia\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Polias ou Roldanas \u00a0 01– Observe que temos duas polias m\u00f3veis\u00a0 —\u00a0 n=2\u00a0 —\u00a0 F =25\/2n\u00a0 —\u00a0 25 = Q\/22\u00a0 —\u00a0\u00a0Q = 100N. 02– a) Uma roldana m\u00f3vel, o homem deve aplicar uma for\u00e7a 2n=21=2 vezes menor, ou seja, de 550\/2=275N. b) Duas vezes menor, ou seja, de 0,5m. 03– O dinam\u00f4metro indica<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1183,"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-1187","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1187","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=1187"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1187\/revisions"}],"predecessor-version":[{"id":10853,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1187\/revisions\/10853"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1183"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1187"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}