{"id":3666,"date":"2018-02-13T14:00:15","date_gmt":"2018-02-13T14:00:15","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3666"},"modified":"2024-08-20T15:59:10","modified_gmt":"2024-08-20T15:59:10","slug":"resolucao-comentada-dinamica-impulsiva","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/dinamica-impulsiva\/resolucao-comentada-dinamica-impulsiva\/","title":{"rendered":"Din\u00e2mica Impulsiva – Resolu\u00e7\u00e3o"},"content":{"rendered":"

 <\/p>\n

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

Din\u00e2mica Impulsiva<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Teorema do Impulso\u00a0 —\u00a0 \u00a0\u201c O impulso da for\u00e7a resultante de um sistema de for\u00e7as que age sobre um corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do corpo\u201d<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a0F.\u00a0\u2206t=m.\u00a0\u2206V\u00a0 —\u00a0 F. \u2206t=m.V\/2\u00a0 —\u00a0 F=mV\/2\u2206t\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 \u00a0m = 0,8 kg\u00a0 — v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 93,6 km\/h = 26 m\/s\u00a0 —\u00a0 \u00a0v = 280,8 km\/h = 78 m\/s\u00a0 —\u00a0 o enunciado afirma que antes da colis\u00e3o a velocidade relativa entre a mola e o capacete \u00e9 V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=78 \u2013 26=52m\/s, e no instante ap\u00f3s a colis\u00e3o a velocidade relativa \u00e9 nula\u00a0 —\u00a0 assim, a quantidade de movimento da mola ap\u00f3s a colis\u00e3o em rela\u00e7\u00e3o ao capacete \u00e9 nula\u00a0 —\u00a0 I=\u0394Q\u00a0 —\u00a0 F.\u0394t=m\u0394V\u00a0 —\u00a0 F.\u0394t=m.(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 F.0,026=0,8.(52 \u2013 0)\u00a0 —\u00a0\u00a0F=1.600N\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em todo gr\u00e1fico F x t, o impulso \u00e9 numericamente igual a \u00e1rea compreendida entre a reta <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

representativa e o eixo t, no caso a \u00e1rea do tri\u00e2ngulo\u00a0 —\u00a0 I=b.h\/2=0,6.8,0\/2\u00a0 —\u00a0 I=2,4N.s.<\/b><\/span><\/span><\/span><\/p>\n

Pelo Teorema do Impulso\u00a0 —\u00a0 \u00a0o impulso da for\u00e7a resultante \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento (\u0394Q)\u00a0 —\u00a0 I = \u0394Q = m\u00a0\u0394v \u00a0— \u00a02,4 = 0,1 (v \u2013 0) \u00a0—\u00a0 \u00a0v = 24 m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quantidade de movimento antes da explos\u00e3o\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 quantidade de movimento do sistema depois da explos\u00e3o\u00a0 — Qd=M<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=234.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.1,0.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=234.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0=234.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=5,85.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m\/s.<\/b><\/span><\/span><\/span><\/p>\n

A varia\u00e7\u00e3o da energia cin\u00e9tica da part\u00edcula fornece o trabalho (W) necess\u00e1rio para par\u00e1-la\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cd<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=234.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>(1,0.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 +4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.(5,85.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 — W=E<\/b><\/span><\/span><\/span>cd<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>= – 6,8.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 mas, o trabalho tamb\u00e9m \u00e9 fornecido por\u00a0 — W=F.d.cos180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u00a0– 6,8.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=F.10.(-1)\u00a0\u00a0—\u00a0 F=6,8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo o esquema do enunciado do exerc\u00edcio antes e depois do fen\u00f4meno (escorregamento da ave)\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quantidade de movimento do sistema (ave + tronco) antes\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=50.(+2) + 10.(-2)=80kg.m\/s\u00a0 —\u00a0 quantidade de movimento do sistema (ave + tronco) depois\u00a0 — \u00a0Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>=50.(V) + 10.(-0,5)=50V \u2013 5\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 80=50V \u2013 5\u00a0 —\u00a0\u00a0V=1,7m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura fornecida\u00a0 —\u00a0 orientando a trajet\u00f3ria como positiva no sentido do movimento da sonda (positivo para a esquerda).<\/b><\/span><\/span><\/span><\/p>\n

Antes \u2013 bola chegando \u00e0 sonda\u00a0 com velocidade de 2.000m\/s e a sonda inicialmente parada\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>=10 x 2.000 + 1.000 x 0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=20.000kg.m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Depois \u2013 bola saindo com velocidade Vb=-2000\/5=- 400m\/s e a sonda se movendo para a esquerda com velocidade +V\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>.(+V)=10.( – 400) + 1.000V\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>= – 4.000 + 1.000V\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 20.000= – 4.000 + 1.000V\u00a0 —\u00a0 V=24m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como a dire\u00e7\u00e3o e o sentido da velocidade coincidem com a dire\u00e7\u00e3o e sentido da quantidade de movimento, e como a quantidade de movimento do sistema antes da explos\u00e3o \u00e9 nula, depois da explos\u00e3o tamb\u00e9m dever\u00e1 ser nula. Assim, a \u00fanica alternativa em que todos os vetores se anulam \u00e9 a\u00a0D\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como os caixotes tem a mesma massa, e os choques s\u00e3o perfeitamente el\u00e1sticos, em todos os choques eles trocam ou conservam suas velocidades conforme voc\u00ea pode observar nos esquemas abaixo:\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Observe que ap\u00f3s o 3<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0choque, V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\"\"0 e V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se de uma colis\u00e3o perfeitamente el\u00e1stica com esferas de mesma massa onde elas trocam suas velocidades\u00a0 —\u00a0 o coeficiente de restitui\u00e7\u00e3o vale e=1\u00a0 —\u00a0 a energia cin\u00e9tica se conserva (as velocidades s\u00e3o iguais antes e depois da colis\u00e3o)\u00a0 —\u00a0 em qualquer tipo de colis\u00e3o a quantidade de movimento sempre se conserva\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 800 kg\u00a0 —\u00a0\u00a0 v<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 90 km\/h = 25 m\/s\u00a0 —\u00a0\u00a0 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 450 kg e v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 120 km\/h =120\/3,6=100\/3 m\/s\u00a0 —\u00a0 lembre-se de que voc\u00ea n\u00e3o deve fazer uma divis\u00e3o que d\u00e1 d\u00edzima no meio da solu\u00e7\u00e3o de um exerc\u00edcio\u00a0 —\u00a0 trabalhe com a fra\u00e7\u00e3o\u00a0 —\u00a0 se na resposta final a d\u00edzima persistir, a\u00ed sim, voc\u00ea faz as contas e os arredondamentos\u00a0 — observe que se fosse feita a divis\u00e3o nessa quest\u00e3o, dividindo 100\/3 e obtendo 33,3 m\/s para v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, voc\u00ea teria muito trabalho e n\u00e3o chegaria \u00e0 resposta exata.\u00a0 —\u00a0 m\u00f3dulo da quantidade de movimento dos dois carros antes da colis\u00e3o\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=800.25\u00a0 — Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=20.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=450.100\/3\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=15.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg.m\/s —\u00a0 quantidade de movimento do <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sistema antes da colis\u00e3o, que s\u00e3o perpendiculares, pois as velocidades tamb\u00e9m s\u00e3o\u00a0 —\u00a0 como quantidade de movimento \u00e9 uma grandeza vetorial, como mostra o esquema, vem\u00a0 —\u00a0\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 aplicando Pit\u00e1goras\u00a0 —\u00a0\u00a0 Q<\/b><\/span><\/span><\/span>Santes<\/b><\/span><\/span><\/sub><\/span>=(20.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (15.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>Santes<\/b><\/span><\/span><\/sub><\/span>=25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg.m\/s.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Sendo a colis\u00e3o inel\u00e1stica eles se movem juntos ap\u00f3s a mesma, na mesma dire\u00e7\u00e3o e sentido que\u00a0\"\"\u00a0com velocidade V e massa total M\u00a0 —\u00a0\u00a0 M=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 M=1.250kg\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>Sdepois<\/b><\/span><\/span><\/sub><\/span>=MV\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>Sdepois<\/b><\/span><\/span><\/sub><\/span>\u00a0=1250V\u00a0 —\u00a0 como o sistema \u00e9 isolado a quantidade de movimento antes e depois do choque \u00e9 a mesma\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>Santes<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>Sdepois<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=1250V\u00a0 —\u00a0 V=20m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Soma vetorial das quantidades de movimento antes dos choques:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Baseado no Princ\u00edpio da Conserva\u00e7\u00e3o da quantidade de Movimento (\"\"), dentre todas as alternativas, a \u00fanica cuja soma vetorial das quantidades de movimento depois do choque tamb\u00e9m \u00e9 vertical e para cima \u00e9 a\u00a0E\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

  Resolu\u00e7\u00e3o Comentada Din\u00e2mica Impulsiva \u00a0   01-\u00a0Teorema do Impulso\u00a0 —\u00a0 \u00a0\u201c O impulso da for\u00e7a resultante de um sistema de for\u00e7as que age sobre um corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do corpo\u201d \u00a0F.\u00a0\u2206t=m.\u00a0\u2206V\u00a0 —\u00a0 F. \u2206t=m.V\/2\u00a0 —\u00a0 F=mV\/2\u2206t\u00a0 —\u00a0\u00a0R- D 02-\u00a0Dados\u00a0 —\u00a0 \u00a0m = 0,8 kg\u00a0 — vo\u00a0= 93,6 km\/h = 26 m\/s\u00a0 —\u00a0 \u00a0v<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3664,"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-3666","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3666","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=3666"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3666\/revisions"}],"predecessor-version":[{"id":10631,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3666\/revisions\/10631"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3664"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3666"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}