{"id":963,"date":"2015-06-29T01:15:53","date_gmt":"2015-06-29T01:15:53","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=963"},"modified":"2024-08-23T12:24:48","modified_gmt":"2024-08-23T12:24:48","slug":"resolucao-dos-exercicios-graficos-de-um-mu","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/graficos-de-um-movimento-uniforme-mu\/resolucao-dos-exercicios-graficos-de-um-mu\/","title":{"rendered":"Resolu\u00e7\u00e3o dos Exerc\u00edcios – Gr\u00e1ficos de um MU"},"content":{"rendered":"

01-<\/b><\/span><\/span>\u00a0<\/span><\/span>Se a velocidade \u00e9 constante, a reta representativa \u00e9 paralela ao eixo dos tempos\u00a0 —<\/span><\/span>\u00a0<\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0<\/span><\/span>Como as retas s\u00e3o paralelas, eles possuem a mesma velocidade, mas, observe que partiram de posi\u00e7\u00f5es iniciais diferentes e que A partiu na frente de B\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span>\u00a0<\/span><\/span>I \u2013 errada \u2013 entre 0 e 4s o movimento foi progressivo<\/span><\/span><\/p>\n

II- Errada \u2013 neste intervalo ele ficou parado no marco 50m<\/span><\/span><\/p>\n

III- V<\/span><\/span>m<\/span><\/sub><\/span>=(S<\/span><\/span>9<\/span><\/sub><\/span>\u00a0<\/span><\/span>\u2013 S<\/span><\/span>4<\/span><\/sub><\/span>)\/(t<\/span><\/span>9<\/span><\/sub><\/span>\u00a0<\/span><\/span>\u2013 t<\/span><\/span>4<\/span><\/sub><\/span>)\u00a0 —\u00a0 V<\/span><\/span>m<\/span><\/sub><\/span>=(60 \u2013 50)\/(9 \u2013 4)\u00a0 —\u00a0 V<\/span><\/span>m<\/span><\/sub><\/span>=10\/5\u00a0 —\u00a0 V<\/span><\/span>m<\/span><\/sub><\/span>=2m\/s\u00a0 —\u00a0 verdadeira\u00a0<\/span><\/span>\u00a0<\/span><\/span>—\u00a0 R- C<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span>\u00a0<\/span><\/span>A part\u00edcula tem velocidade constante (reta inclinada), positiva (reta de inclina\u00e7\u00e3o ascendente) e de valor\u00a0 —\u00a0 V=(S \u2013 S<\/span><\/span>o<\/span><\/sub><\/span>)\/(t \u2013 t<\/span><\/span>o<\/span><\/sub><\/span>)=<\/span><\/span><\/p>\n

(200 \u2013 (-200))\/(20 \u2013 0)\u00a0 —\u00a0 V=400\/20\u00a0 — V=20m\/s\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0<\/span><\/span>Deslocamento entre 0 e 4s\u00a0 —\u00a0 \u0394S=soma alg\u00e9brica das \u00e1reas<\/span><\/span><\/p>\n

\"\"<\/p>\n

Entre 0 e 1h\u00a0 —\u00a0 \u0394S<\/span><\/span>1<\/span><\/sub><\/span>=b.h=1.15\u00a0 —\u00a0 \u0394S<\/span><\/span>1<\/span><\/sub><\/span>=15km\u00a0 —\u00a0 entre 1h e 2h\u00a0 —\u00a0 \u0394S<\/span><\/span>2<\/span><\/sub><\/span>=0 (parado, V=0)\u00a0 —\u00a0 entre 2h e 4h\u00a0 —\u00a0 \u0394S<\/span><\/span>3<\/span><\/sub><\/span>=b.h=2.(-20)\u00a0 —\u00a0 \u0394S<\/span><\/span>3<\/span><\/sub><\/span>=-40km\u00a0 —\u00a0 \u0394S<\/span><\/span>total<\/span><\/sub><\/span>=15 + 0 + (-40)\u00a0 —\u00a0 \u0394S<\/span><\/span>total<\/span><\/sub><\/span>=-25km\u00a0 —\u00a0 \u0394S<\/span><\/span>total<\/span><\/sub><\/span>=S \u2013 S<\/span><\/span>o<\/span><\/sub><\/span>\u00a0 —\u00a0 -25=S \u2013 50\u00a0 —<\/span><\/span>\u00a0<\/span><\/span>\u00a0S= + 25km<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>V<\/span><\/span>A<\/span><\/sub><\/span>=(500-0)\/200=2,5m\/s\u00a0 —\u00a0 V<\/span><\/span>B<\/span><\/sub><\/span>=(500-400)\/200=0,5m\/s\u00a0 —\u00a0 V<\/span><\/span>C<\/span><\/sub><\/span>=(500-540)\/200=0,25m\/s\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0<\/span><\/span>Observe no gr\u00e1fico que o movimento dos dois m\u00f3veis tem in\u00edcio no mesmo instante (t=0) e da mesma posi\u00e7\u00e3o, pois quando t=0, \u0394S<\/span><\/span>1<\/span><\/sub><\/span>=\u0394S<\/span><\/span>2<\/span><\/sub><\/span>=0\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0<\/span><\/span>(01)<\/span><\/span>\u00a0<\/span><\/span>Falsa<\/b><\/span><\/span>\u00a0<\/span><\/span>\u2013 \u00e9 de (18-6)=12h<\/span><\/span><\/p>\n

(02) azul \u2013(16-4)=12h\u00a0 —\u00a0 vermelho\u00a0 —\u00a0 (18-6)=12h\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>Correta<\/b><\/span><\/span><\/p>\n

(04) V<\/span><\/span>mazul<\/span><\/sub><\/span>=(720-0)12=60km\/h\u00a0 —\u00a0 V<\/span><\/span>mvermelho<\/span><\/sub><\/span>=(0-720)\/12=-60km\/h\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>Correta<\/b><\/span><\/span><\/p>\n

(08)<\/span><\/span>\u00a0<\/b><\/span><\/span>Correta \u2013<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>veja o gr\u00e1fico<\/span><\/span><\/p>\n

(16)<\/span><\/span>\u00a0<\/span><\/span>Correta<\/b><\/span><\/span>\u00a0<\/span><\/span>\u2013 veja afirma\u00e7\u00f5es anteriores<\/span><\/span><\/p>\n

(32)<\/span><\/span>\u00a0<\/span><\/span>Correta<\/b><\/span><\/span>\u00a0<\/span><\/span>\u2013 \u00e9 o ponto onde as retas azul e vermelha o se interceptam<\/span><\/span><\/p>\n

S=(02+04+08+16+32)=62<\/span><\/span><\/p>\n

09-<\/b><\/span><\/span>\u00a0<\/span><\/span>S<\/span><\/span>o<\/span><\/sub><\/span>=4m\u00a0 —\u00a0 V=(8-4)\/(8-0)\u00a0 —\u00a0 V=0,5m\/s\u00a0 —\u00a0 S=S<\/span><\/span>o<\/span><\/sub><\/span>\u00a0<\/span><\/span>+ Vt\u00a0 —\u00a0 S=4 + 0,5.4\u00a0 —\u00a0 S=4 + 2\u00a0 —\u00a0 S=6m\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0<\/span><\/span>A onda mais r\u00e1pida \u00e9 a P (reta mais inclinada, maior velocidade) e a diferen\u00e7a de tempo \u00e9 (24 \u2013 16)=8s\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a0<\/span><\/span>Z ultrapassa X quando t=10s e Y quando t=30s\u00a0 —\u00a0 (30 \u2013 10)=20s\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span>\u00a0<\/span><\/span>V=(10 \u2013(-10))\/(10 \u2013 0)=20\/10=2m\/s\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span>\u00a0<\/span><\/span>entre 0 e 10s\u00a0 —\u00a0 V=(50 \u2013 10)\/(10-0)=4ms\u00a0 —\u00a0 entre 10s e 20s (nula, est\u00e1 parado no marco 50m)\u00a0 —\u00a0 entre 20s e 40s\u00a0 —\u00a0 V=(0 \u2013 50)\/(40 \u2013 20)=-2,5m\/s\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span>\u00a0<\/span><\/span>Velocidade menor, menor inclina\u00e7\u00e3o da reta (1 e 4), andou\u00a0 —\u00a0 velocidade maior, maior inclina\u00e7\u00e3o da reta (2), correu\u00a0 — velocidade nula (3), parado\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R= A<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span>\u00a0<\/span><\/span>\u0394S=\u00e1rea entre 0 e 4h\u00a0 —\u00a0 \u0394S=1.4 + 1.(-6) + 1.6\u00a0 —\u00a0 \u0394S=4km\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>\u00a0<\/span><\/span>V<\/span><\/span>A<\/span><\/sub><\/span>=(250 \u2013 0)\/(10 \u2013 0)\u00a0 —\u00a0 V<\/span><\/span>A<\/span><\/sub><\/span>=25m\/s\u00a0 —\u00a0 V<\/span><\/span>B<\/span><\/sub><\/span>=(212,5 \u2013 25)\/(10 \u2013 0)\u00a0 —\u00a0 V<\/span><\/span>B<\/span><\/sub><\/span>=18,75m\/s\u00a0 —\u00a0 ultrapassagem por velocidade relativa\u00a0 —\u00a0 mesmo sentido \u2013 V<\/span><\/span>R<\/span><\/sub><\/span>=25 \u2013 18,75\u00a0 —\u00a0 V<\/span><\/span>R<\/span><\/sub><\/span>=6,25m\/s\u00a0 —\u00a0 V<\/span><\/span>R<\/span><\/sub><\/span>=\u0394S(soma do comprimento das duas carreta)\/\u0394t\u00a0 —\u00a0 6,25=50\/\u0394t\u00a0 —\u00a0 \u0394t=8,0s\u00a0 —<\/span><\/span>\u00a0<\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0<\/span><\/span>R- C<\/span><\/span><\/p>\n

18-<\/b><\/span><\/span>\u00a0<\/span><\/span>gr\u00e1fico 1\u00a0 —\u00a0 S<\/span><\/span>o<\/span><\/sub><\/span>=2m\u00a0 —\u00a0 V=tg<\/span><\/span>\u03b1<\/span><\/span><\/span>=1\/2m\/s pela fun\u00e7\u00e3o hor\u00e1ria)\u00a0 —\u00a0 gr\u00e1fico 2 —\u00a0\u00a0 tg2\u03b1=V\u00a0 —\u00a0 tg2\u03b1=2tg\u03b1\/(1- tg<\/span><\/span>2<\/span><\/sup><\/span>\u03b1)\u00a0 —\u00a0 V=2.1\/2\/(1 \u2013 1\/4 )\u00a0 —\u00a0 V=4\/3m\/s\u00a0 —\u00a0 S=S<\/span><\/span>o<\/span><\/sub><\/span>\u00a0<\/span><\/span>+ Vt\u00a0 —\u00a0 S= 2 +4\/3.t\u00a0 —<\/span><\/span>\u00a0<\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

19-\u00a0<\/b><\/span><\/span>a) Entre t=0 e t=5s o espa\u00e7o varia uniformemente com o tempo e a velocidade \u00e9 constante e vale\u00a0 —\u00a0 V=(S \u2013 S<\/span><\/span>o<\/span><\/sub><\/span>)\/(t \u2013 t<\/span><\/span>o<\/span><\/sub><\/span>)=(40 \u2013 0)\/(10 \u2013 0)\u00a0 —\u00a0\u00a0<\/span><\/span>V=4m\/s<\/b><\/span><\/span><\/p>\n

b) Quando t=20s o homem est\u00e1 em repouso na posi\u00e7\u00e3o 40m\u00a0 —\u00a0\u00a0<\/span><\/span>V=0<\/b><\/span><\/span><\/p>\n

20-\u00a0<\/b><\/span><\/span>No gr<\/span><\/span>\u0644<\/span><\/span><\/span><\/span>fico, a dist\u00e2ncia percorrida \u00e9 d=AO + AB + BC\u00a0 —\u00a0 AO<\/span><\/span>2<\/span><\/sup><\/span>=6<\/span><\/span>2<\/span><\/sup><\/span>\u00a0+ 8<\/span><\/span>2<\/span><\/sup><\/span>=100\u00a0 —\u00a0 AO=10cm\u00a0 —\u00a0 AB=(12 \u2013 6)=6cm\u00a0 —\u00a0 BC=(8-0)=8cm\u00a0 —\u00a0 d=10 + 6 + 8=24cm\u00a0 —\u00a0 de acordo com a escala\u00a0 —\u00a0 d=24×1.000\u00a0 —\u00a0 d=2,4.10<\/span><\/span>2<\/span><\/sup><\/span>m\u00a0 —\u00a0 do gr\u00e1fico voc\u00ea obt\u00e9m o deslocamento vetorial\u00a0 —\u00a0 D=\u2502OC\u2502=12cm, no sentido de O para C\u00a0 —\u00a0 de acordo com a escala\u00a0 —\u00a0 D=12×1.000\u00a0 —\u00a0<\/span><\/span><\/p>\n

D=1,2.10<\/span><\/span>2<\/span><\/sup><\/span>m\u00a0 —\u00a0<\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

21- –\u00a0<\/b><\/span><\/span>C\u00e1lculo da velocidade de cada uma\u00a0 —\u00a0 V<\/span><\/span>T<\/span><\/sub><\/span>=20\/50\u00a0 —\u00a0 V<\/span><\/span>T<\/span><\/sub><\/span>=0,4km\/min\u00a0 —\u00a0 V<\/span><\/span>A<\/span><\/sub><\/span>=15\/50\u00a0 —\u00a0 V<\/span><\/span>A<\/span><\/sub><\/span>=0,3km\/min\u00a0 —\u00a0 observe no gr\u00e1fico que ambas partiram da origem\u00a0 —\u00a0 equa\u00e7\u00e3o de cada uma\u00a0 —\u00a0 S<\/span><\/span>T<\/span><\/sub><\/span>=S<\/span><\/span>o<\/span><\/sub><\/span>\u00a0+ V<\/span><\/span>T.<\/span><\/sub><\/span>t=0 + 0,4t\u00a0 —\u00a0 S<\/span><\/span>T<\/span><\/sub><\/span>=0,4t\u00a0 —\u00a0 S<\/span><\/span>A<\/span><\/sub><\/span>=S<\/span><\/span>o<\/span><\/sub><\/span>\u00a0+ V<\/span><\/span>A.<\/span><\/sub><\/span>t=0 + 0,3t\u00a0 —\u00a0 S<\/span><\/span>A<\/span><\/sub><\/span>=0,3t\u00a0 —\u00a0 quando t=30min, T\u00e2nia est\u00e1 na igreja\u00a0 —\u00a0 S<\/span><\/span>T<\/span><\/sub><\/span>=0,4.30\u00a0 —\u00a0 S<\/span><\/span>T<\/span><\/sub><\/span>=12km\u00a0 —\u00a0 a igreja fica no km12\u00a0 —\u00a0 nesse mesmo<\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0instante \u00c2ngela est\u00e1 na posi\u00e7\u00e3o\u00a0 —\u00a0 S<\/span><\/span>A<\/span><\/sub><\/span>=0,3.30\u00a0 —\u00a0 S<\/span><\/span>A<\/span><\/sub><\/span>=9km\u00a0 —\u00a0 quando \u00c2ngela chega `igreja, j\u00e1 faz 40min que T\u00e2nia est\u00e1 caminhando e est\u00e1 na posi\u00e7\u00e3o\u00a0 —\u00a0 S<\/span><\/span>T<\/span><\/sub><\/span>=0,4.40\u00a0 —\u00a0 S<\/span><\/span>T<\/span><\/sub><\/span>=16km\u00a0 —\u00a0\u00a0<\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

32-\u00a0<\/b><\/span><\/span>V=2\u0394S (ida e volta)\/\u0394t\u00a0 —\u00a0 3.10<\/span><\/span>8<\/span><\/sup><\/span>=2\u0394S\/2,5\u00a0 —\u00a0 \u0394S=3,75.10<\/span><\/span>8<\/span><\/sup><\/span>m\u00a0 —\u00a0<\/span><\/span>\u00a0\u0394S=375.000km<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span>\u00a0Nas figuras abaixo est\u00e3o esquematizadas as duas ondas s\u00edsmicas cujas velocidades de propaga\u00e7\u00e3o s\u00e3o constantes\u00a0 —\u00a0 V<\/span><\/span>P<\/span><\/sub><\/span>=\u0394S\/\u0394t<\/span><\/span>P<\/span><\/sub><\/span>\u00a0 –<\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 \u0394t<\/span><\/span>P<\/span><\/sub><\/span>=1.200\/4\u00a0 —\u00a0 \u0394t<\/span><\/span>P<\/span><\/sub><\/span>=300s\u00a0 —\u00a0 V<\/span><\/span>s<\/span><\/sub><\/span>= \u0394S\/\u0394t<\/span><\/span>s<\/span><\/sub><\/span>\u00a0 —\u00a0 \u0394t<\/span><\/span>s<\/span><\/sub><\/span>=1.200\/6\u00a0 —\u00a0 \u0394t<\/span><\/span>s<\/span><\/sub><\/span>=200s\u00a0 —\u00a0 \u0394t=300 \u2013 200\u00a0 —\u00a0 \u0394t=100s\u00a0 —\u00a0<\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

22-\u00a0<\/b><\/span><\/span>I. Falsa\u00a0 —\u00a0 a velocidade \u00e9 constante\u00a0 — movimento retil\u00edneo progressivo uniforme<\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 ele est\u00e1 em repouso na posi\u00e7\u00e3o 50m.<\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 V=(S \u2013 S<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>)=(60 \u2013 50)\/(9 \u2013 4)\u00a0 —\u00a0 V=2m\/s<\/span><\/span><\/span><\/p>\n

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

23-\u00a0<\/b><\/span><\/span>O gr\u00e1fico sugere\u00a0 — \u00a0movimento progressivo acelerado (corrida para pegar o \u00f4nibus); repouso (espera no ponto); movimento uniforme regressivo (volta para casa); novo repouso (espera pelo t\u00e1xi) e, finalmente, movimento progressivo uniforme (movimento do t\u00e1xi).\u00a0<\/span><\/span><\/p>\n

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

24-\u00a0<\/b><\/span><\/span>Observe a figura abaixo\u00a0 —\u00a0 a catedral se encontra no ponto A(1;1), a prefeitura no ponto B (3;1) e a c\u00e2mara de vereadores\u00a0 no<\/span><\/span><\/p>\n

\"\"<\/p>\n

no ponto C (5;3)\u00a0 —\u00a0 observe que AB=2 e AC<\/span><\/span>2<\/span><\/sup><\/span>=(5 \u2013 1)<\/span><\/span>2<\/span><\/sup><\/span>\u00a0+ (3 \u2013 1)<\/span><\/span>2<\/span><\/sup><\/span>\u00a0 —\u00a0 AC=2\u221a5\u00a0 —\u00a0 como AB=500m\u00a0 —\u00a0 AC=500\u221a5m\u00a0 —<\/span><\/span><\/p>\n

\u00a0R- A<\/b><\/span><\/span><\/p>\n

25<\/b><\/span><\/span>–<\/span><\/span>\u00a0R- A<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/span><\/span><\/p>\n

Voltar para os Exerc\u00edcios<\/a><\/span><\/h2>\n","protected":false},"excerpt":{"rendered":"

01-\u00a0Se a velocidade \u00e9 constante, a reta representativa \u00e9 paralela ao eixo dos tempos\u00a0 —\u00a0\u00a0R- C 02-\u00a0Como as retas s\u00e3o paralelas, eles possuem a mesma velocidade, mas, observe que partiram de posi\u00e7\u00f5es iniciais diferentes e que A partiu na frente de B\u00a0 —\u00a0\u00a0R- C 03-\u00a0I \u2013 errada \u2013 entre 0 e 4s o movimento foi progressivo II- Errada \u2013 neste<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":955,"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-963","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/963","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=963"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/963\/revisions"}],"predecessor-version":[{"id":968,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/963\/revisions\/968"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/955"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=963"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}