{"id":1119,"date":"2015-08-02T16:01:21","date_gmt":"2015-08-02T16:01:21","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1119"},"modified":"2024-08-22T17:45:29","modified_gmt":"2024-08-22T17:45:29","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-cinematica-vetorial","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/cinematica-vetorial\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-cinematica-vetorial\/","title":{"rendered":"Cinem\u00e1tica Vetorial – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Cinem\u00e1tica Vetorial<\/b><\/span><\/span><\/span><\/p>\n

01-\u00a0<\/b><\/span><\/span>Aplicando Pit\u00e1goras no tri\u00e2ngulo hachurado\u00a0 —\u00a0 d=500m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

02-<\/b><\/span><\/span>\u00a0a) Supondo que ele se desloque no sentido anti-hor\u00e1rio indo de M para N, o deslocamento vetorial\u00a0\"\"\u00a0\u00e9 obtido unindo os pontos M\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

e N e a intensidade do vetor deslocamento \u00e9 d=2R=2.5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=10m<\/b><\/span><\/span>\u00a0 —\u00a0 \u0394S=2\u03c0R\/2= \u03c0R=5\u03c0\u00a0 —\u00a0 \u0394S=5\u03c0 m<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- B\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 Quando um corpo sai de um ponto A, desloca-se at\u00e9 um ponto B, e retorna ao ponto A, mesmo que com trajet\u00f3rias diferentes na ida e na volta, seu deslocamento vetorial\u00a0\"\"\u00a0\u00e9 nulo, pois voc\u00ea est\u00e1 somando e subtraindo dois vetores de mesma dire\u00e7\u00e3o, mas de sentidos opostos.<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span>\u00a0a) A varia\u00e7\u00e3o de espa\u00e7o \u0394S representa a dist\u00e2ncia efetivamente (realmente) percorrida pelo corpo (dist\u00e2ncia medida sobre a trajet\u00f3ria) durante o percurso de 3\/4 da mesma\u00a0 —\u00a0\u00a0 \u0394S=3\/4.2\u03c0R=3\/4.2\u03c0.4\u00a0 —\u00a0<\/b><\/span><\/span>\u0394S=6\u03c0 m\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

b) O vetor deslocamento \u00e9 obtido unindo-se os pontos P e Q, com sentido de P para Q<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando Pit\u00e1goras no tri\u00e2ngulo hachurado\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=4<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 4<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=\u221a32\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=4\u221a2 m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>dire\u00e7\u00e3o \u2013\u00a0<\/b><\/span><\/span>obl\u00edqua<\/b><\/span><\/span>\u00a0 —\u00a0 sentido –\u00a0<\/b><\/span><\/span>sudoeste<\/b><\/span><\/span><\/p>\n

c) V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t=6\u03c0\/2\u00a0 —\u00a0 V=3\u03c0m\/s<\/b><\/span><\/span><\/p>\n

d) V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=d\/\u0394t=4\u221a2\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=2\u221a2 m\/s<\/b><\/span><\/span>\u00a0 —\u00a0 dire\u00e7\u00e3o \u2013\u00a0<\/b><\/span><\/span>obl\u00edqua<\/b><\/span><\/span>\u00a0 —\u00a0 sentido –\u00a0<\/b><\/span><\/span>sudoeste<\/b><\/span><\/span><\/p>\n

05- a) (NP)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=4<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 3<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 NP=5cm\u00a0 —\u00a0\u00a0 \u0394S=MN + NP + PQ=4 + 5 + 2\u00a0 —\u00a0\u00a0 \u0394S=11cm\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t=11\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=5,5cm\/s<\/b><\/span><\/span><\/p>\n

b) A velocidade vetorial m\u00e9dia\u00a0\"\"\u00a0\u00e9 obtida unindo M com Q e com sentido de M para Q\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando Pit\u00e1goras no tri\u00e2ngulo hachurado\u00a0 —\u00a0 d=5cm\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=d\/\u0394t=5\/2\u00a0 —<\/b><\/span><\/span>V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=2,5cm\/s \u2013 inclinada – sudeste<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>\u00a0(veja teoria)<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0Quando um corpo sai de um ponto A, desloca-se at\u00e9 um ponto B, e retorna ao ponto A, mesmo que com trajet\u00f3rias diferentes na ida e na volta, seu deslocamento vetorial\u00a0\"\"\u00a0\u00e9 nulo, pois voc\u00ea est\u00e1 somando e subtraindo dois vetores de mesma dire\u00e7\u00e3o, mas de<\/b><\/span><\/span>sentidos opostos. Se\u00a0\"\"\u00a0\u00e9 nulo, a velocidade vetorial m\u00e9dia tamb\u00e9m \u00e9 nula, pois V=d\/t\u00a0 —\u00a0 V=0\/t\u00a0 —\u00a0 V=0\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0R- E \u2013 veja resolu\u00e7\u00e3o do exerc\u00edcio anterior<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span>\u00a0I \u2013 Observe no tri\u00e2ngulo hachurado da figura abaixo que d=300<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 400<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=500m=0,5km\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=d\/t=0,5\/0,5\u00a0 —\u00a0 V=1km\/h –<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

correta<\/b><\/span><\/span><\/p>\n

II- Falsa\u00a0 —\u00a0 a dist\u00e2ncia percorrida vale \u0394S=200 + 100 + 100 + 100 + 200 + 100 + 300\u00a0 —\u00a0 \u0394S=1.100m<\/b><\/span><\/span><\/p>\n

III- Correta \u2013 veja I<\/b><\/span><\/span><\/p>\n

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

10-<\/b><\/span><\/span>\u00a0velocidade escalar\u00a0 —\u00a0 V<\/b><\/span><\/span>me<\/b><\/span><\/span><\/sub>=\u0394S\/\u0394t=6L\/t\u00a0 —\u00a0 velocidade vetorial V<\/b><\/span><\/span>mv<\/b><\/span><\/span><\/sub>=d\/t\u00a0 —\u00a0 Lcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=L.12=L\/2\u00a0 —\u00a0 V<\/b><\/span><\/span>mv<\/b><\/span><\/span><\/sub>=L\/2 + L + L\/2 + L\/2 +<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

L + L\/2=4L\u00a0 —\u00a0 V<\/b><\/span><\/span>mv<\/b><\/span><\/span><\/sub>=4L\/t\u00a0 —\u00a0\u00a0 t \u00e9 o mesmo\u00a0 —\u00a0 V<\/b><\/span><\/span>me<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>mv<\/b><\/span><\/span><\/sub>=(6L\/t)\/(4L\/t)\u00a0 — V<\/b><\/span><\/span>me<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>mv<\/b><\/span><\/span><\/sub>=3\/2\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a0Se a velocidade vetorial \u00e9 constante e n\u00e3o nula, ela tem em todos os pontos a mesma intensidade, mesma dire\u00e7\u00e3o e mesmo sentido e, nesse caso o movimento s\u00f3 pode ser um\u00a0 movimento\u00a0<\/b><\/span><\/span>retil\u00edneo<\/b><\/span><\/span>\u00a0e uniforme\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span>\u00a0a) veja figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Vm=18km\/h=5m\/s\u00a0 —\u00a0 menor caminho\u00a0 —\u00a0 \u0394S=200 + 400 + 200 + 100 =900m\u00a0 —\u00a0 Vm=\u0394S\/\u0394t\u00a0 —\u00a0 5=900\/t\u00a0 —\u00a0 t=180s=3minutos<\/b><\/span><\/span><\/p>\n

b) A intensidade do vetor deslocamento\u00a0\"\"\u00a0\u00e9 obtida aplicando Pit\u00e1goras no tri\u00e2ngulo hachurado da figura abaixo\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 300<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

400<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=500m\u00a0 —\u00a0 o tempo de percurso \u00e9 o mesmo t=180s\u00a0 —\u00a0 V<\/b><\/span><\/span>mv<\/b><\/span><\/span><\/sub>=d\/t=500\/180×3,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>mv<\/b><\/span><\/span><\/sub>=10km\/h\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span>\u00a0\u0394S=20 + 20 + 20 + 20 + 20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394S=100m<\/b><\/span><\/span>\u00a0 —\u00a0 o deslocamento vetorial\u00a0\"\"\u00a0\u00e9 obtido unindo-se A com B, no sentido de A para B<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando Pit\u00e1goras no tri\u00e2ngulo hachurado\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=40<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 20<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d=20\u221a5m<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

14-\u00a0<\/b><\/span><\/span>Como os intervalos de tempo s\u00e3o iguais parara todos os atletas e o deslocamento vetorial tamb\u00e9m \u00e9 o mesmo (obtido, em cada caso, unindo os pontos A e B), o vetor velocidade m\u00e9dia (V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=d\/t), tamb\u00e9m ser\u00e1 o mesmo para todos os atletas\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span>\u00a0(01) Falsa, pois se o movimento for circular e uniforme a velocidade vetorial varia em dire\u00e7\u00e3o e sentido.<\/b><\/span><\/span><\/p>\n

(02) Correta \u2013 veja figuras abaixo<\/b><\/span><\/span><\/p>\n

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

(04) Correta \u2013 todo vetor fica determinado quando se conhece sua intensidade, dire\u00e7\u00e3o e sentido.<\/b><\/span><\/span><\/p>\n

(08) Correta \u2013 veja (02)<\/b><\/span><\/span><\/p>\n

(16) Falsa \u2013 veja (01)<\/b><\/span><\/span><\/p>\n

R- soma (02 + 04 + 08)=<\/b><\/span><\/span>14<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>\u00a0No tri\u00e2ngulo abaixo\u00a0 —\u00a0 sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=cateto oposto\/hipotenusa=x\/d\u00a0 —\u00a0 1\/2 = x\/d\u00a0\u00a0 —\u00a0 x=d\/2\u00a0 — nessas condi\u00e7\u00f5es, o tri\u00e2ngulo do meio<\/b><\/span><\/span><\/p>\n

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

\u00a0eq\u00fcil\u00e1tero de lados d\u00a0 —\u00a0 como tem-se 3 tri\u00e2ngulos eq\u00fcil\u00e1teros de lados d\u00a0 —\u00a0 3d=2,1\u00a0 —\u00a0 d=0,7m \u2013 observe na figura acima que ele percorre \u0394S=6d=6.0,7\u00a0 —\u00a0 \u0394S=4,2m\u00a0 —\u00a0 V= \u0394S\/t\u00a0 —\u00a0 14=4,2\/t\u00a0 —\u00a0 t=0,3s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=30.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>s<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0Sim, pois sempre que um corpo efetua uma curva com qualquer velocidade, mesmo constante, surge sobre ele uma for\u00e7a que o obriga a fazer a curva (no caso a for\u00e7a de atrito), que \u00e9 denominada for\u00e7a resultante centr\u00edpeta, de dire\u00e7\u00e3o radial e dirigida para o centro da curva. Se existe for\u00e7a centr\u00edpeta, existir\u00e1 tamb\u00e9m acelera\u00e7\u00e3o centr\u00edpeta tamb\u00e9m de dire\u00e7\u00e3o radial e dirigida para o centro da curva.<\/b><\/span><\/span><\/p>\n

18-\u00a0<\/b><\/span><\/span>Observe com muita aten\u00e7\u00e3o cada caso esquematizado na figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

19-<\/b><\/span><\/span>\u00a0Como ele est\u00e1 em movimento circular uniforme, a intensidade do vetor velocidade \u00e9 constante e vale V=\u221a2m\/se, ent\u00e3o n\u00e3o existe acelera\u00e7\u00e3o tangencial, mas somente centr\u00edpeta, de intensidade\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=(\u221a2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/0,5=2\/0,5\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span>\u00a0Se a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=0, o movimento \u00e9 uniforme e se a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=constante e n\u00e3o nula ele est\u00e1 em trajet\u00f3ria circular (V e R s\u00e3o constantes). Portanto trata-se de um movimento circular uniforme\u00a0 —<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span>\u00a0Como a velocidade escalar V \u00e9 constante, n\u00e3o existe a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>, somente a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0que \u00e9 fornecida por a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 como V \u00e9 constante, a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0\u00e9 inversamente proporcional a R, ou seja, menor R \u2013 maior a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span>\u00a0No movimento circular uniforme, movimento em que a intensidade (m\u00f3dulo) da velocidade vetorial \u00e9 constante, sua dire\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

e sentido variam, pois\"\"\u00e9 tangente \u00e0 trajet\u00f3ria em cada ponto.\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span>\u00a0Se a indica\u00e7\u00e3o do veloc\u00edmetro \u00e9 constante a acelera\u00e7\u00e3o tangencial \u00e9 nula, restando apenas a acelera\u00e7\u00e3o centr\u00edpeta (a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

1voltas completa\u00a0 —\u00a0 \u0394S=2\u03c0R —\u00a0 \u0394t=2min=2.60\u00a0 —\u00a0 \u0394t=120s\u00a0 —\u00a0 V=\u0394S\/\u0394t=2\u03c0R\/120\u00a0 —\u00a0 V=\u03c0R\/60\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 4=(\u03c0R\/60)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 4=\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/3.600 x 1\/R\u00a0 — 14.400=\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R\u00a0 —\u00a0 R=14.400\/\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=14.400\/9,9596=1.445,9m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span>\u00a0I- Falsa \u2013 como a resist\u00eancia do ar \u00e9 desprezada, a for\u00e7a resultante que age sobre cada bola \u00e9 seu peso (P=mg,vertical e para baixo)<\/b><\/span><\/span><\/p>\n

II- Correta \u2013 sendo a acelera\u00e7\u00e3o tangencial constante (a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=g), a maior altura ser\u00e1 atingida por aquela que tem maior velocidade inicial.<\/b><\/span><\/span><\/p>\n

III_ Correta \u2013 pela acelera\u00e7\u00e3o da gravidade g, vertical e para baixo<\/b><\/span><\/span><\/p>\n

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

25-<\/b><\/span><\/span>\u00a0I- no trecho AB \u00e9 acelerado com a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>, pois partiu do repouso V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 nos trechos BC e DE, n\u00e3o tem a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0(velocidade constante) mas tem a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0(trajet\u00f3ria curva)\u00a0 —\u00a0 no trecho EF \u00e9 retardado com a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>, pois est\u00e1 freando at\u00e9 parar\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>correta\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

II-\u00a0<\/b><\/span><\/span>correta<\/b><\/span><\/span>\u00a0\u2013 num trecho ele acelera e no outro, retarda.<\/b><\/span><\/span><\/p>\n

III-\u00a0<\/b><\/span><\/span>correta<\/b><\/span><\/span>\u00a0\u2013 veja figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

26-<\/b><\/span><\/span>\u00a0O vetor velocidade \u00e9 tangente \u00e0 trajet\u00f3ria e tem o sentido do movimento (horizontal e para a direita)\u00a0 —\u00a0 a acelera\u00e7\u00e3o centr\u00edpeta \u00e9 dirigida para o centro da curva (vertical e para baixo)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R-\u00a0\"\"e\u00a0\"\"<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span>\u00a0De acordo com o enunciado<\/b><\/span><\/span>\u00a0a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=1m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V ap\u00f3s 10s vale V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 V= 0 + 1.10\u00a0 —\u00a0 V=10m\/s\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=1m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

28-<\/b><\/span><\/span>\u00a0Observe a figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0 —\u00a0 a=\u221a2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 sen\u03b2=a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\/a\u00a0 —\u00a0 sen\u03b2=1\/\u221a2\u00a0 —\u00a0 sen<\/b><\/span><\/span>\u03b2<\/b><\/span><\/span><\/span>=\u221a2\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b2<\/b><\/span><\/span><\/span>=45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span>\u00a0\u00a0Movimento uniforme\u00a0 —\u00a0\u00a0\"\"—\u00a0\u00a0\"\"\u00a0e\u00a0\"\"\u00a0s\u00e3o perpendiculares\u00a0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span>\u00a0Entre 2s e 4s ele percorreu um quarto do arco da circunfer\u00eancia\u00a0 —\u00a0 \u0394S=2\u03c0R\/4\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0 \u221a2=(\u03c0R\/2)\/(4-2)\u00a0 —\u00a0 4\u221a2=\u03c0R\u00a0 —\u00a0 R=4\u221a2\/\u03c0\u00a0 —\u00a0 R= 1,8m\u00a0 —\u00a0 sendo a velocidade constante s\u00f3 existe acelera\u00e7\u00e3o centr\u00edpeta, tamb\u00e9m constante\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>= (\u221a2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/1,9=1,0m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span>\u00a0Somente na trajet\u00f3ria II, onde existe acelera\u00e7\u00e3o centr\u00edpeta e consequentemente for\u00e7a centr\u00edpeta e, pelo princ\u00edpio da in\u00e9rcia ele se \u201csentir\u00e1 jogado\u201d, em sentido contr\u00e1rio ao da curva, tentando sair pela tangente\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span>\u00a0Na figura abaixo a acelera\u00e7\u00e3o\u00a0\"\"\u00a0est\u00e1 decomposta em suas parcelas\u00a0\"\"\u00a0e\u00a0\"\"<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

cos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/a\u00a0 —\u00a0 0,5=a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/32\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=16m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 16=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/1\u00a0 —\u00a0 V=4m\/s\u00a0 —\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span>\u00a0Veja esquema abaixo:<\/b><\/span><\/span><\/p>\n

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

34-\u00a0<\/b><\/span><\/span>R- B\u00a0<\/b><\/span><\/span>\u00a0(veja teoria)<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span>\u00a0Se est\u00e1 em MCU, a acelera\u00e7\u00e3o tangencial \u00e9 nula, a velocidade vetorial \u00e9 tangente e a acelera\u00e7\u00e3o \u00e9 a centr\u00edpeta\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span>\u00a0R- C\u00a0 —\u00a0 est\u00e1 em movimento circular uniforme<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span>\u00a0A figura abaixo mostra os vetores\u00a0\"\"\u00a0e\u00a0\"\", decompostos a partir do vetor\u00a0\"\".<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/4\u00a0 —\u00a0 0,87=a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/4\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=3,48m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 3,48=(10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 R=29m\u00a0 —\u00a0 cos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\/4\u00a0 —\u00a0 0,50=a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\/4\u00a0 —\u00a0 a<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=2,0m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/p>\n

Veja figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

39<\/b><\/span><\/span>– Trata-se de uma localiza\u00e7\u00e3o em tr\u00eas dimens\u00f5es<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394V=V \u2013 V<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>=(4i \u2013 2j + k) \u2013 (i -2j + 5k) = 4i -2j +k \u2013i +2j \u2013 5k\u00a0 —\u00a0\u00a0\u0394V=3i \u2013 4k\u00a0 —\u00a0 a=\u00a0\u0394V\/\u0394t=(3i \u2013 4k)\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=1,5i \u2013 2k<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(1,5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (2,0)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a=\u221a6,25\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=2,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

40-\u00a0<\/b><\/span><\/span>a) Como CI parte de A e PA parte de B, o deslocamento de CI deve ser 300m maior que o deslocamento de PA\u00a0 — \u0394S<\/b><\/span><\/span>CI<\/b><\/span><\/span><\/sub>=\u0394S<\/b><\/span><\/span>PA<\/b><\/span><\/span><\/sub>\u00a0+300\u00a0 —\u00a0 como o movimento de CI \u00e9 uniformemente variado e o de PA \u00e9 uniforme\u00a0 —\u00a0 equa\u00e7\u00f5es de cada um com a origem em A\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>CI<\/b><\/span><\/span><\/sub>=gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>PA<\/b><\/span><\/span><\/sub>=Vt + 300\u00a0 —\u00a0 no encontro\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>CI<\/b><\/span><\/span><\/sub>=\u0394S<\/b><\/span><\/span>PA<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=1.t + 300\u00a0 —\u00a0 2,5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 t + 300=0\u00a0 —\u00a0 resolvendo a equa\u00e7\u00e3o\u00a0 —\u00a0 t=600s (tempo do encontro)\u00a0 —\u00a0 deslocamento escalar de PA\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>PA<\/b><\/span><\/span><\/sub>=Vt=1.600\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>PA<\/b><\/span><\/span><\/sub>=600m\u00a0 —\u00a0como PA percorreu 600m at\u00e9 o encontro, ele ocorre no ponto D da figura abaixo\u00a0 —\u00a0 a diagonal do quadrado (BD) \u00e9 o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

deslocamento vetorial (\"\") de PA e tem intensidade\u00a0\u00a0 —\u00a0 d=300\u221a2m\u00a0 —\u00a0 o vetor velocidade m\u00e9dia (\"\") tem intensidade\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=300\u221a2\/600\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0\u2248 0,7m\/s<\/b><\/span><\/span>, dire\u00e7\u00e3o\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>obl\u00edqua\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 sentido\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>NO formando um \u00e2ngulo de 45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0com a horizontal.<\/b><\/span><\/span><\/p>\n

b) CI andar\u00e1 900m at\u00e9 o encontro\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=900\/600\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1,5m\/s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

41-\u00a0<\/b><\/span><\/span>C\u00e1lculo do deslocamento em todo o trajeto\u00a0DS = v.Dt = 24.1 = 24 km\u00a0 —\u00a0 dist\u00e2ncia AB pode ser calculada da seguinte forma\u00a0 —\u00a0DS = 2.AB + 2.BC\u00a0 —\u00a0\u00a0DS\/2 = AB + BC\u00a0 —\u00a0 AB =\u00a0DS\/2 \u2013 BC =24\/2\u2013 3 = 12 \u2013 3 = 9 km\u00a0 —\u00a0 \u00a0tempo total gasto no trecho ABCB<\/b><\/span><\/span><\/p>\n

v =\u00a0DS\/Dt\u00a0 —\u00a0\u00a0Dt =\u00a0DS\/v = (9 + 6)\/20= 15\/20\u00a0 —\u00a0\u00a0Dt = 0,75 h\u00a0 —\u00a0 \u00a0m\u00f3dulo da velocidade vetorial m\u00e9dia 9\/0,75 = 12 km\/h<\/b><\/span><\/span><\/p>\n

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

42-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 largura do rio D = 60 m\u00a0 —\u00a0 \u00a0Dt = 2 min = 120 s\u00a0 —\u00a0 cos\u00a0q\u00a0= 4\/5\u00a0 —\u00a0 sen\u00a0q\u00a0=3\/5\u00a0 —\u00a0 a figura abaixo ilustra as<\/b><\/span><\/span><\/p>\n

velocidades, sendo: v a velocidade de Pedro em rela\u00e7\u00e3o \u00e0 margem; V<\/b><\/span><\/span>p\/ag<\/b><\/span><\/span><\/sub>: a velocidade de Pedro em rela\u00e7\u00e3o \u00e0 \u00e1gua e v<\/b><\/span><\/span>ag<\/b><\/span><\/span><\/sub>\u00a0a velocidade da \u00e1gua.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) V<\/b><\/span><\/span><\/span>p\/ag<\/b><\/span><\/span><\/span><\/sub>=D\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>t=60\/120\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>p\/ag<\/b><\/span><\/span><\/span><\/sub>=0,5m\/s\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) Da figura\u00a0 —\u00a0 cos\u03b8= V<\/b><\/span><\/span>p\/ag<\/b><\/span><\/span><\/sub>\/V\u00a0 —\u00a0 3\/5=0,5\/V\u00a0 —\u00a0 V=2,5\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=0,83m\/s<\/b><\/span><\/span><\/p>\n

c) Da mesma figura\u00a0 —\u00a0 sen\u03b8=V<\/b><\/span><\/span>ag<\/b><\/span><\/span><\/sub>\/V\u00a0 —\u00a0 4\/5=V<\/b><\/span><\/span>ag<\/b><\/span><\/span><\/sub>\/2,5\/3\u00a0 —\u00a0 V<\/b><\/span><\/span>ag<\/b><\/span><\/span><\/sub>=10\/15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>ag<\/b><\/span><\/span><\/sub>=0,67m\/s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

43-\u00a0<\/b><\/span><\/span>O vetor deslocamento (\"\") do carteiro \u00e9 obtido unindo-se os pontos A com B, figura abaixo\u00a0 —\u00a0 o m\u00f3dulo de\u00a0\"\"\u00a0\u00e9 fornecido pelo teorema de Pit\u00e1goras\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=400<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 300<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=500m\u00a0 —\u00a0 dire\u00e7\u00e3o oeste-leste\u00a0 —\u00a0 sentido para o leste\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

44-\u00a0<\/b><\/span><\/span>Dist\u00e2ncia percorrida do sul para o norte\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 20=\u0394S\/(15\/60)\u00a0 —\u00a0 \u0394S=5km\u00a0 —\u00a0 o novo deslocamento de 5km forma um \u00e2ngulo de 60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0com o deslocamento inicial\u00a0 —\u00a0 veja a figura abaixo onde\u00a0\"\"\u00a0\u00e9 o vetor deslocamento pedido\u00a0 —\u00a0 aplicando a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

lei dos cossenos\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=5<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 5<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0+ 2.5.5.cos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=\u221a75\u00a0 —\u00a0 d=5\u221a3m<\/b><\/span><\/span><\/p>\n

 <\/p>\n

45-<\/b><\/span><\/span><\/p>\n

A acelera\u00e7\u00e3o vetorial\u00a0\"\"\u00a0\u00e9 definida como\u00a0\"\", e indica a varia\u00e7\u00e3o de velocidade uma unidade de tempo \u00a0sendo que essa varia\u00e7\u00e3o pode acontecer tanto em dire\u00e7\u00e3o como em sentido.<\/b><\/span><\/span><\/p>\n

Assim, a acelera\u00e7\u00e3o vetorial\u00a0\"\"\u00a0pode ser decomposta em duas; a acelera\u00e7\u00e3o tangencial\u00a0\"\"\u00a0e a acelera\u00e7\u00e3o centr\u00edpeta\u00a0\"\", conforme a figura.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0Caracter\u00edsticas da acelera\u00e7\u00e3o tangencial (\"\"): \u00e9 respons\u00e1vel pela\u00a0varia\u00e7\u00e3o da intensidade (m\u00f3dulo)\u00a0do vetor velocidade, fazendo com que o movimento seja\u00a0acelerado ou\u00a0retardado.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

 <\/p>\n

 <\/p>\n

46-<\/b><\/span><\/span><\/p>\n

Em qualquer movimento circular (uniforme ou n\u00e3o) surge sempre uma acelera\u00e7\u00e3o denominada acelera\u00e7\u00e3o centr\u00edpeta\u00a0\"\"\u00a0respons\u00e1vel pela varia\u00e7\u00e3o da dire\u00e7\u00e3o do vetor velocidade,com as seguintes caracter\u00edsticas: \u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

 <\/p>\n

47-<\/b><\/span><\/span><\/p>\n

Em todo movimento circular existe sempre uma for\u00e7a resultante com dire\u00e7\u00e3o radial e sentido dirigido para o centro da circunfer\u00eancia, de intensidade Fc=mV2\/R, denominada for\u00e7a resultante centr\u00edpeta\u00a0 —<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0R- C<\/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 Cinem\u00e1tica Vetorial 01-\u00a0Aplicando Pit\u00e1goras no tri\u00e2ngulo hachurado\u00a0 —\u00a0 d=500m\u00a0 —\u00a0\u00a0R- D 02-\u00a0a) Supondo que ele se desloque no sentido anti-hor\u00e1rio indo de M para N, o deslocamento vetorial\u00a0\u00a0\u00e9 obtido unindo os pontos M\u00a0 e N e a intensidade do vetor deslocamento \u00e9 d=2R=2.5\u00a0 —\u00a0\u00a0d=10m\u00a0 —\u00a0 \u0394S=2\u03c0R\/2= \u03c0R=5\u03c0\u00a0 —\u00a0 \u0394S=5\u03c0 m 03-\u00a0R- B\u00a0\u00a0—\u00a0 Quando<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1115,"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-1119","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1119","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=1119"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1119\/revisions"}],"predecessor-version":[{"id":10786,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1119\/revisions\/10786"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1115"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1119"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}