{"id":779,"date":"2015-05-13T15:30:43","date_gmt":"2015-05-13T15:30:43","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=779"},"modified":"2024-09-02T17:35:50","modified_gmt":"2024-09-02T17:35:50","slug":"resolucao-comentada-da-ucpel-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-ucpel-2015\/resolucao-comentada-da-ucpel-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da UCPel 2015"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>A for\u00e7a de<\/b><\/span><\/span><\/span> <\/b>atra\u00e7\u00e3o ou de repuls\u00e3o<\/b><\/span><\/span><\/span> <\/b>entre duas cargas<\/b><\/span><\/span><\/span> <\/b>el\u00e9tricas<\/b><\/span><\/span><\/span> <\/b>tem sempre a<\/b><\/span><\/span><\/span> <\/b>mesma intensidade<\/b><\/span><\/span><\/span> <\/b>independente<\/b><\/span><\/span><\/span> <\/b>do sinal e do<\/b><\/span><\/span><\/span> <\/b>valor das cargas.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> =F \u00e9 a intensidade das for\u00e7as sobre as cargas Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> e Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>= F<\/b><\/span><\/span><\/span> <\/b>= m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span>(3m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/4)a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>e<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= F<\/b><\/span><\/span><\/span> <\/b>= m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span>(3m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/4)a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>= m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span>3.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/4 = <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub> \"\" <\/b><\/span><\/span>a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 4\/3.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>Transforma\u00e7\u00e3o isoc\u00f3rica, isom\u00e9trica ou isovolum\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Ocorre a volume constante\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=V\u00a0 —\u00a0 \u0394V=0\u00a0 —\u00a0 W=P.\u0394V\u00a0 —\u00a0 W=P.0\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>W=0<\/b><\/span><\/span><\/span>\u00a0 (trabalho \u00e9 nulo) — \u0394U = Q \u2013 W\u00a0 —\u00a0 \u0394U=Q \u2013 0\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>\u0394U=Q<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Assim, <\/b><\/span><\/span><\/span>todo o calor (Q) recebido pelo sistema \u00e9 igual \u00e0 sua varia\u00e7\u00e3o de energia interna (\u0394U).\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/b><\/span><\/span><\/span>Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 a seguir com algumas \u201cdicas\u201d:<\/b><\/span><\/span><\/span><\/p>\n

–\u00a0P\u00eandulo Simples –<\/b><\/span><\/span><\/span><\/p>\n

Consta de uma <\/b><\/span><\/span><\/span>massa m<\/b><\/span><\/span><\/span>, presa na <\/b><\/span><\/span><\/span>extremidade inferior de um fio ideal<\/b><\/span><\/span><\/span>, fixada verticalmente na sua extremidade superior (figura)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se o <\/b><\/span><\/span><\/span>p\u00eandulo simples oscilar<\/b><\/span><\/span><\/span>, com oscila\u00e7\u00f5es de pequena abertura (no m\u00e1ximo 15<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>), ele descreve um <\/b><\/span><\/span><\/span>movimento circular de raio R=L, sendo L o comprimento do fio.<\/b><\/span><\/span><\/span><\/p>\n

Seu <\/b><\/span><\/span><\/span>per\u00edodo (T)<\/b><\/span><\/span><\/span>, que \u00e9 o tempo que ele demora para efetuar um \u201cvai e vem\u201d completo \u00e9 fornecido pela express\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

onde <\/b><\/span><\/span><\/span>g \u00e9 a acelera\u00e7\u00e3o da gravidade local.<\/b><\/span><\/span><\/span><\/p>\n

– Observe que:<\/b><\/span><\/span><\/span><\/p>\n

* a <\/b><\/span><\/span><\/span>massa pendular m n\u00e3o influi no per\u00edodo T do movimento<\/b><\/span><\/span><\/span>. Assim dois p\u00eandulos de <\/b><\/span><\/span><\/span>mesmo comprimento L<\/b><\/span><\/span><\/span>, mas de <\/b><\/span><\/span><\/span>massas diferentes M e m, apresentam o mesmo per\u00edodo T.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0* O <\/b><\/span><\/span><\/span>per\u00edodo de um p\u00eandulo simples independe da amplitude<\/b><\/span><\/span><\/span>, ou seja, da altura em que m \u00e9 abandonada, assim, os p\u00eandulos da figura abaixo, tanto na situa\u00e7\u00e3o 1 como na 2, demoram o mesmo tempo para ir de A at\u00e9 B, de B at\u00e9 C, de C at\u00e9 B e de B at\u00e9 A.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

* \u00a0O per\u00edodo de um p\u00eandulo simples \u00e9 <\/b><\/span><\/span><\/span>diretamente proporcional \u00e0 raiz quadrada de seu<\/b><\/span><\/span><\/span> <\/b>comprimento L<\/b><\/span><\/span><\/span>. Assim, para dobrar o per\u00edodo T de um p\u00eandulo, seu comprimento L deve ser quadruplicado, etc.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

* O per\u00edodo de um p\u00eandulo simples \u00e9 <\/b><\/span><\/span><\/span>inversamente proporcional \u00e0 raiz quadrada da acelera\u00e7\u00e3o da<\/b><\/span><\/span><\/span> <\/b>gravidade g.<\/b><\/span><\/span><\/span> Assim,\u00a0 quanto maior for a acelera\u00e7\u00e3o da gravidade do local onde est\u00e1 o p\u00eandulo, menor ser\u00e1 o seu per\u00edodo. Uma das aplica\u00e7\u00f5es do p\u00eandulo simples \u00e9 a determina\u00e7\u00e3o da acelera\u00e7\u00e3o da gravidade.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0No caso do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>comprimento final ficou 1\/4 do comprimento inicial<\/b><\/span><\/span><\/span> \"\" o <\/b><\/span><\/span><\/span>per\u00edodo<\/b><\/span><\/span><\/span> de um p\u00eandulo simples \u00e9 <\/b><\/span><\/span><\/span>diretamente proporcional \u00e0 raiz quadrada de seu<\/b><\/span><\/span><\/span> <\/b>comprimento L<\/b><\/span><\/span><\/span>. Assim, para <\/b><\/span><\/span><\/span>dobrar o per\u00edodo T<\/b><\/span><\/span><\/span> de um p\u00eandulo, <\/b><\/span><\/span><\/span>seu comprimento L deve ser quadruplicado<\/b><\/span><\/span><\/span>, e vice versa.<\/b><\/span><\/span><\/span><\/p>\n

Observe na express\u00e3o fornecida que a <\/b><\/span><\/span><\/span>massa pendular m n\u00e3o influi no per\u00edodo T do movimento<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R \u2013 A<\/b><\/span><\/span><\/span><\/p>\n

04- <\/b><\/span><\/span><\/span>Podemos definir o <\/b><\/span><\/span><\/span>Princ\u00edpio Fundamental da Din\u00e2mica ou segunda lei de Newton<\/b><\/span><\/span><\/span> do seguinte modo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim, se essa\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante<\/b><\/span><\/span><\/span>\u00a0for constante<\/b><\/span><\/span><\/span>, a acelera\u00e7\u00e3o (varia\u00e7\u00e3o de velocidade), por ela produzida tamb\u00e9m ser\u00e1 <\/b><\/span><\/span><\/span>constante<\/b><\/span><\/span><\/span> e a for\u00e7a e a acelera\u00e7\u00e3o s\u00e3o grandezas diretamente proporcionais, independente do tipo de trajet\u00f3ria, ou seja:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

05- <\/b><\/span><\/span><\/span>I. Correta<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>O <\/b><\/span><\/span><\/span><\/span>desvio (d) na refra\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/span> corresponde ao <\/b><\/span><\/span><\/span><\/span>\u00e2ngulo <\/b><\/span><\/span><\/span><\/span>entre o prolongamento do raio incidente e o raio refratado.(figura da esquerda)<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

No caso da incid\u00eancia ser <\/b><\/span><\/span><\/span><\/span>normal<\/b><\/span><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span><\/span>raio de luz n\u00e3o sofre desvio e d=0.<\/b><\/span><\/span><\/span><\/span> (figura da direita)<\/b><\/span><\/span><\/span><\/span><\/p>\n

II. Falsa<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Onda n\u00e3o transporta mat\u00e9ria, apenas energia.<\/b><\/span><\/span><\/span><\/p>\n

* O som \u00e9 uma <\/b><\/span><\/span><\/span>onda mec\u00e2nica (precisa de um meio material para se propagar)<\/b><\/span><\/span><\/span> longitudinal (dire\u00e7\u00e3o de propaga\u00e7\u00e3o coincide com a dire\u00e7\u00e3o de oscila\u00e7\u00e3o)<\/b><\/span><\/span><\/span><\/p>\n

* A luz \u00e9 uma <\/b><\/span><\/span><\/span>onda eletromagn\u00e9tica (n\u00e3o precisa de meio material para se propagar)<\/b><\/span><\/span><\/span> transversal (dire\u00e7\u00e3o de propaga\u00e7\u00e3o \u00e9 perpendicular \u00e0 dire\u00e7\u00e3o de oscila\u00e7\u00e3o).<\/b><\/span><\/span><\/span><\/p>\n

III. Correta<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

* Quando duas ou mais ondas sofrem <\/b><\/span><\/span><\/span>superposi\u00e7\u00e3o<\/b><\/span><\/span><\/span> temos o fen\u00f4meno da <\/b><\/span><\/span><\/span>interfer\u00eancia<\/b><\/span><\/span><\/span>, que ocorre tanto para ondas transversais como para ondas longitudinais.<\/b><\/span><\/span><\/span><\/p>\n

* <\/b><\/span><\/span><\/span>Durante a superposi\u00e7\u00e3o, a amplitude da onda resultante \u00e9 a soma alg\u00e9brica das amplitudes de cada onda.<\/b><\/span><\/span><\/span><\/p>\n

* Quando os pulsos <\/b><\/span><\/span><\/span>tiverem a mesma amplitude<\/b><\/span><\/span><\/span> e em oposi\u00e7\u00e3o de fase teremos <\/b><\/span><\/span><\/span>interfer\u00eancia destrutiva\u00a0<\/b><\/span><\/span><\/span>e durante a interfer\u00eancia, quando a corda estiver na horizontal, cada ponto da mesma ter\u00e1 velocidade vertical e horizontal nulas.<\/b><\/span><\/span><\/span><\/p>\n

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

IV. Falsa <\/b><\/span><\/span><\/span> \"\" \u00a0<\/b><\/span><\/span><\/span>Polarizar<\/b><\/span><\/span><\/span> uma onda significa transformar, atrav\u00e9s do polarizador, uma onda <\/b><\/span><\/span><\/span>transversal\u00a0 n\u00e3o polarizada,que vibra em v\u00e1rias dire\u00e7\u00f5es<\/b><\/span><\/span><\/span>, numa onda <\/b><\/span><\/span><\/span>polarizada, que vibra numa \u00fanica dire\u00e7\u00e3o<\/b><\/span><\/span><\/span>.<\/span><\/p>\n

Somente ondas transversais, como a luz, podem ser polarizadas e n\u00e3o se propagam num meio material.<\/b><\/span><\/span><\/span><\/p>\n

O som, que \u00e9 uma onda longitudinal, n\u00e3o pode ser polarizada.<\/b><\/span><\/span><\/span><\/p>\n

V. Correta<\/b><\/span><\/span><\/span> \"\" as <\/b><\/span><\/span><\/span>caracter\u00edsticas do meio<\/b><\/span><\/span><\/span> influem na <\/b><\/span><\/span><\/span>velocidade de propaga\u00e7\u00e3o de qualquer tipo de onda. <\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/b><\/span><\/span><\/span>For\u00e7as que agem sobre a esfera:<\/b><\/span><\/span><\/span><\/p>\n

Peso P<\/b><\/span><\/span><\/span>, vertical, para <\/b><\/span><\/span><\/span>baixo<\/b><\/span><\/span><\/span> e de intensidade <\/b><\/span><\/span><\/span>P = m<\/b><\/span><\/span><\/span>esfera<\/b><\/span><\/span><\/sub><\/span> .g<\/b><\/span><\/span><\/span> = d<\/b><\/span><\/span><\/span>esfera.<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>esfera<\/b><\/span><\/span><\/sub><\/span>.g = 8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>esfera<\/b><\/span><\/span><\/sub><\/span>.10 \"\" <\/b><\/span><\/span><\/span>P = 8.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>V<\/b><\/span><\/span><\/span>esfera<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Empuxo<\/b><\/span><\/span><\/span>, vertical, para <\/b><\/span><\/span><\/span>cima<\/b><\/span><\/span><\/span> e de intensidade <\/b><\/span><\/span><\/span>E = d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span>.g <\/b><\/span><\/span><\/span>= 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>esfera<\/b><\/span><\/span><\/sub><\/span>.10 \"\" <\/b><\/span><\/span><\/span>E = 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>esfera<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P > E<\/b><\/span><\/span><\/span> \"\" a esfera <\/b><\/span><\/span><\/span>afunda<\/b><\/span><\/span><\/span>, for\u00e7a resultante para baixo.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = m.a<\/b><\/span><\/span><\/span> \"\" P \u2013 E = <\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>.a<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> 8.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>V \u2013 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>V = <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>esfera<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>. a \"\" 7.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>V = 8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>V.a \"\" a = 7.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\/8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

a = 0,875.10 \"\" <\/b><\/span><\/span><\/span>a = 8,75 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

A energia potencial gravitacional (E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = mg(h<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> \u2013 h<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>) \"\" a energia potencial depende da <\/b><\/span><\/span><\/span>massa<\/b><\/span><\/span><\/span>, da <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o gravitacional e de sua posi\u00e7\u00e3o vertical inicial e final.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/b><\/span><\/span><\/span>Se o <\/b><\/span><\/span><\/span>\u00f4nibus espacial est\u00e1 em \u00f3rbita em torno da Terra,<\/b><\/span><\/span><\/span> ele <\/b><\/span><\/span><\/span>est\u00e1 dentro do campo gravitacional da Terra e sujeito \u00e0 uma acelera\u00e7\u00e3o da gravidade fornecida por g = G.M<\/b><\/span><\/span><\/span>Terra<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Sendo G e M<\/b><\/span><\/span><\/span>Terra<\/b><\/span><\/span><\/sub><\/span> constantes<\/b><\/span><\/span><\/span>, a <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o da gravidade<\/b><\/span><\/span><\/span> \u00e9 <\/b><\/span><\/span><\/span>inversamente proporcional \u00e0 dist\u00e2ncia d da nave ao centro da Terra.<\/b><\/span><\/span><\/span><\/p>\n

Sendo a <\/b><\/span><\/span><\/span>massa independente da localiza\u00e7\u00e3o de onde o corpo est\u00e1 (sempre a mesma)<\/b><\/span><\/span><\/span>, \u00e0 medida que a <\/b><\/span><\/span><\/span>altura (dist\u00e2ncia d) aumenta,a acelera\u00e7\u00e3o da gravidade g diminuir\u00e1<\/b><\/span><\/span><\/span>, o que implica que <\/b><\/span><\/span><\/span>o peso tamb\u00e9m diminuir\u00e1, pois, P = mg.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/h2>\n

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

RESOLU\u00c7\u00d5ES 01- A for\u00e7a de atra\u00e7\u00e3o ou de repuls\u00e3o entre duas cargas el\u00e9tricas tem sempre a mesma intensidade independente do sinal e do valor das cargas. Assim F1 = F2 =F \u00e9 a intensidade das for\u00e7as sobre as cargas Q1 e Q2. F1 = F = m1.a1 = (3m2\/4)a1 e F2 = F = m2.a2 (3m2\/4)a1 = m2.a2 3.a1\/4 =<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":777,"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-779","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/779","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=779"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/779\/revisions"}],"predecessor-version":[{"id":780,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/779\/revisions\/780"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/777"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=779"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}