{"id":928,"date":"2015-06-29T00:36:40","date_gmt":"2015-06-29T00:36:40","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=928"},"modified":"2024-09-02T17:43:36","modified_gmt":"2024-09-02T17:43:36","slug":"resolucao-comentada-da-uern-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-uern\/resolucao-comentada-da-uern-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da UERN 2015"},"content":{"rendered":"

RESOLU\u00c7\u00c3O DA PROVA DA UERN-2015<\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>Bal\u00e3o <\/b><\/span><\/span><\/span>subindo<\/b><\/span><\/span><\/span> com velocidade constante de <\/b><\/span><\/span><\/span>V = 2m\/s,<\/b><\/span><\/span><\/span> <\/b>percorrendo<\/b><\/span><\/span><\/span> uma altura <\/b><\/span><\/span><\/span>d <\/b><\/span><\/span><\/span>at\u00e9 chegar ao <\/b><\/span><\/span><\/span>teto<\/b><\/span><\/span><\/span> da<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

quadra e demorando um <\/b><\/span><\/span><\/span>intervalo de tempo t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> nessa subida:<\/b><\/span><\/span><\/span><\/p>\n

V = d\/t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>2 =<\/b><\/span><\/span><\/span> <\/b>d\/t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>d = 2t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> (I)<\/b><\/span><\/span><\/span><\/p>\n

O bal\u00e3o <\/b><\/span><\/span><\/span>explode <\/b><\/span><\/span><\/span>no teto e <\/b><\/span><\/span><\/span>o som<\/b><\/span><\/span><\/span> da explos\u00e3o <\/b><\/span><\/span><\/span>desce com velocidade V<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span> = 340 m\/s<\/b><\/span><\/span><\/span>, percorrendo a altura <\/b><\/span><\/span><\/span>d <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>demorando um intervalo de tempo t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> at\u00e9 chegar ao ouvido do garoto:<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span><\/span>som<\/b><\/span><\/span><\/span><\/sub><\/span> = d\/t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span> \"\" 340 = d\/t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>d = 340t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span> (II)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Igualando (I) com (II)<\/b><\/span><\/span><\/span> \"\" 2t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 340t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 170t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> (III)<\/b><\/span><\/span><\/span><\/p>\n

Do enunciado <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 5,13 (IV)<\/b><\/span><\/span><\/span><\/p>\n

(III) e (IV)<\/b><\/span><\/span><\/span> \"\" 170t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> + t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 5,13 \"\" t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 5,13\/171 \"\" <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 0,03 s.<\/b><\/span><\/span><\/span><\/p>\n

Substituindo <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 0,03 s em d = 340t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>d <\/b><\/span><\/span><\/span>= 340×0,03 <\/b><\/span><\/span><\/span>= 10,2 m<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>per\u00edodoT <\/b><\/span><\/span><\/span>(tempo que cada um demora para efetuar uma volta completa) \u00e9 o <\/b><\/span><\/span><\/span>inverso da frequ\u00eancia f<\/b><\/span><\/span><\/span> (n\u00famero de voltas completas efetuadas na unidade de tempo).<\/b><\/span><\/span><\/span><\/p>\n

Primeiro exaustor e\u00f3lico \"\" <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1\/f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1\/2 s.<\/b><\/span><\/span><\/span><\/p>\n

Segundo exaustor e\u00f3lico \"\" <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1\/f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1\/2,5 s.<\/b><\/span><\/span><\/span><\/p>\n

Diferen\u00e7a \u2206T = 1\/2 \u2013 1\/2.5<\/b><\/span><\/span><\/span> = (2,5 \u2013 2)\/5 = 0,5\/5 \"\" <\/b><\/span><\/span><\/span>\u2206T = 0,1 s.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>Calculando esses valores na <\/b><\/span><\/span><\/span>Celsius C<\/b><\/span><\/span><\/span> = K \u2013 273 = 438 \u2013 273 <\/b><\/span><\/span><\/span>= 165<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Varia\u00e7\u00e3o na Celsius<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206C<\/b><\/span><\/span><\/span> = 165 \u2013 120 <\/b><\/span><\/span><\/span>= 45 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

\u00c0 cada <\/b><\/span><\/span><\/span>varia\u00e7\u00e3o de 5 graus na Celsius corresponde uma varia\u00e7\u00e3o de 9 graus na Fahrenheit<\/b><\/span><\/span><\/span> \"\" regra de tr\u00eas \"\" <\/b><\/span><\/span><\/span>\u22065 C \u2013 \u22069 F \"\" 45C – \u2206F<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206F<\/b><\/span><\/span><\/span> = 45×9\/5 <\/b><\/span><\/span><\/span>= 81 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>F.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>Observe na figura abaixo que <\/b><\/span><\/span><\/span>3\u03bb + \u03bb\/2 = 28<\/b><\/span><\/span><\/span> \"\" 7\u03bb\/2 = 28 \"\" <\/b><\/span><\/span><\/span>\u03bb = 8 cm.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

T = 2,5 s <\/b><\/span><\/span><\/span><\/b>f = 1\/T<\/b><\/span><\/span><\/span> = 1\/2,5 <\/b><\/span><\/span><\/span>= 0,4 Hz.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o fundamental da ondulat\u00f3ria \"\" <\/b><\/span><\/span><\/span>V = \u03bbf<\/b><\/span><\/span><\/span> = 8.0,4 <\/b><\/span><\/span><\/span>= 3,2 cm.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Na <\/b><\/span><\/span><\/span>figura abaixo<\/b><\/span><\/span><\/span> est\u00e3o todas as <\/b><\/span><\/span><\/span>for\u00e7as <\/b><\/span><\/span><\/span>que agem sobre cada bloco:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco A<\/b><\/span><\/span><\/span> \"\" desce \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.a \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \u2013 T = m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span> \"\" m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g \u2013 T = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.a \"\" m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.10 \u2013 72 = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.2 \"\" <\/b><\/span><\/span><\/span>8m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span><\/p>\n

72 \"\" m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 9 kg.<\/b><\/span><\/span><\/span><\/p>\n

Bloco B<\/b><\/span><\/span><\/span> \"\" sobe \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.a \"\" <\/b><\/span><\/span><\/span>T – P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span> \"\" T \u2013 m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g = m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.a \"\" 72 \u2013 10m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.2 \"\" <\/b><\/span><\/span><\/span>12m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span><\/p>\n

72 \"\" m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 6 kg.<\/b><\/span><\/span><\/span><\/p>\n

Igualando (I) com (II)<\/b><\/span><\/span><\/span> \"\" 8m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 12m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 1,5m<\/b><\/span><\/span><\/span>B\u00ad\u00ad<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\u00c9 pedida a <\/b><\/span><\/span><\/span>diferen\u00e7a m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \u2013 m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 9 \u2013 6 <\/b><\/span><\/span><\/span>= 3 kg.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/b><\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>\u00fanico caso<\/b><\/span><\/span><\/span> em que um espelho esf\u00e9rico c\u00f4ncavo fornece uma <\/b><\/span><\/span><\/span>imagem virtual<\/b><\/span><\/span><\/span> \u00e9 o apresentado abaixo:<\/b><\/span><\/span><\/span><\/p>\n

Objeto entre o foco F e o v\u00e9rtice V ou entre o foco F e o espelho<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A imagem ter\u00e1:<\/b><\/span><\/span><\/p>\n

Natureza <\/b><\/span><\/span><\/span>\u2013 <\/b><\/span><\/span>virtual<\/b><\/span><\/span><\/span> (obtida na interse\u00e7\u00e3o do prolongamento dos raios de luz )<\/b><\/span><\/span><\/p>\n

Localiza\u00e7\u00e3o \u2013 atr\u00e1s do espelho<\/b><\/span><\/span><\/span><\/p>\n

Tamanho \u2013 maior que o do objeto<\/b><\/span><\/span><\/span><\/p>\n

Orienta\u00e7\u00e3o \u2013 direita em rela\u00e7\u00e3o ao objeto<\/b><\/span><\/span><\/span><\/p>\n

Utilidades:<\/b><\/span><\/span><\/span> s\u00e3o empregados com freq\u00fc\u00eancia quando se deseja obter uma imagem <\/b><\/span><\/span>virtual e ampliada<\/b><\/span><\/span><\/span> de um objeto, como \u00e9 o caso dos <\/b><\/span><\/span>espelhos de barbear, toalete, de dentista, espelho de otorrinolaringologia, etc.<\/b><\/span><\/span><\/span><\/p>\n

R- A\u00a0<\/b><\/span><\/span><\/p>\n

07- <\/b><\/span><\/span><\/span>Empuxo \"\" <\/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>.60.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.10 = 6.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span> <\/b>= 0,6 N.<\/b><\/span><\/span><\/span><\/p>\n

Peso \"\"<\/b><\/span><\/span><\/span>P = m.g <\/b><\/span><\/span><\/span>= 0,4.10 <\/b><\/span><\/span><\/span>= 4 N<\/b><\/span><\/span><\/span><\/p>\n

No fundo e em equil\u00edbrio \"\" E + N = P \"\" 0,6 + N= 4 \"\" N = 3,4 N<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/b><\/span><\/span><\/span>De <\/b><\/span><\/span><\/span>A para B,<\/b><\/span><\/span><\/span> o volume est\u00e1 aumentando \"\" <\/b><\/span><\/span><\/span>W > 0.<\/b><\/span><\/span><\/span><\/p>\n

De <\/b><\/span><\/span><\/span>B para C<\/b><\/span><\/span><\/span>, o volume n\u00e3o varia \"\" <\/b><\/span><\/span><\/span>W = 0.<\/b><\/span><\/span><\/span><\/p>\n

De <\/b><\/span><\/span><\/span>C para D,<\/b><\/span><\/span><\/span> o volume est\u00e1 diminuindo \"\" <\/b><\/span><\/span><\/span>W < 0.<\/b><\/span><\/span><\/span><\/p>\n

De <\/b><\/span><\/span><\/span>D para A<\/b><\/span><\/span><\/span>, o volume n\u00e3o varia \"\" <\/b><\/span><\/span><\/span>W = 0.<\/b><\/span><\/span><\/span><\/p>\n

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

09- <\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>aparente<\/b><\/span><\/span><\/sub><\/span> = f<\/b><\/span><\/span><\/span>fonte<\/b><\/span><\/span><\/sub><\/span>.(V<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span>\/(<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>som \u00ad<\/b><\/span><\/span><\/sub><\/span>– V<\/b><\/span><\/span><\/span>fonte<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> \"\" 1200 = f<\/b><\/span><\/span><\/span>fonte<\/b><\/span><\/span><\/sub><\/span>(340\/(340 \u2013 68) \"\" 1200 = f<\/b><\/span><\/span><\/span>fonte<\/b><\/span><\/span><\/sub><\/span>(340\/272) \"\" 1200 = f<\/b><\/span><\/span><\/span>fonte.<\/b><\/span><\/span><\/sub><\/span>1,25 \"\" <\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>fonte<\/b><\/span><\/span><\/sub><\/span> = 1200\/1,25 <\/b><\/span><\/span><\/span>= 960 Hz.<\/b><\/span><\/span><\/span><\/p>\n

10- <\/b><\/span><\/span><\/span>Uma das conseq\u00fc\u00eancias da <\/b><\/span><\/span><\/span>propaga\u00e7\u00e3o retil\u00ednea da luz<\/b><\/span><\/span><\/span> \u00e9 a observa\u00e7\u00e3o de eclipses, como por exemplo, o eclipse solar:<\/b><\/span><\/span><\/span><\/p>\n

Eclipse solar<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

11- <\/b><\/span><\/span><\/span>Observe na figura abaixo que a <\/b><\/span><\/span><\/span>ddp (tens\u00e3o) U<\/b><\/span><\/span><\/span> no <\/b><\/span><\/span><\/span>ramo <\/b><\/span><\/span><\/span>do resistor de 20 \u2126 que vale <\/b><\/span><\/span><\/span>U <\/b><\/span><\/span><\/span>= R.i = 20.6 <\/b><\/span><\/span><\/span>=120V e que <\/b><\/span><\/span><\/span>\u00e9 a <\/b><\/span><\/span><\/span>mesma <\/b><\/span><\/span><\/span>nos demais ramos pois, todos est\u00e3o em <\/b><\/span><\/span><\/span>paralelo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Corrente el\u00e9trica <\/b><\/span><\/span><\/span>no resistor de <\/b><\/span><\/span><\/span>40 \u2126<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span>R = U\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" 40 = 120\/i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 3 A.<\/b><\/span><\/span><\/span><\/p>\n

i = i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> + i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> \"\" 12 = 3 + 6 + i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 3 A.<\/b><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>\u00faltimo ramo<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = U\/i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> \"\" 10 + R = 120\/3 \"\" R = 40 \u2013 10 \"\" <\/b><\/span><\/span><\/span>R = 30 \u2126<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

RESOLU\u00c7\u00c3O DA PROVA DA UERN-2015 01- Bal\u00e3o subindo com velocidade constante de V = 2m\/s, percorrendo uma altura d at\u00e9 chegar ao teto da quadra e demorando um intervalo de tempo t1 nessa subida: V = d\/t1 2 = d\/t1 d = 2t1 (I) O bal\u00e3o explode no teto e o som da explos\u00e3o desce com velocidade Vsom = 340<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":926,"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-928","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/928","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=928"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/928\/revisions"}],"predecessor-version":[{"id":931,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/928\/revisions\/931"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/926"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=928"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}