{"id":2110,"date":"2016-04-13T02:57:40","date_gmt":"2016-04-13T02:57:40","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2110"},"modified":"2024-09-03T13:43:10","modified_gmt":"2024-09-03T13:43:10","slug":"resolucao-comentada-dos-exercicios-do-enem-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2016\/enem-2015\/resolucao-comentada-dos-exercicios-do-enem-2015\/","title":{"rendered":"ENEM-2015 – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios do ENEM-2015<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>
\n<\/b><\/span><\/span>Voc\u00ea j\u00e1 deve ter notado como \u00e9 <\/b><\/span><\/span><\/span>dif\u00edcil abrir a porta de uma geladeira ou freezer logo ap\u00f3s t\u00ea-la fechado, sendo necess\u00e1rio aguardar alguns<\/b><\/span><\/span><\/span>\"\"<\/p>\n

segundos para abri-la novamente. <\/b><\/span><\/span><\/span><\/p>\n

Isso <\/b><\/span><\/span><\/span>ocorre porque<\/b><\/span><\/span><\/span>, se a <\/b><\/span><\/span><\/span>porta do freezer fica aberta por algum tempo<\/b><\/span><\/span><\/span>, parte do <\/b><\/span><\/span><\/span>ar de seu interior <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>substitu\u00eddo pelo ar mais quente do exterior. <\/b><\/span><\/span><\/span><\/p>\n

Depois que voc\u00ea <\/b><\/span><\/span><\/span>fecha a porta da geladeira,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>temperatura interna do ar em seu interior ir\u00e1 diminui<\/b><\/span><\/span><\/span>r e como se trata de uma <\/b><\/span><\/span><\/span>transforma\u00e7\u00e3o isovolum\u00e9trica (volume constante do interior da geladeira)<\/b><\/span><\/span><\/span>, a <\/b><\/span><\/span><\/span>press\u00e3o tamb\u00e9m diminui (\"\"=constante, s\u00e3o diretamente proporcionais) <\/b><\/span><\/span><\/span>fazendo com que a<\/b><\/span><\/span><\/span>press\u00e3o externa fique maior que a interna, dificultando a abertura da porta. <\/b><\/span><\/span><\/span><\/p>\n

Mas, ap\u00f3s alguns instantes, <\/b><\/span><\/span><\/span>entra ar pelo sistema de veda\u00e7\u00e3o da geladeira, diminuindo a diferen\u00e7a entre as press\u00f5es externa e interna, facilitando, ent\u00e3o, a abertura da porta<\/b><\/span><\/span><\/span>.<\/span><\/p>\n

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

02- <\/span><\/span><\/span>
\n<\/b><\/span><\/span>Veja a teoria sobre o Teorema da energia cin\u00e9tica:<\/b><\/span><\/span><\/span><\/p>\n

Na figura abaixo, o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>carro de massa m<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>tem a intensidade de sua<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>velocidade variando de V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> para V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>num <\/b><\/span><\/span><\/span>intervalo de tempo \u2206t:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Defini\u00e7\u00e3o do Teorema da Energia Cin\u00e9tica: <\/b><\/span><\/span><\/span>\u201cO<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>trabalho da resultante<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>todas<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>as for\u00e7as que agem sobre um corpo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 igual \u00e0 varia\u00e7\u00e3o da energia cin\u00e9tica sofrida pelo corpo\u201d<\/b><\/span><\/span><\/span><\/p>\n

Assim, o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>trabalho da for\u00e7a resultante<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>sobre o carro de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>massa m<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>da figura acima no seu <\/b><\/span><\/span><\/span>deslocamento<\/b><\/span><\/span><\/span><\/p>\n

de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A para B<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 fornecido por:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

As express\u00f5es acima<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>s\u00e3o v\u00e1lidas seja a for\u00e7a resultante conservativa ou n\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

03- <\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>globo ocular pode ser comparado a uma m\u00e1quina fotogr\u00e1fica <\/b><\/span><\/span><\/span>onde o <\/b><\/span><\/span><\/span>cristalino seria a lente (ou sistema de lentes) convergente, a\u00edris seria o diafragma da m\u00e1quina fotogr\u00e1fica<\/b><\/span><\/span><\/span> que abre ou fecha permitindo maior ou menor entrada de luz e a<\/b><\/span><\/span><\/span>retina seria o filme ou sensor onde a imagem real, invertida, reversa e menor \u00e9 formada.<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio<\/b><\/span><\/span><\/span>, a imagem do objeto real est\u00e1 sobre o tecido que corresponde \u00e0 retina localizada no fundo do olho e que <\/b><\/span><\/span><\/span>\u00e9 constitu\u00edda de <\/b><\/span><\/span><\/span>c\u00e9lulas nervosas sens\u00edveis \u00e0 luz<\/b><\/span><\/span><\/span>, que <\/b><\/span><\/span><\/span>transmitem<\/b><\/span><\/span><\/span> ao <\/b><\/span><\/span><\/span>c\u00e9rebro as sensa\u00e7\u00f5es visuais.<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span><\/p>\n

04- <\/b><\/span><\/span><\/span>Se voc\u00ea n\u00e3o domina a teoria, veja as informa\u00e7\u00f5es a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Momento (ou torque) de uma for\u00e7a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

O<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>sinal do momento da for\u00e7a pode ser negativo ou positi<\/b><\/span><\/span><\/span>vo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e, por<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>conven\u00e7\u00e3o,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>vamos adotar o<\/b><\/span><\/span><\/span>sentido hor\u00e1rio de rota\u00e7\u00e3o em torno de O como positivo e anti-hor\u00e1rio como negativo.<\/b><\/span><\/span><\/span><\/p>\n

Condi\u00e7\u00f5es de equil\u00edbrio de um corpo extenso<\/b><\/span><\/span><\/span><\/p>\n

S\u00e3o duas as condi\u00e7\u00f5es para que um corpo extenso r\u00edgido esteja em equil\u00edbrio<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

1<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Equil\u00edbrio de transla\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span>A resultante do sistema de for\u00e7as deve ser nula.<\/b><\/span><\/span><\/span><\/p>\n

2<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Equil\u00edbrio<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>de rota\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0 A<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>soma alg\u00e9brica dos momentos das for\u00e7as<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>que agem sobre o sistema, em rela\u00e7\u00e3o \u00e0 qualquer ponto<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(polo O),deve ser nula.<\/b><\/span><\/span><\/span><\/p>\n

Resolvendo o sistema<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>composto pelas duas equa\u00e7\u00f5es acima voc\u00ea chega<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e0 resolu\u00e7\u00e3o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>do exerc\u00edcio.<\/b><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>caso do exerc\u00edcio<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>vamos<\/b><\/span><\/span><\/span> colocar as for\u00e7as que agem sobre a barra de intensidades:<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>arroz<\/b><\/span><\/span><\/sub><\/span> (peso do arroz) \"\"P<\/b><\/span><\/span><\/span>barra<\/b><\/span><\/span><\/sub><\/span>peso da barra (no meio da barra, pois ela \u00e9 homog\u00eanea) \"\" N (rea\u00e7\u00e3o normal que o apoio aplica na barra)
\n\"\"<\/b><\/span><\/span><\/span><\/p>\n

Momento de cada for\u00e7a em rela\u00e7\u00e3o ao polo 0:<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>barra<\/b><\/span><\/span><\/sub><\/span> = + P<\/b><\/span><\/span><\/span>barra<\/b><\/span><\/span><\/sub><\/span>.1x \"\" M<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> = N.0 = 0 \"\"M<\/b><\/span><\/span><\/span>arroz<\/b><\/span><\/span><\/sub><\/span> = – P<\/b><\/span><\/span><\/span>arroz<\/b><\/span><\/span><\/sub><\/span>.3x.<\/b><\/span><\/span><\/span><\/p>\n


\nCondi\u00e7\u00e3o<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o\"\"<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>soma dos momentos de cada for\u00e7a, em rela\u00e7\u00e3o ao ponto O<\/b><\/span><\/span><\/span>, deve ser <\/b><\/span><\/span><\/span>nula:<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>barra<\/b><\/span><\/span><\/sub><\/span>.1x + 0 – P<\/b><\/span><\/span><\/span>arroz<\/b><\/span><\/span><\/sub><\/span>.3x = 0 \"\"<\/b><\/span><\/span><\/span>\u00a0m<\/b><\/span><\/span><\/span>barra.<\/b><\/span><\/span><\/sub><\/span>g= 3.m<\/b><\/span><\/span><\/span>arroz<\/b><\/span><\/span><\/sub><\/span>.g \"\"<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>arroz<\/b><\/span><\/span><\/sub><\/span> = 5,00 kg <\/b><\/span><\/span><\/span>(veja desenho) \"\"
\nm<\/b><\/span><\/span><\/span>barra<\/b><\/span><\/span><\/sub><\/span>.g = m<\/b><\/span><\/span><\/span>arroz<\/b><\/span><\/span><\/sub><\/span>.g.3\u00a0\"\"m<\/b><\/span><\/span><\/span>barra<\/b><\/span><\/span><\/sub><\/span>\u00a0= 3.5,00\u00a0 \"\"<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>barra\u00a0<\/b><\/span><\/span><\/sub><\/span>= 15,00 kg.<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>R- E
\n<\/b><\/span><\/span><\/span><\/p>\n

05- <\/b><\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>ar \u201cfrio”, mais denso e mais refringente, fica na camada inferior.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00c0 medida que a <\/b><\/span><\/span><\/span>altura aumenta, o ar vai<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>ficando mais quente<\/b><\/span><\/span><\/span>, tornando-se <\/b><\/span><\/span><\/span>menos denso <\/b><\/span><\/span><\/span>e, portanto, <\/b><\/span><\/span><\/span>menos refringente, afastando-se da normal e sofrendo os sucessivos desvios (refra\u00e7\u00f5es) apresentados na figura que, est\u00e1 fora de escala. <\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>
\nCada refra\u00e7\u00e3o <\/b><\/span><\/span><\/span>\u00e9 sempre acompanhada <\/b><\/span><\/span><\/span>de reflex\u00e3o que pode ou n\u00e3o ser reflex\u00e3o total.<\/b><\/span><\/span><\/span><\/p>\n

Assim, para um <\/b><\/span><\/span><\/span>observador, a imagem do iceberg, pode aparecer acima da \u00e1gua, como se estivesse flutuando no ar e, onde estava o iceberg ele pode apenas enxergar \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

Os <\/b><\/span><\/span><\/span>efeitos de Fata Morgana s\u00e3o miragens ditas superiores <\/b><\/span><\/span><\/span>(vistas acima do horizonte), <\/b><\/span><\/span><\/span>diferentes das miragens inferiores <\/b><\/span><\/span><\/span>(vistas abaixo do horizonte), que s\u00e3o mais comuns e criam <\/b><\/span><\/span><\/span>a ilus\u00e3o de lagos de \u00e1gua distantes nos\u00a0desertos<\/a> ou em estradas molhadas com po\u00e7as de \u00e1gua no asfalto muito quente. <\/b><\/span><\/span><\/span><\/p>\n

Pode acontecer tamb\u00e9m que a <\/b><\/span><\/span><\/span>imagem seja <\/b><\/span><\/span><\/span>da pr\u00f3pria superf\u00edcie da \u00e1gua <\/b><\/span><\/span><\/span>que tamb\u00e9m pode aparecer <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

logo acima da superf\u00edcie do mar<\/b><\/span><\/span><\/span>, criando a <\/b><\/span><\/span><\/span>ilus\u00e3o de uma \u201cparede de \u00e1gua”.<\/b><\/span><\/span><\/span><\/p>\n

Essa \u201cparede” poderia ter <\/b><\/span><\/span><\/span>obscurecido a presen\u00e7a do iceberg <\/b><\/span><\/span><\/span>que se chocou com o Titanic da <\/b><\/span><\/span><\/span>vis\u00e3o da tripula\u00e7\u00e3o, o que provocou o seu naufr\u00e1gio.<\/b><\/span><\/span><\/span><\/p>\n

Portanto esse <\/b><\/span><\/span><\/span>fen\u00f4meno \u00e9 resultado da refra\u00e7\u00e3o e posteriormente da eventual reflex\u00e3o total da luz. <\/b><\/span><\/span><\/span>Mas o <\/b><\/span><\/span><\/span>fen\u00f4meno predominante \u00e9 a refra\u00e7\u00e3o da luz.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

07- <\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span>O <\/b><\/span><\/span><\/span>timbre \u00e9 uma qualidade sonora <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>permite distinguir dois sons de mesma altura <\/b><\/span><\/span><\/span> e<\/b><\/span><\/span><\/span>mesma intensidade (mesma nota musical)<\/b><\/span><\/span><\/span>, emitidos <\/b><\/span><\/span><\/span>por instrumentos diferentes tocando a mesma nota <\/b><\/span><\/span><\/span>musical ou acorde.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Assim, <\/b><\/span><\/span><\/span>conseguimos distinguir se a mesma nota musical \u00e9 emitida por um viol\u00e3o ou por um piano <\/b><\/span><\/span><\/span>pelo <\/b><\/span><\/span><\/span>timbre,<\/b><\/span><\/span><\/span> pois ele <\/b><\/span><\/span><\/span>difere nos dois instrumentos e nos fornece sensa\u00e7\u00f5es sonoras diferentes, devido \u00e0s diferentes composi\u00e7\u00f5es de harm\u00f4nicos gerados por cada instrumento.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/b><\/span><\/span><\/span>Sendo\u00a0<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0a dist\u00e2ncia do celular \u00e0 primeira torre T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span>o celular pode estar em <\/b><\/span><\/span><\/span>qualquer ponto da circunfer\u00eancia de centro na primeira torre T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e raio\u00a0R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0(figura I).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo\u00a0<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>a dist\u00e2ncia do celular \u00e0 segunda torre T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, o <\/b><\/span><\/span><\/span>celular pode estar num dos dois pontos <\/b><\/span><\/span><\/span>em que <\/b><\/span><\/span><\/span>as circunfer\u00eancias se interceptam (figura II). <\/b><\/span><\/span><\/span><\/p>\n

Sendo\u00a0<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0a dist\u00e2ncia do celular \u00e0 terceira torre\u00a0T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>intersec\u00e7\u00e3o das tr\u00eas circunfer\u00eancias de raios R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> e R<\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/sub><\/span>ocorre <\/b><\/span><\/span><\/span>num \u00fanico ponto onde deve se localizar o celular (figura III),<\/b><\/span><\/span><\/span> conforme pedido no exerc\u00edcio.<\/b><\/span><\/span><\/span><\/p>\n

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

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

09- <\/b><\/span><\/span><\/span>C\u00e1lculo da <\/b><\/span><\/span><\/span>pot\u00eancia P<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>fornecida ao carro pelo painel fotovoltaico de <\/b><\/span><\/span><\/span>9 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> de \u00e1rea <\/b><\/span><\/span><\/span>por uma <\/b><\/span><\/span><\/span>regra de tr\u00eas:<\/b><\/span><\/span><\/span>
\n\"\"<\/b><\/span><\/span><\/span><\/p>\n

Veja a teoria sobre o Teorema da energia cin\u00e9tica:<\/b><\/span><\/span><\/span><\/p>\n

Na figura abaixo, o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>carro de massa m<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>tem a intensidade de sua<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>velocidade variando de V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> para V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>num <\/b><\/span><\/span><\/span>intervalo de tempo \u2206t:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Defini\u00e7\u00e3o do Teorema da Energia Cin\u00e9tica: <\/b><\/span><\/span><\/span>\u201cO<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>trabalho da resultante<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>todas<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>as for\u00e7as que agem sobre um corpo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 igual \u00e0 varia\u00e7\u00e3o da energia cin\u00e9tica sofrida pelo corpo\u201d<\/b><\/span><\/span><\/span><\/p>\n

Assim, o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>trabalho da for\u00e7a resultante<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>sobre o carro de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>massa m<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>da figura acima no seu <\/b><\/span><\/span><\/span>deslocamento<\/b><\/span><\/span><\/span><\/p>\n

de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A para B<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 fornecido por:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

As express\u00f5es acima<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>s\u00e3o v\u00e1lidas seja a for\u00e7a resultante conservativa ou n\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>caso do exerc\u00edcio \"\" m = 200 kg \"\" V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 (partiu do repouso)\"\" V <\/b><\/span><\/span><\/span>=\"\"<\/b><\/span><\/span><\/span>= 30 m\/s<\/b><\/span><\/span><\/span>
\n\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

Lei de Hooke<\/b><\/span><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>\u00a0 \u2014\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>intensidade da<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>for\u00e7a externa deformadora<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(que comprime ou distende a mola)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Fe<\/b><\/span><\/span>\u00a0 \u2014\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>intensidade da<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>for\u00e7a el\u00e1stica<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>que<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>tende a fazer com que a mola retorne \u00e0 posi\u00e7\u00e3o normal<\/b><\/span><\/span><\/span><\/span><\/p>\n

x<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2014\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>deforma\u00e7\u00e3o da mola em rela\u00e7\u00e3o \u00e0 posi\u00e7\u00e3o normal<\/b><\/span><\/span><\/span><\/span><\/p>\n

K\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u2014\u00a0 constante<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>el\u00e1stica da mola<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>medida no SI em N\/m<\/b><\/span><\/span>, que \u00e9<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>caracter\u00edstica de cada el\u00e1stico ou mola<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>e que depende do material de que a mesma \u00e9 feita, do<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>n\u00famero de espiras<\/b><\/span><\/span>, das dimens\u00f5es, etc.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Hooke <\/b><\/span><\/span>determinou<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>experimentalmente que<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u201cem regime de deforma\u00e7\u00e3o el\u00e1stica, a intensidade da for\u00e7a el\u00e1stica Fe\u00a0\u00e9 diretamente proporcional \u00e0 deforma\u00e7\u00e3o x\u201d,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>ou seja,<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Observa\u00e7\u00e3o:<\/b><\/span><\/span><\/span> O <\/b><\/span><\/span><\/span>enunciado<\/b><\/span><\/span><\/span> <\/b>n\u00e3o mencionou se o lan\u00e7amento \u00e9 horizontal ou obl\u00edquo<\/b><\/span><\/span><\/span>. Se voc\u00ea considerar o <\/b><\/span><\/span><\/span>lan\u00e7amento como horizontal, nenhuma alternativa satisfaz.<\/b><\/span><\/span><\/span> Ent\u00e3o vamos consider\u00e1-lo <\/b><\/span><\/span><\/span>como lan\u00e7amento obl\u00edquo. <\/b><\/span><\/span><\/span><\/p>\n

Rela\u00e7\u00e3o entre o <\/b><\/span><\/span>alcance\u00a0D<\/b><\/span><\/span>, a velocidade <\/b><\/span><\/span>inicial\u00a0V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0e o \u00e2ngulo de <\/b><\/span><\/span>lan\u00e7amento\u00a0\u03b1.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Velocidade vertical<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>da bola no ponto de lan\u00e7amento<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>oy<\/b><\/span><\/sub><\/span> e no ponto de chegada \u2013V<\/b><\/span><\/span>oy<\/b><\/span><\/sub><\/span> <\/b>\u00a0\u00a0\"\"\u00a0 <\/b><\/span><\/span>V<\/b><\/span><\/span>y<\/b><\/span><\/sub><\/span> = V<\/b><\/span><\/span>oy<\/b><\/span><\/sub><\/span> \u00a0\u2013<\/b><\/span><\/span> <\/b>gt\u00a0<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>oy<\/b><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span>oy<\/b><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>2 V<\/b><\/span><\/span>oy<\/b><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span>= gt<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>t=2<\/b><\/span><\/span>.<\/b><\/span><\/span>V<\/b><\/span><\/span>oy<\/b><\/span><\/sub><\/span>\/g<\/b><\/span><\/span>\u00a0\u00a0\"\"<\/b><\/span><\/span>\u00a0 <\/b><\/span><\/span>nesse tempo a bola percorre a dist\u00e2ncia<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>\u0394S =X=Vxt<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>X = V<\/b><\/span><\/span>ox<\/b><\/span><\/sub><\/span>.t\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0 X= V<\/b><\/span><\/span>ox<\/b><\/span><\/sub><\/span>.(2Voy\/g)\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0 X=V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>cos\u03b1.(2V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>sen\u03b1\/g)\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0 X=V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>2<\/b><\/span><\/sup><\/span>.2sen\u03b1.cos\u03b1\/g<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>2sen\u03b1.cos\u03b1 = sen2\u03b1<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0\"\" <\/b><\/span><\/span>X=V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>2<\/b><\/span><\/sup><\/span>sen2\u03b1\/g\u00a0 (I)<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>equa\u00e7\u00e3o do alcance horizontal.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Para a <\/b><\/span><\/span><\/span>mesma intensidade de for\u00e7a F<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>mesmas condi\u00e7\u00f5es de lan\u00e7a mento<\/b><\/span><\/span><\/span> de proj\u00e9teis de <\/b><\/span><\/span><\/span>mesma massa, <\/b><\/span><\/span><\/span>temos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

11- <\/b><\/span><\/span><\/span>Para que as <\/b><\/span><\/span><\/span>tomadas, <\/b><\/span><\/span><\/span>como pede o enunciado<\/b><\/span><\/span><\/span> estejam submetidas \u00e0 mesma diferen\u00e7a de potencial<\/b><\/span><\/span><\/span> <\/b>(ddp), tens\u00e3o ou voltagem U,<\/b><\/span><\/span><\/span> essas duas tomadas <\/b><\/span><\/span><\/span>devem estar ligadas em paralelo j\u00e1 que nesse tipo de associa\u00e7\u00e3o a tens\u00e3o \u00e9 a mesma que a tens\u00e3o nominal da rede el\u00e9trica (U). <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Mas, o <\/b><\/span><\/span><\/span>interruptor deve estar ligado em s\u00e9rie com a l\u00e2mpada<\/b><\/span><\/span><\/span> e, quando <\/b><\/span><\/span><\/span>fechado ele acende a<\/b><\/span><\/span><\/span> <\/b>l\u00e2mpada<\/b><\/span><\/span><\/span> que fica em <\/b><\/span><\/span><\/span>paralelo com as tomadas <\/b><\/span><\/span><\/span>e, portanto, <\/b><\/span><\/span><\/span>submetida \u00e0 tens\u00e3o nominal (U) da rede que as tomadas.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Desligando o interruptor<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>desliga-se a l\u00e2mpada,<\/b><\/span><\/span><\/span> sem alterar a <\/b><\/span><\/span><\/span>liga\u00e7\u00e3o paralelo das tomadas<\/b><\/span><\/span><\/span> com a rede, que <\/b><\/span><\/span><\/span>continuam sujeitas \u00e0 mesma ddp nominal U, da rede. <\/b><\/span><\/span><\/span><\/p>\n

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

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

Princ\u00edpio das trocas de calor<\/b><\/span><\/span><\/span><\/p>\n

Colocando<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>v\u00e1rios corpos a diferentes temperaturas<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>no<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>interior de um recipiente adiab\u00e1tico,<\/b><\/span><\/span><\/span> haver\u00e1 <\/b><\/span><\/span><\/span>trocas de calor entre eles<\/b><\/span><\/span><\/span>, at\u00e9<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>atingirem o equil\u00edbrio t\u00e9rmico.<\/b><\/span><\/span><\/span><\/p>\n

Assim, como o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>recipiente \u00e9 adiab\u00e1tico,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>quantidade de calor cedida pelos corpos mais quentes<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9igual \u00e0 quantidade de calor recebida pelos mais frios.<\/b><\/span><\/span><\/span><\/p>\n

Princ\u00edpio da Conserva\u00e7\u00e3o da energia<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A teoria acima \u00e9<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>v\u00e1lida para mais de um corpo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e, como a quantidade de calor<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>recebida \u00e9 positiva e<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span>cedida \u00e9 negativa, tem-se que<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/span>+ Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0+ \u2026. + Q<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 0 \u00a0ou<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/span>.c<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/span>.(<\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2013 <\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/span>) + m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.( <\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2013 <\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub>) + m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>.( <\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2013 <\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub>) + \u2026\u00a0+ m<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>.(<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2013 <\/b><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub>) = 0.<\/b><\/span><\/span><\/p>\n

Enunciado desse princ\u00edpio das trocas de calor:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Seis horas depois<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>abre-se<\/b><\/span><\/span><\/span> a garrafa e <\/b><\/span><\/span><\/span>mede-se a temperatura da \u00e1gua<\/b><\/span><\/span><\/span>, obtendo-se <\/b><\/span><\/span><\/span>\u03b8 =<\/b><\/span><\/span><\/span> <\/b>16\u00b0C.
\n\"\"<\/b><\/span><\/span><\/span><\/p>\n

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

13- <\/b><\/span><\/span><\/span>C\u00e1lculo dos <\/b><\/span><\/span><\/span>comprimentos de onda das frequ\u00eancias extremas da faixa UV-B.
\nV = c = \u03bb.f<\/b><\/span><\/span><\/span> \"\"\u00a0 \u03bb\u00a0= \"\" \u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u03bb<\/b><\/span><\/span><\/span>min<\/b><\/span><\/span><\/sub><\/span>\u00a0=<\/b><\/span><\/span><\/span> 3,0.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/1,03.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span>\u00a0 \"\" \u00a0\u03bb<\/b><\/span><\/span><\/span>min<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2245<\/b><\/span><\/span><\/span>\u00a02,9.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0m <\/b><\/span><\/span><\/span>= 290 nm.
\n\u03bb<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0=<\/b><\/span><\/span><\/span> 3,0.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/9,34.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0 \"\" \u00a0\u03bb<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2245<\/b><\/span><\/span><\/span>\u00a03,2.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0m <\/b><\/span><\/span><\/span>= 320 nm.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Entre <\/b><\/span><\/span><\/span>290 nm e 320 nm,<\/b><\/span><\/span><\/span> o filtro que apresenta <\/b><\/span><\/span><\/span>absor\u00e7\u00e3o m\u00e1xima \u00e9 o filtro solar IV.
\n<\/b><\/span><\/span><\/span>R- B
\n<\/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 dos exerc\u00edcios do ENEM-2015 01- Voc\u00ea j\u00e1 deve ter notado como \u00e9 dif\u00edcil abrir a porta de uma geladeira ou freezer logo ap\u00f3s t\u00ea-la fechado, sendo necess\u00e1rio aguardar alguns segundos para abri-la novamente. Isso ocorre porque, se a porta do freezer fica aberta por algum tempo, parte do ar de seu interior \u00e9 substitu\u00eddo pelo ar mais quente<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2108,"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-2110","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2110","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=2110"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2110\/revisions"}],"predecessor-version":[{"id":11015,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2110\/revisions\/11015"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2108"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2110"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}