{"id":3396,"date":"2017-11-05T23:29:03","date_gmt":"2017-11-05T23:29:03","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3396"},"modified":"2024-09-04T14:16:08","modified_gmt":"2024-09-04T14:16:08","slug":"resolucao-comentada-vestibulares-recentes-fisica-moderna-2014-2013","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/fisica-moderna\/vestibulares-recentes-fisica-moderna-2014-2013\/resolucao-comentada-vestibulares-recentes-fisica-moderna-2014-2013\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada \u2013 Vestibulares Recentes F\u00cdSICA MODERNA \u2013 2014 – 2013"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u2013 Vestibulares Recentes<\/b><\/span><\/span><\/span><\/p>\n

F\u00cdSICA MODERNA \u2013 2014 – 2013<\/b><\/span><\/span><\/span><\/p>\n

01-<\/span><\/span><\/p>\n

Dilata\u00e7\u00e3o do tempo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 per\u00edodo (tempo que o balan\u00e7o demora para efetuar um \u201cvai e vem\u201d completo medido no interior da nave) — T – per\u00edodo do balan\u00e7o medido por Romeu na plataforma espacial — T=T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/\u221a1 \u2013 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)= T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/\u221a1 \u2013 (0,5)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)= T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/\u221a1 \u2013 0,25)=T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/\u221a( 1 \u2013 1\/4) — T=T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/\u221a3\/4=T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/\u221a3\/2 — T=2T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/\u221a3 =2T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u221a3\/\u221a3.\u00a0\u221a3 —\u00a0T=2T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u221a3\/3=(2\u221a3\/3).T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span><\/p>\n

Dilata\u00e7\u00e3o do comprimento L da nave para Julieta:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=L\/\u221a1 \u2013 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)= L\/\u221a1 \u2013 (0,5)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)= L\/\u221a1 \u2013 0,25)=L\/\u221a( 1 \u2013 1\/4) — L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=L\/\u221a3\/4=L\/\u221a3\/2 — L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=2L\/\u221a3 =2L\u221a3\/\u221a3. \u221a3 —\u00a0L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=2L\u221a3\/3=(2\u221a3\/3).L<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A\u00a0energia m\u00ednima para extrair um el\u00e9tron\u00a0da placa met\u00e1lica \u00e9 denominada\u00a0fun\u00e7\u00e3o trabalho\u00a0e est\u00e1 relacionada com o tipo de metal utilizado.\u00a0Se a energia do f\u00f3ton que incide (h.f) for maior que a fun\u00e7\u00e3o trabalho (W) a energia em excesso ser\u00e1 energia cin\u00e9tica (Ec), de modo que\u00a0 — \u00a0W=h.f \u2013 Ec\u00a0 —\u00a0 denominada\u00a0equa\u00e7\u00e3o fotoel\u00e9trica de Einstein..<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na teoria acima que a frequ\u00eancia m\u00ednima para extra\u00e7\u00e3o dos el\u00e9trons do metal \u00e9 fornecida por fo=W\/h — pelo gr\u00e1fico, os valores m\u00e1ximo e m\u00ednimo para as frequ\u00eancias no espectro vis\u00edvel s\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

Fvioleta\u00a0= 7,50.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0Hz e fvermelho=4,00.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>Hz que, substitu\u00eddos na express\u00e3o f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/h fornecem — violeta — 7.50.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>=W<\/b><\/span><\/span><\/span>violeta<\/b><\/span><\/span><\/sub><\/span>\/ 4,1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0W<\/b><\/span><\/span><\/span>violeta<\/b><\/span><\/span><\/sub><\/span>=30,75.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>=3,075eV\u00a0— vermelha — 4,00.1014=W<\/b><\/span><\/span><\/span>vermelha<\/b><\/span><\/span><\/sub><\/span>\/4,1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>\u00a0—W<\/b><\/span><\/span><\/span>vermelha<\/b><\/span><\/span><\/sub><\/span>=16,4.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>=1,64eV\u00a0— observe na tabela fornecida que o s\u00f3dio \u00e9 o \u00fanico metal cuja W est\u00e1 compreendido entre esses dois valores\u00a0— <\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Correta.<\/b><\/span><\/span><\/span><\/p>\n

02- Correta\u00a0—\u00a0As ondas s\u00e3o fortemente difratadas\u00a0 quando o\u00a0comprimento de onda\u00a0\u00a0tem\u00a0aproximadamente o mesmo tamanho\u00a0 da abertura \u00a0da fenda\u00a0(figura 2)\u00a0ou do tamanho do obst\u00e1culo\u00a0(figura 1).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Figura 1<\/b><\/span><\/span><\/span><\/p>\n

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

figura\u00a0 2<\/b><\/span><\/span><\/span><\/p>\n

Observe na figura da esquerda que a difra\u00e7\u00e3o \u00e9 menos acentuada que na direita.<\/b><\/span><\/span><\/span><\/p>\n

04. Correta. \u00a0<\/b><\/span><\/span><\/span><\/p>\n

08. Falsa\u00a0—\u00a0\u03bb=h\/mV=6,63.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>\/9,1.10<\/b><\/span><\/span><\/span>-31<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=6,63.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>\/27,3.10<\/b><\/span><\/span><\/span>-25<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0\u03bb=2,4.10<\/b><\/span><\/span><\/span>-10<\/b><\/span><\/span><\/sup><\/span> m.<\/b><\/span><\/span><\/span><\/p>\n

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

32. Falsa\u00a0—\u00a0Foi\u00a0Thomas Young\u00a0quem realizou o experimento da dupla fenda.<\/b><\/span><\/span><\/span><\/p>\n

Soma (01 + 02 + 04)=07<\/b><\/span><\/span><\/span><\/p>\n

Efeito fotoel\u00e9trico<\/b><\/span><\/span><\/span><\/p>\n

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

57-<\/b><\/span><\/span><\/span> <\/b>01. Correta\u00a0 —\u00a0 A figura abaixo mostra o gr\u00e1fico da energia cin\u00e9tica do el\u00e9tron extra\u00eddo em fun\u00e7\u00e3o da frequ\u00eancia da radia\u00e7\u00e3o (f\u00f3ton, cor) incidente, para uma mesma placa met\u00e1lica (c\u00e9lula fotoel\u00e9trica).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a freq\u00fc\u00eancia m\u00ednima (frequ\u00eancia de corte) necess\u00e1ria para produzir o efeito fotoel\u00e9trico. Se f=f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0o el\u00e9tron \u00e9 liberado, mas sua energia cin\u00e9tica \u00e9 nula. Para frequ\u00eancias inferiores a f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0o fen\u00f4meno n\u00e3o ocorre. <\/b><\/span><\/span><\/span><\/p>\n

Por\u00e9m, para valores superiores a fo, o n\u00famero de el\u00e9trons arrancados \u00e9 diretamente proporcional \u00e0 intensidade da radia\u00e7\u00e3o eletromagn\u00e9tica incidente, ou seja, aumentando a intensidade da radia\u00e7\u00e3o (frequ\u00eancia, cor) incidente no metal, aumenta-se o n\u00edvel energ\u00e9tico dos f\u00f3tons incidentes, aumentando assim n\u00famero de el\u00e9trons arrancados.<\/b><\/span><\/span><\/span><\/p>\n

A energia m\u00ednima para que ocorra o efeito fotoel\u00e9trico corresponde ao ponto onde a curva intercepta o eixo y\u00a0 —\u00a0 W\u22482,14eV\u00a0 —\u00a0 a frequ\u00eancia m\u00ednima para que o efeito fotoel\u00e9trico ocorra ocorre no ponto onde a curva intercepta o eixo x\u00a0 —\u00a0 f=5,0.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>Hz\u00a0 —\u00a0 W=h.f\u00a0 —\u00a0 2,14=h.5.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 h\u22480,428.10<\/b><\/span><\/span><\/span>-14<\/b><\/span><\/span><\/sup><\/span> \u2248 4,28.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>eVs.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 — \u00a0a\u00a0energia m\u00ednima para extrair um el\u00e9tron\u00a0da placa met\u00e1lica \u00e9 denominada\u00a0fun\u00e7\u00e3o trabalho\u00a0e est\u00e1 relacionada com o tipo de metal utilizado\u00a0 —\u00a0 utilizando o valor 4,28.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>\u00a0eVs da alternativa 01 para constante de Planck, e sendo a frequ\u00eancia de corte fornecida pelo gr\u00e1fico f=5.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0Hz\u00a0 —\u00a0 E = h . f=4,28.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>.5.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E\u22482,14eV.<\/b><\/span><\/span><\/span><\/p>\n

E = 4,43 . 10<\/b><\/span><\/span><\/span>\u201315<\/b><\/span><\/span><\/sup><\/span> . 5 . 10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span> Hz<\/b><\/span><\/span><\/span><\/p>\n

E = 2,215 eV<\/b><\/span><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 fo=5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>Hz \u00e9 a frequ\u00eancia m\u00ednima (de corte) a partir da qual os el\u00e9trons s\u00e3o extra\u00eddos do pot\u00e1ssio, adquirindo velocidade e consequentemente energia cin\u00e9tica.<\/b><\/span><\/span><\/span><\/p>\n

08. Falsa\u00a0 — para f=6.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>\u00a0Hz e h=4,28.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>eVs\u00a0 —\u00a0 E=h.f= 4,28.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>.6.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>=2,528 eV.<\/b><\/span><\/span><\/span><\/p>\n

16. Correta\u00a0 —\u00a0 observe no gr\u00e1fico que, para a mesma placa (mesmo fo), \u00e0 medida que a freq\u00fc\u00eancia (cor, n\u00edvel energ\u00e9tico) da radia\u00e7\u00e3o incidente aumenta, a energia cin\u00e9tica dos el\u00e9trons emitidos tamb\u00e9m aumenta\u00a0 —\u00a0 portanto, quanto mais para a direita a curva cortar o eixo x, maior ser\u00e1 a fun\u00e7\u00e3o trabalho.<\/b><\/span><\/span><\/span><\/p>\n

32. Falsa —\u00a0 aumentando a intensidade da radia\u00e7\u00e3o (freq\u00fc\u00eancia, cor) incidente no metal, aumenta-se o n\u00edvel energ\u00e9tico dos f\u00f3tons incidentes, aumentando assim n\u00famero de el\u00e9trons arrancados\u00a0 —\u00a0 para aumentar a energia cin\u00e9tica de cada el\u00e9tron voc\u00ea deve aumentar a frequ\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

R- ((01 + 02 + 04 +16)=23.<\/b><\/span><\/span><\/span><\/p>\n

58-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Correta.<\/b><\/span><\/span><\/span><\/p>\n

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

04. Falsa\u00a0 —\u00a0 Leis de Bohr\u00a0 —\u00a0 Primeira Lei: os el\u00e9trons podem girar em \u00f3rbita somente a determinadas dist\u00e2ncias permitidas do n\u00facleo\u00a0 —\u00a0 Segunda Lei: um \u00e1tomo irradia energia quando um el\u00e9tron salta de uma \u00f3rbita de maior energia para uma de menor energia.<\/b><\/span><\/span><\/span><\/p>\n

08- Falsa.<\/b><\/span><\/span><\/span><\/p>\n

16- Correta.<\/b><\/span><\/span><\/span><\/p>\n

R- (01, 02, 16).<\/b><\/span><\/span><\/span><\/p>\n

59-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A express\u00e3o da energia de um n\u00edvel n (em el\u00e9tron-volt) \u00e9\u00a0 —\u00a0 E\u00ad<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>= – 13,6\/n<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 primeiro estado estacion\u00e1rio n=1\u00a0 — E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= – 13,6\/1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E1= – 13,6 eV\u00a0 —\u00a0 segundo estado estacion\u00e1rio\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= – 13,6\/2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= – 3,4 eV\u00a0 —\u00a0 energia do f\u00f3ton emitido (liberada nesse decaimento)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>f\u00f3ton<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= – 3,4 \u2013 (- 13,6)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>f\u00f3ton<\/b><\/span><\/span><\/sub><\/span>= 10,2 eV\u00a0 —\u00a0 mas E<\/b><\/span><\/span><\/span>f\u00f3ton<\/b><\/span><\/span><\/sub><\/span>=h.f (I) e V=c=\u03bb.f\u00a0 —\u00a0 f=c\/\u03bb (II)\u00a0 —\u00a0 (II) em (I)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>f\u00f3ton<\/b><\/span><\/span><\/sub><\/span>=h.c\/\u03bb\u00a0 —\u00a0 10,2 = 4,1.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span>.3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/\u03bb\u00a0 —\u00a0 \u03bb = 12,3.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\/10,2\u00a0 —\u00a0 \u039b = 1,2.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>\u00a0m\u00a0 —\u00a0 observe na figura fornecida pelo enunciado que esse f\u00f3ton est\u00e1 na faixa do ultra-violeta\u00a0 —\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

Estrutura at\u00f4mica \u2013 \u00e1tomo de Bohr<\/b><\/span><\/span><\/span><\/p>\n

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

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Autocontra\u00e7\u00e3o gravitacional (potencial gravitacional)\u00a0 —\u00a0 nesse processo, os \u00edons de hidrog\u00eanio (pr\u00f3tons) contidos no seu interior adquirem\u00a0 velocidades muito altas (energia cin\u00e9tica)\u00a0 —\u00a0 o que os leva a atingirem temperaturas da ordem de milh\u00f5es de graus (energia t\u00e9rmica)\u00a0 —\u00a0 com isso, t\u00eam in\u00edcio rea\u00e7\u00f5es exot\u00e9rmicas de fus\u00e3o nuclear, nas quais n\u00facleos de hidrog\u00eanio s\u00e3o fundidos, gerando n\u00facleos de He\u00a0(H\u00e9lio) e propiciando a produ\u00e7\u00e3o da radia\u00e7\u00e3o (energia de massa)\u00a0 —\u00a0 parte dessa radia\u00e7\u00e3o atinge a Terra (sob forma de radia\u00e7\u00e3o eletromagn\u00e9tica) e \u00e9 a principal fonte de toda a energia que utilizamos\u00a0 <\/b><\/span><\/span><\/span>—\u00a0\u00a0R- D.<\/b><\/span><\/span><\/span><\/p>\n

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

49-<\/b><\/span><\/span><\/span> <\/b>R- D. Veja fisicaevestibular.com.br \u2013 F\u00edsica moderna<\/b><\/span><\/span><\/span><\/p>\n

Introdu\u00e7\u00e3o \u00e0 Teoria da Relatividade<\/b><\/span><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como na teoria da relatividade a massa relativ\u00edstica est\u00e1 associada com a velocidade, essa rela\u00e7\u00e3o faz com que a teoria relativ\u00edstica implique tamb\u00e9m com a energia\u00a0 —\u00a0 Einstein contribuiu na sua Teoria Especial da Relatividade com a famosa express\u00e3o matem\u00e1tica:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde E \u00e9 a energia pr\u00f3pria do corpo e a energia emitida na forma de f\u00f3tons pode ser determinada por essa equa\u00e7\u00e3o\u00a0 — E=m.c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2.10<\/b><\/span><\/span><\/span>-26<\/b><\/span><\/span><\/sup><\/span>.(3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>)2=2.10<\/b><\/span><\/span><\/span>-26<\/b><\/span><\/span><\/sup><\/span>.9.10<\/b><\/span><\/span><\/span>16<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E=18.10<\/b><\/span><\/span><\/span>-10<\/b><\/span><\/span><\/sup><\/span>J=1,8.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 essa energia devida \u00e0 aniquila\u00e7\u00e3o dos p\u00f3sitrons e el\u00e9trons, \u00e9 totalmente transformada na energia dos f\u00f3tons, ocorrendo a conserva\u00e7\u00e3o de energia\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Numa explos\u00e3o de supernova foram liberados 3,24.10<\/b><\/span><\/span><\/span>48<\/b><\/span><\/span><\/sup><\/span>\u00a0J , de forma que sua massa foi reduzida para m<\/b><\/span><\/span><\/span>fina<\/b><\/span><\/span><\/sub><\/span>l=4.10<\/b><\/span><\/span><\/span>30<\/b><\/span><\/span><\/sup><\/span>kg. Qual era a massa da estrela antes da explos\u00e3o?<\/b><\/span><\/span><\/span><\/p>\n

a) De acordo com o enunciado s\u00e3o dados\u00a0 —\u00a0 c=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 —\u00a0 \u2206\u03bb=0,0092\u03bb<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 substituindo em express\u00e3o fornecida\u00a0 —\u00a0 v=c.\u2206\u03bb\/\u03bb<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 v=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>.0,092.\u03bb<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/\u03bb<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0,276.10<\/b><\/span><\/span><\/span>18<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 —\u00a0 V=2,76.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 —\u00a0 o valor da constante de Hubble tamb\u00e9m \u00e9 fornecido\u00a0 —\u00a0 H=2,3.10<\/b><\/span><\/span><\/span>-18<\/b><\/span><\/span><\/sup><\/span>s-1\u00a0 —\u00a0 v=H.r\u00a0 —\u00a0 2,76.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>\u00a0= 2,3.10<\/b><\/span><\/span><\/span>-18<\/b><\/span><\/span><\/sup><\/span>.r\u00a0 —\u00a0 r=2,76.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>\/2,3.10<\/b><\/span><\/span><\/span>-18<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0r=1,2.10<\/b><\/span><\/span><\/span>25<\/b><\/span><\/span><\/sup><\/span>m\u00a0(dist\u00e2ncia da Terra at\u00e9 a gal\u00e1xia)<\/b><\/span><\/span><\/span><\/p>\n

b) Dados\u00a0 —\u00a0 E=3,24.10<\/b><\/span><\/span><\/span>48<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 c=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 —\u00a0 massa liberada em forma de energia (massa transformada)\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=?<\/b><\/span><\/span><\/span><\/p>\n

<\/a>E=m<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0\u00a0—\u00a0 3,24.10<\/b><\/span><\/span><\/span>48<\/b><\/span><\/span><\/sup><\/span>=mt.(3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=3,24.10<\/b><\/span><\/span><\/span>48<\/b><\/span><\/span><\/sup><\/span>\/9.10<\/b><\/span><\/span><\/span>16<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 mt=0,36.10<\/b><\/span><\/span><\/span>32<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 mt=36.10<\/b><\/span><\/span><\/span>30<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>final<\/b><\/span><\/span><\/sub><\/span>= m<\/b><\/span><\/span><\/span>inicia<\/b><\/span><\/span><\/sub><\/span>l\u00a0-m<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 dado\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>final<\/b><\/span><\/span><\/sub><\/span>=4.10<\/b><\/span><\/span><\/span>30<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 36.10<\/b><\/span><\/span><\/span>30<\/b><\/span><\/span><\/sup><\/span>\u00a0= m<\/b><\/span><\/span><\/span>inicial<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 4.10<\/b><\/span><\/span><\/span>30<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>inicial<\/b><\/span><\/span><\/sub><\/span>=40.10<\/b><\/span><\/span><\/span>30<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>inicial<\/b><\/span><\/span><\/sub><\/span>=4,0.10<\/b><\/span><\/span><\/span>31<\/b><\/span><\/span><\/sup><\/span>kg.<\/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 \u2013 Vestibulares Recentes F\u00cdSICA MODERNA \u2013 2014 – 2013 01- Dilata\u00e7\u00e3o do tempo: To\u00a0\u2013 per\u00edodo (tempo que o balan\u00e7o demora para efetuar um \u201cvai e vem\u201d completo medido no interior da nave) — T – per\u00edodo do balan\u00e7o medido por Romeu na plataforma espacial — T=To\/\u221a1 \u2013 V2\/c2)= To\/\u221a1 \u2013 (0,5)2c2\/c2)= To\/\u221a1 \u2013 0,25)=To\/\u221a( 1 \u2013 1\/4) —<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3394,"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-3396","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3396","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=3396"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3396\/revisions"}],"predecessor-version":[{"id":3398,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3396\/revisions\/3398"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3394"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3396"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}