{"id":2901,"date":"2017-04-12T05:36:48","date_gmt":"2017-04-12T05:36:48","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2901"},"modified":"2024-09-02T16:29:38","modified_gmt":"2024-09-02T16:29:38","slug":"ondulatoria-resolucao","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/raciocinio-logico\/ondulatoria\/ondulatoria-resolucao\/","title":{"rendered":"Ondulat\u00f3ria – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Ondulat\u00f3ria<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span> <\/b>Densidade volum\u00e9trica=massa\/volume\u00a0— \u03c1=m\/V<\/b><\/span><\/span><\/span>ol<\/b><\/span><\/span><\/sub><\/span>\u00a0— \u03c1=m\/S.L — S=\u03c0.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=\u03c0.(d\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.L —<\/b><\/span><\/span><\/span><\/p>\n

\u03c1=m\/{\u03c0.(d\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.L} —\u00a0\u03c1=4m\/\u03c0d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.L (I).<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Densidade linear=massa\/comprimento\u00a0—\u00a0\u03bc=m\/L\u00a0—\u00a0m=\u03bc.L (II)<\/b><\/span><\/span><\/span><\/p>\n

(II) em (I)\u00a0— \u03c1=4\u03bcL\/\u03c0d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.L — \u03bc=\u03c1\u03c0Ld<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4L —\u00a0\u03bc=\u03c1\u03c0d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4 (III)<\/b><\/span><\/span><\/span><\/p>\n

Velocidade de propaga\u00e7\u00e3o da onda na corda submetida \u00e0 tra\u00e7\u00e3o F\u00a0—\u00a0V=\u221a(F\/\u03bc) (IV)<\/b><\/span><\/span><\/span><\/p>\n

(III) em (VI)\u00a0— V=\u221a({F\/(\u03c1\u03c0d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4) —\u00a0V=(d\/2).\u221a(F\/\u03c1\u03c0)<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se voc\u00ea\u00a0parar o barco, cada crista da onda distante \u03bb=10m uma da outra, se aproximar\u00e1 dele com<\/b><\/span><\/span><\/span>\"\"<\/p>\n

velocidade relativa V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=2 + 8=10m\/s\u00a0e, assim, sua frequ\u00eancia (n\u00famero de oscila\u00e7\u00f5es da canoa por segundo, ser\u00e1 fornecida pela\u00a0equa\u00e7\u00e3o fundamental da ondulat\u00f3ria\u00a0—\u00a0V<\/b><\/span><\/span><\/span>R\u00a0<\/b><\/span><\/span><\/sub><\/span>=\u03bb.f\u00a0— 10=10.f —\u00a0f=10Hz.<\/b><\/span><\/span><\/span><\/p>\n

O\u00a0per\u00edodo T\u00a0pedido corresponde ao\u00a0inverso da frequ\u00eancia\u00a0—\u00a0T=1\/f=1\/1=1s.<\/b><\/span><\/span><\/span><\/p>\n

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

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O\u00a0per\u00edodo\u00a0T\u00a0de um p\u00eandulo simples \u00e9, por defini\u00e7\u00e3o, o\u00a0tempo que ele demora para efetuar um \u201cvai e vem\u201d completo, sair de A e retornar a A —\u00a0T=4s e f=1\/T=1\/4=0,25Hz.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

A\u00a0amplitude\u00a0est\u00e1 mostrada no gr\u00e1fico acima e vale\u00a0AB=AC.<\/b><\/span><\/span><\/span><\/p>\n

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

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Observe na figura abaixo onde foi considerada a onda representada pela\u00a0linha cheia\u00a0que o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

comprimento de onda\u00a0\u03bb\u00a0que corresponde \u00e0\u00a0dist\u00e2ncia percorrida pela onda at\u00e9 come\u00e7ar a repeti\u00e7\u00e3o vale\u00a0\u03bb=4m.<\/b><\/span><\/span><\/span><\/p>\n

b)\u00a0Per\u00edodo T\u00a0\u00e9 o tempo que a onda demora para\u00a0percorrer 1\u03bb (at\u00e9 come\u00e7ar a repeti\u00e7\u00e3o)\u00a0que,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

observado na figura 2 acima nos fornece\u00a0T=8s<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0Observe na figura 1 que a onda com\u00a0curva tracejada est\u00e1 deslocada de um quarto de comprimento de onda em rela\u00e7\u00e3o \u00e0 onda com curva cheia. Portanto a segunda foto (instante t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) foi tirada ( ao menos) um quarto de per\u00edodo depois, ou seja 8\/4=2s. Os valores poss\u00edveis de t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0s\u00e3o\u00a0(0 + 2=2s), (8 + 2=10s), (16 + 2=18s), etc.\u00a0O valor m\u00ednimo \u00e9\u00a02s<\/b><\/span><\/span><\/span>\"\"<\/p>\n

05-<\/b><\/span><\/span><\/span> <\/b>Aparelhos que envolvam a\u00a0emiss\u00e3o ou recep\u00e7\u00e3o de ondas eletromagn\u00e9ticas, como aparelhos celulares podem provocar interfer\u00eancia das ondas eletromagn\u00e9ticas\u00a0nas comunica\u00e7\u00f5es dos pilotos com a torre de controle e,\u00a0esta interfer\u00eancia \u00e9 acentuada quando essas ondas eletromagn\u00e9ticas emitidas pelos aparelhos eletr\u00f4nicos possuem faixas de frequ\u00eancias pr\u00f3ximas \u00e0s do equipamento a bordo da aeronave— R- E<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>O comprimento de onda\u00a0(\u03bb)\u00a0representa a dist\u00e2ncia percorrida pela onda at\u00e9 come\u00e7ar novamente a repeti\u00e7\u00e3o, ou seja, \u00e9 a menor dist\u00e2ncia entre dois pontos consecutivos que est\u00e3o em concord\u00e2ncia de fase, como, por exemplo, a menor dist\u00e2ncia entre duas cristas ou dois vales — o enunciado fornece\u00a0que cada\u00a0per\u00edodo\u00a0de oscila\u00e7\u00e3o cont\u00e9m 16 pessoas,\u00a0que se levantam e sentam organizadamente e\u00a0distanciadas entre si por 80cm — comprimento de onda=15 espa\u00e7osxcomprimento de cada espa\u00e7o — \u03b3=15×0,8=12m — esse comprimento de onda \u03b3=12m corresponde \u00e0 dist\u00e2ncia que a onda percorre no intervalo de tempo de um per\u00edodo\u00a0(T),\u00a0que, por defini\u00e7\u00e3o \u00e9 o tempo que a onda demora a percorrer 1\u03b3=12m com velocidade V=45\/3,6=<\/b><\/span><\/span><\/span>\"\"<\/p>\n

12,5m\/s— c\u00e1lculo do per\u00edodo (T) — V=\u0394S\/\u0394S=\u03b3\/T — 12,5=12\/T — T=12\/12,5=0,96s — o enunciado pede a frequ\u00eancia que representa quantas oscila\u00e7\u00f5es completas a onda efetua em cada unidade de tempo (per\u00edodo T) — lembrando que a frequ\u00eancia \u00e9 o inverso do per\u00edodo —\u00a0f=1\/T=1\/0,96=1,04Hz—\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span> <\/b>Para ondas sofrendo interfer\u00eancia em meios uni, bi ou tridimensionais s\u00e3o v\u00e1lidas sempre as rela\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

Para que um ponto P esteja em interfer\u00eancia\u00a0construtiva (refor\u00e7o), a dist\u00e2ncia da fonte F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0at\u00e9 P (d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>) menos a dist\u00e2ncia da fonte F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0at\u00e9 P (d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>), deve valer um\u00a0n\u00famero par (2n) de meios comprimentos de onda:\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Para que um ponto P esteja\u00a0em interfer\u00eancia destrutiva (anula\u00e7\u00e3o),\u00a0a dist\u00e2ncia da fonte F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0at\u00e9 P (d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>) menos a dist\u00e2ncia da fonte F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0at\u00e9 P (d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>), deve valer um n\u00famero \u00edmpar (2n + 1) de\u00a0meios comprimentos de onda:.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S\u00e3o dados—\u00a0f=440Hz\u00a0—\u00a0V=330m\/s\u00a0— equa\u00e7\u00e3o fundamental da ondulat\u00f3ria — V=\u03bb.f — 330=\u03bb.440 —\u00a0\u03bb=330\/440=0,75m— a dist\u00e2ncia da caixa ac\u00fastica da direita, fonte\u00a0F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, at\u00e9 o Sr. Rubinato (ponto P)\u00a0que vale (L + \u2113) menos a dist\u00e2ncia da caixa ac\u00fastica da esquerda, fonte\u00a0F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, at\u00e9 o Sr. Rubinato (ponto P)\u00a0que vale\u00a0(L)\u00a0deve ser um n\u00famero \u00edmpar, pois o L\u00e1 passou de forte (interfer\u00eancia construtiva) para fraco (interfer\u00eancia destrutiva)— o primeiro \u00edmpar ocorre quando n=0\u00a0— d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(2n + 1).(\u03bb\/2) — (L + \u2113) \u2013 L = (2×0 + 1). 0,75\/2—\u00a0\u2113 =\u00a01×0,75\/2=0,375m=37,5cm\u00a0—\u00a0R- A<\/b><\/span><\/span><\/span>\"\"<\/p>\n

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

I.\u00a0Falsa\u00a0—\u00a0orelha direita\u00a0— observe no gr\u00e1fico que para a orelha direita, sons de\u00a0frequ\u00eancia 6 kHz\u00a0possuem\u00a0intensidade m\u00ednima de 25 dB, ou seja, ele\u00a0n\u00e3o ouve\u00a0sons com intensidade\u00a0abaixo de 25 dB,\u00a0n\u00e3o ouvindo, portanto som de intensidade 20 dB<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

orelha esquerda\u00a0— observe no gr\u00e1fico que para a orelha esquerda, sons de\u00a0frequ\u00eancia 6 kHz\u00a0possuem\u00a0intensidade m\u00ednima de 10 dB, ou seja, ele\u00a0n\u00e3o ouve\u00a0sons com intensidade\u00a0abaixo de 10 dB,\u00a0ouvindo, portanto som de intensidade 20 dB<\/b><\/span><\/span><\/span><\/p>\n

Assim, para a frequ\u00eancia de 6 kHz e n\u00edvel sonoro 20 dB a pessoa\u00a0ouve com a orelha esquerda,\u00a0mas\u00a0n\u00e3o\u00a0com a direita.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta\u00a0— para\u00a0ser ouvido\u00a0um som de\u00a0frequ\u00eancia 0,25 kHz\u00a0deve possuir\u00a0intensidade m\u00ednima de 10 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

dB para as duas orelhas,\u00a0ouvindo, portanto, um sussurro de\u00a015 dB.<\/b><\/span><\/span><\/span><\/p>\n

III. Falsa\u00a0— a transforma\u00e7\u00e3o de\u00a0som em impulso el\u00e9trico\u00a0ocorre na\u00a0orelha interna\u00a0onde est\u00e3o localizados os ossos martelo, bigorna e estribo.<\/b><\/span><\/span><\/span><\/p>\n

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

09-<\/b><\/span><\/span><\/span> <\/b>Na sequ\u00eancia abaixo eu representei as configura\u00e7\u00f5es da corda em cada\u00a0um quarto de per\u00edodo T\u00a0at\u00e9 o instante pedido t=3T\/4\u00a0— a figura 1 representa as deforma\u00e7\u00f5es da onda (I) se propagando <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

para\u00a0a direita, a figura 2 representa as deforma\u00e7\u00f5es da onda (II) se propagando para a esquerda, e a figura 3 representa a configura\u00e7\u00e3o da corda resultante da interfer\u00eancia dessas duas ondas, nos instantes t = T\/2 e t=T3\/4 — observe que no instante pedido t=3T\/4, a corda ter\u00e1 a configura\u00e7\u00e3o reta e horizontal devido, em cada ponto, \u00e0\u00a0interfer\u00eancia destrutiva\u00a0entre as ondas (I) e (II) —\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span> <\/b>Observe na figura que o comprimento de onda destacado vale — \u03bb=450.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>m.<\/b><\/span><\/span><\/span><\/p>\n

Pelo enunciado,\u00a0a\u00a0frequ\u00eancia \u03bd\u00a0\u00e9\u00a0inversamente proporcional ao comprimento de onda \u03bb, sendo a\u00a0constante de proporcionalidade igual \u00e0 velocidade da luz no v\u00e1cuo\u00a0de, aproximadamente,<\/b><\/span><\/span><\/span><\/p>\n

3,0.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\u00a0m\/s, ou seja,\u00a0c=\u03bb.V\u00a0— 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>=450.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>.V —\u00a0V=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/450.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>\u22480,0066.10<\/b><\/span><\/span><\/span>17<\/b><\/span><\/span><\/sup><\/span>=6,6.10<\/b><\/span><\/span><\/span>-14<\/b><\/span><\/span><\/sup><\/span>Hz.<\/b><\/span><\/span><\/span><\/p>\n

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

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I. Correta\u00a0— os raios X s\u00e3o\u00a0ondas eletromagn\u00e9ticas\u00a0e, como elas, est\u00e3o sujeitos \u00e0\u00a0todos os fen\u00f4menos citados.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta\u00a0— Materiais\u00a0densos\u00a0como os\u00a0metais absorvem muito os raios-X, pois tem um n\u00famero at\u00f4mico muito alto. Por outro lado, o\u00a0ar,\u00a0com densidade at\u00f4mica e n\u00famero at\u00f4mico baixos n\u00e3o absorve muito os raios-x. Assim, temos em ordem crescente 5 densidades radiol\u00f3gicas\u00a0 b\u00e1sicas: ar, gordura, \u00e1gua, c\u00e1lcio e metal.\u00a0<\/b><\/span><\/span><\/span>Todas as subst\u00e2ncias s\u00e3o penetradas pelos raios X em maior ou menor grau.<\/b><\/i><\/span><\/span><\/span><\/p>\n

III. Falsa\u00a0—\u00a0veja (II)<\/b><\/span><\/span><\/span><\/p>\n

IV. Correta\u00a0—<\/b><\/span><\/span><\/span><\/p>\n

A luz vis\u00edvel tamb\u00e9m \u00e9 radia\u00e7\u00e3o eletromagn\u00e9tica. A \u00fanica diferen\u00e7a entre a luz e os raios x \u00e9 que a luz tem uma faixa de frequ\u00eancias espec\u00edficas que os nossos olhos conseguem perceber, atrav\u00e9s de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

c\u00e9lulas especializadas localizadas na retina,\u00a0o que nos possibilita v\u00ea-la.\u00a0Os raios x t\u00eam uma faixa de freq\u00fc\u00eancias muito maior, que fica fora do nosso limite de vis\u00e3o. (veja figura acima).<\/b><\/span><\/span><\/span><\/p>\n

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

12-<\/b><\/span><\/span><\/span> <\/b>a) Observe na figura abaixo onde foi considerada a onda representada pela\u00a0linha cheia\u00a0que o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

comprimento de onda\u00a0\u03bb\u00a0que corresponde \u00e0\u00a0dist\u00e2ncia percorrida pela onda at\u00e9 come\u00e7ar a repeti\u00e7\u00e3o vale\u00a0\u03bb=4m.<\/b><\/span><\/span><\/span><\/p>\n

b)\u00a0Per\u00edodo T\u00a0\u00e9 o tempo que a onda demora para\u00a0percorrer 1\u03bb (at\u00e9 come\u00e7ar a repeti\u00e7\u00e3o)\u00a0que,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

observado na figura 2 acima nos fornece\u00a0T=8s<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0Observe na figura 1 que a onda com\u00a0curva tracejada est\u00e1 deslocada de um quarto de comprimento<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

de onda em rela\u00e7\u00e3o \u00e0 onda com curva cheia. Portanto a segunda foto (instante t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) foi tirada ( ao menos) um quarto de per\u00edodo depois, ou seja 8\/4=2s. Os valores poss\u00edveis de t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0s\u00e3o\u00a0(0 + 2=2s), (8 + 2=10s), (16 + 2=18s), etc.\u00a0O valor m\u00ednimo \u00e9\u00a02s<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o metr\u00f4nomo deve ser ajustado para emitir\u00a060 batidas\u00a0por minuto (60s), com ele regulado cada\u00a0batida deve demorar\u00a0t=60\/60 —\u00a0t=1s.<\/b><\/span><\/span><\/span><\/p>\n

Pelo enunciado\u00a08 notas musicais\u00a0devem ser tocadas no\u00a0tempo de 4 batidas\u00a0do metr\u00f4nomo, ou seja,\u00a0as 8\u00a0notas devem ser tocadas em 4s, o tempo pedido de dura\u00e7\u00e3o de\u00a0cada nota\u00a0pode ser obtido por regra de tr\u00eas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

14-<\/b><\/span><\/span><\/span> <\/b>Altura do som\u00a0est\u00e1 relacionada com sua\u00a0freq\u00fc\u00eancia, ou seja, a altura (tom) \u00e9 a qualidade do som que permite ao ouvido distinguir um som grave, de baixa frequ\u00eancia, de um som agudo, de alta freq\u00fc\u00eancia.\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0O som\u00a0mais grave aud\u00edvel\u00a0por um ouvido humano \u00e9 de\u00a0aproximadamente 20 Hz\u00a0e o\u00a0mais agudo\u00a0\u00e9 de\u00a0aproximadamente\u00a020 000 Hz.<\/b><\/span><\/span><\/span><\/p>\n

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

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se de uma onda peri\u00f3dica e o\u00a0per\u00edodo T corresponde ao tempo que ela demora para percorrer um comprimento de onda 1\u03bb\u00a0e, veja na figura acima que\u00a0em 5s ela percorre aproximadamente 7\u03bb.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Regra de tr\u00eas — 5s – 7\u03bb — T s – 1\u03bb — 7\u03bbT=5\u03bb — T=5\/7s.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0frequ\u00eancia f \u00e9 o inverso do per\u00edodo\u00a0—\u00a0f=1\/T=7\/5Hz(rps) e, como o exerc\u00edcio quer em\u00a0rpm\u00a0basta\u00a0multiplicar por 60\u00a0— f=(7\/5)x60=420\/5 —\u00a0f=84rpm.<\/b><\/span><\/span><\/span><\/p>\n

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

16- <\/b><\/span><\/span><\/span>Ru\u00eddo em uma avenida com tr\u00e2nsito congestionado — I=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>w\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— dado — I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— NS=10.log(I\/I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) — NS=10.log(10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>) =10.log10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0— NS\/10=log10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0— 10<\/b><\/span><\/span><\/span>NS\/10<\/b><\/span><\/span><\/sup><\/span>=10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0— NS\/10=9 —\u00a0NS=90dB\u00a0— pela tabela fornecida esse valor corresponde a\u00a04 horas\u00a0de exposi\u00e7\u00e3o —\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

17-\u00a0<\/b><\/span><\/span><\/span>Teoria:\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V\u00a0a velocidade do som,\u00a0V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0a velocidade da fonte,\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0a velocidade do observador,\u00a0f\u00a0a freq\u00fc\u00eancia real emitida pela fonte, afreq\u00fc\u00eancia aparente\u00a0f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0percebida pelo observador ser\u00e1 fornecida pela express\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Orientando a trajet\u00f3ria do observador para a fonte, os sinais de Vo e Vf ser\u00e3o positivos a favor dessa orienta\u00e7\u00e3o e negativos contra essa orienta\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

Fonte se\u00a0aproximando\u00a0da mulher —\u00a0f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=1500(340 + 0)\/(340 – 40)=1500.340\/300 —\u00a0f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=1700Hz.<\/b><\/span><\/span><\/span><\/p>\n

Fonte se\u00a0afastando\u00a0da mulher —\u00a0f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=1500(340 + 0)\/(340 +40)=1500.340\/380 —\u00a0f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=1342,6Hz<\/b><\/span><\/span><\/span><\/p>\n

\u0394f=1700 \u2013 1342=358Hz.<\/b><\/span><\/span><\/span><\/p>\n

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

18-<\/b><\/span><\/span><\/span> <\/b>1. Falsa\u00a0 —\u00a0 no ar\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>=d\/t\u00a0 —\u00a0 319=6380\/t\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>=20s\u00a0 —\u00a0 no solo\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>solo<\/b><\/span><\/span><\/sub><\/span>=d\/t\u00a0 —\u00a0 5720=5720\/t\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>solo<\/b><\/span><\/span><\/sub><\/span>=1s\u00a0 —\u00a0 \u00e1gua<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=d\/t\u00a0 —\u00a0 1320=660\/t\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=0,5s\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=1,5s\u00a0 —\u00a0 diferen\u00e7a 18,5s\u00a0 —\u00a0 Os \u00edndios ouvir\u00e3o o sinal de ataque da cavalaria 18,5s depois de colocarem o ouvido no ch\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

2. Falsa\u00a0 —\u00a0 a frequ\u00eancia \u00e9 a mesma na \u00e1gua e no solo\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 f=V\/\u03bb\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\/\u03bb<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>solo<\/b><\/span><\/span><\/sub><\/span>\/\u03bb<\/b><\/span><\/span><\/span>solo<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1320\/3 =5720\/\u03bb<\/b><\/span><\/span><\/span>solo<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u03bb<\/b><\/span><\/span><\/span>solo<\/b><\/span><\/span><\/sub><\/span>=13m.<\/b><\/span><\/span><\/span><\/p>\n

3. Falsa\u00a0 —\u00a0 Na aproxima\u00e7\u00e3o entre fonte (cavalaria) e observador (\u00edndios), os \u00edndios perceber\u00e3o o som emitido pela cavalaria mais agudo (maior freq\u00fc\u00eancia, recebe maior n\u00famero de frentes de onda na unidade de tempo) do que perceberia se fonte e observador estivessem parados. Nesse caso, o comprimento de onda aparente percebido pelo observador ser\u00e1 menor que o comprimento da onda emitido pela fonte.<\/b><\/span><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 o som n\u00e3o de propaga no v\u00e1cuo.<\/b><\/span><\/span><\/span><\/p>\n

R- (F,F,F,F).<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando uma\u00a0onda\u00a0de superf\u00edcie se\u00a0aproxima da costa e encontra \u00e1guas menos profundas\u00a0do que metade do seu comprimento de onda,\u00a0s\u00f3 o seu per\u00edodo continua o mesmo. A\u00a0sua velocidade e comprimento de onda diminuem e a altura aumenta<\/b><\/span><\/span><\/span><\/p>\n

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

20-<\/b><\/span><\/span><\/span> <\/b>Qualquer sistema f\u00edsico\u00a0possui uma ou mais\u00a0freq\u00fc\u00eancias naturais de vibra\u00e7\u00e3o. Quando ele \u00e9 \u201cexcitado\u201d por algum agente externo, agindo no ritmo de uma dessas freq\u00fc\u00eancias, surge o fen\u00f4meno da\u00a0resson\u00e2ncia,\u00a0ou seja, ele\u00a0come\u00e7a a oscilar gradativamente at\u00e9 atingir uma dessas freq\u00fc\u00eancias, onde sua amplitude \u00e9 m\u00e1xima.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Um est\u00e1dio de futebol deve ser constru\u00eddo levando-se em conta a\u00a0resson\u00e2ncia,\u00a0pois com a<\/b><\/span><\/span><\/span><\/p>\n

empolga\u00e7\u00e3o da torcida ao bater os p\u00e9s e pular, pode ocorrer a\u00a0resson\u00e2ncia.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Ondulat\u00f3ria 01- Densidade volum\u00e9trica=massa\/volume\u00a0— \u03c1=m\/Vol\u00a0— \u03c1=m\/S.L — S=\u03c0.R2=\u03c0.(d\/2)2.L — \u03c1=m\/{\u03c0.(d\/2)2.L} —\u00a0\u03c1=4m\/\u03c0d2.L (I). Densidade linear=massa\/comprimento\u00a0—\u00a0\u03bc=m\/L\u00a0—\u00a0m=\u03bc.L (II) (II) em (I)\u00a0— \u03c1=4\u03bcL\/\u03c0d2.L — \u03bc=\u03c1\u03c0Ld2\/4L —\u00a0\u03bc=\u03c1\u03c0d2\/4 (III) Velocidade de propaga\u00e7\u00e3o da onda na corda submetida \u00e0 tra\u00e7\u00e3o F\u00a0—\u00a0V=\u221a(F\/\u03bc) (IV) (III) em (VI)\u00a0— V=\u221a({F\/(\u03c1\u03c0d2\/4) —\u00a0V=(d\/2).\u221a(F\/\u03c1\u03c0) R- A 02-\u00a0Se voc\u00ea\u00a0parar o barco, cada crista da onda distante \u03bb=10m uma<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2899,"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-2901","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2901","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=2901"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2901\/revisions"}],"predecessor-version":[{"id":10984,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2901\/revisions\/10984"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2899"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2901"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}