{"id":2331,"date":"2016-07-09T23:56:13","date_gmt":"2016-07-09T23:56:13","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2331"},"modified":"2024-08-23T14:56:04","modified_gmt":"2024-08-23T14:56:04","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-ondas-sonoras","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/ondulatoria\/acustica\/ondas-sonoras\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-ondas-sonoras\/","title":{"rendered":"Ondas Sonoras – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Ondas Sonoras<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>a) V=\u0394S\/\u0394t\u00a0 —\u00a0 340=\u0394S\/3\u00a0 —\u00a0\u00a0\u0394S=1020m (ida e volta)\u00a0 —\u00a0\u0394S=510m<\/b><\/span><\/span><\/span><\/p>\n

b) V=\u0394S\/\u0394t\u00a0 —\u00a0 340=\u0394S\/0,1\u00a0 —\u00a0\u00a0\u0394S=34m (ida e volta)\u00a0 —\u00a0\u00a0\u0394S=17m<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span> <\/b>V=\u0394S\/\u0394t\u00a0 —\u00a0 350=\u0394S\/0,100\u00a0 —\u00a0\u00a0\u0394S=35m (ida e volta)\u00a0 —\u00a0\u0394S=35\/2=17,5m<\/b><\/span><\/span><\/span><\/p>\n

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

04-<\/b><\/span><\/span><\/span> <\/b>T = 8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>s\u00a0 —\u00a0 T=1\/f\u00a0 —\u00a0 8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=1\/f\u00a0 —\u00a0 f=125Hz<\/b><\/span><\/span><\/span><\/p>\n

05\u2013<\/b><\/span><\/span><\/span> <\/b>C\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>C\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span> <\/b>Como a freq\u00fc\u00eancia (f) dobra e o per\u00edodo (T) \u00e9 o inverso da freq\u00fc\u00eancia, ele ser\u00e1 reduzido \u00e0 metade, o que ocorre na figura C.<\/b><\/span><\/span><\/span><\/p>\n

Ou, sendo V= \u03bbf\u00a0 —\u00a0 se f dobrar, sendo V constante (mesmo meio), o comprimento de onda\u00a0\u03bb\u00a0dever\u00e1 ser reduzido \u00e0 metade, passando de 40cm para 20cm.<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span> <\/b>a) V=<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>S\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t\u00a0 —\u00a0 1540=0,2\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t=1,3.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>s<\/b><\/span><\/span><\/span><\/p>\n

b) V=<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>S\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t\u00a0 —\u00a0 1540=<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>S\/2,5.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>S=3,85cm\u00a0 —\u00a0 at\u00e9 o ponto A\u00a0 —\u00a0 d=10-3,85\u00a0 —\u00a0\u00a0d=6,15cm<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span> <\/b>V= \u03bbf\u00a0—\u00a0 340= \u03bb1000\u00a0 —\u00a0\u00a0\u03bb =0,34m=34cm\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span>a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>=330m\/s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 a 40<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0 V=330+0,59X40\u00a0 —\u00a0 V=353,6m\/s<\/b><\/span><\/span><\/span><\/p>\n

a raz\u00e3o V\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 353,6\/330=1,07 e \u00e9 linear (reta)\u00a0\u00a0\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span> <\/b>a) A\u00a0 —\u00a0\u00a0\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=1,5m<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0\u00a0 B\u00a0 —\u00a0\u00a0\u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=0,5m<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\u00a0\u00a0 C\u00a0 —\u00a0\u00a0\u03bb<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=0,3m<\/b><\/span><\/span><\/span><\/p>\n

b)\u00a0\u03bb<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=1,6-0,1=1,5m<\/b><\/span><\/span><\/span><\/p>\n

c) a velocidade de propaga\u00e7\u00e3o \u00e9 a mesma para as tr\u00eas ondas A, B e C e vale 340m\/s.<\/b><\/span><\/span><\/span><\/p>\n

c\u00e1lculo da frequ\u00eancia de cada uma\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 f=V\/\u03bb<\/b><\/span><\/span><\/span><\/p>\n

f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V\/\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=340\/1,5=226,7Hz<\/b><\/span><\/span><\/span><\/p>\n

f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=V\/\u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=340\/0,5=680Hz<\/b><\/span><\/span><\/span><\/p>\n

f<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=V\/\u03bb<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=340\/0,3=1133.3Hz<\/b><\/span><\/span><\/span><\/p>\n

– O menor comprimento de onda que o exerc\u00edcio manda chamar de\u00a0f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 o da A\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=226,7Hz<\/b><\/span><\/span><\/span><\/p>\n

Segundo o enunciado \u03bb=A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=16 ua<\/b><\/span><\/span><\/span><\/p>\n

– A maior freq\u00fc\u00eancia \u00e9 a da onda C e vale\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1133,3\/226,7=5\u00a0 —\u00a0\u00a0f<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=5f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span><\/p>\n

\u03bb<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=(A<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03bb<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u03bb<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=1 ua<\/b><\/span><\/span><\/span><\/p>\n

Colocando-os no gr\u00e1fico, temos:<\/b><\/span><\/span><\/span><\/p>\n

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

13- \u00a0a) V=\u0394S\/\u0394t\u00a0 —\u00a0 1400=\u0394S\/0,8\u00a0 —\u00a0\u00a0\u0394S=1120m (ida e volta)\u00a0 —\u00a0\u00a0\u0394S=560m<\/b><\/span><\/span><\/span><\/p>\n

b) a frequ\u00eancia no ar e na \u00e1gua \u00e9 a mesma (mesma fonte) e vale f=4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>Hz<\/b><\/span><\/span><\/span><\/p>\n

\u00a0ar\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 370=\u03bb<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>.4.10<\/b><\/span><\/span><\/span>4\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0\u03bb<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>=9,25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span><\/p>\n

\u00e1gua\u00a0 — V=\u03bb<\/b><\/span><\/span><\/span>agua<\/b><\/span><\/span><\/sub><\/span>.f\u00a0 —\u00a0 1400=\u00a0\u03bb<\/b><\/span><\/span><\/span>agua<\/b><\/span><\/span><\/sub><\/span>.4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u03bb<\/b><\/span><\/span><\/span>agua<\/b><\/span><\/span><\/sub><\/span>=35.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span><\/p>\n

\u03bb<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>\/\u03bb<\/b><\/span><\/span><\/span>agua<\/b><\/span><\/span><\/sub><\/span>=9,25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/35.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=0,264 <\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span> <\/b>a) 2001, Uma odiss\u00e9ia no espa\u00e7o, pois o som n\u00e3o se propaga no v\u00e1cuo.<\/b><\/span><\/span><\/span><\/p>\n

b) Os dois est\u00e3o corretos, pois a luz se propaga no v\u00e1cuo<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span> <\/b>Quando, novamente, som e imagem forem simult\u00e2neos, a pessoa estar\u00e1 ouvindo o som da pancada anterior, que estar\u00e1 percorrendo a dist\u00e2ncia de 172m em 0,5s.<\/b><\/span><\/span><\/span><\/p>\n

V=\u0394S\/\u0394t\u00a0 —\u00a0 V=172\/0,5\u00a0 —\u00a0\u00a0V=344m\/s<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span> <\/b>a) dist\u00e2ncia at\u00e9 o pared\u00e3o d<\/b><\/span><\/span><\/span><\/p>\n

\u00a0emerso\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>=2d\/0,731\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 imerso\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=2d\/0,170\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>=2d\/0,170X0,731\/2d=4,3<\/b><\/span><\/span><\/span><\/p>\n

b) a frequ\u00eancia f \u00e9 a mesma (mesma fonte)<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>=\u03bb<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.f\/\u03bb<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>.f\u00a0 —\u00a0\u00a04,3=\u03bb<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u03bbl<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span> <\/b>Per\u00edodo T=20.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>s\u00a0 — f=1\/T\u00a0 —\u00a0 f=1\/20.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 f=5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>Hz\u00a0 —\u00a0 f=50.000Hz\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span> <\/b>Tempo que o som demora para percorrer\u00a0\u0394S=3X340=1.020m, com V=340m\/s\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 340=1020\/\u0394t\u00a0 —\u00a0\u00a0\u0394t=3s.<\/b><\/span><\/span><\/span><\/p>\n

Nesse tempo, o corredor com V=10m\/s, percorreu\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 10=\u0394S\/3\u00a0 —\u00a0\u0394S=AX=30m<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span><\/span> <\/b>V=\u0394S\/\u0394t\u00a0 —\u00a0 1520=\u0394S\/3\u00a0 —\u00a0\u00a0\u0394S=4.560m<\/b><\/span><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span>Para anular o som original, o computador deve enviar ondas que provoqueminterfer\u00eancia\u00a0destrutiva. R- A<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span> <\/b>O som chega ao observador com atraso e som e imagem ser\u00e3o simult\u00e2neos quando o atraso for um m\u00faltiplo do per\u00edodo das marteladas, ou seja, a cada 0,9s, 1,8s, 2,7s, etc. No caso, a menor dist\u00e2ncia ocorre com um atraso de\u00a0\u0394t=0,9s.<\/b><\/span><\/span><\/span><\/p>\n

V=\u0394S\/\u0394t\u00a0 —\u00a0 330=\u0394S\/0,9\u00a0 —\u00a0\u00a0\u0394S=297m<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span> <\/b>Calculo do instante em que o alto-falante est\u00e1 emitindo o som de freq\u00fc\u00eancia F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1.080Hz<\/b><\/span><\/span><\/span><\/p>\n

F(t)=1000+200t\u00a0 —\u00a0 1080=1000+200t\u00a0 —\u00a0 t=0,4s<\/b><\/span><\/span><\/span><\/p>\n

Calculo do intervalo de tempo que qualquer som emitido pela fonte demora\u00a0 para chegar \u00e0 pessoa P que est\u00e1 a \u0394=34m da fonte.<\/b><\/span><\/span><\/span><\/p>\n

V=\u0394S\/\u0394t\u00a0 —\u00a0 340=34\/Dt\u00a0 —\u00a0\u00a0\u0394t=0,1s<\/b><\/span><\/span><\/span><\/p>\n

O som que a pessoa percebe tem um atraso de 0,1s em rela\u00e7\u00e3o ao som emitido pela fonte. Assim, t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,4s \u2013 0,1s=0,3s<\/b><\/span><\/span><\/span><\/p>\n

F(2)=1000+200.0,3\u00a0 —\u00a0\u00a0F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1.060Hz<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span> <\/b>atrav\u00e9s do trilho\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 3400=L\/t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=L\/3400<\/b><\/span><\/span><\/span><\/p>\n

– pelo ar\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 340=L\/t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=L\/340<\/b><\/span><\/span><\/span><\/p>\n

Como a viagem pelo ar demora mais, temos que t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,18\u00a0 —\u00a0 L\/340-L\/3400=0,18\u00a0 —\u00a0L=68m<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

As ondas eletromagn\u00e9ticas emitidas na regi\u00e3o litor\u00e2nea do Brasil atingem a regi\u00e3o amaz\u00f4nica ap\u00f3s sofrer reflex\u00e3o total na ionosfera\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A dist\u00e2ncia percorrida pelo som at\u00e9 que Patr\u00edcia ou\u00e7a o eco de sua voz direta \u00e9 igual a 2d (ida e volta)\u00a0 —\u00a0 v=\u2206S\/\u2206t\u00a0 —\u00a0 v = 2d\/\u2206t\u00a0 — 340=2d\/0,4\u00a0 —\u00a0 d=136\/2\u00a0 —\u00a0 d=68m\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>01. Falsa\u00a0 —\u00a0 a frequ\u00eancia de uma onda n\u00e3o se altera quando ela reflete ou refrata, permanecendo constante, igual a frequ\u00eancia com que foi emitida pela fonte.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 trata-se da afirma a 1\u00aa lei da reflex\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 desde que a onda seja mec\u00e2nica\u00a0 —\u00a0 as ondas eletromagn\u00e9ticas tamb\u00e9m s\u00e3o transversais, mas n\u00e3o provocam vibra\u00e7\u00f5es nas part\u00edculas do meio\u00a0 —\u00a0 a vibra\u00e7\u00e3o d\u00e1-se nas intensidades dos campos el\u00e9trico e magn\u00e9tico que se propagam pelo meio.<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 a velocidade de uma onda depende das condi\u00e7\u00f5es do meio\u00a0 —\u00a0 por exemplo, no ar, a velocidade do som depende da umidade e da temperatura e na \u00e1gua depende da profundidade.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (02 + 04 + 08) = 14<\/b><\/span><\/span><\/span><\/p>\n

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

27-\u00a0<\/b><\/span><\/span><\/span>Como o som vai e volta ele percorre \u2206S=2h\u00a0 —\u00a0 V=\u2206S\/\u2206t\u00a0 —1,6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0= 2h\/0,8\u00a0 —\u00a0 h=1,6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.0,4\u00a0 —\u00a0 h=640m\u00a0 —\u00a0 equa\u00e7\u00e3o fundamental da ondulat\u00f3ria\u00a0 —\u00a0 V=\u03bb.f\u00a0 —\u00a0 1,6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>= \u03bb.4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03bb=1,6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03bb=0,4.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>=4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

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

28-<\/b><\/span><\/span><\/span> <\/b>Raio, rel\u00e2mpago e trov\u00e3o s\u00e3o fen\u00f4menos atmosf\u00e9ricos que ocorrem antes ou durante tempestades, causados por descargas el\u00e9tricas entre nuvens ( rel\u00e2mpago ) ou entre uma nuvem e o solo ( raio ), acompanhados de um estrondo ( trov\u00e3o )\u00a0 —\u00a0 o raio ocorre quando uma nuvem, carregada de eletricidade est\u00e1tica, atinge um potencial eletrost\u00e1tico t\u00e3o elevado que a camada de ar existente entre ela e o solo deixa de ser isolante, tornando-se condutor, permitindo a movimenta\u00e7\u00e3o das cargas el\u00e9tricas at\u00e9 o solo\u00a0 —\u00a0 com o rel\u00e2mpago ocorre a mesma coisa s\u00f3 que \u00e9 entre as nuvens\u00a0 —\u00a0\u00a0 essas descargas el\u00e9tricas produzem uma grande luminosidade devido ao aquecimento do ar, que pode ser vista bem distante\u00a0 —\u00a0\u00a0 o trov\u00e3o \u00e9 o som que acompanha a violenta expans\u00e3o do ar quando \u00e9 rapidamente aquecido\u00a0 —\u00a0 devido \u00e0 diferen\u00e7a de velocidades da luz e do som (luz, 3.108m\/s e som, 340m\/s), quando o raio ou rel\u00e2mpago ocorre a grande dist\u00e2ncia do local de observa\u00e7\u00e3o, o trov\u00e3o \u00e9 ouvido alguns segundos depois do clar\u00e3o\u00a0 —\u00a0\u00a0R- E.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Press\u00e3o=for\u00e7a\/\u00e1rea\u00a0 —\u00a0 Pr=F\/S\u00a0 —\u00a0 densidade=massa\/volume\u00a0 —\u00a0 \u03c1=m\/Vo\u00a0 —\u00a0 Vo=S.\u2113\u00a0 —\u00a0 \u03c1=m\/S.\u2113\u00a0 —\u00a0 Pr=F\/S\u00a0 —F=m.a\u00a0 —\u00a0 Pr=m.a\/S\u00a0 — multiplicando o numerador e denominador por \u2113\u00a0 —\u00a0 Pr=m.a.\u2113\/S.\u2113\u00a0 —\u00a0 Pr=m.a.\u2113\/Vo\u00a0 — mas, m\/Vo=densidade=\u03c1\u00a0 — Pr=\u03c1.a.\u2113\u00a0 —\u00a0 acelera\u00e7\u00e3o a=ve\/t\u00a0 —\u00a0 Pr= \u03c1.(Ve\/t).\u2113\u00a0 —\u00a0 Ve= \u2113\/t\u00a0 —\u00a0 Pr=\u03c1.Ve.Ve\u00a0 —\u00a0 Pr=\u03c1.Ve2\u00a0 —<\/b><\/span><\/span><\/p>\n

Ve=\u221a(Pr\/\u03c1) ou Ve=(Pr\/\u03c1)1\/2\u00a0 —\u00a0\u00a0R- C.<\/b><\/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 de vestibulares sobre Ondas Sonoras 01-\u00a0 a) V=\u0394S\/\u0394t\u00a0 —\u00a0 340=\u0394S\/3\u00a0 —\u00a0\u00a0\u0394S=1020m (ida e volta)\u00a0 —\u00a0\u0394S=510m b) V=\u0394S\/\u0394t\u00a0 —\u00a0 340=\u0394S\/0,1\u00a0 —\u00a0\u00a0\u0394S=34m (ida e volta)\u00a0 —\u00a0\u00a0\u0394S=17m 02- V=\u0394S\/\u0394t\u00a0 —\u00a0 350=\u0394S\/0,100\u00a0 —\u00a0\u00a0\u0394S=35m (ida e volta)\u00a0 —\u00a0\u0394S=35\/2=17,5m 03- E 04- T = 8.10-3s\u00a0 —\u00a0 T=1\/f\u00a0 —\u00a0 8.10-3=1\/f\u00a0 —\u00a0 f=125Hz 05\u2013 C\u00a0\u00a0\u00a0\u00a0\u00a0 06- C\u00a0\u00a0\u00a0\u00a0 07- E 08- Como a freq\u00fc\u00eancia (f)<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2327,"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-2331","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2331","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=2331"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2331\/revisions"}],"predecessor-version":[{"id":10906,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2331\/revisions\/10906"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2327"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2331"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}