{"id":2706,"date":"2017-02-07T03:30:03","date_gmt":"2017-02-07T03:30:03","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2706"},"modified":"2024-09-03T17:37:19","modified_gmt":"2024-09-03T17:37:19","slug":"resolucao-comentada-unifesp-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/unifesp-2017\/resolucao-comentada-unifesp-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada \u2013 UNIFESP 2017"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada<\/b><\/span><\/span><\/span> \u2013 UNIFESP017 \u2013 1<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span> fase<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>a)<\/b><\/span><\/span> <\/b>Se a <\/b><\/span><\/span>freq\u00fc\u00eancia da h\u00e9lice \u00e9 de f = 4 H<\/b><\/span><\/span>z<\/b><\/span><\/sub><\/span>, o per\u00edodo vale <\/b><\/span><\/span>T = \"\"<\/b><\/span><\/span> =<\/b><\/span><\/span><\/span> \"\" \"\" <\/b><\/span><\/span><\/span>T = 0,25 s (tempo que demora para efetuar uma volta completa). <\/b><\/span><\/span><\/span><\/span><\/p>\n

Se, <\/b><\/span><\/span>para efetuar uma volta completa a h\u00e9lice demora T = 0,25 s,<\/b><\/span><\/span> para <\/b><\/span><\/span>efetuar 12 voltas completas<\/b><\/span><\/span> ela demorar\u00e1 \"\"<\/b><\/span><\/span> <\/b>= 12×0,25 <\/b><\/span><\/span>= 3 s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como o avi\u00e3o se desloca \"\" com velocidade escalar constante V = 2 m\/s em \"\"<\/b><\/span><\/span> <\/b>= 3 s \"\" V = \"\"<\/b><\/span><\/span> \"\"<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

2 =<\/b><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206S = 6 m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span><\/span><\/p>\n

Velocidade angular (W) de um MCU<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

\u00a0<\/span>Considere um<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>m\u00f3vel em MCU<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>de modo que no<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>instante t<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span>sua posi\u00e7\u00e3o seja determinada pelo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00e2ngulo<\/b><\/span><\/span> <\/b>de fase \u03c6<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0e num<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>outro instante t\u00a0(t > t<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>)\u00a0pelo \u00e2ngulo de fase \u03c6.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, observa-se que,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>no intervalo de tempo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u0394t = t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>raio<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>que<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>acompanha<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>o m\u00f3vel em seu movimento<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>descreveu \u201cvarreu\u201d um \u00e2ngulo \u0394\u03c6 = \u03c6 \u2013 \u03c6<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

C\u00e1lculo da <\/b><\/span><\/span>velocidade V\u2019 de rota\u00e7\u00e3o <\/b><\/span><\/span>de um ponto da <\/b><\/span><\/span>extremidade R = 1m de uma das p\u00e1s <\/b><\/span><\/span>da h\u00e9lice \"\" <\/b><\/span><\/span>W = \"\"<\/b><\/span><\/span> <\/b> \"\"<\/b><\/span><\/span><\/span> <\/b>24 = \"\"<\/b><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>V\u2019 = 24 m\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

Acontece que a <\/b><\/span><\/span>velocidade vetorial, em rela\u00e7\u00e3o ao solo, \u00e9 a <\/b><\/span><\/span>soma vetorial<\/b><\/span><\/span> <\/b>da <\/b><\/span><\/span>velocidade <\/b><\/span><\/span>devida \u00e0 <\/b><\/span><\/span>rota\u00e7\u00e3o <\/b><\/span><\/span>na extremidade da h\u00e9lice <\/b><\/span><\/span>(V\u2019 = 24 m\/s) <\/b><\/span><\/span>com a <\/b><\/span><\/span>velocidade de transla\u00e7\u00e3o <\/b><\/span><\/span>do avi\u00e3o <\/b><\/span><\/span>(V = 2 m\/s), pois ambas s\u00e3o perpendiculares.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/a> V<\/b><\/span><\/span>R<\/b><\/span><\/sub><\/span>2<\/b><\/span><\/sup><\/span> = V<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> + V\u2019<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> <\/b>= 2<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> + 24<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> = 4 + 576 = 580 \"\" V<\/b><\/span><\/span>R<\/b><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span>R<\/b><\/span><\/sub><\/span> \"\" 24,1 m\/s<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

02- <\/span><\/span><\/span>a) <\/b><\/span><\/span>Considerando o sistema \u201cproj\u00e9til, saco de areia e carrinho\u201d<\/b><\/span><\/span> isolado de <\/b><\/span><\/span>for\u00e7as externas<\/b><\/span><\/span> voc\u00ea pode aplicar o <\/b><\/span><\/span>teorema da conserva\u00e7\u00e3o da quantidade de movimento.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span>sa<\/b><\/span><\/sub><\/span> = Q<\/b><\/span><\/span>sd<\/b><\/span><\/sub><\/span> \"\" 10 = 1,6 + 100V \"\" V = \"\"<\/b><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>V = 8,4.10<\/b><\/span><\/span>-2<\/b><\/span><\/sup><\/span> m\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span><\/span><\/p>\n

Teorema da energia cin\u00e9tica:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u201c\u00a0O<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>trabalho da resultant<\/b><\/span><\/span>e<\/b><\/span>\u00a0<\/b><\/span><\/span><\/span>de<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>todas as for\u00e7as<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>que agem sobre um corpo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 igual \u00e0 varia\u00e7\u00e3o da energia cin\u00e9tica sofrida pelo corpo \u201c<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W<\/b><\/span><\/span>FR<\/b><\/span><\/sub><\/span> = \u2206E<\/b><\/span><\/span>c<\/b><\/span><\/sub><\/span> = E<\/b><\/span><\/span>cd<\/b><\/span><\/sub><\/span> \u2013 E<\/b><\/span><\/span>ca<\/b><\/span><\/sub><\/span> = m<\/b><\/span><\/span>p<\/b><\/span><\/sub><\/span>.(V<\/b><\/span><\/span>pd<\/b><\/span><\/sub><\/span>)<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> 2 – m<\/b><\/span><\/span>p<\/b><\/span><\/sub><\/span>.(V<\/b><\/span><\/span>pa<\/b><\/span><\/sub><\/span>)<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> 2<\/b><\/span><\/span> \"\" W<\/b><\/span><\/span>FR<\/b><\/span><\/sub><\/span> = 20.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>.80<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> 2 – 20.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>.500<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> 2 = 64 \u2013 2500 \"\" <\/b><\/span><\/span>W<\/b><\/span><\/span>FR<\/b><\/span><\/sub><\/span> = – 2435 J<\/b><\/span><\/span> (o sinal negativo significa que a energia cin\u00e9tica do proj\u00e9til foi dissipada).<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

03- <\/span><\/span><\/span>a) <\/span><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de Clapeyron ou Equa\u00e7\u00e3o de um g\u00e1s ideal<\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Pelo gr\u00e1fico:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ponto A \"\" P<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span> = 2.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span> N\/m<\/b><\/span><\/span>2 \"\" <\/b><\/span><\/sup><\/span>V<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span> = 2.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span> m<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span> \"\" T<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span><\/span><\/span><\/p>\n

Ponto B \"\" P<\/b><\/span><\/span>B<\/b><\/span><\/sub><\/span> = 5.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span> N\/m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> \"\" V<\/b><\/span><\/span>B<\/b><\/span><\/sub><\/span> = 6.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span> m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> \"\" T<\/b><\/span><\/span>B<\/b><\/span><\/sub><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) <\/b><\/span><\/span><\/span><\/span><\/p>\n

Primeiro Princ\u00edpio da Termodin\u00e2mica ou Princ\u00edpio da Conserva\u00e7\u00e3o da energia<\/span><\/span><\/span><\/span><\/p>\n

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

O<\/span><\/span>\u00a0<\/b><\/span><\/span>Primeiro Princ\u00edpio da Termodin\u00e2mica (Princ\u00edpio da Conserva\u00e7\u00e3o da Energia)<\/span><\/span> afirma que:<\/span><\/span>\u00a0<\/b><\/span><\/span>\u201cA energia n\u00e3o pode ser criada nem destru\u00edda, mas apenas transformada\u201d<\/span><\/span><\/span><\/span><\/p>\n

Exemplo num\u00e9rico:<\/span><\/span><\/span><\/span><\/p>\n

Suponha que um<\/span><\/span> sistema isolado receba Q = 300 J de calor. <\/span><\/span>Se, por exemplo,<\/span><\/span> \u2206U = 50 J dessa energia forem absorvidos pelo sistema aumentando sua energia interna de 50 J, <\/span><\/span>ent\u00e3o a parte restante,<\/span><\/span> W = 250 J, ser\u00e1 fornecida ao ambiente sob forma de trabalho.<\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ent\u00e3o, Q = W + \u2206U<\/b><\/span><\/span> ou <\/span><\/span>\u2206U = Q \u2013 W.<\/span><\/span><\/span><\/span><\/p>\n

Como nas transforma\u00e7\u00f5es <\/b><\/span><\/span>ACDEB e AFB os estados inicial (A, de temperatura T<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span>)) e final (B, de<\/b><\/span><\/span> <\/b>temperatura T<\/b><\/span><\/span>B<\/b><\/span><\/sub><\/span>) coincidem,<\/b><\/span><\/span> a <\/b><\/span><\/span>varia\u00e7\u00e3o de energia interna \u2206U ser\u00e1 a mesma nas duas transforma\u00e7\u00f5es<\/b><\/span><\/span> <\/b><\/span><\/span><\/p>\n

\"\" \u2206U<\/b><\/span><\/span>ACDEB<\/b><\/span><\/sub><\/span> = \u2206U<\/b><\/span><\/span>AFB<\/b><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u2206U<\/b><\/span><\/span>ACDEB<\/b><\/span><\/sub><\/span> = Q<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> \u2013 W<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> e \u2206U<\/b><\/span><\/span>AFB<\/b><\/span><\/sub><\/span> = Q<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> \u2013 W<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> \"\" Q<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> \u2013 Q<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> = W<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> \u2013 W<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Mas, <\/b><\/span><\/span>em<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>toda transforma\u00e7\u00e3o <\/b><\/span><\/span>representada no<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>diagrama PxV, o trabalho realizado<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 numericamente igual \u00e0 \u00e1rea.<\/b><\/span><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do W<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> no trecho ACDEB:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do W<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> no trecho AFB:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> \u2013 Q<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> = W<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> \u2013 W<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> = 17.10<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> \u2013 8.0<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> \u2013 Q<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> = 9.10<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span> J<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

 <\/p>\n

04- <\/span><\/span><\/span>a) <\/b><\/span><\/span>O <\/b><\/span><\/span>\u00edndice de refra\u00e7\u00e3o absoluto<\/b><\/span><\/span> aplicado ao <\/b><\/span><\/span>material <\/b><\/span><\/span>do <\/b><\/span><\/span>cilindro<\/b><\/span><\/span> constitu\u00eddo de um <\/b><\/span><\/span>meio homog\u00eaneo e transparente \u00e9 <\/b><\/span><\/span>fornecido por:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

n = 1,6 e c = 3.10<\/b><\/span><\/span>8<\/b><\/span><\/sup><\/span> m\/s \"\" n =<\/b><\/span><\/span> \"\" \"\" <\/b><\/span><\/span><\/span>1,6 =<\/b><\/span><\/span> \"\" \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span> <\/b>=<\/b><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>V = 1,875 m\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Aplicando a <\/b><\/span><\/span>lei de Snell-Descartes<\/b><\/span><\/span> na refra\u00e7\u00e3o do feixe de luz em A, <\/b><\/span><\/span>do ar para o cilindro <\/b><\/span><\/span> \"\" <\/b><\/span><\/span>n<\/b><\/span><\/span>ar<\/b><\/span><\/sub><\/span>.seni = n<\/b><\/span><\/span>.<\/b><\/span><\/sub><\/span>senr<\/b><\/span><\/span> \"\" 1.sen53<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span> = 1,6.senr \"\" 1.0,8 = 1,6.senr \"\" senr = \"\"<\/b><\/span><\/span> <\/b>= 0,5 \"\" <\/b><\/span><\/span>r = 30<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe atentamente a figura abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se o<\/b><\/span><\/span> \u00e2ngulo de incid\u00eancia <\/b><\/span><\/span>(com a normal N)<\/b><\/span><\/span> em A \u00e9 de 53<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>, <\/b><\/span><\/span>o <\/b><\/span><\/span>\u00e2ngulo de emerg\u00eancia em B <\/b><\/span><\/span>(com a<\/b><\/span><\/span> <\/b>normal N)<\/b><\/span><\/span> tamb\u00e9m ser\u00e1 de 53<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>, <\/b><\/span><\/span>pois os meios<\/b><\/span><\/span> s\u00e3o os mesmos.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como o tri\u00e2ngulo AOB \u00e9 is\u00f3sceles (dois lados iguais R) ent\u00e3o seus \u00e2ngulos opostos s\u00e3o iguais (30<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span>).<\/b><\/span><\/span><\/span><\/span><\/p>\n

No<\/b><\/span><\/span> tri\u00e2ngulo ABC o \u00e2ngulo <\/b><\/span><\/span>\u03b1<\/b><\/span><\/span><\/span> \u00e9 externo <\/b><\/span><\/span>o que implica que<\/b><\/span><\/span> <\/b>\u03b1 <\/b><\/span><\/span><\/span>= 2×23 <\/b><\/span><\/span>= 46<\/b><\/span><\/span>o<\/b><\/span><\/sup><\/span><\/span><\/span><\/p>\n

 <\/p>\n

05- <\/span><\/span><\/span>a)<\/b><\/span><\/span> <\/b>C\u00e1lculo da resist\u00eancia equivalente R<\/b><\/span><\/span>eqA<\/b><\/span><\/sub><\/span> quando a chave est\u00e1 <\/b><\/span><\/span>ligada<\/b><\/span><\/span> no <\/b><\/span><\/span>ponto A<\/b><\/span><\/span> e a corrente <\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

\"\"el\u00e9trica passando apenas pelos dois resistores que est\u00e3o em paralelo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

b) <\/b><\/span><\/span>Se a<\/b><\/span><\/span> densidade da \u00e1gua \u00e9 d = 1 kg\/L, <\/b><\/span><\/span>isso significa que o <\/b><\/span><\/span>volume de 1 L de \u00e1gua possui massa de m = 1 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, o<\/b><\/span><\/span> volume de 4L <\/b><\/span><\/span>de \u00e1gua equivale \u00e0<\/b><\/span><\/span> massa de m = 4 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Da equa\u00e7\u00e3o fundamental da calorimetria \"\" Q = m.c.\u2206\u03b8 = 4.1.(80 \u2013 20) = 4.60 \"\" Q = W = 240 kcal.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Com a chave em A \"\" P<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span> = \"\"= \"\"<\/b><\/span><\/span>= \"\"<\/b><\/span><\/span><\/span> \"\" P<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span> = 400 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>2.40 =<\/b><\/span><\/span> \"\" \"\"<\/b><\/span><\/span><\/span> R =<\/b><\/span><\/span> \"\" \"\" <\/b><\/span><\/span><\/span>R =<\/b><\/span><\/span> <\/b>18 \u2126<\/b><\/span><\/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 UNIFESP017 \u2013 1a fase 01- a) Se a freq\u00fc\u00eancia da h\u00e9lice \u00e9 de f = 4 Hz, o per\u00edodo vale T = = T = 0,25 s (tempo que demora para efetuar uma volta completa). Se, para efetuar uma volta completa a h\u00e9lice demora T = 0,25 s, para efetuar 12 voltas completas ela demorar\u00e1 = 12×0,25<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2703,"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-2706","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2706","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=2706"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2706\/revisions"}],"predecessor-version":[{"id":2708,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2706\/revisions\/2708"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2703"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2706"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}