{"id":4779,"date":"2019-11-07T02:59:47","date_gmt":"2019-11-07T02:59:47","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4779"},"modified":"2024-07-10T13:05:23","modified_gmt":"2024-07-10T13:05:23","slug":"cederj-meio-do-ano-2019-2020","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2019\/cederj-meio-do-ano-2019-2020\/","title":{"rendered":"CEDERJ \u2013 meio do ano \u2013 2019\/2020"},"content":{"rendered":"

CEDERJ \u2013 meio do ano \u2013 2019\/2020<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>Funda\u00e7\u00e3o Centro de Educa\u00e7\u00e3o a Dist\u00e2ncia do Estado do Rio de Janeiro<\/b><\/span><\/span><\/span>\u00a0(<\/b><\/span><\/span><\/span>Cecierj<\/b><\/span><\/span><\/span>) desenvolve projetos nas \u00e1reas de educa\u00e7\u00e3o superior a dist\u00e2ncia e divulga\u00e7\u00e3o cient\u00edfica, atingindo diretamente mais de 60 mil pessoas\/ano residentes nos 92 munic\u00edpios do Estado do Rio de Janeiro, com a oferta de:<\/b><\/span><\/span><\/span><\/p>\n

    \n
  1. Cursos de gradua\u00e7\u00e3o a dist\u00e2ncia por interm\u00e9dio do\u00a0Cons\u00f3rcio Cederj<\/a>, parceria formada entre o\u00a0Governo do Estado do Rio de Janeiro<\/a>\u00a0e as seguintes institui\u00e7\u00f5es de ensino:\u00a0Centro Federal de Educa\u00e7\u00e3o Tecnol\u00f3gica Celso Suckow da Fonseca<\/a>\u00a0(CEFET-RJ);\u00a0Universidade Federal do Rio de Janeiro<\/a>\u00a0(UFRJ);\u00a0Universidade do Estado do Rio de Janeiro<\/a>\u00a0(UERJ);\u00a0Universidade Estadual do Norte Fluminense Darcy Ribeiro<\/a>\u00a0(UENF);\u00a0Universidade Federal Fluminense<\/a>\u00a0(<\/b><\/span><\/span><\/span>UFF<\/b><\/span><\/span><\/span>);\u00a0Universidade Federal do Estado do Rio de Janeiro<\/a>\u00a0(UNIRIO); e\u00a0Universidade Federal Rural do Rio de Janeiro<\/a>\u00a0(UFRRJ);<\/b><\/span><\/span><\/span><\/li>\n
  2. Pr\u00e9-Vestibular Social<\/a>\u00a0(PVS), projeto dirigido \u00e0queles que j\u00e1 conclu\u00edram ou que frequentam o \u00faltimo ano do Ensino M\u00e9dio e desejam realizar as provas para ingresso nas universidades, mas n\u00e3o t\u00eam condi\u00e7\u00f5es de arcar com os custos dos cursos preparat\u00f3rios particulares;<\/b><\/span><\/span><\/span><\/li>\n
  3. Forma\u00e7\u00e3o continuada de professores da educa\u00e7\u00e3o b\u00e1sica, com a oferta de cursos modulares em diversas \u00e1reas: Antropologia, Arte e Comunica\u00e7\u00e3o, Biologia, Educa\u00e7\u00e3o em Ci\u00eancias, Educa\u00e7\u00e3o Especial e Inclusiva, F\u00edsica, Inform\u00e1tica Educativa, Geoci\u00eancias, Governan\u00e7a: Gest\u00e3o, Auditoria e TI, Letras, Matem\u00e1tica e Qu\u00edmica, sob responsabilidade da\u00a0Diretoria de Extens\u00e3o<\/a>; e<\/b><\/span><\/span><\/span><\/li>\n
  4. Projetos de\u00a0Divulga\u00e7\u00e3o Cient\u00edfica<\/a>, como Pra\u00e7a da Ci\u00eancia Itinerante, Jovens Talentos para Ci\u00eancia, Lona da Ci\u00eancia, Feira de Ci\u00eancia, Tecnologia e Inova\u00e7\u00e3o (FECTI), entre outros, que visam popularizar o conhecimento cient\u00edfico no Estado do Rio de Janeiro.<\/b><\/span><\/span><\/span><\/li>\n<\/ol>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    01- (CEDERJ \u2013 RJ \u2013 meio do ano \u2013 019\/029)<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Uma part\u00edcula e um detector movimentam-se sobre uma linha reta, <\/b><\/span><\/span><\/span>que define <\/b><\/span><\/span><\/span>o eixo de coordenadas x. <\/b><\/span><\/span><\/span><\/p>\n

    Quan\u00addo a part\u00edcula encontra-se na posi\u00e7\u00e3o x<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span><\/span><\/span> <\/b>= 0 m, <\/b><\/span><\/span><\/span>com uma <\/b><\/span><\/span><\/span>velocidade positiva de 10 m\/s, o<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    detector est\u00e1 na posi\u00e7\u00e3o x = 24m com velocidade, tamb\u00e9m positiva, de 4 m\/s. <\/b><\/span><\/span><\/span><\/p>\n

    As <\/b><\/span><\/span><\/span>com\u00adponentes (x) da posi\u00e7\u00e3o e (v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span><\/span><\/span>) da velocidade da part\u00edcula ao longo do eixo x, medidas em rela\u00e7\u00e3o ao detector, <\/b><\/span><\/span><\/span>s\u00e3o dadas por: <\/b><\/span><\/span><\/span><\/p>\n

    (A) <\/span><\/span><\/span><\/b><\/span><\/span>x = – 24 m, v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span><\/span><\/span> <\/b>= 14 m\/s <\/b><\/span><\/span><\/span><\/p>\n

    (B) <\/span><\/span><\/span><\/b><\/span><\/span>x = – 24 m, v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span><\/span><\/span> <\/b>= 6 m\/s <\/b><\/span><\/span><\/span><\/p>\n

    (C) <\/span><\/span><\/span><\/b><\/span><\/span>x = 24 m, v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span><\/span><\/span> = – 1 4 m\/s <\/b><\/span><\/span><\/span><\/p>\n

    (D) <\/span><\/span><\/span><\/b><\/span><\/span>x = 24 m, v<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span><\/span><\/span> = – 6 m\/s <\/b><\/span><\/span><\/span><\/p>\n

    02- (CEDERJ \u2013 RJ \u2013 meio do ano \u2013 019\/029)<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Um objeto se afasta de um observador em alta ve\u00adlocidade. <\/b><\/span><\/span><\/span>Ele <\/b><\/span><\/span><\/span>emite uma luz amarela<\/b><\/span><\/span><\/span> que, <\/b><\/span><\/span><\/span>no entanto, che\u00adga ao observador em tom de vermelho, devido ao efeito Doppler.<\/b><\/span><\/span><\/span><\/p>\n

    Esse efeito se deve \u00e0 mudan\u00e7a na <\/b><\/span><\/span><\/span><\/p>\n

    (A) <\/span><\/span><\/span><\/b><\/span><\/span>frequ\u00eancia e na velocidade da onda recebida. <\/b><\/span><\/span><\/span><\/p>\n

    (B) <\/span><\/span><\/span><\/b><\/span><\/span>amplitude e na velocidade da onda recebida. <\/b><\/span><\/span><\/span><\/p>\n

    (C) <\/span><\/span><\/span><\/b><\/span><\/span>amplitude e no comprimento da onda recebida. <\/b><\/span><\/span><\/span><\/p>\n

    (D) <\/span><\/span><\/span><\/b><\/span><\/span>frequ\u00eancia e no comprimento da onda recebida. <\/b><\/span><\/span><\/span><\/p>\n

    03- (CEDERJ \u2013 RJ \u2013 meio do ano \u2013 019\/029)<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Uma <\/b><\/span><\/span><\/span>amostra de g\u00e1s ideal est\u00e1 contida em um cilindro met\u00e1lico<\/b><\/span><\/span><\/span> que <\/b><\/span><\/span><\/span>est\u00e1 imerso em uma grande quantidade de \u00e1gua a temperatura constante.<\/b><\/span><\/span><\/span> O <\/b><\/span><\/span><\/span>g\u00e1s \u00e9 comprimido lenta\u00admente numa transforma\u00e7\u00e3o isot\u00e9rmica. <\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    A <\/b><\/span><\/span><\/span>varia\u00e7\u00e3o da ener\u00adgia interna do g\u00e1s (\u0394U) e o trabalho W realizado pelo g\u00e1s nesse processo s\u00e3o: <\/b><\/span><\/span><\/span><\/p>\n

    (A) <\/span><\/span><\/span><\/span><\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>U = 0; W > 0 <\/b><\/span><\/span><\/span><\/span><\/p>\n

    (B) <\/span><\/span><\/span><\/span><\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>U = 0; W < 0 <\/b><\/span><\/span><\/span><\/span><\/p>\n

    (C) <\/span><\/span><\/span><\/span><\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>U > 0; W > 0 <\/b><\/span><\/span><\/span><\/span><\/p>\n

    (D) <\/span><\/span><\/span><\/b><\/span><\/span>\u0394U < 0; W < 0 <\/b><\/span><\/span><\/span><\/p>\n

    04- (CEDERJ \u2013 RJ \u2013 meio do ano \u2013 019\/029)<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Um <\/b><\/span><\/span><\/span>cubo de gelo encontra-se em repouso flutuando na \u00e1gua. <\/b><\/span><\/span><\/span>Considere que <\/b><\/span><\/span><\/span>P e E representam,<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    respectivamen\u00adte, os <\/b><\/span><\/span><\/span>m\u00f3dulos <\/b><\/span><\/span><\/span>do peso do cubo de gelo e do empuxo que a \u00e1gua exerce sobre o gelo.<\/b><\/span><\/span><\/span> <\/b>F denomina o <\/b><\/span><\/span><\/span>m\u00f3dulo <\/b><\/span><\/span><\/span>da resul\u00adtante das for\u00e7as que atuam no cubo de gelo.<\/b><\/span><\/span><\/span> Ent\u00e3o: <\/b><\/span><\/span><\/span><\/p>\n

    (A) <\/span><\/span><\/span><\/b><\/span><\/span>F = (E + P) > 0 <\/b><\/span><\/span><\/span><\/p>\n

    (B) <\/span><\/span><\/span><\/b><\/span><\/span>F= P; E = 0 <\/b><\/span><\/span><\/span><\/p>\n

    (C) <\/span><\/span><\/span><\/b><\/span><\/span>F = 0; E = P <\/b><\/span><\/span><\/span><\/p>\n

    (D) <\/span><\/span><\/span><\/b><\/span><\/span>F < P < E<\/b><\/span><\/span><\/span><\/p>\n

    05- (CEDERJ \u2013 RJ \u2013 meio do ano \u2013 019\/029)<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Um <\/b><\/span><\/span><\/span>foguete de massa M viaja pelo espa\u00e7o sideral com velocidade <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span> Em <\/b><\/span><\/span><\/span>um dado instante, um peda\u00e7o do foguete, com um d\u00e9cimo de sua massa, desprende-se e continua movimentando-se com a mesma velocidade origi\u00adnal <\/b><\/span><\/span><\/span>V <\/b><\/span><\/span><\/span>do foguete.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

    <\/p>\n

    Ap\u00f3s o desprendimento desse peda\u00e7o, o que restou do foguete viaja com velocidade <\/b><\/span><\/span><\/span><\/p>\n

    (A) <\/span><\/span><\/span><\/b><\/span><\/span>v<\/b><\/span><\/span><\/span>\u2019 = <\/b><\/span><\/span><\/span>V <\/b><\/span><\/span><\/span><\/p>\n

    (B) <\/span><\/span><\/span><\/span><\/b><\/span><\/span>v<\/b><\/span><\/span><\/span><\/span>\u2019 = (9\/10)<\/b><\/span><\/span><\/span><\/span>V <\/b><\/span><\/span><\/span><\/span><\/p>\n

    (C) <\/span><\/span><\/span><\/span><\/b><\/span><\/span>v<\/b><\/span><\/span><\/span><\/span>\u2019 = (10\/9)<\/b><\/span><\/span><\/span><\/span>V <\/b><\/span><\/span><\/span><\/span><\/p>\n

    (D) <\/span><\/span><\/span><\/span><\/b><\/span><\/span>v<\/b><\/span><\/span><\/span><\/span>\u2019 = (1\/10)<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span><\/p>\n

    06- (CEDERJ \u2013 RJ \u2013 meio do ano \u2013 019\/029)<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Uma <\/b><\/span><\/span><\/span>bilha de a\u00e7o com massa m = 3,0×10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span> <\/b>kg encontra-se em repouso sobre uma mola de massa des\u00adprez\u00edvel <\/b><\/span><\/span><\/span>que est\u00e1 <\/b><\/span><\/span><\/span>comprimida de 5,0×10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span> <\/b>m.<\/b><\/span><\/span><\/span> <\/b>A mola \u00e9 libera\u00adda impulsionando verticalmente a bilha para cima.<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Considerando que a <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o da gravidade vale g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span> <\/b>e sabendo que a <\/b><\/span><\/span><\/span>constante el\u00e1stica da mola \u00e9 k = 4,8 x 10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span> <\/b>N\/m, <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>altura m\u00e1xima atingida pela bilha a partir do repouso <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span><\/p>\n

    (A) <\/span><\/span><\/span><\/b><\/span><\/span>2,0 m <\/b><\/span><\/span><\/span>(B) <\/span><\/span><\/span><\/b><\/span><\/span>4,0 m <\/b><\/span><\/span><\/span>(C) <\/span><\/span><\/span><\/b><\/span><\/span>6,0 m <\/b><\/span><\/span><\/span>(D) <\/span><\/span><\/span><\/b><\/span><\/span>8,0 m <\/b><\/span><\/span><\/span><\/p>\n

    07- (CEDERJ \u2013 RJ \u2013 meio do ano \u2013 019\/029)<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Um <\/b><\/span><\/span><\/span>pedreiro utiliza uma roldana fixa (ideal)<\/b><\/span><\/span><\/span> e uma <\/b><\/span><\/span><\/span>corda de massa desprez\u00edvel <\/b><\/span><\/span><\/span>para <\/b><\/span><\/span><\/span>descer verticalmente um balde de concreto de massa m = 20 kg.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

    \"CEDERJ\"<\/p>\n

    O <\/b><\/span><\/span><\/span>pedreiro controla a corda de modo a que o balde des\u00e7a com uma acelera\u00ad\u00e7\u00e3o constante a = g\/20<\/b><\/span><\/span><\/span>, onde <\/b><\/span><\/span><\/span>g representa a acelera\u00e7\u00e3o da gravidade, <\/b><\/span><\/span><\/span>cujo <\/b><\/span><\/span><\/span>valor no local \u00e9 de 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span>Nesse caso, a <\/b><\/span><\/span><\/span>intensidade da tens\u00e3o T na corda <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span><\/p>\n

    (A) <\/span><\/span><\/span><\/b><\/span><\/span>1,0 x 10<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span>N <\/b><\/span><\/span><\/span><\/p>\n

    (B) <\/span><\/span><\/span><\/b><\/span><\/span>1,9 x 10<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span>N <\/b><\/span><\/span><\/span><\/p>\n

    (C) <\/span><\/span><\/span><\/b><\/span><\/span>2,0 x 10<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span>N <\/b><\/span><\/span><\/span><\/p>\n

    (D) <\/span><\/span><\/span><\/b><\/span><\/span>2,1 x 10<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span>N <\/b><\/span><\/span><\/span><\/p>\n

    08- (CEDERJ \u2013 RJ \u2013 meio do ano \u2013 019\/029)<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Disjuntor \u00e9 um dispositivo de seguran\u00e7a que impede a passagem de corrente el\u00e9trica em um circuito quando ela atinge um valor superior ao recomendado. <\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Um <\/b><\/span><\/span><\/span>chuveiro el\u00e9\u00adtrico, cuja pot\u00eancia \u00e9 de 5,5×10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span><\/span><\/span><\/span> <\/b>W, <\/b><\/span><\/span><\/span>necessita ser <\/b><\/span><\/span><\/span>instalado em uma resid\u00eancia onde a tens\u00e3o \u00e9 de 220 V.<\/b><\/span><\/span><\/span> O <\/b><\/span><\/span><\/span>disjuntor adequado para a instala\u00e7\u00e3o desse chuveiro deve ser <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span><\/p>\n

    (A) <\/span><\/span><\/span><\/b><\/span><\/span>10 A <\/b><\/span><\/span><\/span>(B) <\/span><\/span><\/span><\/b><\/span><\/span>15 A <\/b><\/span><\/span><\/span>(C) <\/span><\/span><\/span><\/b><\/span><\/span>20 A <\/b><\/span><\/span><\/span>(D) <\/span><\/span><\/span><\/b><\/span><\/span>30 A <\/b><\/span><\/span><\/span><\/p>\n

    Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica da CEDERJ realizada e Julho\/2019<\/span><\/span><\/span><\/p>\n

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

    Orientando a trajet\u00f3ria para direita, ambas as velocidades que s\u00e3o positivas ser\u00e3o para a direita <\/b><\/span><\/span><\/span>(movimento progressivo, a favor dos marcos crescentes da trajet\u00f3ria) <\/b><\/span><\/span><\/span>a velocidade <\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    relativa de aproxima\u00e7\u00e3o entre eles ser\u00e1 o m\u00f3dulo da diferen\u00e7a entre as velocidades<\/b><\/span><\/span><\/span> (se movem no mesmo sentido, se fosse em sentido contr\u00e1rio seria a soma). <\/b><\/span><\/span><\/span><\/p>\n

    Assim V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> <\/b>= 10 \u2013 4 <\/b><\/span><\/span><\/span>= 6 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

    Como a part\u00edcula se encontra 24 m \u00e0 esquerda do detetor, sua posi\u00e7\u00e3o em rela\u00e7\u00e3o \u00e0 ele ser\u00e1 X = – 24 m.<\/b><\/span><\/span><\/span><\/p>\n

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

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

    Efeito Doppler<\/b><\/span><\/span><\/span><\/p>\n

    Refere-se \u00e0\u00a0<\/b><\/span><\/span><\/span>varia\u00e7\u00e3o da freq\u00fc\u00eancia notada<\/b><\/span><\/span><\/span>\u00a0por um\u00a0<\/b><\/span><\/span><\/span>observador<\/b><\/span><\/span><\/span>\u00a0quando a\u00a0<\/b><\/span><\/span><\/span>dist\u00e2ncia entre ele e uma fonte de ondas est\u00e1 aumentando ou diminuindo.<\/b><\/span><\/span><\/span><\/p>\n

    Na\u00a0<\/b><\/span><\/span><\/span>aproxima\u00e7\u00e3o entre fonte e observador,\u00a0<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/span>mesmo perceber\u00e1 o som emitido pela fonte mais agudo\u00a0<\/b><\/span><\/span><\/span>(maior freq\u00fc\u00eancia, recebe maior n\u00famero de frentes de onda na unidade de tempo) do que\u00a0<\/b><\/span><\/span><\/span>perceberia se fonte e observador estivessem parados.<\/b><\/span><\/span><\/span><\/p>\n

    Nesse caso, o\u00a0<\/b><\/span><\/span><\/span>comprimento de onda aparente percebido pelo observador ser\u00e1 menor\u00a0<\/b><\/span><\/span><\/span>que o\u00a0<\/b><\/span><\/span><\/span>comprimento da onda emitido pela fonte (observador O1\u00a0das figuras abaixo).<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    No\u00a0<\/b><\/span><\/span><\/span>afastamento entre fonte e observador,\u00a0<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/span>mesmo perceber\u00e1 o som emitido pela fonte mais grave\u00a0<\/b><\/span><\/span><\/span>(menor freq\u00fc\u00eancia, recebe menor n\u00famero de frentes de onda na unidade de tempo) do que\u00a0<\/b><\/span><\/span><\/span>perceberia se fonte e observador estivessem parados.<\/b><\/span><\/span><\/span><\/p>\n

    Nesse caso, o\u00a0<\/b><\/span><\/span><\/span>comprimento de onda aparente percebido pelo observador ser\u00e1 maior que o comprimento da onda emitido pela fonte (observador O2\u00a0das figuras acima).<\/b><\/span><\/span><\/span><\/p>\n

    Observe que o\u00a0<\/b><\/span><\/span><\/span>motorista da ambul\u00e2ncia n\u00e3o percebe nenhuma altera\u00e7\u00e3o no som emitido pela sirene, pois eles se movem juntos.<\/b><\/span><\/span><\/span><\/p>\n

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

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

    Transforma\u00e7\u00e3o isot\u00e9rmica<\/b><\/span><\/span><\/span><\/p>\n

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

    Como a\u00a0<\/b><\/span><\/span><\/span>temperatura (T) e a energia interna (U) de um sistema\u00a0<\/b><\/span><\/span><\/span>est\u00e3o associadas \u00e0\u00a0<\/b><\/span><\/span><\/span>energia cin\u00e9tica das mol\u00e9culas,\u00a0<\/b><\/span><\/span><\/span>se a\u00a0<\/b><\/span><\/span><\/span>temperatura for constante (transforma\u00e7\u00e3o isot\u00e9rmica), a energia interna (U) do sistema tamb\u00e9m ser\u00e1 constante.<\/b><\/span><\/span><\/span><\/p>\n

    Assim, a\u00a0<\/b><\/span><\/span><\/span>varia\u00e7\u00e3o de energia interna (\u0394U) do sistema ser\u00e1 nula.<\/b><\/span><\/span><\/span><\/p>\n

    Transforma\u00e7\u00e3o isot\u00e9rmica\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394U = 0\u00a0<\/b><\/span><\/span><\/span>\u00a0 \u0394U = Q \u2013 W\u00a0\u00a0\u00a0 0 = Q \u2013 W\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>Q = W.<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Q = W<\/b><\/span><\/span><\/span>\u00a0\u00a0Portanto, numa\u00a0<\/b><\/span><\/span><\/span>transforma\u00e7\u00e3o isot\u00e9rmica\u00a0<\/b><\/span><\/span><\/span>(temperatura constante)\u00a0<\/b><\/span><\/span><\/span>todo calor Q recebido pelo sistema \u00e9 totalmente transformado em trabalho.<\/b><\/span><\/span><\/span><\/p>\n

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

    Como o <\/b><\/span><\/span><\/span>g\u00e1s \u00e9 comprimido <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>trabalho realizado \u00e9 negativo. (W < 0).<\/b><\/span><\/span><\/span><\/p>\n

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

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

    Teorema de Arquimedes \u2013 Empuxo<\/b><\/span><\/span><\/span><\/p>\n

    Enunciado do princ\u00edpio de Arquimedes<\/b><\/span><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Express\u00e3o matem\u00e1tica do Empuxo<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    <\/p>\n

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

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

    S\u00e3o duas partes, m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 9M\/10 e m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 1M\/10. <\/b><\/span><\/span><\/span><\/p>\n

    Quantidade de movimento do sistema antes do peda\u00e7o se desprender Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = M.V.<\/b><\/span><\/span><\/span><\/p>\n

    Quantidade de movimento do sistema depois onde o peda\u00e7o de massa m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 1.M\/10 se desprende com velocidade V e o outro de massa <\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 9m\/10 continua com velocidade V\u2019.<\/b><\/span><\/span><\/span><\/p>\n

    Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = (1M\/10).V + (9M\/10).V\u2019.<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Teorema da conserva\u00e7\u00e3o da quantidade de movimento Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> M.V = (1M\/10).V + (9M\/10).V\u2019 <\/b><\/span><\/span><\/span> MV \u2013 1MV\/10 = 9MV\u2019\/10 (10MV \u2013 1MV)\/10 = 9MV\u2019\/10 9V\/10 = 9V\u2019\/10 <\/b><\/span><\/span><\/span> V\u2019 = V<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

    <\/a> R- A
    \n<\/b><\/span><\/span><\/span>06-<\/span><\/span><\/span><\/p>\n

    Colocando o n\u00edvel zero de altura no ponto A e aplicando o teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica nos pontos A e B.<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    No ponto A V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 0<\/b><\/span><\/span><\/span> (partiu do repouso) e <\/b><\/span><\/span><\/span>no ponto B V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 0 <\/b><\/span><\/span><\/span>(altura m\u00e1xima).<\/b><\/span><\/span><\/span><\/p>\n

    Ponto A E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>pgA<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>peA<\/b><\/span><\/span><\/sub><\/span> <\/b>= m.g.h<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> + m.V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + k.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = m.g.0 + m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> <\/b>+ 4,8.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>. (5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = 0 + 0 + 4,8.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.25.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\/2 <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = 60.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> J<\/b><\/span><\/span><\/span><\/p>\n

    Ponto B E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>pgB<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>peB<\/b><\/span><\/span><\/sub><\/span> <\/b>= m.g.h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> + m.V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + k.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 3.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.10.h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> + m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> <\/b>+ 0 E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = 30.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> + 0 + 0 <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = 30.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.h<\/b><\/span><\/span><\/span>B.<\/b><\/span><\/span><\/sub><\/span><\/p>\n

    E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span> 60.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> = 30.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> =60\/30 <\/b><\/span><\/span><\/span>= 2 m.<\/b><\/span><\/span><\/span><\/p>\n

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

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

    A <\/b><\/span><\/span><\/span>for\u00e7a que o pedreiro aplica na corda \u00e9 a for\u00e7a de tens\u00e3o<\/b><\/span><\/span><\/span> (tra\u00e7\u00e3o) <\/b><\/span><\/span><\/span>T na mesma que \u00e9 transmitida at\u00e9 o balde.<\/b><\/span><\/span><\/span><\/p>\n

    As <\/b><\/span><\/span><\/span>for\u00e7as que agem sobre o balde na descida s\u00e3o T<\/b><\/span><\/span><\/span> (vertical e para cima) e <\/b><\/span><\/span><\/span>o peso P <\/b><\/span><\/span><\/span>(vertical e para baixo).<\/b><\/span><\/span><\/span><\/p>\n

    <\/p>\n

    Como o <\/b><\/span><\/span><\/span>balde desce com acelera\u00e7\u00e3o a<\/b><\/span><\/span><\/span> = 9\/20 = 10\/<\/b><\/span><\/span><\/span>20 = 0,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>a <\/b><\/span><\/span><\/span>for\u00e7a resultante F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> sobre o mesmo \u00e9 para baixo e P > T.<\/b><\/span><\/span><\/span><\/p>\n

    Segunda lei de Newton F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = ma P \u2013 T = m.a <\/b><\/span><\/span><\/span> m.g \u2013 T = m.a 20.10 – T = 20.0,5 <\/b><\/span><\/span><\/span><\/p>\n

    200 \u2013 T = 10 T = 200 \u2013 10 <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span> = 190 N <\/b><\/span><\/span><\/span>= 1,9.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> N.<\/b><\/span><\/span><\/span><\/p>\n

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

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

    S\u00e3o dados: P <\/b><\/span><\/span><\/span>= 5,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> W; <\/b><\/span><\/span><\/span>U <\/b><\/span><\/span><\/span>= 220 V e <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span> = ?<\/b><\/span><\/span><\/span><\/p>\n

    P = i.U <\/b><\/span><\/span><\/span> 5,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> = i.220 i = 5500\/220 <\/b><\/span><\/span><\/span>i = 25 A.<\/b><\/span><\/span><\/span><\/p>\n

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

     <\/p>\n

    Clique aqui para ver sobre o tema!<\/a><\/p>\n

    Confira as v\u00eddeo aulas!<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

    CEDERJ \u2013 meio do ano \u2013 2019\/2020 A\u00a0Funda\u00e7\u00e3o Centro de Educa\u00e7\u00e3o a Dist\u00e2ncia do Estado do Rio de Janeiro\u00a0(Cecierj) desenvolve projetos nas \u00e1reas de educa\u00e7\u00e3o superior a dist\u00e2ncia e divulga\u00e7\u00e3o cient\u00edfica, atingindo diretamente mais de 60 mil pessoas\/ano residentes nos 92 munic\u00edpios do Estado do Rio de Janeiro, com a oferta de: Cursos de gradua\u00e7\u00e3o a dist\u00e2ncia por interm\u00e9dio do\u00a0Cons\u00f3rcio<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4337,"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-4779","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4779","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=4779"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4779\/revisions"}],"predecessor-version":[{"id":10147,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4779\/revisions\/10147"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4337"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4779"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}