{"id":3149,"date":"2017-08-19T03:04:16","date_gmt":"2017-08-19T03:04:16","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3149"},"modified":"2024-09-03T17:04:06","modified_gmt":"2024-09-03T17:04:06","slug":"resolucao-comentada-das-questoes-de-fisica-da-ufpr-pr-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/ufpr-2017\/resolucao-comentada-das-questoes-de-fisica-da-ufpr-pr-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UFPR-PR-2017"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da <\/b><\/span><\/span><\/span><\/p>\n

UFPR-PR-2017<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>Massa do corpo \"\"<\/b><\/span><\/span><\/span> <\/b>m<\/b><\/span><\/span><\/span> = 2000g <\/b><\/span><\/span><\/span>= 2kg<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Peso do corpo \"\"<\/b><\/span><\/span><\/span> <\/b>P = mg <\/b><\/span><\/span><\/span>= 2.10 <\/b><\/span><\/span><\/span>= 20 N.<\/b><\/span><\/span><\/span><\/p>\n

Nas <\/b><\/span><\/span><\/span>figuras abaixo <\/b><\/span><\/span><\/span>est\u00e3o <\/b><\/span><\/span><\/span>colocadas as for\u00e7as <\/b><\/span><\/span><\/span>que agem sobre o sistema e a <\/b><\/span><\/span><\/span>soma vetorial dessas for\u00e7as<\/b><\/span><\/span><\/span> pelo <\/b><\/span><\/span><\/span>m\u00e9todo da linha poligonal <\/b><\/span><\/span><\/span>deve <\/b><\/span><\/span><\/span>fornecer um tri\u00e2ngulo fechado <\/b><\/span><\/span><\/span>(equil\u00edbrio, for\u00e7a resultante nula).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o <\/b><\/span><\/span><\/span>tri\u00e2ngulo da direita \u00e9 ret<\/b><\/span><\/span><\/span>\u00e2ngulo \"\" <\/b><\/span><\/span><\/span>sen 30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 0,5 = \"\" \"\" <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10 N.<\/b><\/span><\/span><\/span><\/p>\n

Observe que a <\/b><\/span><\/span><\/span>intensidade da for\u00e7a de tra\u00e7\u00e3o T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10 N<\/b><\/span><\/span><\/span> \u00e9 a <\/b><\/span><\/span><\/span>for\u00e7a el\u00e1stica <\/b><\/span><\/span><\/span>que deforma a mola <\/b><\/span><\/span><\/span>fornecida por F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = k.x <\/b><\/span><\/span><\/span>onde <\/b><\/span><\/span><\/span>k<\/b><\/span><\/span><\/span> \u00e9 a <\/b><\/span><\/span><\/span>constante el\u00e1stica <\/b><\/span><\/span><\/span>da mola <\/b><\/span><\/span><\/span>k = 20 N\/cm <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>x a deforma\u00e7\u00e3o da mola.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = kx \"\"<\/b><\/span><\/span><\/span> 10 N = 20 N\/cm.x \"\" x = \"\" \"\" <\/b><\/span><\/span><\/span>x = 0,5 cm.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>Dist\u00e2ncia entre o sat\u00e9lite e o receptor \"\" d =<\/b><\/span><\/span><\/span> 30 000 km <\/b><\/span><\/span><\/span>= 3.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span> m \"\"<\/b><\/span><\/span><\/span> <\/b>V =<\/b><\/span><\/span><\/span> 300 000 km\/s <\/b><\/span><\/span><\/span>= 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s \"\"<\/b><\/span><\/span><\/span> <\/b>\u2206t = ?.<\/b><\/span><\/span><\/span><\/p>\n

V = \"\"<\/b><\/span><\/span><\/span> \"\" 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> = \"\" \"\" \"\"3.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>\/ 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> \"\" \"\"= 10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span> <\/b>= 0,1 s.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/b><\/span><\/span><\/span>Observe nas <\/b><\/span><\/span><\/span>sequ\u00eancias das figuras abaixo<\/b><\/span><\/span><\/span> que, tanto na <\/b><\/span><\/span><\/span>Situa\u00e7\u00e3o 1 quanto na Situa\u00e7\u00e3o <\/b><\/span><\/span><\/span>2 as <\/b><\/span><\/span><\/span>resist\u00eancias el\u00e9tricas equivalentes s\u00e3o iguais <\/b><\/span><\/span><\/span>o que implica que a <\/b><\/span><\/span><\/span>intensidade de corrente el\u00e9trica no gerador \u00e9 a mesma nas duas situa\u00e7\u00f5es.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

N\u00e3o existe diferen\u00e7a de potencial entre A e B <\/b><\/span><\/span><\/span>estando os fios <\/b><\/span><\/span><\/span>conectados ou n\u00e3o, <\/b><\/span><\/span><\/span>pois os <\/b><\/span><\/span><\/span>pontos A e B possuem o mesmo potencial devido \u00e0 simetria do circuito.<\/b><\/span><\/span><\/span><\/p>\n

Pelas informa\u00e7\u00f5es acima a \u00fanica alternativa correta \u00e9 a primeira.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/b><\/span><\/span><\/span>S\u00e3o dados:<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> <\/b>= 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>press\u00e3o da bolha ao atingir a superf\u00edcie \u00e9 a press\u00e3o atmosf\u00e9rica P = 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> Pa<\/b><\/span><\/span><\/span>(N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) e, na <\/b><\/span><\/span><\/span>parte inferior P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>volume V na superf\u00edcie \u00e9 V = 50%<\/b><\/span><\/span><\/span> do seu volume em <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o ao volume que tinha quando do in\u00edcio do movimento de subida, ou seja, V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + 0,5V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" V = 1,5V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como <\/b><\/span><\/span><\/span>n\u00e3o houve varia\u00e7\u00e3o significativa de temperatura <\/b><\/span><\/span><\/span>durante a subida <\/b><\/span><\/span><\/span>trata-se <\/b><\/span><\/span><\/span>de uma <\/b><\/span><\/span><\/span>transforma\u00e7\u00e3o isot\u00e9rmica com T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = T. <\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o geral dos gases Perfeit<\/b><\/span><\/span><\/span>os \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V\u00ad<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = P.V\/<\/b><\/span><\/span><\/span>T \"\" P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V\u00ad<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/T = P.V\/T \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = P.V \"\"<\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.1,5V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 1,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> Pa.<\/b><\/span><\/span><\/span><\/p>\n

Teorema de Stevin \"\"<\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = P + d.g.h \"\"<\/b><\/span><\/span><\/span> 1,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> = 1,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> + 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.h \"\" h = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>h =<\/b><\/span><\/span><\/span> 0,5.10<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sup><\/span> = <\/b><\/span><\/span><\/span><\/p>\n

5 m<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/b><\/span><\/span><\/span>Pode-se\u00a0<\/b><\/span><\/span><\/span>transformar uma indica\u00e7\u00e3o de uma escala para outra\u00a0<\/b><\/span><\/span><\/span>conforme o procedimento a seguir, de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

acordo com a\u00a0<\/b><\/span><\/span><\/span>rela\u00e7\u00e3o matem\u00e1tica baseada no teorema de Thales:<\/b><\/span><\/span><\/span><\/p>\n

\"\" = \"\"<\/b><\/span><\/span><\/span> \"\" \"\" = \"\" \"\" \"\" = \"\"<\/b><\/span><\/span><\/span> \"\" 9C = 211,5 \"\" C = \"\" \"\" <\/b><\/span><\/span><\/span>C = 23,5 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 apresentada a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Diagrama de estado \u2013 ponto triplo<\/b><\/span><\/span><\/span><\/p>\n

\u00a0As\u00a0<\/b><\/span><\/span><\/span>curvas do gr\u00e1fico recebem o nome de diagrama de estado ou diagrama de fases. Nele, para uma dada subst\u00e2ncia:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/span>Curva CB\u00a0\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>\u00a0 fornece a\u00a0<\/b><\/span><\/span><\/span>temperatura e a press\u00e3o, de fus\u00e3o ou solidifica\u00e7\u00e3o<\/b><\/span><\/span><\/span>\u00a0e essa\u00a0<\/b><\/span><\/span><\/span>curva representa a curva de fus\u00e3o onde os estados s\u00f3lido e l\u00edquido est\u00e3o em equil\u00edbrio ou seja, em qualquer ponto desta curva a subst\u00e2ncia est\u00e1 em um estado de coexist\u00eancia de estados s\u00f3lido e l\u00edquido.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Curva CD\u00a0\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>fornece a temperatura e a press\u00e3o, de vaporiza\u00e7\u00e3o ou condensa\u00e7\u00e3o<\/b><\/span><\/span><\/span>\u00a0e essa\u00a0<\/b><\/span><\/span><\/span>curva representa a curva de vaporiza\u00e7\u00e3o onde os estados l\u00edquido e vapor est\u00e3o em equil\u00edbrio<\/b><\/span><\/span><\/span>, ou seja, em\u00a0<\/b><\/span><\/span><\/span>qualquer ponto desta curva a subst\u00e2ncia\u00a0<\/b><\/span><\/span><\/span>est\u00e1 em um estado de\u00a0<\/b><\/span><\/span><\/span>coexist\u00eancia de estados vapor e l\u00edquido.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Curva AC\u00a0\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>fornece a temperatura e press\u00e3o, de sublima\u00e7\u00e3o ou cristaliza\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>e essa curva representa a\u00a0<\/b><\/span><\/span><\/span>curva de sublima\u00e7\u00e3o onde os estados s\u00f3lido e vapor est\u00e3o em equil\u00edbrio, ou seja, em qualquer ponto desta curva a subst\u00e2ncia est\u00e1 em um estado de coexist\u00eancia de estados s\u00f3lido e vapor.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

Massa do corpo s\u00f3lido \"\"<\/b><\/span><\/span><\/span> <\/b>m<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span> = 600g \"\"<\/b><\/span><\/span><\/span> <\/b>V<\/b><\/span><\/span><\/span>s\u00f3lido<\/b><\/span><\/span><\/sub><\/span> = 1 L <\/b><\/span><\/span>= 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>densidade do corpo s\u00f3lid<\/b><\/span><\/span><\/span>o \"\" <\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> <\/b> \"\" <\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span> = 0,6 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Volume total =<\/b><\/span><\/span><\/span> <\/b>1,0.10 <\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span><\/p>\n

Volume imerso <\/b><\/span><\/span><\/span>= 80% do V<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span> = 0,8.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\" =<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/sup><\/span> \"\"<\/b><\/span><\/span><\/sup> <\/b>=<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>l\u00edquido <\/b><\/span><\/span><\/sub><\/span> <\/b>= \"\"<\/b><\/span><\/span><\/span> <\/b>= 0,75 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/b><\/span><\/span><\/span>a. Falsa \"\"<\/b><\/span><\/span><\/span> O <\/b><\/span><\/span><\/span>efeito fotoel\u00e9trico ocorre quando<\/b><\/span><\/span><\/span> a luz\u00a0(onda eletromagn\u00e9tica, radia\u00e7\u00e3o eletromagn\u00e9tica)\u00a0de frequ\u00eancia suficientemente alta incide sobre a superf\u00edcie de um metal, ela pode retirar el\u00e9trons do mesmo.<\/b><\/span><\/span><\/span><\/p>\n

b. Falsa \"\"<\/b><\/span><\/span><\/span> Nas l\u00e2mpadas <\/b><\/span><\/span><\/span>incandescentes a emiss\u00e3o de luz ocorre devido ao aquecimento de seu filamento<\/b><\/span><\/span><\/span> por <\/b><\/span><\/span><\/span>efeito Joule <\/b><\/span><\/span><\/span>(choque dos el\u00e9trons da corrente el\u00e9trica com as part\u00edculas do filamento)<\/b><\/span><\/span><\/span><\/p>\n

c. Falsa \"\"<\/b><\/span><\/span><\/span> <\/b>N\u00e3o \u00e9 a carga el\u00e9trica <\/b><\/span><\/span><\/span>do f\u00f3ton que \u00e9 <\/b><\/span><\/span><\/span>absorvida pelo metal, \u00e9 a energia do mesmo.<\/b><\/span><\/span><\/span><\/p>\n

d. Falsa \"\"<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span><\/span>fa\u00edsca<\/b><\/span><\/span><\/span> ocorre quando o <\/b><\/span><\/span><\/span>meio que era isolante se torna condutor <\/b><\/span><\/span><\/span>devido \u00e0 uma <\/b><\/span><\/span><\/span>diferen\u00e7a de potencial muito elevada.<\/b><\/span><\/span><\/span><\/p>\n

e. Correta \"\"<\/b><\/span><\/span><\/span> <\/b>O\u00a0efeito fotoel\u00e9trico s\u00f3 surge se o metal receber um feixe de radia\u00e7\u00e3o com\u00a0energia superior\u00a0\u00e0\u00a0energia m\u00ednima de remo\u00e7\u00e3o dos el\u00e9trons\u00a0do metal, provocando a\u00a0sua sa\u00edda das \u00f3rbitas\u00a0o que\u00a0pode ocorrer sem energia cin\u00e9tica\u00a0(se a energia da radia\u00e7\u00e3o for igual \u00e0 energia de remo\u00e7\u00e3o) ou\u00a0com energia cin\u00e9tica, (se a energia da radia\u00e7\u00e3o exceder a energia de remo\u00e7\u00e3o dos el\u00e9trons).<\/b><\/span><\/span><\/span><\/p>\n

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

09- <\/b><\/span><\/span><\/span>a) <\/b><\/span><\/span><\/span>Observe o <\/b><\/span><\/span><\/span>c\u00e1lculo da resist\u00eancia el\u00e9trica do resistor equivalente (R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>da associa\u00e7\u00e3o de resistores na <\/b><\/span><\/span><\/span>sequ\u00eancia das figuras abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Colocando na associa\u00e7\u00e3o <\/b><\/span><\/span><\/span>as correntes <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> e i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> <\/b>e lembrando que, pelo <\/b><\/span><\/span><\/span>enunciado, i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 0,5 A <\/b><\/span><\/span><\/span>vamos calcular a <\/b><\/span><\/span><\/span>tens\u00e3o nos terminais do resistor de R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 6 \u2126 \"\"<\/b><\/span><\/span><\/span> <\/b>U<\/b><\/span><\/span><\/span>R2<\/b><\/span><\/span><\/sub><\/span> = R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 6×0,5 \"\"<\/b><\/span><\/span><\/span> U<\/b><\/span><\/span><\/span>R2<\/b><\/span><\/span><\/sub><\/span> = 3 V.<\/b><\/span><\/span><\/span><\/p>\n

Mas, observe que os <\/b><\/span><\/span><\/span>resistores R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> e R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> est\u00e3o em paralelo <\/b><\/span><\/span><\/span>e, portanto, sujeitos \u00e0 <\/b><\/span><\/span><\/span>mesma tens\u00e3o de<\/b><\/span><\/span><\/span><\/p>\n

U = 3 V.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo de i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 3 = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 1 A.<\/b><\/span><\/span><\/span><\/p>\n

Veja na <\/b><\/span><\/span><\/span>figura<\/b><\/span><\/span><\/span> que <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> + i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 0,5 + 1,0 =1,5 A.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ent\u00e3o, a <\/b><\/span><\/span><\/span>tens\u00e3o U nos terminais da associa\u00e7\u00e3o valer\u00e1 U = U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + 3 <\/b><\/span><\/span><\/span>= <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + 3 = 2×1,5 + 3 \"\"<\/b><\/span><\/span><\/span> U = 6,0 V.<\/b><\/span><\/span><\/span><\/p>\n

c) <\/b><\/span><\/span><\/span>Se o <\/b><\/span><\/span><\/span>gerador tem for\u00e7a eletromotriz E = 7,5 V<\/b><\/span><\/span><\/span> e ele fornece ao <\/b><\/span><\/span><\/span>circuito externo U = 6,0 V <\/b><\/span><\/span><\/span>ent\u00e3o a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

queda de tens\u00e3o U\u2019 no interior do gerador deve valer U\u2019 =<\/b><\/span><\/span><\/span> 7,5 \u2013 6,0 <\/b><\/span><\/span><\/span>= 1,5V <\/b><\/span><\/span><\/span>e \u00e9 devida \u00e0 <\/b><\/span><\/span><\/span>resist\u00eancia interna do gerador <\/b><\/span><\/span><\/span>e fornecida <\/b><\/span><\/span><\/span>por U\u2019 = r.i \"\"<\/b><\/span><\/span><\/span> 1,5 = r.1,5 \"\" <\/b><\/span><\/span><\/span>r = 1 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

10- <\/b><\/span><\/span><\/span>a) <\/b><\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>foco F <\/b><\/span><\/span><\/span>est\u00e1 localizado na <\/b><\/span><\/span><\/span>dist\u00e2ncia m\u00e9dia entre o centro de curvatura C e o espelho.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

f <\/b><\/span><\/span><\/span>= R\/2 = 10\/2 <\/b><\/span><\/span><\/span>= 5 cm <\/b><\/span><\/span><\/span>(positivo, espelho c\u00f4ncavo).<\/b><\/span><\/span><\/span><\/p>\n

P = 3 cm<\/b><\/span><\/span><\/span><\/p>\n

P\u2019 = ?<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o dos pontos conjugados \"\" \"\"<\/b><\/span><\/span><\/span> = \"\" <\/b><\/span><\/span><\/span>+<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>=<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>+<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\" \"\"<\/b><\/span><\/span><\/span> <\/b>–<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\" \"\"<\/b><\/span><\/span><\/span> <\/b>=<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>P\u2019 = – 7,5 cm <\/b><\/span><\/span><\/span>(P\u2019 < 0, imagem virtual)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

i=<\/b><\/span><\/span><\/span> ?<\/b><\/span><\/span><\/span><\/p>\n

o = <\/b><\/span><\/span><\/span>– 2cm (invertida)<\/b><\/span><\/span><\/span><\/p>\n

P =<\/b><\/span><\/span><\/span> 3 cm<\/b><\/span><\/span><\/span><\/p>\n

P\u2019 =<\/b><\/span><\/span><\/span> – 7,5 cm<\/b><\/span><\/span><\/span><\/p>\n

\"\" =<\/b><\/span><\/span><\/span> – \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> = – \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>i =<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>= 5 cm <\/b><\/span><\/span><\/span>( a imagem \u00e9 direita em rela\u00e7\u00e3o ao objeto e tem 5cm de altura).<\/b><\/span><\/span><\/span><\/p>\n

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

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UFPR-PR-2017 01- Massa do corpo m = 2000g = 2kg Peso do corpo P = mg = 2.10 = 20 N. Nas figuras abaixo est\u00e3o colocadas as for\u00e7as que agem sobre o sistema e a soma vetorial dessas for\u00e7as pelo m\u00e9todo da linha poligonal deve fornecer um tri\u00e2ngulo fechado (equil\u00edbrio, for\u00e7a resultante nula).<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3147,"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-3149","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3149","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=3149"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3149\/revisions"}],"predecessor-version":[{"id":3150,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3149\/revisions\/3150"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3147"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3149"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}