{"id":3063,"date":"2017-06-04T17:46:48","date_gmt":"2017-06-04T17:46:48","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3063"},"modified":"2024-09-03T17:42:10","modified_gmt":"2024-09-03T17:42:10","slug":"resolucao-comentada-das-questoes-de-fisica-da-unitau-medicina-sp-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/unitau-medicina-2017\/resolucao-comentada-das-questoes-de-fisica-da-unitau-medicina-sp-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UNITAU MEDICINA \u2013 SP \u2013 2017"},"content":{"rendered":"

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

UNITAU MEDICINA \u2013 SP \u2013 2017<\/span><\/span><\/span><\/p>\n

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

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

a) Errado \"\" <\/b><\/span><\/span>A posi\u00e7\u00e3o <\/b><\/span><\/span>do objeto \u00e9 dada por x(t) = (30 cos30\u00ba)t e y(t) = (30 sen30\u00ba)t-5t<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>, o que <\/b><\/span><\/span>caracteriza<\/b><\/span><\/span> um movimento parab\u00f3lico. <\/b><\/span><\/span>O eixo x <\/b><\/span><\/span>representa a dist\u00e2ncia <\/b><\/span><\/span>horizontal<\/b><\/span><\/span> percorrida pelo objeto e o eixo y <\/b><\/span><\/span>representa<\/b><\/span><\/span> a dist\u00e2ncia vertical percorrida pelo objeto. Note que \u00e9 <\/b><\/span><\/span>o termo <\/b><\/span><\/span>quadr\u00e1tico da fun\u00e7\u00e3o y(t) que <\/b><\/span><\/span>caracteriza<\/b><\/span><\/span> a par\u00e1bola e <\/b><\/span><\/span>indica que <\/b><\/span><\/span>o objeto tem uma fase de <\/b><\/span><\/span>subida<\/b><\/span><\/span> (primeira metade da par\u00e1bola) e uma fase de <\/b><\/span><\/span>descida<\/b><\/span><\/span> (segunda metade da par\u00e1bola).<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

b) Errado \"\" Como <\/b><\/span><\/span>explicado <\/b><\/span><\/span>no item a, as fun\u00e7\u00f5es de <\/b><\/span><\/span>posi\u00e7\u00e3o <\/b><\/span><\/span>do objeto, x(t) e y(t), caracterizam um movimento <\/b><\/span><\/span>curvil\u00edneo<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

c) Errada \"\" Utilizando a equa\u00e7\u00e3o de <\/b><\/span><\/span>Torricelli<\/b><\/span><\/span>, podemos obter a altura m\u00e1xima que o <\/b><\/span><\/span>objeto<\/b><\/span><\/span> atinge, em rela\u00e7\u00e3o ao solo \"\" \"\"<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span>velocidade<\/b><\/span><\/span> V<\/b><\/span><\/span>y<\/b><\/span><\/sub><\/span> \u00e9 a componente vertical da velocidade no <\/b><\/span><\/span>ponto mais alto <\/b><\/span><\/span>da trajet\u00f3ria, que vale zero (<\/b><\/span><\/span>momento<\/b><\/span><\/span> em que o objeto est\u00e1 <\/b><\/span><\/span>na transi\u00e7\u00e3o <\/b><\/span><\/span>da fase subida para a fase de <\/b><\/span><\/span>descida<\/b><\/span><\/span>).<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Podemos descobrir a <\/b><\/span><\/span>velocidade<\/b><\/span><\/span> inicial V<\/b><\/span><\/span>oy<\/b><\/span><\/sub><\/span> das equa\u00e7\u00f5es de <\/b><\/span><\/span>posi\u00e7\u00e3o<\/b><\/span><\/span> dadas no enunciado da quest\u00e2o:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

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

Temos ent\u00e3o que:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

A <\/b><\/span><\/span>altura m\u00e1xima <\/b><\/span><\/span>que o objeto atinge, portanto, \u00e9 11,25 metros.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\u00c9 importante saber <\/b><\/span><\/span>diferenciar<\/b><\/span><\/span> o movimento em x do <\/b><\/span><\/span>movimento<\/b><\/span><\/span> em y, pois podem ter valores <\/b><\/span><\/span>distintos<\/b><\/span><\/span>, como \u00e9 este o caso. Tamb\u00e9m devemos <\/b><\/span><\/span>notar<\/b><\/span><\/span> que a dist\u00e2ncia <\/b><\/span><\/span>\u0394S na <\/b><\/span><\/span>equa\u00e7\u00e3o<\/b><\/span><\/span> de Torricelli \u00e9 relativa ao movimento <\/b><\/span><\/span>vertical<\/b><\/span><\/span>, pois \u00e9 somente nele que <\/b><\/span><\/span>consideramos<\/b><\/span><\/span> a atua\u00e7\u00e3o da gravidade.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

d) Errada \"\" <\/b><\/span><\/span>Podemos calcular <\/b><\/span><\/span>o tempo em que o objeto atinge o ponto mais alto e <\/b><\/span><\/span>chega ao solo <\/b><\/span><\/span>novamente, atrav\u00e9s da <\/b><\/span><\/span>equa\u00e7\u00e3o dada <\/b><\/span><\/span>no enunciado:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

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

Resolvendo essa <\/b><\/span><\/span>equa\u00e7\u00e3o<\/b><\/span><\/span> de segundo grau encontramos que o <\/b><\/span><\/span>tempo de subida <\/b><\/span><\/span>\u00e9 de 1,5 segundos. Como o <\/b><\/span><\/span>tempo de descida <\/b><\/span><\/span>\u00e9 igual ao tempo de subida, o tempo total \u00e9 3 <\/b><\/span><\/span>segundos<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

e) Correto \"\" Podemos <\/b><\/span><\/span>calcular<\/b><\/span><\/span> a dist\u00e2ncia horizontal atrav\u00e9s da equa\u00e7\u00e3o em x, dada no enunciado, <\/b><\/span><\/span>utilizando o tempo <\/b><\/span><\/span>calculado no item d, j\u00e1 que \u00e9 o mesmo.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

R-<\/b><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa E.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

Para <\/b><\/span><\/span>encontrar<\/b><\/span><\/span> o m\u00f3dulo da acelera\u00e7\u00e3o de descida do objeto, \u00e9 <\/b><\/span><\/span>necess\u00e1rio<\/b><\/span><\/span> fazer o somat\u00f3rio de <\/b><\/span><\/span>for\u00e7as<\/b><\/span><\/span> sobre o corpo e assim obter a equa\u00e7\u00e3o <\/b><\/span><\/span>din\u00e2mica<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

Sabendo que \"\" m = 8 Kg<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

A = \u00b5.N<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

N = P.cos\u03f4<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

<\/b>Podemos<\/b><\/span><\/span> escrever a seguinte <\/b><\/span><\/span>equa\u00e7\u00e3o<\/b><\/span><\/span> da resultante das <\/b><\/span><\/span>for\u00e7as<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

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

a) <\/b><\/span><\/span>Coeficiente<\/b><\/span><\/span> de atrito cin\u00e9tico igual a <\/b><\/span><\/span>0,3 <\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

b) Coeficiente de <\/b><\/span><\/span>atrito<\/b><\/span><\/span> cin\u00e9tico igual a <\/b><\/span><\/span>0,1<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

c) <\/b><\/span><\/span>Coeficiente<\/b><\/span><\/span> de atrito cin\u00e9tico igual a <\/b><\/span><\/span>0,2<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

d) Coeficiente de atrito <\/b><\/span><\/span>cin\u00e9tico<\/b><\/span><\/span> igual a <\/b><\/span><\/span>0,4<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

e) Coeficiente <\/b><\/span><\/span>de atrito <\/b><\/span><\/span>cin\u00e9tico igual a <\/b><\/span><\/span>0,7<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

R-<\/b><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa a.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

A <\/b><\/span><\/span>energia dissipada <\/b><\/span><\/span>pode ser calculada pela varia\u00e7\u00e3o da energia <\/b><\/span><\/span>mec\u00e2nica<\/b><\/span><\/span>, isto \u00e9, se n\u00e3o <\/b><\/span><\/span>houvesse<\/b><\/span><\/span> perdas durante o <\/b><\/span><\/span>movimento<\/b><\/span><\/span> do objeto, toda energia <\/b><\/span><\/span>cin\u00e9tica <\/b><\/span><\/span>adquirida durante a <\/b><\/span><\/span>subida deveria <\/b><\/span><\/span>converter-se em energia <\/b><\/span><\/span>potencial<\/b><\/span><\/span> gravitacional, mas <\/b><\/span><\/span>como isso <\/b><\/span><\/span>n\u00e3o ocorre, a diferen\u00e7a <\/b><\/span><\/span>entre elas <\/b><\/span><\/span>ser\u00e1 a parcela de energia <\/b><\/span><\/span>dissipada <\/b><\/span><\/span>pelo atrito com o ar durante a subida.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

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

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

<\/b>A energia <\/b><\/span><\/span>dissipada durante a <\/b><\/span><\/span>subida<\/b><\/span><\/span> ser\u00e1 de <\/b><\/span><\/span>2 joules<\/b><\/span><\/span>, enquanto que se n\u00e3o houvesse a <\/b><\/span><\/span>resist\u00eancia<\/b><\/span><\/span> do atrito, a energia obtida no ponto mais alto da subida seria:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

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

Portanto, o <\/b><\/span><\/span>percentual<\/b><\/span><\/span> de energia perdida ser\u00e1 a <\/b><\/span><\/span>parcela dissipada<\/b><\/span><\/span> pelo total que poderia ser obtido:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

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

R-<\/b><\/span><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa B.<\/b><\/span><\/span><\/span><\/p>\n

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

Podemos <\/b><\/span><\/span><\/span>resolver<\/b><\/span><\/span><\/span> este problema, atrav\u00e9s da <\/b><\/span><\/span><\/span>conserva\u00e7\u00e3o<\/b><\/span><\/span><\/span> da energia mec\u00e2nica, dada pela rela\u00e7\u00e3o \"\" \"\"\"\"\"\"<\/b><\/span><\/span><\/span><\/p>\n

Sabendo que <\/b><\/span><\/span><\/span>no in\u00edcio <\/b><\/span><\/span><\/span>n\u00e3o havia energia cin\u00e9tica, pois o objeto estava em repouso, e que quando <\/b><\/span><\/span><\/span>atinge o solo <\/b><\/span><\/span><\/span>n\u00e3o h\u00e1 mais energia <\/b><\/span><\/span><\/span>potencial<\/b><\/span><\/span><\/span> gravitacional, pois n\u00e3o h\u00e1 mais <\/b><\/span><\/span><\/span>altura<\/b><\/span><\/span><\/span>, podemos reescrever a <\/b><\/span><\/span><\/span>express\u00e3o<\/b><\/span><\/span><\/span> de conserva\u00e7\u00e3o da energia \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" <\/b><\/span><\/span><\/span><\/p>\n

R-<\/b><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa D.<\/b><\/span><\/span><\/p>\n

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

Podemos<\/b><\/span><\/span><\/span> resolver esta <\/b><\/span><\/span><\/span>quest\u00e3o<\/b><\/span><\/span><\/span> atrav\u00e9s da Lei de <\/b><\/span><\/span><\/span>Snell-Descartes \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Temos, ent\u00e3o, que \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\" \"\"<\/b><\/span><\/span><\/span><\/p>\n

R-<\/b><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa A.<\/b><\/span><\/span><\/p>\n

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

a) Correto \"\" Podemos <\/b><\/span><\/span><\/span>calcular a carga <\/b><\/span><\/span><\/span>nos capacitores atrav\u00e9s da <\/b><\/span><\/span><\/span>express\u00e3o<\/b><\/span><\/span><\/span> \"\" \"\" \"\"\"\"<\/b><\/span><\/span><\/span><\/p>\n

Sabendo<\/b><\/span><\/span><\/span> que a tens\u00e3o \u00e9 dividida pela <\/b><\/span><\/span><\/span>metade<\/b><\/span><\/span><\/span> entre os dois capacitores, por estarem em s\u00e9rie e terem <\/b><\/span><\/span><\/span>capacit\u00e2ncias<\/b><\/span><\/span><\/span> iguais \"\" \"\"<\/b><\/span><\/span><\/span><\/p>\n

Temos, ent\u00e3o, que \"\" \"\" \"\" \"\" \"\" \"\"<\/b><\/span><\/span><\/span><\/p>\n

b) Errado \"\" <\/b><\/span><\/span><\/span>As cargas <\/b><\/span><\/span><\/span>em cada um s\u00e3o iguais pois est\u00e3o em <\/b><\/span><\/span><\/span>s\u00e9rie<\/b><\/span><\/span><\/span> e possuem mesmo valor de <\/b><\/span><\/span><\/span>capacit\u00e2ncia<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

c) Errado \"\" A <\/b><\/span><\/span><\/span>diferen\u00e7a<\/b><\/span><\/span><\/span> de potencial que sai da fonte \u00e9 <\/b><\/span><\/span><\/span>dividida<\/b><\/span><\/span><\/span> pela metade em cada <\/b><\/span><\/span><\/span>capacitor<\/b><\/span><\/span><\/span> por estarem em s\u00e9rie e terem <\/b><\/span><\/span><\/span>mesmo valor <\/b><\/span><\/span><\/span>de capacit\u00e2ncia, dessa forma vai <\/b><\/span><\/span><\/span>50 V <\/b><\/span><\/span><\/span>para cada um.<\/b><\/span><\/span><\/span><\/p>\n

d) Errado \"\" Assim como <\/b><\/span><\/span><\/span>explicado<\/b><\/span><\/span><\/span> no item c, a diferen\u00e7a de <\/b><\/span><\/span><\/span>potencial<\/b><\/span><\/span><\/span> em cada um dos <\/b><\/span><\/span><\/span>capacitores<\/b><\/span><\/span><\/span> s\u00e3o iguais e valem <\/b><\/span><\/span><\/span>50 V<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

e) Errado \"\" Como foi <\/b><\/span><\/span><\/span>explicado<\/b><\/span><\/span><\/span> nos itens a e c, as cargas e as <\/b><\/span><\/span><\/span>tens\u00f5es<\/b><\/span><\/span><\/span> em cada um s\u00e3o iguais.<\/b><\/span><\/span><\/span><\/p>\n

R-<\/b><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa A.<\/b><\/span><\/span><\/p>\n

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

Para <\/b><\/span><\/span><\/span>compreender<\/b><\/span><\/span><\/span> melhor a quest\u00e3o podemos <\/b><\/span><\/span><\/span>marcar <\/b><\/span><\/span><\/span>a polaridade dos <\/b><\/span><\/span><\/span>elementos<\/b><\/span><\/span><\/span> de carga da seguinte forma:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) Errado \"\" De <\/b><\/span><\/span><\/span>acordo<\/b><\/span><\/span><\/span> com a teoria de <\/b><\/span><\/span><\/span>circuitos<\/b><\/span><\/span><\/span> el\u00e9tricos, em ramos paralelos a <\/b><\/span><\/span><\/span>tens\u00e3o<\/b><\/span><\/span><\/span> \u00e9 a mesma desde que n\u00e3o exista algum <\/b><\/span><\/span><\/span>elemento<\/b><\/span><\/span><\/span> entre eles. Neste caso, parte da tens\u00e3o que sai <\/b><\/span><\/span><\/span>da fonte <\/b><\/span><\/span><\/span>de 36 V, \u00e9 retida pelo <\/b><\/span><\/span><\/span>resistor<\/b><\/span><\/span><\/span> de 4 Ohms, <\/b><\/span><\/span><\/span>dessa forma<\/b><\/span><\/span><\/span>, a tens\u00e3o <\/b><\/span><\/span><\/span>que chega <\/b><\/span><\/span><\/span>no capacitor \u00e9 \"\" \"\"<\/b><\/span><\/span><\/span> \"\" Portanto, \u00e9 menor que 36 V.<\/b><\/span><\/span><\/span><\/p>\n

b) Errado \"\" <\/b><\/span><\/span><\/span>pelo mesmo <\/b><\/span><\/span><\/span>motivo do item a. O <\/b><\/span><\/span><\/span>resistor<\/b><\/span><\/span><\/span> de 4 Ohms pega parte dos 36 V que <\/b><\/span><\/span><\/span>saem da fonte<\/b><\/span><\/span><\/span>. A tens\u00e3o no resistor de 8 Ohms \u00e9 a mesma que no <\/b><\/span><\/span><\/span>capacitor<\/b><\/span><\/span><\/span>, por estarem em <\/b><\/span><\/span><\/span>ramos paralelos <\/b><\/span><\/span><\/span>e sem algum outro elemento entre eles. Portanto \"\" \"\"<\/b><\/span><\/span><\/span><\/p>\n

c) Errado \"\" Como o <\/b><\/span><\/span><\/span>circuito<\/b><\/span><\/span><\/span> est\u00e1 em <\/b><\/span><\/span><\/span>regime estacion\u00e1rio<\/b><\/span><\/span><\/span>, podemos concluir que o capacitor est\u00e1 <\/b><\/span><\/span><\/span>totalmente <\/b><\/span><\/span><\/span>carregado e se comporta como um <\/b><\/span><\/span><\/span>circuito aberto<\/b><\/span><\/span><\/span>, veja a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Dessa forma<\/b><\/span><\/span><\/span>, a corrente el\u00e9trica n\u00e3o passa mais pelo <\/b><\/span><\/span><\/span>capacitor<\/b><\/span><\/span><\/span>. Portanto a corrente entre os resistores de <\/b><\/span><\/span><\/span>8 e 4 Ohms <\/b><\/span><\/span><\/span>pode ser calculada como \"\" \"\"<\/b><\/span><\/span><\/span><\/p>\n

Note que nesta nova <\/b><\/span><\/span><\/span>configura\u00e7\u00e3o<\/b><\/span><\/span><\/span>, a resist\u00eancia de 4 Ohms est\u00e1 <\/b><\/span><\/span><\/span>em s\u00e9rie <\/b><\/span><\/span><\/span>com a de 8 Ohms, portanto a resist\u00eancia <\/b><\/span><\/span><\/span>equivalente<\/b><\/span><\/span><\/span> \u00e9 a soma das duas.<\/b><\/span><\/span><\/span><\/p>\n

d) Errado \"\" Como o circuito est\u00e1 em <\/b><\/span><\/span><\/span>regime estacion\u00e1rio<\/b><\/span><\/span><\/span>, admitimos que o <\/b><\/span><\/span><\/span>capacitor<\/b><\/span><\/span><\/span> est\u00e1 totalmente <\/b><\/span><\/span><\/span>carregado<\/b><\/span><\/span><\/span> e se comporta como um circuito aberto, ou seja, nenhuma <\/b><\/span><\/span><\/span>corrente<\/b><\/span><\/span><\/span> passa por ele.<\/b><\/span><\/span><\/span><\/p>\n

e) Correto \"\" A <\/b><\/span><\/span><\/span>energia<\/b><\/span><\/span><\/span> armazenada em um <\/b><\/span><\/span><\/span>capacitor<\/b><\/span><\/span><\/span> pode ser calculada pela <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o<\/b><\/span><\/span><\/span> entre a quantidade de carga e a <\/b><\/span><\/span><\/span>tens\u00e3o<\/b><\/span><\/span><\/span> em cima dele, veja o gr\u00e1fico abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A energia <\/b><\/span><\/span><\/span>potencial<\/b><\/span><\/span><\/span> el\u00e9trica E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>, armazenada no <\/b><\/span><\/span><\/span>capacitor<\/b><\/span><\/span><\/span>, \u00e9 numericamente igual a \u00e1rea sob a <\/b><\/span><\/span><\/span>curva<\/b><\/span><\/span><\/span> na figura. Sabendo que \"\" \"\"<\/b><\/span><\/span><\/span> \"\" \"\"<\/b><\/span><\/span><\/span><\/p>\n

Sendo \"\" E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> = a <\/b><\/span><\/span><\/span>energia potencial <\/b><\/span><\/span><\/span>el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

Q = a <\/b><\/span><\/span><\/span>quantidade de carga <\/b><\/span><\/span><\/span>no capacitor<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span> = a <\/b><\/span><\/span><\/span>tens\u00e3o el\u00e9trica<\/b><\/span><\/span><\/span> no capacitor<\/b><\/span><\/span><\/span><\/p>\n

C = a <\/b><\/span><\/span><\/span>capacit\u00e2ncia<\/b><\/span><\/span><\/span><\/p>\n

Podemos <\/b><\/span><\/span><\/span>calcular<\/b><\/span><\/span><\/span> a energia <\/b><\/span><\/span><\/span>armazenada <\/b><\/span><\/span><\/span>no capacitor da seguinte forma:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

R-<\/b><\/span><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa E.<\/b><\/span><\/span><\/span><\/p>\n

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

Neste problema, \u00e9 importante notar que n\u00e3o somente o vidro do recipiente dilata mas tamb\u00e9m o l\u00edquido dentro dele. A varia\u00e7\u00e3o do volume do l\u00edquido \u00e9 dada pela soma da varia\u00e7\u00e3o do volume do vidro mais a varia\u00e7\u00e3o do volume de l\u00edquido vazado, conforme a seguinte express\u00e3o \"\" \"\" (I)<\/b><\/span><\/span><\/span><\/p>\n

Podemos calcular a varia\u00e7\u00e3o do volume atrav\u00e9s da seguinte express\u00e3o de dilata\u00e7\u00e3o \"\" \"\" \"\" \"\" (II)<\/b><\/span><\/span><\/span><\/p>\n

Substituindo a express\u00e3o (II) em (I), temos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

R-<\/b><\/span><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa A.<\/b><\/span><\/span><\/span><\/p>\n

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

a) Errado \"\" Isoc\u00f3rica significa uma transforma\u00e7\u00e3o a volume constante, e de A para B o volume varia de 10 para 20 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

b) Errado \"\" Isob\u00e1rica significa uma transforma\u00e7\u00e3o a press\u00e3o constante, e de B para C a press\u00e3o varia de 1 para 3 atm.<\/b><\/span><\/span><\/span><\/p>\n

c) Errado \"\" Podemos usar o modelo de g\u00e1s ideal no qual \"\" \"\" \"\" <\/b><\/span><\/span><\/span><\/p>\n

\"\" \"\" \"\" \"\" A temperatura no ponto B \u00e9 duas vezes a temperatura no ponto A.<\/b><\/span><\/span><\/span><\/p>\n

d) Correto \"\" \"\" \"\" \"\" \"\" \"\"<\/b><\/span><\/span><\/span><\/p>\n

e) Errado \"\" Como mostrado no item d, \"\".<\/b><\/span><\/span><\/span><\/p>\n

R-<\/b><\/span><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa D.<\/b><\/span><\/span><\/span><\/p>\n

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

a) Errado \"\" <\/b><\/span><\/span><\/span>Nas usinas <\/b><\/span><\/span><\/span>nucleares a energia \u00e9 gerada a <\/b><\/span><\/span><\/span>partir <\/b><\/span><\/span><\/span>da rea\u00e7\u00e3o nuclear de um <\/b><\/span><\/span><\/span>elemento<\/b><\/span><\/span><\/span> quando \u00e9 <\/b><\/span><\/span><\/span>transformado<\/b><\/span><\/span><\/span> em um de seus is\u00f3topos.<\/b><\/span><\/span><\/span><\/p>\n

b) Correto \"\" Nas usinas <\/b><\/span><\/span><\/span>hidrel\u00e9tricas<\/b><\/span><\/span><\/span> a \u00e1gua proveniente de um <\/b><\/span><\/span><\/span>rio<\/b><\/span><\/span><\/span> \u00e9 represada e <\/b><\/span><\/span><\/span>ent\u00e3o sofre <\/b><\/span><\/span><\/span>uma queda, <\/b><\/span><\/span><\/span>caindo<\/b><\/span><\/span><\/span> sobre as p\u00e1s de uma <\/b><\/span><\/span><\/span>turbina<\/b><\/span><\/span><\/span> convertendo a energia potencial em <\/b><\/span><\/span><\/span>energia cin\u00e9tica <\/b><\/span><\/span><\/span>e depois em <\/b><\/span><\/span><\/span>energia<\/b><\/span><\/span><\/span> el\u00e9trica atrav\u00e9s de geradores.<\/b><\/span><\/span><\/span><\/p>\n

c) Errado \"\" Nas usinas <\/b><\/span><\/span><\/span>termel\u00e9tricas<\/b><\/span><\/span><\/span> a energia \u00e9 <\/b><\/span><\/span><\/span>gerada<\/b><\/span><\/span><\/span> a partir da queima de combust\u00edveis <\/b><\/span><\/span><\/span>vegetais<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

d) Errado \"\" Nos <\/b><\/span><\/span><\/span>parques<\/b><\/span><\/span><\/span> e\u00f3licos a energia <\/b><\/span><\/span><\/span>\u00e9 gerada <\/b><\/span><\/span><\/span>atrav\u00e9s do fluxo <\/b><\/span><\/span><\/span>de vento<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

e) Errado \"\" Nas usinas <\/b><\/span><\/span><\/span>solares<\/b><\/span><\/span><\/span> a energia \u00e9 <\/b><\/span><\/span><\/span>gerada a partir <\/b><\/span><\/span><\/span>de c\u00e9lulas solares que <\/b><\/span><\/span><\/span>captam a<\/b><\/span><\/span><\/span> energia dos raios <\/b><\/span><\/span><\/span>de sol<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

R-<\/b><\/span><\/span><\/span> Portanto, a resposta correta \u00e9 a alternativa B.<\/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 UNITAU MEDICINA \u2013 SP \u2013 2017 01- a) Errado A posi\u00e7\u00e3o do objeto \u00e9 dada por x(t) = (30 cos30\u00ba)t e y(t) = (30 sen30\u00ba)t-5t2, o que caracteriza um movimento parab\u00f3lico. O eixo x representa a dist\u00e2ncia horizontal percorrida pelo objeto e o eixo y representa a dist\u00e2ncia vertical percorrida pelo objeto. Note<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3061,"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-3063","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3063","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=3063"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3063\/revisions"}],"predecessor-version":[{"id":11055,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3063\/revisions\/11055"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3061"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3063"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}