{"id":1501,"date":"2015-11-02T22:22:24","date_gmt":"2015-11-02T22:22:24","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1501"},"modified":"2024-09-03T13:44:34","modified_gmt":"2024-09-03T13:44:34","slug":"1501-2","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2016\/escola-naval\/1501-2\/","title":{"rendered":"Resolu\u00e7\u00e3o da Escola naval – 2015\/2016"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios da<\/b><\/span><\/span><\/span><\/p>\n

Escola Naval – 015\/016<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

 <\/p>\n

\"\"\u00a0<\/span>O<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>trabalho das for\u00e7as\u00a0conservativas<\/b><\/span><\/span><\/span>,\u00a0como, por exemplo, as<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>for\u00e7as peso, da el\u00e1stica , da<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>el\u00e9trica e da<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>magn\u00e9tica<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>n\u00e3o dependem da trajet\u00f3ria<\/b><\/span><\/span><\/span>, mas apenas das<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>posi\u00e7\u00f5es inicial e final da mesma.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

\"\"\u00a0<\/span>Gr\u00e1ficos das<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>energias cin\u00e9tica, potencial gravitacional e mec\u00e2nica<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>para um corpo de<\/b><\/span><\/span><\/span>massa<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>quando<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>lan\u00e7ado verticalmente para cima<\/b><\/span><\/span><\/span>, a partir do<\/b><\/span><\/span><\/span>ponto de lan\u00e7amento, tomado como<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>referencial e desprezando-se as for\u00e7as resistivas, em<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>fun\u00e7\u00e3o do tempo de subida e descida.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Durante todo o movimento<\/b><\/span><\/span><\/span>, conclui-se que, \u00e0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>diminui\u00e7\u00e3o de energia cin\u00e9tica<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>corresponde um <\/b><\/span><\/span><\/span>aumento de energia potencial gravitacional e vice-versa,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>mantendo-se constante a totalidade da energia mec\u00e2nica.<\/b><\/span><\/span><\/span><\/p>\n

Observe que as<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>representa\u00e7\u00f5es gr\u00e1ficas das energias cin\u00e9tica e potencial gravitacional <\/b><\/span><\/span><\/span>correspondem \u00e0 duas par\u00e1bolas invertidas de modo que, em cada ponto, a<\/b><\/span><\/span><\/span>soma dessas duas energias<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>corresponda \u00e0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>energia mec\u00e2nica<\/b><\/span><\/span><\/span>, que \u00e9<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>constante.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe tamb\u00e9m que o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>tempo de subida \u00e9 igual ao tempo de descida,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>pois as<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>for\u00e7as dissipativas s\u00e3o desprezadas.<\/b><\/span><\/span><\/span><\/p>\n

Aplicando a <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o trigonom\u00e9trica<\/b><\/span><\/span><\/span> temos que:<\/b><\/span><\/span><\/span><\/p>\n

sen2x = sen x . senx E<\/b><\/span><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub><\/span> = 4.sen<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>(2\/3<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>t – <\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2) <\/b><\/span><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub><\/span> = 4.sen(2\/3<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>t – <\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2).sen(2\/3<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>t –<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2)<\/b><\/span><\/span><\/span><\/span><\/p>\n

para t=0 \"\" Ec = 4.sen(0 – <\/b><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2).sen(0 – <\/b><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2<\/b><\/span><\/span><\/span>)<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Ec = 4.(-1).(-1)<\/b><\/span><\/span><\/span> \"\" E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 4 J (valor m\u00e1ximo),<\/b><\/span><\/span><\/span> <\/b>portanto <\/b><\/span><\/span><\/span>Ep = 0,<\/b><\/span><\/span><\/span> pois, pelo <\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da energia E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>= constante.<\/b><\/span><\/span><\/span><\/p>\n

para t=1,5s \"\" Ec = 4.sen(<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span> – <\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2).sen(<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span> – <\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2) \"\" <\/b><\/span><\/span><\/span><\/span>Ec = 4.sen(<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2).sen(<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2)<\/b><\/span><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>Ec = 4.1.1 \"\"<\/b><\/span><\/span><\/span><\/p>\n

Ec = 4 J<\/b><\/span><\/span><\/span> <\/b>(valor m\u00e1ximo),<\/b><\/span><\/span><\/span> <\/b>portanto <\/b><\/span><\/span><\/span>Ep = 0,<\/b><\/span><\/span><\/span> pois, pelo <\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da energia E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = constante.<\/b><\/span><\/span><\/span><\/p>\n

para t = 3 s \"\" Ec = 4.sen(2<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span> – <\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2).sen(2<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span> – <\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2)<\/b><\/span><\/span><\/span><\/span> <\/b>Ec = 4.sen(2<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span> – <\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2).sen(2<\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span> – <\/b><\/span><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2) \"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ec = 4.sen(3<\/b><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2).sen(3<\/b><\/span><\/span><\/span>\uf070<\/b><\/span><\/span><\/span>\/2)<\/b><\/span><\/span><\/span> \"\" Ec = 4.(-1).(-1) \"\" <\/b><\/span><\/span>Ec = 4 J<\/b><\/span><\/span><\/span> <\/b>(valor m\u00e1ximo),<\/b><\/span><\/span><\/span> <\/b>portanto <\/b><\/span><\/span><\/span>Ep = 0,<\/b><\/span><\/span><\/span> pois, pelo <\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da energia E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = constante.<\/b><\/span><\/span><\/span><\/p>\n

Logo, o <\/b><\/span><\/span><\/span>\u00fanico gr\u00e1fico que corresponde aos valores obtidos<\/b><\/span><\/span><\/span> \u00e9 o primeiro.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

02- <\/b><\/span><\/span><\/span>Quando voc\u00ea <\/b><\/span><\/span><\/span>comprime a mola com for\u00e7as de intensidade 60N<\/b><\/span><\/span><\/span>, a extremidade <\/b><\/span><\/span><\/span>esquerda<\/b><\/span><\/span><\/span> da mola<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

estar\u00e1 empurrando o <\/b><\/span><\/span><\/span>bloco A para a esquerda com F = 60N e a extremidade direita da mola estar\u00e1 empurrando o bloco B para a direita com F = 60N.<\/b><\/span><\/span><\/span><\/p>\n

Analisando primeiramente o <\/b><\/span><\/span><\/span>bloco A <\/b><\/span><\/span><\/span>temos:<\/b><\/span><\/span><\/span><\/p>\n

Fat A = <\/b><\/span><\/span><\/span><\/span>\uf06d<\/b><\/span><\/span><\/span><\/span>.N<\/b><\/span><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Fat A = <\/b><\/span><\/span><\/span><\/span>\uf06d<\/b><\/span><\/span><\/span><\/span>.m.g \"\" Fat A = 0,4.20.10<\/b><\/span><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Fat A = 80 N<\/b><\/span><\/span><\/span><\/span><\/p>\n

Retirando simultaneamente ambas as for\u00e7as<\/b><\/span><\/span><\/span>, o <\/b><\/span><\/span><\/span>bloco A<\/b><\/span><\/span><\/span> ficar\u00e1 sujeito a uma for\u00e7a <\/b><\/span><\/span><\/span>de 60N para a esquerda e uma for\u00e7a de atrito de 80 N para a direita.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como a for\u00e7a<\/b><\/span><\/span><\/span> <\/b>F vale 60 N e<\/b><\/span><\/span><\/span> <\/b>a <\/b><\/span><\/span><\/span>for\u00e7a de atrito vale 80 N,<\/b><\/span><\/span><\/span> <\/b>chegamos a conclus\u00e3o que<\/b><\/span><\/span><\/span> <\/b>o bloco A fica parado, uma vez que o atrito impede seu movimento.<\/b><\/span><\/span><\/span><\/p>\n

Analisando agora o <\/b><\/span><\/span><\/span>bloco B:<\/b><\/span><\/span><\/span><\/p>\n

Fat B = <\/b><\/span><\/span><\/span><\/span>\uf06d<\/b><\/span><\/span><\/span><\/span>.N<\/b><\/span><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Fat B = <\/b><\/span><\/span><\/span><\/span>\uf06d<\/b><\/span><\/span><\/span><\/span>.m.g \"\" Fat B = 0,4.10.10 \"\" <\/b><\/span><\/span><\/span><\/span>Fat B = 40 N<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como a <\/b><\/span><\/span><\/span>for\u00e7a F vale 60 N e a for\u00e7a de atrito vale 40 N,<\/b><\/span><\/span><\/span> chegamos \u00e0 conclus\u00e3o que o <\/b><\/span><\/span><\/span>bloco B entra em<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

movimento para a direita,<\/b><\/span><\/span><\/span> uma vez que o atrito agora somente dificulta o movimento do bloco, que acaba sendo <\/b><\/span><\/span><\/span>impulsionado a sua direita<\/b><\/span><\/span><\/span> pela a\u00e7\u00e3o da mola, sujeito a uma <\/b><\/span><\/span><\/span>for\u00e7a resultante de F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = 60 \u2013 40 = 20N..<\/b><\/span><\/span><\/span><\/p>\n

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

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

Atrav\u00e9s da figura verificamos que<\/b><\/span><\/span><\/span> se trata de uma <\/b><\/span><\/span>liga\u00e7\u00e3o em paralelo<\/b><\/span><\/span><\/span><\/p>\n

Associa\u00e7\u00e3o paralelo, caracter\u00edsticas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0Os <\/b><\/span><\/span><\/span>resistores s\u00e3o associados pelos seus terminais<\/b><\/span><\/span><\/span>, ou seja, todos <\/b><\/span><\/span><\/span>saem de um mesmo ponto e todos chegam a um mesmo ponto.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0A <\/b><\/span><\/span><\/span>diferen\u00e7a de potencial (tens\u00e3o) U<\/b><\/span><\/span><\/span> de toda a associa\u00e7\u00e3o (entre A e B) \u00e9 <\/b><\/span><\/span><\/span>a mesma para todos os resistores<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0A <\/b><\/span><\/span><\/span>corrente total i<\/b><\/span><\/span><\/span> \u00e9 a soma das correntes parciais, ou seja,\u00a0<\/b><\/span><\/span><\/span>i = i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

04- <\/span><\/span><\/span>Campo el\u00e9trico gerado por uma carga pontual Q<\/b><\/span><\/span><\/p>\n

\"\" Uma <\/b><\/span><\/span><\/span>carga puntiforme (Q)<\/b><\/span><\/span><\/span> geradora de campo <\/b><\/span><\/span><\/span>provoca num ponto P<\/b><\/span><\/span><\/span>, distante\u00a0<\/b><\/span><\/span><\/span>d\u00a0<\/b><\/span><\/span><\/span>da carga, um <\/b><\/span><\/span><\/span>vetor<\/b><\/span><\/span><\/span> <\/b>campo el\u00e9trico\u00a0\"\"\u00a0que faz surgir sobre uma carga de prova q a\u00ed colocada uma for\u00e7a el\u00e9trica\u00a0\"\"\u00a0de intensidade F=KQq\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, que substitu\u00edda na equa\u00e7\u00e3o E=F\/q fornece:<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a0Observe <\/b><\/span><\/span><\/span>atentamente as figuras abaixo<\/b><\/span><\/span><\/span> onde a carga geradora <\/b><\/span><\/span><\/span>Q > 0<\/b><\/span><\/span><\/span> provoca em <\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>< 0<\/b><\/span><\/span><\/span> localizado em M uma for\u00e7a de atra\u00e7\u00e3o\u00a0\"\"\u00a0e, como <\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 negativa, campo e for\u00e7a tem mesma dire\u00e7\u00e3o e sentidos opostos, estando\"\"\u00a0em M se <\/b><\/span><\/span><\/span>afastando<\/b><\/span><\/span><\/span> de Q > 0.<\/b><\/span><\/span><\/span><\/p>\n

Verifique agora que a <\/b><\/span><\/span><\/span>carga geradora Q > 0 provoca em q<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>> 0<\/b><\/span><\/span><\/span> localizado em <\/b><\/span><\/span><\/span>N <\/b><\/span><\/span><\/span>uma for\u00e7a de
\n<\/b><\/span><\/span><\/span>repuls\u00e3o\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0e, como <\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 positiva,<\/b><\/span><\/span><\/span> campo e for\u00e7a t\u00eam <\/b><\/span><\/span><\/span>mesma dire\u00e7\u00e3o e mesmo sentido, estando<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>em N se afastando de Q > 0<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Generalizando: <\/b><\/span><\/span><\/span>em qualquer ponto do\u00a0campo gerado por Q > 0\u00a0colocando-se cargas de prova q positivas ou negativas, o campogerado\u00a0ser\u00e1 sempre de afastamento.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Analogamente, se a\u00a0carga geradora fosse negativa Q < 0, em\u00a0todos os pontos o campo el\u00e9trico gerado seria de aproxima\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o <\/b><\/span><\/span><\/span>el\u00e9tron (carga de prova negativa)<\/b><\/span><\/span><\/span> se desloca para a <\/b><\/span><\/span><\/span>direita a for\u00e7a sobre ele tamb\u00e9m \u00e9 para a direita e o campo el\u00e9trico \u00e9 para a esquerda.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Considere uma <\/b><\/span><\/span><\/span>determinada m\u00e1quina realizando certo trabalho<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span> A <\/b><\/span><\/span><\/span>pot\u00eancia \u00fatil (P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> \u00e9 a pot\u00eancia que a m\u00e1quina utiliza na <\/b><\/span><\/span><\/span>realiza\u00e7\u00e3o de um <\/b><\/span><\/span>trabalho externo;<\/b><\/span><\/span><\/span> a<\/b><\/span><\/span><\/span>pot\u00eancia dissipada (P<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> corresponde \u00e0 <\/b><\/span><\/span><\/span>pot\u00eancia n\u00e3o aproveitada, transformada no interior da m\u00e1quina em energia t\u00e9rmica (calor).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Para poder realizar o <\/b><\/span><\/span><\/span>trabalho \u00fatil (externo)<\/b><\/span><\/span>, a m\u00e1quina deve <\/b><\/span><\/span><\/span>receber<\/b><\/span><\/span> uma <\/b><\/span><\/span><\/span>pot\u00eancia total (P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>),<\/b><\/span><\/span> que deve valer:\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span>t <\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>u<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

Portanto, da <\/b><\/span><\/span><\/span>pot\u00eancia total<\/b><\/span><\/span> fornecida \u00e0 m\u00e1quina <\/b><\/span><\/span><\/span>s\u00f3 uma porcentagem, <\/b><\/span><\/span>pot\u00eancia \u00fatil \u00e9 aproveitada,<\/b><\/span><\/span><\/span> <\/b>pois parte dela \u00e9 perdida (pot\u00eancia dissipada). Assim, <\/b><\/span><\/span><\/span>rendimento (\u00a0\u03b7\u00a0\u2013 letra grega\u00a0eta)<\/b><\/span><\/span><\/span> de uma m\u00e1quina \u00e9 <\/b><\/span><\/span><\/span>sua capacidade de realiza\u00e7\u00e3o de determinado trabalho<\/b><\/span><\/span> e \u00e9 \u00a0definido como sendo a <\/b><\/span><\/span><\/span>raz\u00e3o entre a <\/b><\/span><\/span>pot\u00eancia \u00fatil (P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>) e a pot\u00eancia total (P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ent\u00e3o, por exemplo, se o <\/b><\/span><\/span><\/span>motor de um carro<\/b><\/span><\/span><\/span> estiver <\/b><\/span><\/span><\/span>bem regulado<\/b><\/span><\/span><\/span> ele apresentar\u00e1 <\/b><\/span><\/span><\/span>maior rendimento<\/b><\/span><\/span><\/span>, percorrendo uma <\/b><\/span><\/span><\/span>distancia maior com a mesma quantidade de combust\u00edvel<\/b><\/span><\/span><\/span> que <\/b><\/span><\/span><\/span>outro carro<\/b><\/span><\/span><\/span> de mesmas caracter\u00edsticas, mas com <\/b><\/span><\/span><\/span>o motor desregulado.<\/b><\/span><\/span><\/span><\/p>\n

Como o <\/b><\/span><\/span><\/span>rendimento \u00e9 uma rela\u00e7\u00e3o entre duas grandezas de mesmas unidades<\/b><\/span><\/span><\/span>, elas <\/b><\/span><\/span><\/span>se cancelam<\/b><\/span><\/span><\/span> e ele <\/b><\/span><\/span><\/span>n\u00e3o ter\u00e1 unidade (grandeza adimensional).<\/b><\/span><\/span><\/span><\/p>\n

Sendo P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span>\u00a0sempre menor que P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>,\u00a0<\/b><\/span><\/span><\/span>\u00a0\u03b7\u00a0<\/b><\/span><\/span><\/span>\u00a0sempre ser\u00e1 <\/b><\/span><\/span><\/span>menor que 1<\/b><\/span><\/span><\/span>, que normalmente \u00e9 <\/b><\/span><\/span><\/span>multiplicado por 100, sendo assim expresso em porcentagem.<\/b><\/span><\/span><\/span><\/p>\n

Podemos agora calcular a <\/b><\/span><\/span><\/span>Pot\u00eancia total da turbina:<\/b><\/span><\/span><\/span><\/p>\n

\u03b7<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>= Pu\/P<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>15\/100 = 40.10<\/b><\/span><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup><\/span> \/ P<\/b><\/span><\/span><\/span><\/span>t <\/b><\/span><\/span><\/span><\/sub><\/span>\"\" <\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span> = 40.10<\/b><\/span><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup><\/span>\/15\/100<\/b><\/span><\/span><\/span><\/span> <\/b>\"\" P<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span> = 4000.10<\/b><\/span><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup><\/span>\/15 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A seguir,<\/b><\/span><\/span><\/span> <\/b>encontramos<\/b><\/span><\/span><\/span> a Energia total consumida em 3 minutos (180s):<\/b><\/span><\/span><\/span><\/p>\n

E = P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span>\uf044<\/b><\/span><\/span><\/span>t \"\" E = (4000.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\/15).180 \"\" E = 4000. 10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>.12 \"\" E = 48000. 10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span> J = 48.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span> J. <\/b><\/span><\/span><\/span>Sendo a<\/b><\/span><\/span><\/span> varia\u00e7\u00e3o de entropia vapor- turbinas (1\/12) GJ\/K,<\/b><\/span><\/span><\/span>conseguimos encontrar<\/b><\/span><\/span><\/span> a temperatura do vapor por uma regra de tr\u00eas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"Utilizando a<\/b><\/span><\/span><\/span> rela\u00e7\u00e3o entre as escala Celsius e Kelvin<\/b><\/span><\/span><\/span>, temos<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

K = C + 273 \"\" <\/b><\/span><\/span><\/span>576 = C + 273<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>C = 576 \u2013 273<\/b><\/span><\/span><\/span> \"\" C = 303 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>Em<\/b><\/span><\/span> <\/b>todo gr\u00e1fico VXt<\/b><\/span><\/span><\/span>,\u00a0 a<\/b><\/span><\/span> <\/b>\u00e1rea <\/b><\/span><\/span><\/span>compreendida entre a reta<\/b><\/span><\/span> <\/b>representativa e o eixo do tempo \u00e9<\/b><\/span><\/span> <\/b>numericamente igual ao espa\u00e7o (\u0394S) percorrido pelo m\u00f3vel.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A<\/b><\/span><\/span><\/span> <\/b>\u00e1rea <\/b><\/span><\/span><\/span>do ret\u00e2ngulo hachurado acima vale<\/b><\/span><\/span><\/span> <\/b>baseXaltura = (t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)XV = \u0394t.V<\/b><\/span><\/span><\/span>\u00a0\"\" \u00e1rea = V.\u0394t\u00a0\"\"\u00a0 <\/b><\/span><\/span><\/span>\u0394S=V.\u0394t\u00a0 \"\"\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

Sendo assim, <\/b><\/span><\/span><\/span>podemos determinar a dist\u00e2ncia percorrida em cada etapa, calculando cada \u00e1rea :<\/b><\/span><\/span><\/span><\/p>\n

1\u00aa etapa<\/b><\/span><\/span><\/span> \"\" \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=b.h = V.\u0394t\u00a0\"\" \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= 55,5.t km<\/b><\/span><\/span><\/span><\/p>\n

2\u00aa etapa<\/b><\/span><\/span><\/span> \"\" \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 0.(t + 2 – t) \"\" \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 0.2 \"\" \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= 0 km<\/b><\/span><\/span><\/span><\/p>\n

3\u00aa etapa<\/b><\/span><\/span><\/span> \"\" \u0394S<\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/sub><\/span>= 72.(t + 6 – t – 2) \"\" \u0394S<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>= 72.4 \"\" \u0394S<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>= 288 km<\/b><\/span><\/span><\/span><\/p>\n

Como<\/b><\/span><\/span><\/span> o trajeto <\/b><\/span><\/span><\/span>entre as cidades<\/b><\/span><\/span><\/span> vale 510 km, <\/b><\/span><\/span><\/span>podemos encontrar<\/b><\/span><\/span><\/span> o valor de t \"\" \u0394S = \u0394S<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>+ \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> + \u0394S<\/b><\/span><\/span><\/span>3 \"\" <\/b><\/span><\/span><\/sub><\/span>510 = 55,5t + 0 + 288 \"\" 510 – 288 = 55,5t \"\" t = 222\/55,5\"\" t = 4 h.<\/b><\/span><\/span><\/span><\/p>\n

Logo,<\/b><\/span><\/span><\/span> o tempo total vale \"\" \u0394t = t + 6 = 4 + 6 \"\" \u0394t = 10 h.<\/b><\/span><\/span><\/span><\/p>\n

Agora resta apenas<\/b><\/span><\/span><\/span> determinar a velocidade m\u00e9dia:<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = \u0394S\/ \u0394t \"\" <\/b><\/span><\/span><\/span>Vm = 510\/10 \"\" <\/b><\/span><\/span><\/span>Vm = 51 km\/h.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/span><\/span><\/span>Na<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>coordenada cartesiana<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>tridimensional,<\/b><\/span><\/span><\/span> <\/b>os elementos <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span>,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>j<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>k<\/b><\/span><\/span>\u00a0 s\u00e3o<\/b><\/span><\/span><\/span> <\/b>vetores unit\u00e1rios<\/b><\/span><\/span><\/span> nas <\/b><\/span><\/span>dire\u00e7\u00f5es<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

dos eixos<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>x<\/b><\/i><\/span><\/span><\/span>,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>y<\/b><\/i><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>z<\/b><\/i><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>respectivamente.<\/b><\/span><\/span><\/span><\/p>\n

No nosso caso, a <\/b><\/span><\/span><\/span>coordenada cartesiana \u00e9 bidimensional e a dist\u00e2ncia horizontal (eixo x) x \u00e9<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

representada pelos valores de i e os do eixo y, pelos valores de j.<\/b><\/span><\/span><\/span><\/p>\n

Como<\/b><\/span><\/span><\/span> j representa o eixo y, <\/b><\/span><\/span><\/span>que corresponde a altura<\/b><\/span><\/span><\/span>, atrav\u00e9s da equa\u00e7\u00e3o podemos determinar a altura no instante t = 1s, usando <\/b><\/span><\/span><\/span>apenas <\/b><\/span><\/span><\/span>os valores de j:<\/b><\/span><\/span><\/span><\/p>\n

r = – 5,0t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2,0t + 8,4 \"\" y = -5,0.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2,0.1 + 8,4 \"\" y = -5 + 2 + 8,4 \"\" y = h = 5,4 m.<\/b><\/span><\/span><\/span><\/p>\n

Na horizontal x, quando t = 1s voc\u00ea ter\u00e1 \"\" x = 6,0t + 2,5 = 6.1 + 2,5 \"\" x = 8,5 m.<\/b><\/span><\/span><\/span><\/p>\n

Assim, ap\u00f3s 1s, a <\/b><\/span><\/span><\/span>part\u00edcula se deslocou<\/b><\/span><\/span><\/span> simultaneamente <\/b><\/span><\/span><\/span>5,4m para cima e 8,5m para a direita e o m\u00f3dulo de seu <\/b><\/span><\/span><\/span>deslocamento d<\/b><\/span><\/span><\/span> foi<\/b><\/span><\/span><\/span> \"\" d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 5,4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 8,5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 29,16 + 72,25 \"\" <\/b><\/span><\/span><\/span>d <\/b><\/span><\/span><\/span>= \u221a(101,41) = 10,07 <\/b><\/span><\/span><\/span>= 10m<\/b><\/span><\/span><\/span><\/p>\n

Velocidade \"\" V = d\/t = 10\/1 \"\" V = 10 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/span><\/span><\/span>Potencial el\u00e9trico<\/b><\/span><\/span><\/span> de um <\/b><\/span><\/span><\/span>condutor esf\u00e9rico eletrizado<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>com carga Q<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"C\u00e1lculo do m\u00f3dulo da carga Q utilizando o potencial el\u00e9trico fornecido no ponto P, V = 10V, distante d = (0,04+1,56) = 1,6 m do centro da carga \"\" V = K.Q\/d \"\" 10 = k.Q\/(0,04 + 1,56) \"\" 10 = k.Q\/1,6 \"\"k.Q = 16<\/b><\/span><\/span><\/span><\/p>\n

Agora, conseguimos obter o valor do <\/b><\/span><\/span><\/span>potencial para o raio da esfera 4 vezes menor<\/b><\/span><\/span><\/span>, no interior da esfera, que \u00e9 <\/b><\/span><\/span>o mesmo que da superf\u00edcie e de intensidade<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = k.Q\/R<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 16\/0,01<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span><\/p>\n

1600 V.<\/b><\/span><\/span><\/span><\/p>\n

Em seguida,<\/b><\/span><\/span><\/span> <\/b>determinamos<\/b><\/span><\/span><\/span> <\/b>tamb\u00e9m <\/b><\/span><\/span><\/span>o potencial externo \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub> = k.Q\/d \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 16\/(0,01+1,56) \"\"<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 16\/1,57 \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 10,2 V<\/b><\/span><\/span><\/span><\/p>\n

Finalmente, conseguimos obter a<\/b><\/span><\/span><\/span> <\/b>ddp entre o centro da casca e o ponto P \"\" ddp = V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> \u2013 V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

ddp = 1600 – 10,2 \"\" ddp = 1589,8 V ~ 1,59 kV<\/b><\/span><\/span><\/span><\/p>\n

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

09- <\/span><\/span><\/span>Um dos <\/b><\/span><\/span>processos pr\u00e1ticos<\/b><\/span><\/span><\/span> para se <\/b><\/span><\/span>determinar a dire\u00e7\u00e3o e o sentido do vetor indu\u00e7\u00e3o magn\u00e9tica\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0ou <\/b><\/span><\/span>vetor campo magn\u00e9tico\"\", \u00e9 a regra da m\u00e3o direita.<\/b><\/span><\/span><\/span><\/b><\/p>\n

Esse <\/b><\/span><\/span>sentido de\"\"\u00a0depende do sentido da corrente que o origina.<\/b><\/span><\/span><\/span><\/p>\n

Voc\u00ea <\/b><\/span><\/span>coloca o polegar no sentido da corrente<\/b><\/span><\/span><\/span> com a <\/b><\/span><\/span>m\u00e3o espalmada (primeira figura), em seguida<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

voc\u00ea fecha a m\u00e3o para pegar o<\/b><\/span><\/span><\/span> <\/b>fio (segunda figura)<\/b><\/span><\/span><\/span> e o <\/b><\/span><\/span>sentid<\/b><\/span><\/span><\/span>o da \u201cfechada\u201d de m\u00e3o \u00e9 o <\/b><\/span><\/span>sentido do vetor\"\"\u00a0(terceira figura). Observe na terceira figura que\u00a0\"\"\u00a0\u00e9 sempre tangente \u00e0s linhas de indu\u00e7\u00e3o em cada ponto.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0Comprova-se <\/b><\/span><\/span>experimentalment<\/b><\/span><\/span><\/span>e que a <\/b><\/span><\/span>intensidade do campo magn\u00e9tico\"\"\u00a0depende da intensidade da corrente el\u00e9trica i, da dist\u00e2ncia r do fio at\u00e9 o ponto (P) onde se quer o campo<\/b><\/span><\/span><\/span> <\/b>magn\u00e9tico e do meio onde o condutor se encontra.<\/b><\/span><\/span><\/span><\/b><\/p>\n

Essa <\/b><\/span><\/span>depend\u00eancia de\"\"\u00a0com o meio \u00e9 fornecida pela constante \u03bc que recebe o nome de permeabilidade magn\u00e9tica do meio e no v\u00e1cuo ela vale \u00a0\u03bc<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sub><\/span>= 4\u03c0.10-7T.m\/A. Matematicamente:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na <\/b><\/span><\/span>express\u00e3o acima<\/b><\/span><\/span><\/span> que, se as correntes <\/b><\/span><\/span>tiverem o mesmo m\u00f3dulo I,<\/b><\/span><\/span><\/span> os <\/b><\/span><\/span>campos magn\u00e9ticos<\/b><\/span><\/span><\/span> <\/b>originados por cada fio<\/b><\/span><\/span><\/span> ter\u00e3o o <\/b><\/span><\/span>mesmo m\u00f3dulo B.<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea <\/b><\/span><\/span>aplicar a regra da m\u00e3o direita para cada fio<\/b><\/span><\/span><\/span>, com as correntes tendo <\/b><\/span><\/span>sentidos contr\u00e1rios,<\/b><\/span><\/span><\/span><\/b><\/p>\n

\"\"<\/p>\n

observar\u00e1<\/b><\/span><\/span><\/span> que, nas <\/b><\/span><\/span>regi\u00f5es superiores e inferiores, os campos magn\u00e9ticos criados por cada fio se anulam (sentidos opostos, entrando e saindo da folha) e de mesmo m\u00f3dulo.<\/b><\/span><\/span><\/span><\/p>\n

Sendo assim, o <\/b><\/span><\/span>campo magn\u00e9tico medido pelo sensor do alicate amper\u00edmetro ser\u00e1 ZERO.<\/b><\/span><\/span><\/span><\/p>\n

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

10- \"\"<\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>sinal do momento<\/b><\/span><\/span><\/span> da for\u00e7a pode ser <\/b><\/span><\/span><\/span>negativo ou positivo<\/b><\/span><\/span><\/span>e, por <\/b><\/span><\/span><\/span>conven\u00e7\u00e3o<\/b><\/span><\/span><\/span>, vamos <\/b><\/span><\/span><\/span>adotar o sentido hor\u00e1rio de rota\u00e7\u00e3o em torno de O como positivo e anti-hor\u00e1rio como negativo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Condi\u00e7\u00f5es de equil\u00edbrio de um corpo extenso<\/b><\/span><\/span><\/span><\/p>\n

S\u00e3o <\/b><\/span><\/span><\/span>duas as condi\u00e7\u00f5es para que um corpo extenso r\u00edgido esteja em equil\u00edbrio:<\/b><\/span><\/span><\/span><\/p>\n

1<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span> <\/b>\u2013 Equil\u00edbrio de transla\u00e7\u00e3o\u00a0 —\u00a0 A <\/b><\/span><\/span><\/span>resultante <\/b><\/span><\/span><\/span>do sistema de for\u00e7as deve ser nula<\/b><\/span><\/span><\/span><\/p>\n

2<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span> \u2013 Equil\u00edbrio de rota\u00e7\u00e3o\u00a0 —\u00a0 A <\/b><\/span><\/span><\/span>soma alg\u00e9brica dos momentos das for\u00e7as<\/b><\/span><\/span><\/span> que agem sobre o sistema, em rela\u00e7\u00e3o \u00e0 qualquer ponto <\/b><\/span><\/span><\/span>(p\u00f3lo O),<\/b><\/span><\/span><\/span>deve ser nula.<\/b><\/span><\/span><\/span><\/p>\n

Resolvendo o <\/b><\/span><\/span>sistema composto<\/b><\/span><\/span><\/span> pelas <\/b><\/span><\/span>duas equa\u00e7\u00f5es acima<\/b><\/span><\/span><\/span> voc\u00ea chega \u00e0 resolu\u00e7\u00e3o do exerc\u00edcio.<\/b><\/span><\/span><\/p>\n

No caso do <\/b><\/span><\/span>nosso exerc\u00edcio, colocando o <\/b><\/span><\/span><\/span>p\u00f3lo (eixo de rota\u00e7\u00e3o) na dobradi\u00e7a, em 0<\/b><\/span><\/span><\/span>, estabelecendo o <\/b><\/span><\/span><\/span>sentido hor\u00e1rio <\/b><\/span><\/span><\/span>de rota\u00e7\u00e3o como <\/b><\/span><\/span><\/span>positivo e aplicando a 2\u00aa condi\u00e7\u00e3o (equil\u00edbrio de rota\u00e7\u00e3o)<\/b><\/span><\/span><\/span>, temos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Agora, <\/b><\/span><\/span>aplicando a <\/b><\/span><\/span><\/span>1\u00aa condi\u00e7\u00e3o<\/b><\/span><\/span><\/span> (equil\u00edbrio de transla\u00e7\u00e3o), temos:<\/b><\/span><\/span><\/span><\/p>\n

T + D = F<\/b><\/span><\/span><\/span><\/p>\n

12 + D = 36<\/b><\/span><\/span><\/span><\/p>\n

D = 36 – 12<\/b><\/span><\/span><\/span><\/p>\n

D = 24 N<\/b><\/span><\/span><\/span><\/p>\n

R – C<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios da Escola Naval – 015\/016 01-   \u00a0O\u00a0trabalho das for\u00e7as\u00a0conservativas,\u00a0como, por exemplo, as\u00a0for\u00e7as peso, da el\u00e1stica , da\u00a0el\u00e9trica e da\u00a0magn\u00e9tica\u00a0n\u00e3o dependem da trajet\u00f3ria, mas apenas das\u00a0posi\u00e7\u00f5es inicial e final da mesma.   \u00a0Gr\u00e1ficos das\u00a0energias cin\u00e9tica, potencial gravitacional e mec\u00e2nica\u00a0para um corpo demassa\u00a0m,\u00a0quando\u00a0lan\u00e7ado verticalmente para cima, a partir doponto de lan\u00e7amento, tomado como\u00a0referencial e desprezando-se as for\u00e7as<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1499,"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-1501","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1501","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=1501"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1501\/revisions"}],"predecessor-version":[{"id":1508,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1501\/revisions\/1508"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1499"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1501"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}