{"id":2787,"date":"2017-03-15T05:04:52","date_gmt":"2017-03-15T05:04:52","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2787"},"modified":"2024-09-03T14:20:48","modified_gmt":"2024-09-03T14:20:48","slug":"vestibular-da-espcex-aman-2016-2017-resolucao","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/espcex-aman-20162017\/vestibular-da-espcex-aman-2016-2017-resolucao\/","title":{"rendered":"Vestibular da EsPCEx-Aman-2016\/2017 – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica do vestibular da <\/span><\/span><\/span><\/p>\n

<\/a> EsPCEx-Aman-2016\/2017<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>Veja na figura da esquerda os \u00e2ngulos que confirmam que o \u00e2ngulo de incid\u00eancia do raio de luz na interface ar – l\u00edquido (ponto A) vale i = 60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Aplicando a lei de <\/b><\/span><\/span><\/span>Snell \u2013 Descartes <\/b><\/span><\/span><\/span>na figura da direita acima no <\/b><\/span><\/span><\/span>ponto A <\/b><\/span><\/span><\/span>da <\/b><\/span><\/span><\/span>interface ar \u2013 l\u00edquido<\/b><\/span><\/span><\/span><\/p>\n

N<\/b><\/span><\/span><\/span>ar<\/b><\/span><\/span><\/sub><\/span>.seni = n<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.senr \"\" <\/b><\/span><\/span><\/span>1.sen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = n<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> \"\" 1.\"\"<\/b><\/span><\/span><\/span> = n<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span> =\"\"<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

Trem de comprimento (x = 150 m) <\/b><\/span><\/span><\/span>atravessando totalmente um <\/b><\/span><\/span><\/span>t\u00fanel de comprimento (c=<\/b><\/span><\/span><\/span>?) \"\" observe que para <\/b><\/span><\/span><\/span>atravessar completamente o t\u00fanel<\/b><\/span><\/span><\/span>, um <\/b><\/span><\/span><\/span>ponto P fixo em qualquer parte do trem <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

deve <\/b><\/span><\/span><\/span>percorrer uma dist\u00e2ncia \u0394S = 150 + c,<\/b><\/span><\/span><\/span> com <\/b><\/span><\/span><\/span>velocidade V num intervalo de tempo \u0394t, tal que \"\"<\/b><\/span><\/span><\/span><\/p>\n

V = \"\" \"\"<\/b><\/span><\/span><\/span> 16 = \"\" \"\" 150 + c = 800 \"\" c = 800 \u2013 150 \"\" <\/b><\/span><\/span><\/span>c = 650 m<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

No <\/b><\/span><\/span><\/span>equil\u00edbrio de transla\u00e7\u00e3o, <\/b><\/span><\/span><\/span>para que ocorra equil\u00edbrio na <\/b><\/span><\/span><\/span>vertical a resultante das for\u00e7as <\/b><\/span><\/span><\/span>deve ser <\/b><\/span><\/span><\/span>nula,<\/b><\/span><\/span><\/span> ou seja, <\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> + N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = N<\/b><\/span><\/span><\/span>N <\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> 75 + 150 = N<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> N<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> = 225 N<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>intensidade da for\u00e7a F aplicada na polia A,<\/b><\/span><\/span><\/span> chega ao cubo com <\/b><\/span><\/span><\/span>intensidade \"\"<\/b><\/span><\/span><\/span>,<\/b><\/span><\/span><\/span> dire\u00e7\u00e3o <\/b><\/span><\/span><\/span>vertical<\/b><\/span><\/span><\/span> e sentido <\/b><\/span><\/span><\/span>para cima <\/b><\/span><\/span><\/span>(veja figura)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A <\/b><\/span><\/span><\/span>intensidade do empuxo <\/b><\/span><\/span><\/span>(for\u00e7a vertical e para cima que equivale ao peso do volume de l\u00edquido deslocado) \u00e9 <\/b><\/span><\/span><\/span>fornecida por:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o <\/b><\/span><\/span><\/span>cubo est\u00e1 totalmente imerso o volume de l\u00edquido deslocado corresponde ao pr\u00f3prio volume do cubo, <\/b><\/span><\/span><\/span>sendo <\/b><\/span><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> a densidade do l\u00edquido e V o volume do cubo \"\"<\/b><\/span><\/span><\/span> <\/b>E = \u03c1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V.g.<\/b><\/span><\/span><\/span><\/p>\n

Peso do cubo <\/b><\/span><\/span><\/span>(vertical e para baixo) \"\" <\/b><\/span><\/span><\/span>P = mg <\/b><\/span><\/span><\/span>=d<\/b><\/span><\/span><\/span>cubo<\/b><\/span><\/span><\/sub><\/span>.V.d \"\" <\/b><\/span><\/span><\/span>P = \u03c1.V.g.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>mola de constante el\u00e1stica K <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>deformada de X <\/b><\/span><\/span><\/span>puxa o cubo <\/b><\/span><\/span><\/span>para baixo com uma for\u00e7a el\u00e1stica <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>intensidade F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = KX.<\/b><\/span><\/span><\/span><\/p>\n

Estando, pelo enunciado, o <\/b><\/span><\/span><\/span>cubo em equil\u00edbrio, <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>for\u00e7a resultante sobre ele deve ser nula \"\" \"\" + E = P + F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>+<\/b><\/span><\/span><\/span> <\/b>\u03c1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V.g = \u03c1.V.g + KX \"\" F = 2\u03c1.V.g + 2KX – 2\u03c1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V.g \"\" <\/b><\/span><\/span><\/span>F = 2[Vg(\u03c1 \u2013 \u03c1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) + KX].<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Observe que o dispositivo <\/b><\/span><\/span><\/span>de 8V (maior valor) \u00e9 o gerador <\/b><\/span><\/span><\/span>(a corrente sai de seu polo positivo) e o de <\/b><\/span><\/span><\/span>6 V \u00e9 o receptor <\/b><\/span><\/span><\/span>(a corrente sai de seu polo negativo) \"\" <\/b><\/span><\/span><\/span>U <\/b><\/span><\/span><\/span>= 8 \u2013 6 <\/b><\/span><\/span><\/span>= 2V.<\/b><\/span><\/span><\/span><\/p>\n

Os <\/b><\/span><\/span><\/span>resistores est\u00e3o associados em s\u00e9rie \"\" R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> <\/b>= 3 + 3 + 4 <\/b><\/span><\/span><\/span>= 10 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>10 =<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>i = 0,2 A (corrente em cada resistor) \"\" P<\/b><\/span><\/span><\/span>4\u2126<\/b><\/span><\/span><\/sub><\/span> = R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>= 4.(0,2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 4.0,04 \"\" <\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>4\u2126<\/b><\/span><\/span><\/sub><\/span> = 0,16 W.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>S\u00e3o dados \"\" m =<\/b><\/span><\/span><\/span> 4 kg \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> 0 (inicialmente em repouso) \"\" <\/b><\/span><\/span><\/span>t =<\/b><\/span><\/span><\/span> 3 s \"\" <\/b><\/span><\/span><\/span>\u2206S = <\/b><\/span><\/span><\/span>9m. <\/b><\/span><\/span><\/span>Equa\u00e7\u00e3o hor\u00e1ria <\/b><\/span><\/span><\/span>dos <\/b><\/span><\/span><\/span>espa\u00e7os do MUV <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span> \"\" S = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + \"\"<\/b><\/span><\/span><\/span> \"\" S \u2013 S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206S = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>9 = 0.3 + \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>18 = 9.a \"\" <\/b><\/span><\/span><\/span>a = 2 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Lei fundamental da din\u00e2mica <\/b><\/span><\/span><\/span>(segunda lei de Newton) \"\" <\/b><\/span><\/span><\/span>F = m.a <\/b><\/span><\/span><\/span>= 4,2 \"\" <\/b><\/span><\/span><\/span>F = 8 N.<\/b><\/span><\/span><\/span><\/p>\n

Velocidade V no instante t = 3 s \"\" <\/b><\/span><\/span><\/span> <\/b>V<\/b><\/span><\/span><\/span> <\/b>= V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + a.t <\/b><\/span><\/span><\/span>= 0 + 2.3 \"\" <\/b><\/span><\/span><\/span>V = 6 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Impulso \"\" <\/b><\/span><\/span><\/span> <\/b>I = F.\u2206t <\/b><\/span><\/span><\/span>= 8.3 <\/b><\/span><\/span><\/span>I = 24 N.s.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de movimento Q <\/b><\/span><\/span><\/span>no instante t = 3 s quando sua <\/b><\/span><\/span><\/span>velocidade \u00e9 V = 6 m\/<\/b><\/span><\/span><\/span>s \"\" <\/b><\/span><\/span><\/span>Q = m.V <\/b><\/span><\/span><\/span>= 4.6<\/b><\/span><\/span><\/span><\/p>\n

Q = 24 kg.m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Dire\u00e7\u00e3o e sentido do campo magn\u00e9tico<\/b><\/span><\/span>\u00a0\"\" originado por um fio retil\u00edneo percorrido por corrente el\u00e9trica i<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0Um dos<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>processos pr\u00e1ticos<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>para se determinar a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>dire\u00e7\u00e3o e o sentido do vetor indu\u00e7\u00e3o magn\u00e9tica\u00a0\"\"\u00a0ou vetor campo magn\u00e9tico<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>, \u00e9 a regra da m\u00e3o direita<\/b><\/span><\/span>. Esse<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>sentido de<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0depende do sentido da corrente que o origina. <\/b><\/span><\/span><\/span><\/span><\/p>\n

Voc\u00ea<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>coloca o polegar no sentido da corrente com a m\u00e3o espalmada (primeira figura),<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>em seguida<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

voc\u00ea<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>fecha a m\u00e3o para pegar o\u00a0fio (segunda figura)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>e o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>sentido da \u201cfechada\u201d de m\u00e3o \u00e9 o sentido do vetor<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0(terceira figura).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe na<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>terceira figura que \u00a0\"\"\u00a0\u00e9 sempre tangente \u00e0s linhas de indu\u00e7\u00e3o em cada ponto.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Intensidade do campo magn\u00e9tico<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Comprova-se<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>experimentalmente que a intensidade do campo magn\u00e9tico<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>depende da<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>intensidade da corrente el\u00e9trica i,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>da<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>dist\u00e2ncia r do fio at\u00e9 o ponto (P) onde se quer o campo magn\u00e9tico e do meio onde o condutor se encontra.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Essa<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>depend\u00eancia de<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0com o meio<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00e9 fornecida pela<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>constante \u03bc que recebe o nome de permeabilidade magn\u00e9tica do meio e no v\u00e1cuo ela vale \u03bc<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 4\u03c0.10<\/b><\/span><\/span>-7<\/b><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span>T.m\/A. Matematicamente:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Agora vamos calcular o campo el\u00e9trico no interior da espira circular:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Dire\u00e7\u00e3o e sentido do vetor campo magn\u00e9tico\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>no interior de uma espira circular<\/b><\/span><\/span><\/span><\/span><\/p>\n

Na<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>figura da esquerda<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>voc\u00ea observa um<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>condutor sob forma de espira circular com centro O e raio R<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>sendo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>percorrido por uma corrente el\u00e9trica de intensidade i<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Em<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>torno da espira (figura da direita)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>surge um<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>campo magn\u00e9tico<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>cuja <\/b><\/span><\/span>dire\u00e7\u00e3o e sentido<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00e9 fornecido pela<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>regra da m\u00e3o direita<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(voc\u00ea coloca o polegar no sentido da corrente com a m\u00e3o espalmada, em seguida voc\u00ea fecha a m\u00e3o como se fosse pegar o pegar o fio e o sentido da \u201cfechada\u201d de m\u00e3o \u00e9 o sentido do vetor<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>),\u00a0<\/b><\/span><\/span>considerando cada pequeno trecho da circunfer\u00eancia como sendo um condutor retil\u00edneo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe na figura da direita que no<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>plano do c\u00edrculo,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>todas as<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>linhas de for\u00e7a t\u00eam sentidos coincidentes e que no centro da espira, em qualquer posi\u00e7\u00e3o que voc\u00ea use a regra da m\u00e3o direita<\/b><\/span><\/span>, o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>campo magn\u00e9tico<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\u00e9 perpendicular ao plano do papel e, nesse exemplo, entrando nele.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Intensidade de \"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

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

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

Transforma\u00e7\u00e3o adiab\u00e1tica<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Nela, o\u00a0<\/b><\/span><\/span><\/span>sistema n\u00e3o troca calor com o meio externo (Q = 0) e isso ocorre porque o g\u00e1s est\u00e1 termicamente isolado ou porque a transforma\u00e7\u00e3o \u00e9 muito r\u00e1pida de modo que qualquer troca de calor com possa ser considerada.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Se Q = 0\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394U = Q \u2013 W\u00a0\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0 \u0394U = 0 \u2013 W\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394U = \u2013 W.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

09- <\/span><\/span><\/span>Como <\/b><\/span><\/span><\/span>o elevador est\u00e1 subindo com velocidade constante, <\/b><\/span><\/span><\/span>ele est\u00e1 em <\/b><\/span><\/span><\/span>MU e a for\u00e7a F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> <\/b>aplicada pelo motor do elevador para <\/b><\/span><\/span><\/span>elev\u00e1-lo de h = 20 m \u00e9 o seu peso P = m.g<\/b><\/span><\/span><\/span> = 6.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10 \"\" <\/b><\/span><\/span><\/span>P = 6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> N.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>energia dissipada pelo elevador <\/b><\/span><\/span><\/span>nessa subida <\/b><\/span><\/span><\/span>(trabalho W<\/b><\/span><\/span><\/span>) \u00e9 a energia <\/b><\/span><\/span><\/span>potencial gravitacional \"\" W =<\/b><\/span><\/span><\/span> <\/b>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = P.h <\/b><\/span><\/span><\/span>= 6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.20 \"\" <\/b><\/span><\/span><\/span>W = 12.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> J.<\/b><\/span><\/span><\/span><\/p>\n

P = \"\"<\/b><\/span><\/span><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span> = 12.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> = 120.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> W <\/b><\/span><\/span><\/span>= 120 kW.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

Observe que a alternativa correta \u00e9 a C, cuja an\u00e1lise \u00e9 feita a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span><\/span>sequ\u00eancia abaixo <\/b><\/span><\/span><\/span>\u00e9 fornecido o <\/b><\/span><\/span><\/span>c\u00e1lculo da corrente el\u00e9trica total i.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

11- <\/span><\/span><\/span>Chamando de <\/b><\/span><\/span><\/span>A o ponto onde a altura em rela\u00e7\u00e3o ao solo \u00e9 de h = 4m<\/b><\/span><\/span><\/span> e onde a <\/b><\/span><\/span><\/span>esfera de massa m = 0,8 kg \u00e9 abandonada (V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 0),<\/b><\/span><\/span><\/span> vamos calcular a <\/b><\/span><\/span><\/span>energia mec\u00e2nica total nesse ponto que vale E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>pgA<\/b><\/span><\/span><\/sub><\/span> = \"\" + mgh <\/b><\/span><\/span><\/span>= 0,8.\"\" + 0,8.10.4 \"\" E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = 0 + 32 \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = 32 J.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

C\u00e1lculo da <\/b><\/span><\/span><\/span>energia mec\u00e2nica <\/b><\/span><\/span><\/span>da esfera no <\/b><\/span><\/span><\/span>ponto B<\/b><\/span><\/span><\/span>, quando <\/b><\/span><\/span><\/span>a mola <\/b><\/span><\/span><\/span>j\u00e1 est\u00e1 <\/b><\/span><\/span><\/span>comprimida de X, m\u00e1xima<\/b><\/span><\/span><\/span> <\/b>deforma\u00e7\u00e3o <\/b><\/span><\/span><\/span>que ocorre quando a <\/b><\/span><\/span><\/span>velocidade da esfera \u00e9 nula (E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span> = 0).<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Nesse ponto a <\/b><\/span><\/span><\/span>energia potencial gravitacional tamb\u00e9m \u00e9 nula (h = 0). <\/b><\/span><\/span><\/span>Portanto no <\/b><\/span><\/span><\/span>ponto B <\/b><\/span><\/span><\/span>(compress\u00e3o m\u00e1xima da mola deformada de X) a <\/b><\/span><\/span><\/span>energia do sistema \u00e9 apenas a energia potencial el\u00e1stica <\/b><\/span><\/span><\/span>da mola de <\/b><\/span><\/span><\/span>constante el\u00e1stica k = 400 N\/m \"\" E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span> = k.\"\"<\/b><\/span><\/span><\/span> = 400.\"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB <\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/sub><\/span> <\/b>E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span> = 200X<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Supondo o <\/b><\/span><\/span><\/span>sistema conservativo <\/b><\/span><\/span><\/span>pode-se aplicar o <\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da energia mec\u00e2nica igualando E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> com E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> 32 = 200X<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" X = \"\" \"\" <\/b><\/span><\/span><\/span>X = <\/b><\/span><\/span><\/span>0,4 m <\/b><\/span><\/span><\/span>= 40 cm.<\/b><\/span><\/span><\/span><\/p>\n

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

12- <\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>for\u00e7a peso <\/b><\/span><\/span><\/span>que age <\/b><\/span><\/span><\/span>sobre a part\u00edcula <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>vertical e para baixo <\/b><\/span><\/span><\/span>e tem <\/b><\/span><\/span><\/span>intensidade<\/b><\/span><\/span><\/span> <\/b>P = m.g<\/b><\/span><\/span><\/span> =10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.10 \"\"<\/b><\/span><\/span><\/span><\/p>\n

P = 10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span> N.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/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 das quest\u00f5es de F\u00edsica do vestibular da EsPCEx-Aman-2016\/2017 01- Veja na figura da esquerda os \u00e2ngulos que confirmam que o \u00e2ngulo de incid\u00eancia do raio de luz na interface ar – l\u00edquido (ponto A) vale i = 60o. Aplicando a lei de Snell \u2013 Descartes na figura da direita acima no ponto A da interface ar \u2013 l\u00edquido<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2785,"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-2787","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2787","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=2787"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2787\/revisions"}],"predecessor-version":[{"id":11038,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2787\/revisions\/11038"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2785"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2787"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}