{"id":4716,"date":"2019-09-10T00:58:50","date_gmt":"2019-09-10T00:58:50","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4716"},"modified":"2024-07-11T13:59:34","modified_gmt":"2024-07-11T13:59:34","slug":"unicap-medicina-019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2019\/unicap-medicina-019\/","title":{"rendered":"UNICAP \u2013 MEDICINA – 019"},"content":{"rendered":"

UNICAP \u2013 MEDICINA – 019<\/b><\/span><\/span><\/span><\/p>\n

<\/a> \"UNICAP\"<\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>Universidade Cat\u00f3lica de Pernambuco<\/b><\/span><\/span><\/span>\u00a0(<\/b><\/span><\/span><\/span>UNICAP)<\/b><\/span><\/span><\/span>, popularmente conhecida apenas como a\u00a0<\/b><\/span><\/span><\/span>Cat\u00f3lica<\/b><\/span><\/span><\/span>) \u00e9 uma institui\u00e7\u00e3o de ensino superior\u00a0privada<\/a>,\u00a0filantr\u00f3pica<\/a>\u00a0e\u00a0confessional<\/a>, localizada em\u00a0Recife<\/a>.<\/b><\/span><\/span><\/span><\/p>\n

A institui\u00e7\u00e3o oferece, atualmente, 36 cursos de gradua\u00e7\u00e3o, sendo o curso de Direito o com mais alunos matriculados e o de Ci\u00eancias Econ\u00f4micas o mais antigo, criado em 1943.<\/b><\/span><\/span><\/span><\/p>\n

Constitui-se no maior complexo educacional dos jesu\u00edtas no Brasil,\u00a0com aproximadamente quinze mil estudantes. <\/b><\/span><\/span><\/span><\/p>\n

Na \u00e1rea da gradua\u00e7\u00e3o, registram-se, nos diversos cursos, cerca de cinquenta mil diplomados pela institui\u00e7\u00e3o, ao longo dos sessenta anos de atividades ininterruptas.<\/b><\/span><\/span><\/span><\/p>\n

A Unicap recebeu 33 estrelas de certifica\u00e7\u00e3o de qualidade. Um reconhecimento nacional pelo trabalho da institui\u00e7\u00e3o no ramo de ensino.<\/b><\/span><\/span><\/span><\/p>\n

A Universidade Cat\u00f3lica de Pernambuco comemora o excelente desempenho na primeira edi\u00e7\u00e3o da Avalia\u00e7\u00e3o Nacional Seriada dos Estudantes de Medicina.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

01-(MEDICINA – Universidade Cat\u00f3lica de Pernambuco \u2013 UNICAP \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

0 \u2013 0<\/b><\/span><\/span><\/span> <\/b><\/span><\/span><\/span>Duas part\u00edculas A e B se movem na mesma trajet\u00f3ria retil\u00ednea,<\/b><\/span><\/span><\/span> com as <\/b><\/span><\/span><\/span>seguintes equa\u00e7\u00f5es: <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/sub><\/span> = 5 + 5<\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>(<\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>) e <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/sub><\/span> = 15 + 10<\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>(<\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>).<\/b><\/span><\/span><\/span> <\/b><\/p>\n

No <\/b><\/span><\/span><\/span>instante <\/b><\/span><\/span><\/span>t = 2s<\/b><\/i><\/span><\/span><\/span>, a posi\u00e7\u00e3o da part\u00edcula <\/b><\/span><\/span><\/span>B <\/b><\/i><\/span><\/span><\/span>relativa \u00e0 part\u00edcula <\/b><\/span><\/span><\/span>A <\/b><\/i><\/span><\/span><\/span>\u00e9 30m.<\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 <\/b><\/span><\/span><\/span> Uma <\/b><\/span><\/span><\/span>part\u00edcula se move com movimento retil\u00edneo uniformemente variado. <\/b><\/span><\/span><\/span>Sabe-se que, <\/b><\/span><\/span><\/span>em <\/b><\/span><\/span><\/span>t = 5s<\/b><\/i><\/span><\/span><\/span>, sua velocidade \u00e9 <\/b><\/span><\/span><\/span>10m\/s <\/b><\/i><\/span><\/span><\/span>e que, em <\/b><\/span><\/span><\/span>t = 15s<\/b><\/i><\/span><\/span><\/span>, a sua velocidade \u00e9 <\/b><\/span><\/span><\/span>30m\/s<\/b><\/i><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Sua velocidade inicial \u00e9 zero.<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 <\/b><\/span><\/span><\/span>Um <\/b><\/span><\/span><\/span>avi\u00e3o voa horizontalmente com uma velocidade de <\/b><\/span><\/span><\/span>100m\/s<\/b><\/i><\/span><\/span><\/span>, quando abandona uma bomba, que atinge o solo em <\/b><\/span><\/span><\/span>10s<\/b><\/i><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>altura do avi\u00e3o no instante em que abandonou a bomba \u00e9 <\/b><\/span><\/span><\/span>500m<\/b><\/i><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 <\/b><\/span><\/span><\/span> Um <\/b><\/span><\/span><\/span>movimento retardado pode ser progressivo.<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 4 <\/b><\/span><\/span><\/span> <\/b>Duas polias de raios R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10 cm e R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 20 cm est\u00e3o ligadas entre si por uma correia <\/b><\/span><\/span><\/span>(observe a ilustra\u00e7\u00e3o abaixo). <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Se a <\/b><\/span><\/span><\/span>polia de raio R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> executa 25 r.p.m, a polia de raio R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> executa 50 r.p.m.<\/b><\/span><\/span><\/span><\/p>\n

02-(MEDICINA – Universidade Cat\u00f3lica de Pernambuco \u2013 UNICAP \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

0 \u2013 0 Aten\u00e7\u00e3o para a figura que segue:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

No <\/b><\/span><\/span><\/span>sistema da figura, a corda e a roldana s\u00e3o ideais. <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>for\u00e7a tem m\u00f3dulo 30N.<\/b><\/span><\/span><\/span><\/p>\n

Sendo <\/b><\/span><\/span><\/span>m<\/b><\/i><\/span><\/span><\/span>A<\/b><\/i><\/span><\/span><\/sub><\/span> = 2Kg<\/b><\/i><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>m<\/b><\/i><\/span><\/span><\/span>B<\/b><\/i><\/span><\/span><\/sub><\/span> = 4Kg <\/b><\/i><\/span><\/span><\/span>e o atrito desprez\u00edvel, podemos afirmar que o bloco A se move com acelera\u00e7\u00e3o de m\u00f3dulo 15m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 <\/b><\/span><\/span><\/span>Uma <\/b><\/span><\/span><\/span>for\u00e7a de m\u00f3dulo 10N empurra um corpo contra uma parede vertical.<\/b><\/span><\/span><\/span><\/p>\n

Sendo <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/sub><\/span> <\/b>= 0,2 <\/b><\/span><\/span><\/span>e sabendo que o <\/b><\/span><\/span><\/span>corpo se encontra em repouso, o peso do corpo \u00e9 2N.<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 <\/b><\/span><\/span><\/span>Um <\/b><\/span><\/span><\/span>bloco se move com velocidade de m\u00f3dulo V, <\/b><\/span><\/span><\/span>quando encontra uma <\/b><\/span><\/span><\/span>regi\u00e3o com atrito de coeficiente cin\u00e9tico <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span> .<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>bloco chega ao repouso depois de percorrer uma dist\u00e2ncia igual a <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span>\u0001\u0001.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 <\/b><\/span><\/span><\/span>Uma <\/b><\/span><\/span><\/span>for\u00e7a \u00e9 aplicada a um bloco de massa 4Kg<\/b><\/span><\/span><\/span> (veja a figura ilustrativa abaixo).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

O <\/b><\/span><\/span><\/span>bloco sobe o plano inclinado com uma acelera\u00e7\u00e3o de m\u00f3dulo 2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, partindo do repouso no ponto A. <\/b><\/span><\/span><\/span><\/p>\n

A pot\u00eancia m\u00e9dia da for\u00e7a em (plano inclinado) \u00e9 200W.<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 4 <\/b><\/span><\/span><\/span> Com <\/b><\/span><\/span><\/span>refer\u00eancia ao item anterior, a pot\u00eancia da for\u00e7a <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span> <\/b>\u20d7<\/b><\/span><\/span><\/span> no instante correspondente ao ponto B \u00e9 400W.<\/b><\/span><\/span><\/span><\/p>\n

03-(MEDICINA – Universidade Cat\u00f3lica de Pernambuco \u2013 UNICAP \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

0 \u2013 0 <\/b><\/span><\/span><\/span>Um <\/b><\/span><\/span><\/span>cilindro de madeira flutua verticalmente em \u00e1gua com metade de seu comprimento<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

submerso. <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>densidade da madeira \u00e9 metade da densidade da \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>temperatura de uma subst\u00e2ncia aumenta de 20 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>. Na <\/b><\/span><\/span><\/span>escala Kelvin, esse<\/b><\/span><\/span><\/span><\/p>\n

aumento \u00e9 igual a 20K.<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 <\/b><\/span><\/span><\/span> <\/b>540g de gelo a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>misturado com 540g de \u00e1gua a 80<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C. <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>temperatura final da mistura \u00e9 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 <\/b><\/span><\/span><\/span> A <\/b><\/span><\/span><\/span>quantidade de calor necess\u00e1ria para converter 2g de gelo a \u201310 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C em \u00e1gua a 80 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C \u00e9 600 cal.<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 4 <\/b><\/span><\/span><\/span> Um <\/b><\/span><\/span><\/span>sistema absorve 2Kcal de calor<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>realiza um trabalho de 500J<\/b><\/span><\/span><\/span>; a <\/b><\/span><\/span><\/span>varia\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

de sua energia interna \u00e9 de 7900J.<\/b><\/span><\/span><\/span><\/p>\n

04-(MEDICINA – Universidade Cat\u00f3lica de Pernambuco \u2013 UNICAP \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

0 \u2013 0 <\/b><\/span><\/span><\/span> Uma <\/b><\/span><\/span><\/span>part\u00edcula de carga igual a 2<\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>c \u00e9 abandonada em repouso em um campo el\u00e9trico uniforme de m\u00f3dulo 4 N\/C.<\/b><\/span><\/span><\/span><\/p>\n

Desprezando a a\u00e7\u00e3o da gravidade, o trabalho realizado pelo campo el\u00e9trico em um descolamento espont\u00e2neo de 10m \u00e9 80 <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>J.<\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 <\/b><\/span><\/span><\/span> <\/b>Quatro cargas el\u00e9tricas q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10 <\/b><\/span><\/span><\/span>\u0001C<\/b><\/span><\/span><\/span>, q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 5<\/b><\/span><\/span><\/span> \u0001C<\/b><\/span><\/span><\/span>, q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = \u20133<\/b><\/span><\/span><\/span> \u0001C<\/b><\/span><\/span><\/span> e q<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span> = 8<\/b><\/span><\/span><\/span> \u0001C<\/b><\/span><\/span><\/span> s\u00e3o colocadas nos v\u00e9rtices de um quadrado <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>lado \u221a2m. <\/b><\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>potencial el\u00e9trico relativo ao infinito, no centro do<\/b><\/span><\/span><\/span><\/p>\n

quadrado, \u00e9 1,8.10<\/b><\/span><\/span><\/span>5 <\/b><\/span><\/span><\/sup><\/span>V.<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 <\/b><\/span><\/span><\/span> O <\/b><\/span><\/span><\/span>circuito da figura abaixo \u00e9 ideal.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>d.d.P (V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) entre os pontos A e B \u00e9 4\/3 V.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 <\/b><\/span><\/span><\/span> Com <\/b><\/span><\/span><\/span>refer\u00eancia ao item anterior, a pot\u00eancia total dissipada no circuito \u00e9 10 W.<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 4 <\/b><\/span><\/span><\/span> A <\/b><\/span><\/span><\/span>curva caracter\u00edstica de um motor el\u00e9trico \u00e9 representada na figura que segue<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A f A <\/b><\/span><\/span><\/span>for\u00e7a contraeletromotriz do motor \u00e9 12V.<\/b><\/span><\/span><\/span><\/p>\n

05-(MEDICINA – Universidade Cat\u00f3lica de Pernambuco \u2013 UNICAP \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

0 \u2013 0 <\/b><\/span><\/span><\/span> O <\/b><\/span><\/span><\/span>tempo necess\u00e1rio para a luz passar atrav\u00e9s de uma placa de vidro, <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>\u00edndice de<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

refra\u00e7\u00e3o 1,5 e espessura 2mm, \u00e9 10<\/b><\/span><\/span><\/span>\u201310<\/b><\/span><\/span><\/sup><\/span> s.<\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 <\/b><\/span><\/span><\/span> O <\/b><\/span><\/span><\/span>comprimento de onda da luz em dois meios <\/b><\/span><\/span><\/span>A <\/b><\/i><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>B <\/b><\/i><\/span><\/span><\/span>s\u00e3o 3.500 A<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> e 7000 A<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>, respectivamente. O \u00e2ngulo cr\u00edtico \u00e9 45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 <\/b><\/span><\/span><\/span> Um <\/b><\/span><\/span><\/span>espelho c\u00f4ncavo, de dist\u00e2ncia focal igual a 50cm, <\/b><\/span><\/span><\/span>produz uma <\/b><\/span><\/span><\/span>imagem invertida e duas vezes maior que o objeto.<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span><\/span>dist\u00e2ncia do objeto ao espelho \u00e9 75 cm.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 <\/b><\/span><\/span><\/span> Uma <\/b><\/span><\/span><\/span>part\u00edcula carregada penetra em uma regi\u00e3o onde existe um campo magn\u00e9tico uniforme, paralelamente \u00e0s linhas de campo. Seu movimento \u00e9 circular uniforme<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 4<\/b><\/span><\/span><\/span> Em uma <\/b><\/span><\/span><\/span>onda eletromagn\u00e9tica, os campos el\u00e9tricos e magn\u00e9ticos s\u00e3o perpendiculares entre si, e oscilam com a mesma frequ\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica da <\/span><\/span><\/span>Universidade Cat\u00f3lica de Pernambuco \u2013 UNICAP \u2013 Medicina \u2013 2019<\/b><\/span><\/span><\/span><\/p>\n

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

0 \u2013 0 Falsa (II).<\/b><\/span><\/span><\/span><\/p>\n

<\/span> quando t = 2 s a posi\u00e7\u00e3o de A vale S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 5 + 5t <\/b><\/span><\/span><\/span>= 5 + 5.2 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 5 + 10 <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 15 m.<\/b><\/span><\/span><\/span><\/p>\n

quando t = 2 s a posi\u00e7\u00e3o de B vale S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 15 + 10t <\/b><\/span><\/span><\/span>= 15 + 10.2 S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 15 + 20 <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 35 m.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>posi\u00e7\u00e3o de B em rela\u00e7\u00e3o a de A \u00e9 S<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> <\/b>= 35 \u2013 15 <\/b><\/span><\/span><\/span>= 20 m<\/b><\/span><\/span><\/span><\/p>\n

1 – 1 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

quando t = 5 s <\/b><\/span><\/span><\/span>sua <\/b><\/span><\/span><\/span>velocidade vale V = 10 m\/s V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + at <\/b><\/span><\/span><\/span> 10 = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + a.5 <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 10 \u2013 5.a <\/b><\/span><\/span><\/span><\/p>\n

quando t = 15 s <\/b><\/span><\/span><\/span>sua <\/b><\/span><\/span><\/span>velocidade vale V = 30 m\/s V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + at <\/b><\/span><\/span><\/span> 30 = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + a.15 <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 30 \u2013 15.a<\/b><\/span><\/span><\/span><\/p>\n

Igualando as duas equa\u00e7\u00f5es acima 10 \u2013 5.a = 30 \u2013 15.a <\/b><\/span><\/span><\/span> 10.a = 20 <\/b><\/span><\/span><\/span>a = 2 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Substituindo <\/b><\/span><\/span><\/span>a = 2 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> em cada uma das duas equa\u00e7\u00f5es acima V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> <\/b>= 10 \u2013 5.2 = 30 \u2013 15.2 <\/b><\/span><\/span><\/span>= 0<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

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

Lan\u00e7amento horizontal<\/b><\/span><\/span><\/span><\/span><\/p>\n

Colocando-se a\u00a0origem do sistema de refer\u00eancia<\/b><\/span><\/span>\u00a0no\u00a0<\/b><\/span><\/span>ponto de lan\u00e7amento, orienta-se<\/b><\/span><\/span>, por exemplo, o\u00a0<\/b><\/span><\/span>eixo X para a direita e o eixo Y para baixo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Decomp\u00f5e-se, em cada instante\u00a0<\/b><\/span><\/span>o movimento em\u00a0<\/b><\/span><\/span>duas parcelas:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Segundo o eixo X\u00a0<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>trata-se de um\u00a0movimento horizontal uniforme\u00a0<\/b><\/span><\/span>com\u00a0<\/b><\/span><\/span>velocidade constante\u00a0<\/b><\/span><\/span>de<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>intensidade\u00a0V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>,\u00a0<\/b><\/span><\/span>que \u00e9 a\u00a0<\/b><\/span><\/span>velocidade de lan\u00e7amento<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>S = S<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0+ V.t<\/b><\/span><\/span>\u00a0\u00a0X= 0 + V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.t\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

X = V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Segundo o eixo Y\u00a0<\/b><\/span><\/span>\u00a0 trata-se de um\u00a0<\/b><\/span><\/span>movimento uniformemente variado\u00a0<\/b><\/span><\/span>com\u00a0<\/b><\/span><\/span>velocidade vertical\u00a0<\/b><\/span><\/span>inicia nula<\/b><\/span><\/span>\u00a0(<\/b><\/span><\/span>V<\/b><\/span><\/span>oy\u00a0<\/b><\/span><\/sub><\/span>= 0),<\/b><\/span><\/span>\u00a0ou seja, \u00e9 uma\u00a0<\/b><\/span><\/span>queda livre com o corpo abandonado da origem<\/b><\/span><\/span>,\u00a0sujeito\u00a0<\/b><\/span><\/span>apenas \u00e0 acelera\u00e7\u00e3o da<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>gravidade, de intensidade\u00a0g, dire\u00e7\u00e3o vertical e sentido para baixo<\/b><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

S = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>.t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Y= 0 + 0.t + gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Y = g.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Vy= 0 + g.t\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0= g.t<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.\u0394S\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.\u0394h<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.\u0394h<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 2.g.\u0394h\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>caso do exerc\u00edcio na vertical, o tempo t que ela demora para chegar ao solo \u00e9 t = 10 s <\/b><\/span><\/span><\/span><\/p>\n

Y = h = gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 <\/b><\/span><\/span><\/span>= 10.(10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2) = 10.(100\/2) = 10.50 <\/b><\/span><\/span><\/span>h = 500 m.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

Se o <\/b><\/span><\/span><\/span>movimento \u00e9 retardado <\/b><\/span><\/span><\/span>o m\u00f3vel est\u00e1 <\/b><\/span><\/span><\/span>freando <\/b><\/span><\/span><\/span>(velocidade diminuindo em m\u00f3dulo) e se <\/b><\/span><\/span><\/span>progressivo<\/b><\/span><\/span><\/span> (se movimenta no sentido dos marcos crescentes da trajet\u00f3ria).<\/b><\/span><\/span><\/span><\/p>\n

Pode ocorrer sim com o <\/b><\/span><\/span><\/span>m\u00f3vel freando e se movendo no sentido dos marcos crescentes.<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 4 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

Acoplamento de polias e engrenagens<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

f<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>.R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> = f<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>.R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> <\/b><\/span><\/span>se R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> = 10 cm; R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> = 20 cm e se f<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> = 50 rpm, f<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> valer\u00e1 <\/b><\/span><\/span> <\/b>f<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>.R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span> = f<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>.R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span><\/span><\/p>\n

50.10 = f<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>.20 f<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> = 500\/20 <\/b><\/span><\/span>f<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> = 25 rpm.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- (0-0) \u2013 II: (1-1)- I: (2-2)- I: (3-3)- I e (4-4)- I.<\/b><\/span><\/span><\/span><\/p>\n

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

0 \u2013 0<\/b><\/span><\/span><\/span> Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

Observe <\/b><\/span><\/span><\/span>na figura que nessas condi\u00e7\u00f5es com a for\u00e7a de 30 N empurrando A para a direita, <\/b><\/span><\/span><\/span>sua <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o \u00e9 maior que a acelera\u00e7\u00e3o de ca\u00edda do bloco B <\/b><\/span><\/span><\/span>e, assim a <\/b><\/span><\/span><\/span>corda n\u00e3o est\u00e1 tracionada <\/b><\/span><\/span><\/span>e a <\/b><\/span><\/span><\/span>\u00fanica for\u00e7a que age sobre A na dire\u00e7\u00e3o do movimento \u00e9 a de 30 N <\/b><\/span><\/span><\/span>de modo que <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = F = m.<\/b><\/span><\/span><\/span>a 30 = 2.a <\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>= 30\/2 <\/b><\/span><\/span><\/span>= 15 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 Falsa (II)<\/b><\/span><\/span><\/span><\/p>\n

Equil\u00edbrio na vertical <\/b><\/span><\/span><\/span> veja na <\/b><\/span><\/span><\/span>figura abaixo <\/b><\/span><\/span><\/span>que, como o corpo <\/b><\/span><\/span><\/span>n\u00e3o escorrega na vertical F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> = \u00b5<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>.N = 0,2.N <\/b><\/span><\/span><\/span>(N \u00e9 a normal) <\/b><\/span><\/span><\/span>\u00e9 igual ao peso P na situa\u00e7\u00e3o de F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> m\u00e1ximo P = 0,2N.(I)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Equil\u00edbrio na horizontal <\/b><\/span><\/span><\/span> F = N <\/b><\/span><\/span><\/span>10 = N (II)<\/b><\/span><\/span><\/span><\/p>\n

(II) em (I) <\/b><\/span><\/span><\/span> P = 0,2N P = 0,2.10 P = 2 N.<\/b><\/span><\/span><\/span><\/p>\n

Para <\/b><\/span><\/span><\/span>valor de peso acima de 2 N o bloco n\u00e3o fica em equil\u00edbrio, ele desce<\/b><\/span><\/span><\/span> mas, para <\/b><\/span><\/span><\/span>pesos inferiores a 2N ele permanece em repouso<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 Falsa (II)<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>atc<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> <\/b>\u0001<\/b><\/span><\/span><\/span>.N = <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>.P = <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>.m.g<\/b><\/span><\/span><\/span> <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>for\u00e7a resultante sobre ele \u00e9 a pr\u00f3pria for\u00e7a de atrito <\/b><\/span><\/span><\/span>tal que <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span>atc<\/b><\/span><\/span><\/sub><\/span> = m.a<\/b><\/span><\/span><\/span> <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>.m.g =<\/b><\/span><\/span><\/span> <\/b>m.a <\/b><\/span><\/span><\/span>a = <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>.g.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de Torricelli com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = V, V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = o (para) e a = – <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>.g <\/b><\/span><\/span><\/span>(movimento retardado)<\/b><\/span><\/span><\/span> <\/b><\/span><\/span><\/span> <\/b>V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>\u00ado<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2ad<\/b><\/span><\/span><\/span><\/p>\n

0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2.<\/b><\/span><\/span><\/span> \u0001<\/b><\/span><\/span><\/span>.g.d <\/b><\/span><\/span><\/span> d = V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2.<\/b><\/span><\/span><\/span> \u0001<\/b><\/span><\/span><\/span>.g.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 Falsa (II)<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da dist\u00e2ncia d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> que ele percorre sobre o plano inclinado sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> <\/b>= cateto oposto\/hipotenusa 0,5 = 10\/d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 20 m<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Trabalho W realizado pela for\u00e7a de intensidade F = 4 N do deslocamento d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 20 m.<\/b><\/span><\/span><\/span><\/p>\n

W = F.d.cos0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> <\/b>= 4.20.1 <\/b><\/span><\/span><\/span>W = 80 J.<\/b><\/span><\/span><\/span><\/p>\n

Determinando o <\/b><\/span><\/span><\/span>tempo t que ele demora para percorrer os 20m com acelera\u00e7\u00e3o de 2 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> partindo do repouso <\/b><\/span><\/span><\/span> <\/b>d = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + a.(t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2) <\/b><\/span><\/span><\/span> 20 = 0.t + 2.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 <\/b><\/span><\/span><\/span>t <\/b><\/span><\/span><\/span>= <\/b><\/span><\/span><\/span> s <\/b><\/span><\/span><\/span>= 4,47 s.<\/b><\/span><\/span><\/span><\/p>\n

P = W\/t <\/b><\/span><\/span><\/span>= 80\/4,47 <\/b><\/span><\/span><\/span>= 17,89 W<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 4 Falsa (II)<\/b><\/span><\/span><\/span><\/p>\n

Velocidade quando chega ao ponto B V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + a.t <\/b><\/span><\/span><\/span>= 0 + 2.4,47 <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 8,94 s.<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = F.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> <\/b>= 4.8,94 <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 35,76 W<\/b><\/span><\/span><\/span><\/p>\n

R- (0-0) \u2013 I: (1-1)- II: (2-2)- II: (3-3)- II e (4-4)- II.<\/b><\/span><\/span><\/span><\/p>\n

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

0 \u2013 0 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>\/2 <\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span> = 2V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>madeira<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span> d<\/b><\/span><\/span><\/span>madeira<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span>\/2V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>madeira<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> = 1\/2 <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>madeira<\/b><\/span><\/span><\/sub><\/span> = d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\/2.<\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

Os <\/b><\/span><\/span><\/span>intervalos entre os pontos fixos nas escalas Celsius <\/b><\/span><\/span><\/span>(0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a 100<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C) e <\/b><\/span><\/span><\/span>na Kelvin <\/b><\/span><\/span><\/span>(273 K e 373 K) <\/b><\/span><\/span><\/span>s\u00e3o os mesmos (100 graus). <\/b><\/span><\/span><\/span>Assim,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>varia\u00e7\u00e3o de temperatura entre as duas escalas \u00e9 a mesma.<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de calor cedida pela \u00e1gua para ir de \u00e1gua a 80<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C at\u00e9 <\/b><\/span><\/span><\/span>\u00e1gua a 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span> <\/b><\/span><\/span><\/span> <\/b>Q = m.c (t \u2013 t<\/b><\/span><\/span><\/span>O<\/b><\/span><\/span><\/sub><\/span>) =540.1.(80 \u2013 0) <\/b><\/span><\/span><\/span>= 43200 cal.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de calor para fundir o gelo a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>( (passar de gelo a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a <\/b><\/span><\/span><\/span>\u00e1gua a 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>) <\/b><\/span><\/span><\/span>Q = m.L <\/b><\/span><\/span><\/span>= 540.80 <\/b><\/span><\/span><\/span>= 43200 cal<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Portanto no <\/b><\/span><\/span><\/span>final teremos \u00e1gua a 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 Falsa (Ii)<\/b><\/span><\/span><\/span><\/p>\n

De gelo a \u2013 10 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C at\u00e9 gelo a 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C Q = m.c.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) = 2.0,5.(0 \u2013 (-10)) = 1.10 <\/b><\/span><\/span><\/span>= 10 cal<\/b><\/span><\/span><\/span><\/p>\n

De gelo a 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C at\u00e9 \u00e1gua a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C Q = m.L <\/b><\/span><\/span><\/span>= 2.80 <\/b><\/span><\/span><\/span>= 160 cal<\/b><\/span><\/span><\/span><\/p>\n

De \u00e1gua a 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C at\u00e9 \u00e1gua a 80 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C Q = m.c.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> = 2.1.(80 \u2013 0) = 2.80 <\/b><\/span><\/span><\/span>= 160 cal.<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span> = 10 + 160 + 160 <\/b><\/span><\/span><\/span>= 330 cal<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 4 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span> = 2 Kcal = 2000 calx4,2 <\/b><\/span><\/span><\/span>= 8400 J<\/b><\/span><\/span><\/span>; <\/b><\/span><\/span><\/span>W = 500 J e \u2206U = ?.<\/b><\/span><\/span><\/span><\/p>\n

\u2206<\/span>U = Q \u2013 W <\/b><\/span><\/span><\/span>= 8400 \u2013 500 <\/b><\/span><\/span><\/span>= 7900 J<\/b><\/span><\/span><\/span><\/p>\n

R- (0-0) \u2013 I: (1-1)- I: (2-2)- I: (3-3)- II e (4-4)- I.<\/b><\/span><\/span><\/span><\/p>\n

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

0 \u2013 0 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

U = E.d <\/b><\/span><\/span><\/span>= 4.10 <\/b><\/span><\/span><\/span> U = 40 V.<\/b><\/span><\/span><\/span><\/p>\n

W = q.U <\/b><\/span><\/span><\/span>= 2.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span>.40 <\/b><\/span><\/span><\/span> W = <\/b><\/span><\/span><\/span>80.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span> <\/b>= 80 nJ<\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

C C\u00e1lculo da dist\u00e2ncia d<\/b><\/span><\/span><\/span> entre <\/b><\/span><\/span><\/span>cada carga e o ponto C no centro do quadrad<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>Pit\u00e1goras<\/b><\/span><\/span><\/span> no <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

Potencial el\u00e9trico (V) de uma carga fonte (Q)<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Potencial el\u00e9trico originado num ponto P por v\u00e1rias cargas fonte<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

0 2 \u20132 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da corrente i entre A e C <\/b><\/span><\/span><\/span> <\/b>R<\/b><\/span><\/span><\/span>eqAC<\/b><\/span><\/span><\/sub><\/span> = 5 + 5 + 5 <\/b><\/span><\/span><\/span>= 15 \u2126 <\/b><\/span><\/span><\/span> <\/b>U<\/b><\/span><\/span><\/span>AC<\/b><\/span><\/span><\/sub><\/span> = 2 V <\/b><\/span><\/span><\/span> <\/b>R<\/b><\/span><\/span><\/span>eqAC<\/b><\/span><\/span><\/sub><\/span> = U<\/b><\/span><\/span><\/span>AC<\/b><\/span><\/span><\/sub><\/span>\/i<\/b><\/span><\/span><\/span>AC<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

15 = 2\/i<\/b><\/span><\/span><\/span>AC<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>AC<\/b><\/span><\/span><\/sub><\/span> = 2\/15 A<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

No <\/b><\/span><\/span>trecho AB a corrente \u00e9 a mesma que no trecho AC <\/b><\/span><\/span>(est\u00e3o em s\u00e9rie) <\/b><\/span><\/span>i<\/b><\/span><\/span>AB<\/b><\/span><\/sub><\/span> <\/b>= i<\/b><\/span><\/span>AC<\/b><\/span><\/sub><\/span> <\/b>= 2\/15 A.<\/b><\/span><\/span><\/p>\n

<\/p>\n

R<\/b><\/span><\/span>eqAB<\/b><\/span><\/sub><\/span> = 5 + 5 <\/b><\/span><\/span>= 10 \u2126 <\/b><\/span><\/span> <\/b>R<\/b><\/span><\/span><\/span>eqAB<\/b><\/span><\/span><\/sub><\/span> = U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/i<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span>10 = U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>\/(2\/15<\/b><\/span><\/span><\/span>) U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>20\/15 <\/b><\/span><\/span><\/span>= 4\/3 V.<\/b><\/span><\/span><\/span><\/p>\n

2 3 \u2013 3 Errada (II)<\/b><\/span><\/span><\/span><\/p>\n

Pot\u00eancia no trecho AC pela direita P<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = U.i<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 2.(2\/15) P<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 4\/15 W mas a essa pot\u00eancia \u00e9 a m mesma que a do trecho AC pela esquerda P<\/b><\/span><\/span><\/span>total <\/b><\/span><\/span><\/sub><\/span>= 4\/15 + 4\/15 = 8\/15 W <\/b><\/span><\/span><\/span><\/p>\n

3 4 \u2013 4 Errada (II)<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o do receptor<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Pelo gr\u00e1fico quando i = 6 A, U = 18 V <\/b><\/span><\/span><\/span> U = E\u2019 \u2013 r\u2019.i <\/b><\/span><\/span><\/span>18 = E\u2019<\/b><\/span><\/span><\/span> <\/b>\u2013 r\u2019.6 (I)<\/b><\/span><\/span><\/span><\/p>\n

Pelo gr\u00e1fico quando i = 12 A, U = 30 V <\/b><\/span><\/span><\/span> U = E\u2019 \u2013 r\u2019.i <\/b><\/span><\/span><\/span>30 = E\u2019<\/b><\/span><\/span><\/span> <\/b>\u2013 r\u2019.12 (II)<\/b><\/span><\/span><\/span><\/p>\n

Resolvendo (I) com (II) r\u2019 = 2 \u2126 e E\u2019 = 54 V.<\/b><\/span><\/span><\/span><\/p>\n

R- (0-0) \u2013 I: (1-1)- I: (2-2)- I: (3-3)- II e (4-4)- II.<\/b><\/span><\/span><\/span><\/p>\n

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

0 0 – 0 Falsa (II)<\/b><\/span><\/span><\/span><\/p>\n

n = c\/V<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span> 1<\/b><\/span><\/span><\/span>,5 = 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/V<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span> V<\/b><\/span><\/span><\/span>vidro<\/b><\/span><\/span><\/sub><\/span> = 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/1,5 <\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>vidro <\/b><\/span><\/span><\/sub><\/span>= V = 2.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s.<\/b><\/span><\/span><\/span><\/p>\n

V = d\/t <\/b><\/span><\/span><\/span>2.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> = 2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/t t = 2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/2.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> t = 10<\/b><\/span><\/span><\/span>-11<\/b><\/span><\/span><\/sup><\/span> s.<\/b><\/span><\/span><\/span><\/p>\n

1 \u2013 1 Falsa (II)<\/b><\/span><\/span><\/span><\/p>\n

sen L = n<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/n<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= \u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/ \u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span> senL = n<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span>\/n<\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/sub><\/span>\/ \u03bb<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span> <\/b>= 3500\/7000 = 1\/2 <\/b><\/span><\/span><\/span> <\/b><\/span><\/span><\/span>sen L = 1\/2 <\/b><\/span><\/span><\/span><\/p>\n

L = 30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 2 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

I = – 2.o f = 50 cm P = ?<\/b><\/span><\/span><\/span><\/p>\n

i\/o = – P\u2019\/P <\/b><\/span><\/span><\/span> i\/o = – P\u2019\/P – 2.o\/o = – P\u2019\/P <\/b><\/span><\/span><\/span>P\u2019 = 2P.<\/b><\/span><\/span><\/span><\/p>\n

1\/f = 1\/P + 1\/P\u2019 <\/b><\/span><\/span><\/span> <\/b>1\/50 = 1\/P + 1\/2P <\/b><\/span><\/span><\/span> 1\/50 = (2 +1)\/2P 1\/50 = 3\/2P 2P = 150 <\/b><\/span><\/span><\/span>P = 75 cm<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 3 Falsa (II)<\/b><\/span><\/span><\/span><\/p>\n

Carga em repouso (V = 0) ou lan\u00e7ada com velocidade\u00a0paralelamente \u00e0s linhas de indu\u00e7\u00e3o de um campo magn\u00e9tico uniforme\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Est\u00e1 em movimento retil\u00edneo uniforme.<\/span><\/span><\/span><\/p>\n

4 \u2013 4 Correta (I)<\/b><\/span><\/span><\/span><\/p>\n

Caracter\u00edsticas das ondas eletromagn\u00e9ticas<\/b><\/span><\/span><\/span><\/p>\n