{"id":4488,"date":"2019-06-30T23:06:45","date_gmt":"2019-06-30T23:06:45","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4488"},"modified":"2024-07-11T11:57:24","modified_gmt":"2024-07-11T11:57:24","slug":"acafe-medicina-2019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2019\/acafe-medicina-2019\/","title":{"rendered":"ACAFE \u2013 MEDICINA – 2019"},"content":{"rendered":"

ACAFE \u2013 MEDICINA – 2019<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Com o objetivo de desenvolver o ensino superior em Santa Catarina, 16 Funda\u00e7\u00f5es Educacionais criadas pelos munic\u00edpios se uniram em torno da constitui\u00e7\u00e3o de um \u00f3rg\u00e3o com as atribui\u00e7\u00f5es de planejar, articular e coordenar a\u00e7\u00f5es integradas entre elas. Nasce ent\u00e3o, em 2 de maio de 1974, a Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013 ACAFE, que vem se fortalecendo at\u00e9 os dias atuais.<\/b><\/span><\/span><\/span><\/p>\n

Das 15 Funda\u00e7\u00f5es iniciais, mantenedoras de Institui\u00e7\u00f5es de Ensino Superior \u2013 IES isoladas, o Sistema ACAFE evoluiu para configurar, hoje, um conjunto de 16 IES, sendo 11 Universidades e 5 Centros Universit\u00e1rios. A ACAFE \u00e9 uma sociedade civil sem fins lucrativos que congrega as funda\u00e7\u00f5es educacionais criadas no Estado de Santa Catarina por lei dos poderes p\u00fablicos estaduais e municipais.<\/b><\/span><\/span><\/span><\/p>\n

O objetivo desta uni\u00e3o \u00e9 promover o interc\u00e2mbio administrativo, t\u00e9cnico e cient\u00edfico entre as Institui\u00e7\u00f5es de Educa\u00e7\u00e3o Superior, na busca de solu\u00e7\u00f5es para problemas comuns nas \u00e1reas de ensino, pesquisa, extens\u00e3o e administra\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

01- (ACAFE – <\/b><\/span><\/span><\/span>Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013Medicina – 2019.<\/span><\/span><\/span><\/p>\n

<\/p>\n

O <\/b><\/span><\/span><\/span>gr\u00e1fico da figura abaixo<\/b><\/span><\/span><\/span> mostra o <\/b><\/span><\/span><\/span>comportamento da velocidade (v) de um ve\u00edculo variando <\/b><\/span><\/span><\/span>em<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

fun\u00e7\u00e3o do tempo (t) em uma trajet\u00f3ria retil\u00ednea. <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>dist\u00e2ncia, em metros, percorrida por esse ve\u00edculo durante o tempo total de movimento (4s) <\/b><\/span><\/span><\/span>ser\u00e1:<\/b><\/span><\/span><\/span><\/p>\n

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

02- (ACAFE – <\/b><\/span><\/span><\/span>Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013Medicina – 2019.<\/span><\/span><\/span><\/p>\n

<\/p>\n

Um <\/b><\/span><\/span><\/span>drone<\/b><\/span><\/span><\/span> <\/b>eleva uma caixa de 5,0 kg com velocidade vertical constante de m\u00f3dulo 2 m\/s, <\/b><\/span><\/span><\/span>como mostra a <\/b><\/span><\/span><\/span>figura abaixo. <\/b><\/span><\/span><\/span>Em <\/b><\/span><\/span><\/span>certa altura, o fio que prende a caixa ao drone arrebenta<\/b><\/span><\/span><\/span> e o <\/b><\/span><\/span><\/span>drone passa a subir sozinho.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

<\/p>\n

Considere o intervalo de tempo entre o instante em que a caixa se solta do drone e o instante em que a caixa come\u00e7a a descer para marcar com V as afirma\u00e7\u00f5es verdadeiras e com F as falsas.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Considere o <\/b><\/span><\/span><\/span>sistema conservativo. <\/b><\/span><\/span><\/span><\/p>\n

( ) O trabalho realizado pela for\u00e7a peso sobre a caixa \u00e9 -10 J. <\/b><\/span><\/span><\/span><\/p>\n

( ) A quantidade de movimento da caixa permanece constante. <\/b><\/span><\/span><\/span><\/p>\n

( ) O impulso aplicado pela for\u00e7a peso sobre a caixa \u00e9 -10 Ns. <\/b><\/span><\/span><\/span><\/p>\n

( ) A energia cin\u00e9tica da caixa na altura m\u00e1xima alcan\u00e7ada \u00e9 nula. <\/b><\/span><\/span><\/span><\/p>\n

( ) A energia potencial gravitacional do drone permanece constante. <\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>sequ\u00eancia correta, de cima para baixo<\/b><\/span><\/span><\/span>, \u00e9: <\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span> V – F – V – V \u2013 F<\/b><\/span><\/span><\/span><\/p>\n

B<\/b><\/span><\/span><\/span>\u00a0 V – V – F – F – F <\/b><\/span><\/span><\/span><\/p>\n

C\u00a0<\/b><\/span><\/span><\/span> F – V – F – V \u2013 V<\/b><\/span><\/span><\/span><\/p>\n

D<\/b><\/span><\/span><\/span>\u00a0 F – F – V – F – V<\/b><\/span><\/span><\/span><\/p>\n

03- (ACAFE – <\/b><\/span><\/span><\/span>Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013Medicina – 2019.<\/span><\/span><\/span><\/p>\n

<\/p>\n

Brinquedo das \u201cantigas\u201d, <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>carrinho de rolim\u00e3 \u00e9 o nome dado a um carrinho, <\/b><\/span><\/span><\/span>geralmente <\/b><\/span><\/span><\/span>constru\u00eddo de madeira com um eixo m\u00f3vel montado com rolamentos de a\u00e7o <\/b><\/span><\/span><\/span>(dispensados por mec\u00e2nicas de<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

autom\u00f3veis), <\/b><\/span><\/span><\/span>utilizado para controlar o carrinho enquanto este desce pela rua.<\/b><\/span><\/span><\/span> Ao <\/b><\/span><\/span><\/span>construir devemos encaixar firmemente os rolamentos no eixo cil\u00edndrico de determinado metal <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>di\u00e2metro um pouco maior que o di\u00e2metro interno do rolamento de a\u00e7o. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Para esse <\/b><\/span><\/span><\/span>procedimento aquecemos ambos para o encaixe e depois resfriamos. <\/b><\/span><\/span><\/span>Sendo assim, o <\/b><\/span><\/span><\/span>coeficiente de dilata\u00e7\u00e3o do metal utilizado em rela\u00e7\u00e3o ao coeficiente de dilata\u00e7\u00e3o do a\u00e7o deve ser: <\/b><\/span><\/span><\/span><\/p>\n

A<\/b><\/span><\/span><\/span>\u00a0 igual ou maior
\n<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span>\u00a0 maior
\n<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span>\u00a0 igual <\/b><\/span><\/span><\/span>
\nD\u00a0<\/b><\/span><\/span><\/span> menor <\/b><\/span><\/span><\/span><\/p>\n

04- (ACAFE – <\/b><\/span><\/span><\/span>Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013Medicina – 2019.<\/span><\/span><\/span><\/p>\n

<\/p>\n

Seja um <\/b><\/span><\/span><\/span>bal\u00e3o de l\u00e1tex que tem forma esf\u00e9rica<\/b><\/span><\/span><\/span> e est\u00e1 <\/b><\/span><\/span><\/span>eletrizado com carga negativa distribu\u00edda uniformemente em sua superf\u00edcie externa.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

<\/p>\n

Considerando o <\/b><\/span><\/span><\/span>exposto acima <\/b><\/span><\/span><\/span>e que <\/b><\/span><\/span><\/span>inicialmente o bal\u00e3o est\u00e1 semi-inflado, assinale a alternativa correta que completa as lacunas das frases a seguir<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Em <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o <\/b><\/span><\/span><\/span>ao <\/b><\/span><\/span><\/span>estado inicial, enquanto o bal\u00e3o estiver sendo inflado, constatamos<\/b><\/span><\/span><\/span> que em <\/b><\/span><\/span><\/span>pontos interiores a superf\u00edcie do mesmo, o campo el\u00e9trico ___________. <\/b><\/span><\/span><\/span><\/p>\n

J\u00e1 em <\/b><\/span><\/span><\/span>pontos externos na superf\u00edcie, o campo el\u00e9trico ____________ e o potencial ______________.<\/b><\/span><\/span><\/span><\/p>\n

A ser\u00e1 sempre nulo – aumenta – aumenta. <\/b><\/span><\/span><\/span><\/p>\n

B<\/b><\/span><\/span><\/span> diminui – diminui – ser\u00e1 sempre igual. <\/b><\/span><\/span><\/span><\/p>\n

C<\/b><\/span><\/span><\/span> ser\u00e1 sempre nulo – diminui – diminui.<\/b><\/span><\/span><\/span><\/p>\n

D<\/b><\/span><\/span><\/span> aumenta – diminui – aumenta.<\/b><\/span><\/span><\/span><\/p>\n

05- (ACAFE – <\/b><\/span><\/span><\/span>Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013Medicina – 2019.<\/span><\/span><\/span><\/p>\n

<\/p>\n

Em uma <\/b><\/span><\/span><\/span>loja tr\u00eas pe\u00e7as de roupas est\u00e3o em uma arara <\/b><\/span><\/span><\/span>(suporte para pendurar roupas), conforme <\/b><\/span><\/span><\/span>mostra a figura. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>arara \u00e9 constitu\u00edda por tr\u00eas partes, duas verticais (parte A e B) e uma na horizontal (parte C),<\/b><\/span><\/span><\/span> <\/b>todas<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span><\/span>mesma massa (m = 1,00kg). <\/b><\/span><\/span><\/span><\/p>\n

Cada pe\u00e7a de roup<\/b><\/span><\/span><\/span>a e seu <\/b><\/span><\/span><\/span>cabide formam um conjunto, ent\u00e3o temos o conjunto 1 (m = 1,00kg) <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>est\u00e1 0,10m da parte A, o conjunto 2 (m = 0,50kg) que est\u00e1 a 0,20m do conjunto 1 e o conjunto 3 (m = 1,50kg) que est\u00e1 a 0,20m da parte B. <\/b><\/span><\/span><\/span><\/p>\n

Considerando todas as partes <\/b><\/span><\/span><\/span>da arara <\/b><\/span><\/span><\/span>homog\u00eaneas<\/b><\/span><\/span><\/span> e o m\u00f3dulo da <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o da gravidade igual a 10 m\/s2 , assinale a alternativa correta que apresenta os m\u00f3dulos das for\u00e7as, em newtons,<\/b><\/span><\/span><\/span> que a <\/b><\/span><\/span><\/span>parte C aplica sobre a parte A e B,<\/b><\/span><\/span><\/span> respectivamente.<\/b><\/span><\/span><\/span><\/p>\n

A<\/b><\/span><\/span><\/span>\u00a0 20,5 e 19,5
\n<\/b><\/span><\/span><\/span> B <\/b><\/span><\/span><\/span>20,0 e 20,0
\n<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span> 24,5 e 15,5
\n<\/b><\/span><\/span><\/span> D <\/b><\/span><\/span><\/span>29,5 e 10,5 <\/b><\/span><\/span><\/span><\/p>\n

<\/a> 06- (ACAFE – <\/b><\/span><\/span><\/span>Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013Medicina – 2019.<\/span><\/span><\/span><\/p>\n

<\/p>\n

O <\/b><\/span><\/span><\/span>quadro abaixo apresenta algumas informa\u00e7\u00f5es <\/b><\/span><\/span><\/span>de uma <\/b><\/span><\/span><\/span>fatura da conta de energia el\u00e9trica <\/b><\/span><\/span><\/span>de uma <\/b><\/span><\/span><\/span>resid\u00eancia por um per\u00edodo de 30 dias. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Sabe-se que uma <\/b><\/span><\/span><\/span>chaleira el\u00e9trica \u00e9 utilizada todos os dias por quinze minutos <\/b><\/span><\/span><\/span>e que <\/b><\/span><\/span><\/span>a energia gasta por ela, em 30 dias, representa 2% do consumo de energia da casa. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>alternativa correta que apresenta o valor da pot\u00eancia dessa chaleira, em watt, <\/b><\/span><\/span><\/span>\u00e9: <\/b><\/span><\/span><\/span><\/p>\n

A<\/b><\/span><\/span><\/span> 1500
\n<\/b><\/span><\/span><\/span> B <\/b><\/span><\/span><\/span>1200
\n<\/b><\/span><\/span><\/span> C <\/b><\/span><\/span><\/span>1000
\n<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span> 800<\/b><\/span><\/span><\/span><\/p>\n

07- (ACAFE – <\/b><\/span><\/span><\/span>Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013Medicina – 2019.<\/span><\/span><\/span><\/p>\n

<\/p>\n

Em tempos <\/b><\/span><\/span><\/span>de crise econ\u00f4mica, uma pessoa deseja empreender montando uma pequena loja de roupas.<\/b><\/span><\/span><\/span> <\/b>Um dos itens essenciais \u00e9 colocar um espelho em uma parede vertical, <\/b><\/span><\/span><\/span>de modo que <\/b><\/span><\/span><\/span>qualquer cliente de m\u00e9dia altura (h) possa se ver inteiro nesse espelho a certa dist\u00e2ncia horizontal<\/b><\/span><\/span><\/span><\/p>\n

(d). <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Mas <\/b><\/span><\/span><\/span>para economizar, o espelho dever\u00e1 ter a menor altura poss\u00edvel.<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span><\/span>alternativa correta que indica a altura desse espelho<\/b><\/span><\/span><\/span> \u00e9: <\/b><\/span><\/span><\/span><\/p>\n

A<\/b><\/span><\/span><\/span> h\/3
\n<\/b><\/span><\/span><\/span> B <\/b><\/span><\/span><\/span>h\/d
\n<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span> 2h\/3
\n<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span> h\/2 <\/b><\/span><\/span><\/span><\/p>\n

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica da ACAFE \u2013 2019<\/span><\/span><\/span><\/p>\n

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

Rela\u00e7\u00e3o entre o deslocamento \u2206S e o gr\u00e1fico da velocidade x tempo de um MUV.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Em\u00a0<\/b><\/span><\/span>todo\u00a0gr\u00e1fico VXt\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>\u00e1rea\u00a0<\/b><\/span><\/span>entre a\u00a0<\/b><\/span><\/span>reta representativa\u00a0e o\u00a0eixo dos tempos<\/b><\/span><\/span>\u00a0\u00e9 <\/b><\/span><\/span>numericamente<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

igual\u00a0\u00e0\u00a0varia\u00e7\u00e3o de espa\u00e7o \u0394S<\/b><\/span><\/span>,\u00a0entre dois\u00a0<\/b><\/span><\/span>instantes quaisquer t<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0e t<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio dist\u00e2ncia percorrida \u00e9 numericamente igual a \u00e1rea no gr\u00e1fico v X t, como o gr\u00e1fico de 0 a 4 s \u00e9 metade de um c\u00edrculo de raio R = 2, temos A = \u2206S = \u03c0.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = \u03c0.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \/2 = 4\u03c0\/2 = 2\u03c0.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Primeira – Verdadeira \u2013 <\/b><\/span><\/span><\/span>colocando a <\/b><\/span><\/span><\/span>origem da altura no ponto em que a caixa se solta, <\/b><\/span><\/span><\/span>ela tem <\/b><\/span><\/span><\/span>velocidade de V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 2 m\/s <\/b><\/span><\/span><\/span>(mesma que a do drone) e <\/b><\/span><\/span><\/span>energia potencial gravitacional nula <\/b><\/span><\/span><\/span>(h = 0) portanto <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>pi<\/b><\/span><\/span><\/sub><\/span> = 0.<\/b><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>ponto de altura m\u00e1xima quando V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 0, <\/b><\/span><\/span><\/span>sua <\/b><\/span><\/span><\/span>altura ser\u00e1 <\/b><\/span><\/span><\/span> equa\u00e7\u00e3o de Torricelli <\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2.g.h<\/b><\/span><\/span><\/span> 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2.10.h <\/b><\/span><\/span><\/span> h = 4\/20 <\/b><\/span><\/span><\/span> <\/b>h = 0,2 m.<\/b><\/span><\/span><\/span><\/p>\n

Trabalho como varia\u00e7\u00e3o de energia potencial gravitacional <\/b><\/span><\/span><\/span> W = E<\/b><\/span><\/span><\/span>pi<\/b><\/span><\/span><\/sub><\/span> \u2013 E<\/b><\/span><\/span><\/span>pf<\/b><\/span><\/span><\/sub><\/span> <\/b>= 0 \u2013 mgh = -5.10.0,2 <\/b><\/span><\/span><\/span> <\/b><\/p>\n

W = – 10 J.<\/b><\/span><\/span><\/span><\/p>\n

Segunda – Falsa <\/b><\/span><\/span><\/span>\u2013 a <\/b><\/span><\/span><\/span>quantidade de movimento \u00e9 fornecida por Q = m.V <\/b><\/span><\/span><\/span>e, como <\/b><\/span><\/span><\/span>V varia de 2 m\/s para zero, Q tamb\u00e9m variar\u00e1 de Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = m.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 5.2 <\/b><\/span><\/span><\/span>= 10 kg.m\/s <\/b><\/span><\/span><\/span>para <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = m.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> <\/b>= 5.0 <\/b><\/span><\/span><\/span>= 0.<\/b><\/span><\/span><\/span><\/p>\n

Terceira \u2013 Verdadeira \u2013<\/b><\/span><\/span><\/span> veja a anterior <\/b><\/span><\/span><\/span> <\/b>I = Q<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> \u2013 Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 0 \u2013 10 = <\/b><\/span><\/span><\/span>– 10 kg.m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Quarta – Verdadeira \u2013 <\/b><\/span><\/span><\/span>na <\/b><\/span><\/span><\/span>altura m\u00e1xima V = 0 e ent\u00e3o E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> = 5.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 <\/b><\/span><\/span><\/span>= 0<\/b><\/span><\/span><\/span><\/p>\n

Quinta \u2013 Falsa \u2013 veja a primeira justificativa.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Dilata\u00e7\u00e3o volum\u00e9trica dos s\u00f3lidos<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe na <\/b><\/span><\/span>express\u00e3o acima que, quanto maior o valor do coeficiente de dilata\u00e7\u00e3o, mais o corpo se dilata quando aquecido e mais se contrai quando resfriado.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como o <\/b><\/span><\/span><\/span>di\u00e2metro do eixo cil\u00edndrico de metal \u00e9 um pouco maior <\/b><\/span><\/span><\/span>que o do <\/b><\/span><\/span><\/span>rolamento de a\u00e7o, no encaixe eles ficam folgados. <\/b><\/span><\/span><\/span><\/p>\n

Quando <\/b><\/span><\/span><\/span>voc\u00ea aquece os dois, o di\u00e2metro do a\u00e7o se dilata mais ficando encaixado no rolamento do metal.<\/b><\/span><\/span><\/span><\/p>\n

Depois <\/b><\/span><\/span><\/span>quando resfriados o rolamento de metal se contrai mais que o do a\u00e7o, fazendo com que fiquem mais presos.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Campo e potencial el\u00e9trico de um condutor esf\u00e9rico em equil\u00edbrio eletrost\u00e1tico<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe nas <\/b><\/span><\/span><\/span>informa\u00e7\u00f5es acima <\/b><\/span><\/span><\/span>que em <\/b><\/span><\/span><\/span>pontos internos de um condutor em equil\u00edbrio eletrost\u00e1tico <\/b><\/span><\/span><\/span>(pelo enunciado a distribui\u00e7\u00e3o \u00e9 uniforme) o <\/b><\/span><\/span><\/span>campo el\u00e9trico \u00e9 nulo<\/b><\/span><\/span><\/span>, estando ele inflado ou n\u00e3o e, em <\/b><\/span><\/span><\/span>pontos externos ele diminui<\/b><\/span><\/span><\/span> <\/b>(inversamente proporcional ao quadrado do raio da esfera \u2013 E = KQ\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>potencial el\u00e9trico, para pontos externos diminui<\/b><\/span><\/span><\/span> <\/b>tamb\u00e9m, pois para <\/b><\/span><\/span><\/span>pontos internos vale V = KQ\/R e para pontos externos V = KQ\/d com d > R.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Colocando as for\u00e7as que agem sobre cada ponto <\/b><\/span><\/span><\/span>da arara:<\/b><\/span><\/span><\/span><\/p>\n

Peso da arara colocado no centro de massa da mesma localizado a d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = 0,5 m de A e 0,5 m de B <\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = m.g <\/b><\/span><\/span><\/span>= 1.10 <\/b><\/span><\/span><\/span>= 10 N<\/b><\/span><\/span><\/span><\/p>\n

Peso do conjunto 1 <\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.g <\/b><\/span><\/span><\/span>= 1.10 <\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10 N com d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 0,1 m de A<\/b><\/span><\/span><\/span><\/p>\n

Peso do conjunto 2 <\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.g <\/b><\/span><\/span><\/span>= 0,5.10 <\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 5 N com d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 0,3 m de A<\/b><\/span><\/span><\/span><\/p>\n

Peso do conjunto 3 <\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.g <\/b><\/span><\/span><\/span>= 1,5.10 <\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 15 N com d<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 0,8 m de A<\/b><\/span><\/span><\/span><\/p>\n

N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> <\/b> rea\u00e7\u00e3o normal em A com d<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 0<\/b><\/span><\/span><\/span><\/p>\n

N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> <\/b> rea\u00e7\u00e3o normal em B com d<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 1m de A.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Colocando o polo<\/b><\/span><\/span><\/span> (eixo de rota\u00e7\u00e3o) <\/b><\/span><\/span><\/span>em A, calculando o momento de cada for\u00e7a em rela\u00e7\u00e3o ao polo <\/b><\/span><\/span><\/span>e estabelecendo o <\/b><\/span><\/span><\/span>sentido hor\u00e1rio de rota\u00e7\u00e3o em torno de A como positivo:<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>NA<\/b><\/span><\/span><\/sub><\/span> <\/b>= N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.0 <\/b><\/span><\/span><\/span>= 0<\/b><\/span><\/span><\/span> <\/b><\/p>\n

M<\/b><\/span><\/span><\/span>P1<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>= 10.0,1 <\/b><\/span><\/span><\/span>= 1 N.m <\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>P2<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 5.0,3 <\/b><\/span><\/span><\/span>= 1,5 N.m<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> <\/b>= 10.0,5 <\/b><\/span><\/span><\/span>= 5 N.m <\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>P3<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> <\/b>= 15.0,8 <\/b><\/span><\/span><\/span>= 12 N.m<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>NB<\/b><\/span><\/span><\/sub><\/span> = – N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.1 = – N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o a soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao polo deve ser nula <\/b><\/span><\/span><\/span> <\/p>\n

0 + 1,0 + 1,5 + 5,0 + 12 \u2013 N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 0 <\/b><\/span><\/span><\/span> <\/b>N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 19,5 N<\/b><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>equil\u00edbrio de transla\u00e7\u00e3o<\/b><\/span><\/span><\/span> a soma das for\u00e7as para cima de anular a soma das for\u00e7as para baixo <\/b><\/span><\/span><\/span> <\/p>\n

10 + 5 + 10 + 15 = N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> + N<\/b><\/span><\/span><\/span>B <\/b><\/span><\/span><\/sub><\/span> <\/b>40 = N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> + 19,5 <\/b><\/span><\/span><\/span> N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 40 \u2013 19,5 <\/b><\/span><\/span><\/span> <\/b>N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 20,5 N<\/b><\/span><\/span><\/span><\/p>\n

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

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

Veja na <\/b><\/span><\/span><\/span>tabela que a energia total consumida em 1 m\u00eas (30 dias) foi de Wt = 375 kWh e <\/b><\/span><\/span><\/span>como a <\/b><\/span><\/span><\/span>chaleira consome 2% dessa energia em 30 dias ela consumir\u00e1 W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>= 0,02×375 <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 7,5 kWh (em 30 dias).<\/b><\/span><\/span><\/span><\/p>\n

Em <\/b><\/span><\/span><\/span>1 dia ela consumir\u00e1 W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> 7,5\/30 <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 0,25 kWh (energia gasta pela chaleira em 1 dia).<\/b><\/span><\/span><\/span><\/p>\n

Mas o <\/b><\/span><\/span><\/span>tempo de funcionamento dela por dia \u00e9 t <\/b><\/span><\/span><\/span>= 15 min = 15\/60 = 1\/4 <\/b><\/span><\/span><\/span>= 0,25 h.<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/t <\/b><\/span><\/span><\/span>= 0,25\/0,25 <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> 1 kW <\/b><\/span><\/span><\/span>= 1000 W<\/b><\/span><\/span><\/span><\/p>\n

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

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

C\u00e1lculo da altura de um espelho plano vertical para que, a partir do ch\u00e3o, uma pessoa possa ver-se de corpo inteiro, desde a cabe\u00e7a at\u00e9 os p\u00e9s.<\/b><\/span><\/span><\/span><\/p>\n

Localizar a imagem\u00a0<\/b><\/span><\/span><\/span>da pessoa que fica\u00a0<\/b><\/span><\/span><\/span>atr\u00e1s do espelho pelos seus pontos extremos AA\u2019<\/b><\/span><\/span><\/span>(superior) e<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

BB\u2019<\/b><\/span><\/span><\/span>\u00a0(inferior).<\/b><\/span><\/span><\/span><\/p>\n

Em seguida\u00a0<\/b><\/span><\/span><\/span>ligar, com linha pontilhada, A\u2019 e B\u2019 ao olho da pessoa objeto<\/b><\/span><\/span><\/span>, que\u00a0<\/b><\/span><\/span><\/span>interceptam <\/b><\/span><\/span><\/span>o espelho<\/b><\/span><\/span><\/span><\/p>\n

nos\u00a0<\/b><\/span><\/span><\/span>pontos M<\/b><\/span><\/span><\/span>\u00a0(inferior) e\u00a0<\/b><\/span><\/span><\/span>N\u00a0<\/b><\/span><\/span><\/span>(superior), que\u00a0<\/b><\/span><\/span><\/span>delimitam o tamanho m\u00ednimo do espelho\u00a0<\/b><\/span><\/span><\/span>para que a\u00a0<\/b><\/span><\/span><\/span>pessoa possa ver-se de corpo inteiro no mesmo.<\/b><\/span><\/span><\/span>\"ACAFE\"<\/p>\n

Observe que os\u00a0<\/b><\/span><\/span><\/span>tri\u00e2ngulos OMN e OB\u2019A\u2019\u00a0<\/b><\/span><\/span><\/span>s\u00e3o\u00a0<\/b><\/span><\/span><\/span>semelhantes<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>dessa semelhan\u00e7a tiramos o tamanho m\u00ednimo do espelho MN\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>H\/MN = 2d\/d\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>MN = H\/2\u00a0<\/b><\/span><\/span><\/span>\u00a0(o\u00a0<\/b><\/span><\/span><\/span>tamanho m\u00ednimo do espelho\u00a0<\/b><\/span><\/span><\/span>para que a\u00a0<\/b><\/span><\/span><\/span>pessoa se veja de corpo inteiro deve ter a metade da altura da pessoa).\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Importante:<\/b><\/span><\/span><\/span>\u00a0Observe que a\u00a0<\/b><\/span><\/span><\/span>altura m\u00ednima do espelho \u00e9 sempre a mesma independente do fato de a pessoa estar a uma dist\u00e2ncia d, 2d, 3d, etc. do espelho<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Assim, a\u00a0<\/b><\/span><\/span><\/span>imagem da pessoa encontra-se ajustada ao tamanho do espelho independente da dist\u00e2ncia a que ela se encontra do mesmo<\/b><\/span><\/span><\/span>, mas, \u00e0\u00a0<\/b><\/span><\/span><\/span>medida que a pessoa se afasta do espelho<\/b><\/span><\/span><\/span>, sua\u00a0<\/b><\/span><\/span><\/span>imagem<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>tamb\u00e9m se afasta dando a impress\u00e3o<\/b><\/span><\/span><\/span>, devido ao\u00a0<\/b><\/span><\/span><\/span>\u00e2ngulo visual, que ela parece menor, mas continua sempre ajustada ao tamanho do espelho.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

Clique aqui para ver sobre o tema!<\/a><\/p>\n

Confira as v\u00eddeo aulas!<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

ACAFE \u2013 MEDICINA – 2019 Com o objetivo de desenvolver o ensino superior em Santa Catarina, 16 Funda\u00e7\u00f5es Educacionais criadas pelos munic\u00edpios se uniram em torno da constitui\u00e7\u00e3o de um \u00f3rg\u00e3o com as atribui\u00e7\u00f5es de planejar, articular e coordenar a\u00e7\u00f5es integradas entre elas. Nasce ent\u00e3o, em 2 de maio de 1974, a Associa\u00e7\u00e3o Catarinense das Funda\u00e7\u00f5es Educacionais \u2013 ACAFE, que<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4337,"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-4488","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4488","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=4488"}],"version-history":[{"count":5,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4488\/revisions"}],"predecessor-version":[{"id":10298,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4488\/revisions\/10298"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4337"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4488"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}