{"id":4726,"date":"2019-09-10T01:14:47","date_gmt":"2019-09-10T01:14:47","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4726"},"modified":"2024-07-11T11:58:28","modified_gmt":"2024-07-11T11:58:28","slug":"puc-rj-2019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2019\/puc-rj-2019\/","title":{"rendered":"PUC RJ 2019"},"content":{"rendered":"

Puc \u2013 RJ – 2019<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>Pontif\u00edcia Universidade Cat\u00f3lica do Rio de Janeiro<\/b><\/span><\/span><\/span>\u00a0(<\/b><\/span><\/span><\/span>PUC-Rio<\/b><\/span><\/span><\/span>) \u00e9 uma\u00a0institui\u00e7\u00e3o de ensino superior<\/a><\/b><\/span><\/span><\/span> <\/b>comunit\u00e1ria, filantr\u00f3pica e sem fins lucrativos da Secretaria de Regula\u00e7\u00e3o e Supervis\u00e3o da Educa\u00e7\u00e3o Superior. Sediada na\u00a0G\u00e1vea<\/a>,\u00a0Rio de Janeiro<\/a>, no\u00a0Brasil<\/a>.\u00a0Nos tempos atuais, \u00e9 reconhecida como uma das melhores e mais prestigiadas\u00a0universidades da Am\u00e9rica Latina<\/a>.<\/b><\/span><\/span><\/span><\/p>\n

A PUC-Rio adota duas formas de ingresso para gradua\u00e7\u00e3o: o\u00a0Enem<\/a>\u00a0e seu vestibular pr\u00f3prio. Por meio dessas provas, distribuem-se bolsas acad\u00eamicas aos estudantes.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0PUC-Rio foi uma das universidades pioneiras no ensino do\u00a0empreendedorismo<\/a>\u00a0nos cursos de\u00a0gradua\u00e7\u00e3o<\/a>\u00a0no Brasil.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

01-(PUC \u2013 RIO \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>partir do solo, uma bola \u00e9 lan\u00e7ada verticalmente para cima <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>alcan\u00e7a uma altura de 3,2 metros. Quanto tempo, em segundos, a bola leva para subir e retornar ao n\u00edvel do solo?<\/b><\/span><\/span><\/span><\/p>\n

Despreze a resist\u00eancia do ar. Dado: g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

A) 0,32 B) 0,64 <\/b><\/span><\/span><\/span>C) <\/b><\/span><\/span><\/span>0,80 <\/b><\/span><\/span><\/span>D) <\/b><\/span><\/span><\/span>1,6 <\/b><\/span><\/span><\/span>E) <\/b><\/span><\/span><\/span>3,2 <\/b><\/span><\/span><\/span><\/p>\n

02-(PUC \u2013 RIO \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Uma <\/b><\/span><\/span><\/span>diferen\u00e7a de potencial V \u00e9 aplicada a um resistor de resist\u00eancia R. <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>pot\u00eancia dissipada nesse resistor \u00e9 P. <\/b><\/span><\/span><\/span><\/p>\n

Ao dobrar a resist\u00eancia e triplicar a diferen\u00e7a de potencial, a nova pot\u00eancia dissipada <\/b><\/span><\/span><\/span>ser\u00e1: <\/b><\/span><\/span><\/span><\/p>\n

A) <\/b><\/span><\/span><\/span>6P <\/b><\/span><\/span><\/span>B) <\/b><\/span><\/span><\/span>3P <\/b><\/span><\/span><\/span>C) <\/b><\/span><\/span><\/span>P <\/b><\/span><\/span><\/span>D) <\/b><\/span><\/span><\/span>2P <\/b><\/span><\/span><\/span>E) <\/b><\/span><\/span><\/span>9P\/2 <\/b><\/span><\/span><\/span><\/p>\n

03-(PUC \u2013 RIO \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Dentro de um calor\u00edmetro perfeito, de capacidade C = 40 cal\/<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>temperatura inicial 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C, <\/b><\/span><\/span><\/span>colocam-se <\/b><\/span><\/span><\/span>100 g de um material de calor espec\u00edfico 0,50 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a uma temperatura de 90 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>, e <\/b><\/span><\/span><\/span>uma massa de 10 g de gelo a 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Calcule, em <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C,<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>valor da temperatura final de equil\u00edbrio do sistema.<\/b><\/span><\/span><\/span><\/p>\n

Dados: C<\/b><\/span><\/span><\/span>\u00c1GUA<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1,0 cal\/go C <\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/span>FUSAO GELO<\/b><\/span><\/span><\/sub><\/span> <\/b>= 80 cal\/g<\/b><\/span><\/span><\/span><\/p>\n

A) <\/b><\/span><\/span><\/span>40 <\/b><\/span><\/span><\/span>B) <\/b><\/span><\/span><\/span>39 <\/b><\/span><\/span><\/span>C) <\/b><\/span><\/span><\/span>38 <\/b><\/span><\/span><\/span>D) <\/b><\/span><\/span><\/span>37 <\/b><\/span><\/span><\/span>E) <\/b><\/span><\/span><\/span>36 <\/b><\/span><\/span><\/span><\/p>\n

04-(PUC \u2013 RIO \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Nas f\u00e9rias, <\/b><\/span><\/span><\/span>um casal faz uma <\/b><\/span><\/span><\/span>viagem de carro, saindo do Rio de Janeiro rumo a Tiradentes<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>primeiro trecho, na BR-101 at\u00e9 Barbacena<\/b><\/span><\/span><\/span>, \u00e9 <\/b><\/span><\/span><\/span>feito <\/b><\/span><\/span><\/span>com uma <\/b><\/span><\/span><\/span>velocidade escalar m\u00e9dia de 70 km\/h. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

O <\/b><\/span><\/span><\/span>casal faz uma parada para um lanche por meia hora <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>prossegue a viagem tomando a BR-265 para Tiradentes<\/b><\/span><\/span><\/span>, um <\/b><\/span><\/span><\/span>trecho de 60 km que eles percorrem em 60 min<\/b><\/span><\/span><\/span>. O <\/b><\/span><\/span><\/span>tempo total de viagem \u00e9 de 5h30min. Calcule, em km\/h,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>velocidade m\u00e9dia do carro no percurso total Rio-Tiradentes.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

A) <\/b><\/span><\/span><\/span>60 <\/b><\/span><\/span><\/span>B) <\/b><\/span><\/span><\/span>62 <\/b><\/span><\/span><\/span>C) <\/b><\/span><\/span><\/span>65 <\/b><\/span><\/span><\/span>D) <\/b><\/span><\/span><\/span>68 <\/b><\/span><\/span><\/span>E) <\/b><\/span><\/span><\/span>70 <\/b><\/span><\/span><\/span><\/p>\n

05-(PUC \u2013 RIO \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Uma <\/b><\/span><\/span><\/span>caixa de 20 kg est\u00e1 na borda da ca\u00e7amba de um caminh\u00e3o, <\/b><\/span><\/span><\/span>a uma <\/b><\/span><\/span><\/span>altura de 1,0 m do piso. <\/b><\/span><\/span><\/span>Um rapaz <\/b><\/span><\/span><\/span>coloca uma t\u00e1bua de madeira, de 2,0 m de comprimento, desde a borda da ca\u00e7amba at\u00e9 o piso.<\/b><\/span><\/span><\/span> Ele d\u00e1 um <\/b><\/span><\/span><\/span>pequeno empurr\u00e3o na caixa e a mesma desliza pela t\u00e1bua com velocidade constante at\u00e9 chegar ao piso.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

<\/p>\n

Calcule, em J, o m\u00f3dulo do trabalho realizado pela for\u00e7a de atrito exercida pela t\u00e1bua sobre a caixa durante a descida<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

A) <\/b><\/span><\/span><\/span>20 <\/b><\/span><\/span><\/span>B) <\/b><\/span><\/span><\/span>40 <\/b><\/span><\/span><\/span>C) <\/b><\/span><\/span><\/span>100 <\/b><\/span><\/span><\/span>D) <\/b><\/span><\/span><\/span>200 <\/b><\/span><\/span><\/span>E) <\/b><\/span><\/span><\/span>400 <\/b><\/span><\/span><\/span><\/p>\n

06-(PUC \u2013 RIO \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Tr\u00eas corpos, 1, 2 e 3, <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>massas m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10 kg, m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 15 kg e m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 25 kg, <\/b><\/span><\/span><\/span>se <\/b><\/span><\/span><\/span>movem horizontalmente sobre um trilho no eixo infinito x, sem nenhuma resist\u00eancia ou atrito, <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>velocidades iniciais v<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 4,0 m\/s, v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = -2,0 m\/s e v<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 0,0 m\/s, <\/b><\/span><\/span><\/span>respectivamente. <\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>dist\u00e2ncia inicial entre os blocos 1 e 2 \u00e9 1,0 m <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>entre os blocos 2 e 3 \u00e9 2,0m, <\/b><\/span><\/span><\/span>como <\/b><\/span><\/span><\/span>mostrado na figura.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Os <\/b><\/span><\/span><\/span>corpos 1 e 2 sofrem uma colis\u00e3o completamente inel\u00e1stica,<\/b><\/span><\/span><\/span> ou seja, eles <\/b><\/span><\/span><\/span>grudam um no outro ap\u00f3s colidir.<\/b><\/span><\/span><\/span> Esse <\/b><\/span><\/span><\/span>conjunto ent\u00e3o colide elasticamente com o corpo 3. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Calcule o m\u00f3dulo da velocidade do corpo 3, em m\/s, ap\u00f3s 153 s a partir do instante inicial.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

A) <\/b><\/span><\/span><\/span>4,00 <\/b><\/span><\/span><\/span>B) <\/b><\/span><\/span><\/span>2,00 <\/b><\/span><\/span><\/span>C) <\/b><\/span><\/span><\/span>0,40 <\/b><\/span><\/span><\/span>D) <\/b><\/span><\/span><\/span>0,20 <\/b><\/span><\/span><\/span>E) <\/b><\/span><\/span><\/span>0,00 <\/b><\/span><\/span><\/span><\/p>\n

07-(PUC \u2013 RIO \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Uma <\/b><\/span><\/span><\/span>bateria fornece uma voltagem V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> e possui uma resist\u00eancia interna r.<\/b><\/span><\/span><\/span> <\/b>Colocando a bateria em contato com um volt\u00edmetro, a medida \u00e9 12 V<\/b><\/span><\/span><\/span>. Colocando a seguir <\/b><\/span><\/span><\/span>um amper\u00edmetro em s\u00e9rie com a bateria e com um resistor R = 1,0 k\u03a9, a medida \u00e9 8 mA. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>partir dessas medidas, os valores da voltagem V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> e da resist\u00eancia interna r, em V e k\u03a9, s\u00e3o<\/b><\/span><\/span><\/span> respectivamente: <\/b><\/span><\/span><\/span><\/p>\n

A) <\/b><\/span><\/span><\/span>12 e 0,50 <\/b><\/span><\/span><\/span><\/p>\n

B) <\/b><\/span><\/span><\/span>24 e 1,0 <\/b><\/span><\/span><\/span><\/p>\n

C) <\/b><\/span><\/span><\/span>12 e 1,0 <\/b><\/span><\/span><\/span><\/p>\n

D) <\/b><\/span><\/span><\/span>24 e 0,50 <\/b><\/span><\/span><\/span><\/p>\n

E) <\/b><\/span><\/span><\/span>12 e 2,0 <\/b><\/span><\/span><\/span><\/p>\n

08-(PUC \u2013 RIO \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Sejam as seguintes afirma\u00e7\u00f5es: <\/b><\/span><\/span><\/span><\/p>\n

I. Quanto maior a frequ\u00eancia de uma onda de luz no v\u00e1cuo, maior a velocidade de propaga\u00e7\u00e3o dessa onda. <\/b><\/span><\/span><\/span><\/p>\n

II. Em uma corda com seus dois extremos fixos, ondas estacion\u00e1rias somente poder\u00e3o ser produzidas se o comprimento da corda for um m\u00faltiplo do comprimento de onda. <\/b><\/span><\/span><\/span><\/p>\n

III. Ondas sonoras precisam de um meio material para se propagarem e s\u00e3o longitudinais no ar. <\/b><\/span><\/span><\/span>Marque a <\/b><\/span><\/span><\/span>op\u00e7\u00e3o correta:<\/b><\/span><\/span><\/span><\/p>\n

A) <\/b><\/span><\/span><\/span>Somente a afirma\u00e7\u00e3o I \u00e9 verdadeira. <\/b><\/span><\/span><\/span><\/p>\n

B) <\/b><\/span><\/span><\/span>Somente a afirma\u00e7\u00e3o II \u00e9 verdadeira. <\/b><\/span><\/span><\/span><\/p>\n

C) <\/b><\/span><\/span><\/span>Somente a afirma\u00e7\u00e3o III \u00e9 verdadeira. <\/b><\/span><\/span><\/span><\/p>\n

D) <\/b><\/span><\/span><\/span>Somente as afirma\u00e7\u00f5es I e II s\u00e3o verdadeiras. <\/b><\/span><\/span><\/span><\/p>\n

E) <\/b><\/span><\/span><\/span>Somente as afirma\u00e7\u00f5es I e III s\u00e3o verdadeiras. 2 <\/b><\/span><\/span><\/span><\/p>\n

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

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

Lan\u00e7amento vertical para cima<\/b><\/span><\/span><\/span><\/span><\/p>\n

Considere um corpo\u00a0<\/b><\/span><\/span>lan\u00e7ado verticalmente para cima<\/b><\/span><\/span>, a partir de um ponto A (origem), com\u00a0<\/b><\/span><\/span>velocidade escalar inicial\u00a0<\/b><\/span><\/span>Vo<\/b><\/span><\/span>.<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0O\u00a0<\/b><\/span><\/span>tempo subida \u00e9 igual ao tempo de descida.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0A\u00a0<\/b><\/span><\/span>velocidade (Vo)\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span>lan\u00e7amento\u00a0na origem\u00a0<\/b><\/span><\/span>\u00e9\u00a0<\/b><\/span>igual \u00e0\u00a0mesma velocidade de\u00a0chegada \u00e0 origem<\/b><\/span><\/span>,\u00a0mas de\u00a0<\/b><\/span>sinal contr\u00e1rio (-Vo).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0Em\u00a0<\/b><\/span><\/span>qualquer ponto da trajet\u00f3ria\u00a0<\/b><\/span><\/span>o corpo\u00a0<\/b><\/span>tem duas velocidades de mesmo m\u00f3dulo,\u00a0<\/b><\/span><\/span>uma<\/b><\/span>positiva na subida\u00a0<\/b><\/span><\/span>e uma\u00a0<\/b><\/span>negativa na descida.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0Se um\u00a0<\/b><\/span><\/span>m\u00f3vel A partir um tempo x antes de um m\u00f3vel B,\u00a0<\/b><\/span><\/span>t\u00eam-se: tA\u00a0\u2013 tB\u00a0= x\u00a0\u00a0<\/b><\/span>tA\u00a0= tB\u00a0+ x,<\/b><\/span><\/span>que\u00a0<\/b><\/span>deve-se substituir nas fun\u00e7\u00f5es hor\u00e1rias do espa\u00e7o ou da velocidade para continuar a resolu\u00e7\u00e3o do exerc\u00edcio.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio, h = <\/b><\/span><\/span><\/span>3,2 m e <\/b><\/span><\/span><\/span>g = <\/b><\/span><\/span><\/span>10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b><\/span><\/span><\/span>na altura m\u00e1xima V = 0 <\/b><\/span><\/span><\/span> c\u00e1lculo da <\/b><\/span><\/span><\/span>velocidade de lan\u00e7amento por Torricelli V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>o<\/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> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2.10.3,2 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 64 <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 8 m\/s. <\/b><\/span><\/span><\/span><\/p>\n

Tempo que demora para atingir a altura m\u00e1xima V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> – a.t <\/b><\/span><\/span><\/span> 0 = 8 \u2013 10t t = 8\/10 <\/b><\/span><\/span><\/span>t = 0,8 s.<\/b><\/span><\/span><\/span><\/p>\n

Como <\/b><\/span><\/span><\/span>o atrito \u00e9 desprezado, o tempo de subida \u00e9 igual ao tempo de descida <\/b><\/span><\/span><\/span> t<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span> = 0,8 + 0,8<\/b><\/span><\/span><\/span><\/p>\n

T<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span> = 1,6 s.<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

<\/p>\n

Se <\/b><\/span><\/span><\/span>voc\u00ea dobrar o valor da resist\u00eancia a pot\u00eancia ficar\u00e1 2 vezes menor <\/b><\/span><\/span><\/span>e se voc\u00ea <\/b><\/span><\/span><\/span>triplicar a ddp a pot\u00eancia ficar\u00e1 3<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 9 vezes maior P\u2019 = 9P\/2. <\/b><\/span><\/span><\/span><\/p>\n

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

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

Recipiente passando de t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C at\u00e9 t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = C.(t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> = 40.(t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \u2013 0) <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 40t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> <\/b><\/p>\n

Material passando de to = 90 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C at\u00e9 t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> m.c .(t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) = 100.0,5.(t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \u2013 90) <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 50t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \u2013 4500<\/b><\/span><\/span><\/span><\/p>\n

Gelo a 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C se fundindo <\/b><\/span><\/span><\/span>(passando 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) <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = m.L <\/b><\/span><\/span><\/span>= 10.80 <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 800 cal<\/b><\/span><\/span><\/span><\/p>\n

\u00c1gua passando de t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C at\u00e9 t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> Q<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span> = m.c.(t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> = 10.1.(t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \u2013 0) <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span> = 10t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Como o <\/b><\/span><\/span><\/span>calor\u00edmetro \u00e9 perfeito <\/b><\/span><\/span><\/span>n\u00e3o h\u00e0 trocas de calor com o meio exterior e a <\/b><\/span><\/span><\/span>soma das quantidades de calor trocadas entre eles deve ser nula 40t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> + 50t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> \u2013 4500 + 800 + 10t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> <\/b>= 0 100t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 3700 <\/b><\/span><\/span><\/span><\/p>\n

t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 3700\/100 <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 37 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

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

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

O <\/b><\/span><\/span><\/span>segundo trecho de Barbacena at\u00e9 Tiradentes <\/b><\/span><\/span><\/span>tem <\/b><\/span><\/span><\/span>dist\u00e2ncia de d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 60 km <\/b><\/span><\/span><\/span>e \u00e9 percorrido em <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 60 min <\/b><\/span><\/span><\/span>= 1 h.<\/b><\/span><\/span><\/span><\/p>\n

Como ele <\/b><\/span><\/span><\/span>parou durante 0,5 h no restaurante, <\/b><\/span><\/span><\/span>e o <\/b><\/span><\/span><\/span>tempo total de viajem foi de <\/b><\/span><\/span><\/span>5h30min = <\/b><\/span><\/span><\/span>5,5 h<\/b><\/span><\/span><\/span>, o <\/b><\/span><\/span><\/span>primeiro trecho do percurso foi efetuado em t<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>= 5,5 h (tempo total) \u2013 (1 h \u201cprimeiro trecho\u201d + 0,5 h \u201crestaurante\u201d) = 5,5 \u2013 1,5 t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 4 h (tempo para percorrer o primeiro trecho).<\/b><\/span><\/span><\/span><\/p>\n

Primeiro trecho V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span> 70 = d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/4 <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 280 km.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Velocidade m\u00e9dia total = dist\u00e2ncia total\/tempo total (inclusive parad<\/b><\/span><\/span><\/span>a) V<\/b><\/span><\/span><\/span>mt<\/b><\/span><\/span><\/sub><\/span> = (280 + 60)\/5,5 = 340\/5,5 <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>mt<\/b><\/span><\/span><\/sub><\/span> = 61,8 km\/h.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Como a <\/b><\/span><\/span><\/span>caixa desce com velocidade constante a energia cin\u00e9tica (E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2) \u00e9 a mesma <\/b><\/span><\/span><\/span>durante todo o trajeto e o <\/b><\/span><\/span><\/span>trabalho realizado pela for\u00e7a de atrito corresponde \u00e0 varia\u00e7\u00e3o de energia potencial gravitacional<\/b><\/span><\/span><\/span>, j\u00e1 que a <\/b><\/span><\/span><\/span>normal n\u00e3o realiza trabalho <\/b><\/span><\/span><\/span>(W<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> = F.d.cos90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = F.d.0 = 0).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

Primeira colis\u00e3o, inel\u00e1stica entre 1 e 2 que se movem juntos com velocidade comum V ap\u00f3s a colis\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

Como os <\/b><\/span><\/span><\/span>atritos s\u00e3o desprezados eles se movem com velocidades constantes. <\/b><\/span><\/span><\/span><\/p>\n

Utilizando o <\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Quantidade de movimento do sistema antes do choque Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 10.4 + 15.(-2) <\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = 10 kg.m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de movimento do sistema depois do choque quando se movem unidos com velocidade comum V Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V <\/b><\/span><\/span><\/span>= 10.V + 15.(V) <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = 25V.<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> 10 = 25V<\/b><\/span><\/span><\/span> V = 10\/25 <\/b><\/span><\/span><\/span>V = 0,4 m\/s <\/b><\/span><\/span><\/span>(velocidade comum dos blocos 1 e 2 ap\u00f3s o choque).<\/b><\/span><\/span><\/span><\/p>\n

Segunda colis\u00e3o entre os blocos 1 e 2, considerados como um \u00fanico bloco de massa m<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sub><\/span> = (10 + 15) = 25 kg e o bloco 3 de massa m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 25 kg.<\/b><\/span><\/span><\/span><\/p>\n

Como a <\/b><\/span><\/span><\/span>colis\u00e3o \u00e9 el\u00e1stica<\/b><\/span><\/span><\/span> <\/b>(coeficiente de restitui\u00e7\u00e3o \u00e9 1) e <\/b><\/span><\/span><\/span>eles <\/b><\/span><\/span><\/span>possuem a mesma massa <\/b><\/span><\/span><\/span>eles<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

trocam suas velocidades ap\u00f3s o choque <\/b><\/span><\/span><\/span>e, assim o bloco de massa m<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sub><\/span> fica parado (V = 0) e o bloco 3 se desloca com velocidade constante de V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = 0,4 m\/s. <\/b><\/span><\/span><\/span><\/p>\n

Sendo os atritos desprezados, ap\u00f3s qualquer instante (inclusive 153 s) essa velocidade ser\u00e1 a mesma.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Quando o <\/b><\/span><\/span><\/span>volt\u00edmetro \u00e9 colocado em contato com os polos da bateria<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>indicar 12 V <\/b><\/span><\/span><\/span>ele <\/b><\/span><\/span><\/span>estar\u00e1 fornecendo o valor da for\u00e7a eletromotriz E da bateria E = 12 V.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>ddp, tens\u00e3o ou voltagem U nos terminais do resistor de R = 1 k\u2126 = 1.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> \u2126, percorrido por i = 8 mA = 8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> A <\/b><\/span><\/span><\/span>(indica\u00e7\u00e3o do amper\u00edmetro<\/b><\/span><\/span><\/span>) ser\u00e1 R = U\/i <\/b><\/span><\/span><\/span> 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> = U\/8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> <\/b><\/span><\/span><\/span> U = 8 V.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o do gerador U = E \u2013 r.i <\/b><\/span><\/span><\/span> 8 = 12 \u2013 r.8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> r = 4\/8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> <\/b><\/span><\/span><\/span>r <\/b><\/span><\/span><\/span>= 0,5 .10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> <\/b>= 0,5 k\u2126.<\/b><\/span><\/span><\/span><\/p>\n

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

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

I. Falsa <\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>velocidade da luz <\/b><\/span><\/span><\/span>(e de outra qualquer onda eletromagn\u00e9tica) <\/b><\/span><\/span><\/span>no v\u00e1cuo \u00e9 constante e vale V = c = 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s, independente da frequ\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

II. Falsa <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Devem ter o mesmo comprimento de onda.<\/b><\/span><\/span><\/span><\/p>\n

III. Verdadeira <\/b><\/span><\/span><\/span><\/p>\n

\"PUC<\/p>\n

R- C<\/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":"

Puc \u2013 RJ – 2019 A\u00a0Pontif\u00edcia Universidade Cat\u00f3lica do Rio de Janeiro\u00a0(PUC-Rio) \u00e9 uma\u00a0institui\u00e7\u00e3o de ensino superior comunit\u00e1ria, filantr\u00f3pica e sem fins lucrativos da Secretaria de Regula\u00e7\u00e3o e Supervis\u00e3o da Educa\u00e7\u00e3o Superior. Sediada na\u00a0G\u00e1vea,\u00a0Rio de Janeiro, no\u00a0Brasil.\u00a0Nos tempos atuais, \u00e9 reconhecida como uma das melhores e mais prestigiadas\u00a0universidades da Am\u00e9rica Latina. A PUC-Rio adota duas formas de ingresso para gradua\u00e7\u00e3o:<\/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-4726","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4726","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=4726"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4726\/revisions"}],"predecessor-version":[{"id":10299,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4726\/revisions\/10299"}],"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=4726"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}