{"id":4741,"date":"2019-09-10T11:13:11","date_gmt":"2019-09-10T11:13:11","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4741"},"modified":"2024-07-11T03:03:24","modified_gmt":"2024-07-11T03:03:24","slug":"unitau-2019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2019\/unitau-2019\/","title":{"rendered":"UNITAU 2019"},"content":{"rendered":"

UNITAU \u2013 MEDICINA \u2013 2019<\/b><\/span><\/span><\/span><\/p>\n

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Faculdade de Medicina de Taubat\u00e9 (FMT)<\/b><\/span><\/span><\/span>, ligada hoje \u00e0\u00a0Universidade de Taubat\u00e9<\/a>, \u00e9 uma\u00a0escola<\/a>\u00a0de\u00a0medicina<\/a>do\u00a0Interior de S\u00e3o Paulo<\/a>, localizada no Campus do Bom Conselho, em\u00a0Taubat\u00e9<\/a>. \u00c9 a \u00fanica escola de medicina do\u00a0Vale do Para\u00edba<\/a>\u00a0e uma das mais antigas do interior do\u00a0Estado de S\u00e3o Paulo<\/a>.<\/b><\/span><\/span><\/span><\/p>\n

Por pertencer a uma\u00a0universidade<\/a>\u00a0de car\u00e1ter\u00a0municipal<\/a>\u00a0e filantr\u00f3pica, a faculdade possui diversos programas de bolsas de estudos que variam de 20% at\u00e9 100% de desconto nas mensalidades. A Unitau disponibiliza cerca de 1,8 milh\u00f5es de reais por ano para bolsas Simube (al\u00e9m de outros programas de bolsas e descontos oferecidos pela universidade). Por essa mesma raz\u00e3o, a\u00a0<\/b><\/span><\/span><\/span>FMT possui uma das menores mensalidades dentre as faculdades de medicina do estado de S\u00e3o Paulo.<\/b><\/span><\/span><\/span><\/p>\n

A Faculdade ainda disponibiliza bolsas de inicia\u00e7\u00e3o cient\u00edfica e monitoria da pr\u00f3pria universidade (PIC, PIBIC), do CNPq (governo federal) e\u00a0FAPESP<\/a>(governo de S\u00e3o Paulo).<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>figura abaixo mostra um objeto, de massa M, em equil\u00edbrio est\u00e1tico, suspenso<\/b><\/span><\/span><\/span> por <\/b><\/span><\/span><\/span>cabos cujas extremidades est\u00e3o presas ao teto <\/b><\/span><\/span><\/span>de uma sala. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Considere o <\/b><\/span><\/span><\/span>m\u00f3dulo da acelera\u00e7\u00e3o gravitacional terrestre g = 10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> . <\/b><\/span><\/span><\/span>Sabendo que o <\/b><\/span><\/span><\/span>\u00e2ngulo \u03b8 \u00e9 de 30<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span> , e que a tens\u00e3o m\u00e1xima suportada por cada um dos cabos utilizados \u00e9 de 2 kN, qual \u00e9 o valor m\u00e1ximo de M suportado pelo sistema?<\/b><\/span><\/span><\/span> <\/b>Dados: sen(30<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span>) <\/b><\/span><\/span><\/span>= 0,5 e <\/b><\/span><\/span><\/span>cos(30<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span>) <\/b><\/span><\/span><\/span>= 0,87 <\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>0,2 kg <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span>10 kg <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>100 kg <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>200 kg <\/b><\/span><\/span><\/span>e) <\/b><\/span><\/span><\/span>2000 kg <\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

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Considere <\/b><\/span><\/span><\/span>um objeto inicialmente em repouso e apoiado sobre uma superf\u00edcie horizontal e plana.<\/b><\/span><\/span><\/span> Esse objeto foi <\/b><\/span><\/span><\/span>arrastado de modo a descrever um movimento de transla\u00e7\u00e3o, com trajet\u00f3ria linear de 100 m,<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>sempre apoiado sobre a superf\u00edcie, at\u00e9 atingir a velocidade de 30 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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Desconsidere todos os tipos de atrito que poderiam atuar sobre o sistema. <\/b><\/span><\/span><\/span>Sabendo que a <\/b><\/span><\/span><\/span>massa do objeto \u00e9 de 90 kg, calcule o m\u00f3dulo da for\u00e7a resultante que atuou sobre o objeto ao longo da trajet\u00f3ria at\u00e9 atingir a velocidade de 30 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>102 N <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span>205 N <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>405 N <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>810 N <\/b><\/span><\/span><\/span>e) <\/b><\/span><\/span><\/span>1215 N<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

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Uma part\u00edcula, que \u00e9 um objeto cujas dimens\u00f5es s\u00e3o desprez\u00edveis em rela\u00e7\u00e3o \u00e0s demais dimens\u00f5es envolvidas no problema, deve percorrer uma dist\u00e2ncia linear<\/b><\/span><\/span><\/span> (movimento em linha reta) <\/b><\/span><\/span><\/span>de 320 m. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>dist\u00e2ncia ser\u00e1 percorrida com um movimento retil\u00edneo, uniformemente acelerado, sendo nula a velocidade inicial.<\/b><\/span><\/span><\/span> <\/b>Desconsidere todos os tipos de atrito<\/b><\/span><\/span><\/span> que poderiam atuar sobre o sistema. <\/b><\/span><\/span><\/span>Calcule o m\u00f3dulo da acelera\u00e7\u00e3o do movimento para que a velocidade m\u00e9dia ao longo dos 320 m seja de 8 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>0,2 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>b) <\/b><\/span><\/span><\/span>0,4 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>c) <\/b><\/span><\/span><\/span>0,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>d) <\/b><\/span><\/span><\/span>2,0 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>e) <\/b><\/span><\/span><\/span>4,0 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

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Um <\/b><\/span><\/span><\/span>p\u00eandulo simples \u00e9 composto por um objeto, cujas dimens\u00f5es s\u00e3o desprez\u00edveis, preso a uma linha inextens\u00edvel, sendo o comprimento da linha muito maior do que as dimens\u00f5es do objeto.<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span><\/span>outra extremidade da linha \u00e9 fixada em um suporte, ou mesmo ao teto, como mostra a figura.<\/b><\/span><\/span><\/span><\/p>\n

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A <\/b><\/span><\/span><\/span>posi\u00e7\u00e3o de equil\u00edbrio est\u00e1tico do p\u00eandulo corresponde ao ponto B. <\/b><\/span><\/span><\/span>Entretanto, <\/b><\/span><\/span><\/span>ao ser deslocado at\u00e9 o ponto A e, posteriormente, ao ser abandonado, a partir do repouso, o p\u00eandulo passa a desenvolver um movimento peri\u00f3dico do ponto A at\u00e9 o ponto C, passando por B, e, depois, fazendo o movimento inverso.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

No <\/b><\/span><\/span><\/span>sistema representado na figura, a massa do objeto pendurado na extremidade da linha \u00e9 de 10 g; <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>comprimento da linha inextens\u00edvel \u00e9 de 2,0 m; <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>m\u00f3dulo da acelera\u00e7\u00e3o gravitacional terrestre <\/b><\/span><\/span><\/span>pode ser adotado <\/b><\/span><\/span><\/span>como 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>; <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>massa do fio inextens\u00edvel \u00e9 desprez\u00edvel, e todos os tipos de atrito podem ser desprezados <\/b><\/span><\/span><\/span>para o movimento desse p\u00eandulo. <\/b><\/span><\/span><\/span>Sabendo que a energia cin\u00e9tica da massa m, ao passar pelo ponto B, \u00e9 de 0,1 J, o cosseno do \u00e2ngulo de abertura inicial do p\u00eandulo (cos \u03b8) <\/b><\/span><\/span><\/span>\u00e9 igual a <\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>0,98 <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span>0,87 <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>0,65 <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>0,<\/b><\/span><\/span><\/span>71 e) <\/b><\/span><\/span><\/span>0,50<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

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Considere <\/b><\/span><\/span><\/span>um objeto cujas dimens\u00f5es s\u00e3o desprez\u00edveis. <\/b><\/span><\/span><\/span>No <\/b><\/span><\/span><\/span>instante de tempo t = 0 s, esse objeto \u00e9 abandonado, a partir do repouso, de uma altura de H = 80 m, medida em rela\u00e7\u00e3o ao solo.<\/b><\/span><\/span><\/span> No <\/b><\/span><\/span><\/span>mesmo instante em que o objeto \u00e9 abandonado (t=0 s), uma pessoa, situada a uma dist\u00e2ncia linear R = 40 m do ponto em que o objeto ir\u00e1 tocar o solo, come\u00e7a a se deslocar em movimento retil\u00edneo uniforme em dire\u00e7\u00e3o ao ponto de colis\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Sobre o movimento descrito, \u00e9 totalmente CORRETO <\/b><\/span><\/span><\/span>afirmar: <\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>Se a velocidade m\u00e9dia de deslocamento da pessoa for igual a 5 m\/s, ela estar\u00e1 a uma dist\u00e2ncia de 20 m do ponto de colis\u00e3o quando o objeto atingir o solo. <\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>Se a velocidade m\u00e9dia de deslocamento da pessoa for igual a 8,0 m\/s, ela percorrer\u00e1 a dist\u00e2ncia R no mesmo intervalo de tempo que o objeto levar\u00e1 para atingir o solo. <\/b><\/span><\/span><\/span><\/p>\n

c) <\/b><\/span><\/span><\/span>Se a velocidade m\u00e9dia de deslocamento da pessoa for igual a 9,5 m\/s, ela estar\u00e1 a 8 m de dist\u00e2ncia do ponto de colis\u00e3o quando o objeto atingir o solo. <\/b><\/span><\/span><\/span><\/p>\n

d) <\/b><\/span><\/span><\/span>Se a velocidade m\u00e9dia de deslocamento da pessoa for igual a 12,0 m\/s, ela passar\u00e1 pelo ponto de colis\u00e3o dois segundos antes de o objeto atingir o solo. <\/b><\/span><\/span><\/span><\/p>\n

e) <\/b><\/span><\/span><\/span>Se a velocidade m\u00e9dia de deslocamento da pessoa for igual a 8,0 m\/s, ela chegar\u00e1 ao ponto de colis\u00e3o dois segundos ap\u00f3s o objeto atingir o solo.<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Uma <\/b><\/span><\/span><\/span>part\u00edcula com carga positiva \u00e9 injetada com velocidade V numa regi\u00e3o com campos el\u00e9trico e magn\u00e9tico cruzados e passa a se movimentar em uma trajet\u00f3ria retil\u00ednea. <\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>campo magn\u00e9tico \u00e9 um vetor que sai perpendicularmente \u00e0 folha do papel, enquanto o campo el\u00e9trico \u00e9 paralelo \u00e0 folha, conforme apresentado <\/b><\/span><\/span><\/span>na figura abaixo. <\/b><\/span><\/span><\/span><\/p>\n

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Sabendo que <\/b><\/span><\/span><\/span>E e B s\u00e3o os m\u00f3dulos dos campos el\u00e9trico e magn\u00e9tico, determine o m\u00f3dulo da velocidade V nessa regi\u00e3o e assinale a alternativa CORRETA.<\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>V = E\/B <\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>V = B\/E <\/b><\/span><\/span><\/span><\/p>\n

c) <\/b><\/span><\/span><\/span>V = EB <\/b><\/span><\/span><\/span><\/p>\n

d) <\/b><\/span><\/span><\/span>V = E\/B2 <\/b><\/span><\/span><\/span><\/p>\n

e) <\/b><\/span><\/span><\/span>V = B\/E2<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

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Considere o circuito representado abaixo. Ap\u00f3s fecharmos a chave faca F, o gerador n\u00e3o ideal alimenta o resistor R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> de 100 ohms.<\/b><\/span><\/span><\/span><\/p>\n

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Sabendo que a <\/b><\/span><\/span><\/span>resist\u00eancia interna do gerador \u00e9 de r = 20 ohms,<\/b><\/span><\/span><\/span> e que a <\/b><\/span><\/span><\/span>for\u00e7a eletromotriz do gerador \u00e9 de 120 volts,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>pot\u00eancia dissipada no gerador, medida em watts, <\/b><\/span><\/span><\/span>\u00e9 de <\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>100 <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span>50 <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>40 <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>30 <\/b><\/span><\/span><\/span>e) <\/b><\/span><\/span><\/span>20<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Um <\/b><\/span><\/span><\/span>agitador \u00e9 usado para aquecer uma certa quantidade de \u00e1gua,<\/b><\/span><\/span><\/span> contida num <\/b><\/span><\/span><\/span>recipiente termicamente isolado.<\/b><\/span><\/span><\/span><\/p>\n

Considere que o <\/b><\/span><\/span><\/span>calor espec\u00edfico da \u00e1gua \u00e9 1 cal\/(g.<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C).<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Admitindo que <\/b><\/span><\/span><\/span>1 cal = 4,2 J, determine a varia\u00e7\u00e3o de temperatura sofrida por uma massa de \u00e1gua de 400 g que recebe desse agitador 4,2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> J de energia.<\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>25,0 graus Celsius <\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>52,0 graus Celsius <\/b><\/span><\/span><\/span><\/p>\n

c) <\/b><\/span><\/span><\/span>22,0 graus Celsius <\/b><\/span><\/span><\/span><\/p>\n

d) <\/b><\/span><\/span><\/span>55,0 graus Celsius <\/b><\/span><\/span><\/span><\/p>\n

e) <\/b><\/span><\/span><\/span>2,50 graus Celsius<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span> (Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Qual quantidade de energia um agitador deve fornecer a 200 g de \u00e1gua, que est\u00e1 inicialmente no estado l\u00edquido <\/b><\/span><\/span><\/span>e a <\/b><\/span><\/span><\/span>zero graus Celsius, para levar essa \u00e1gua at\u00e9 a temperatura de vapor (100 \u00baC)?<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Sabe-se que o <\/b><\/span><\/span><\/span>calor espec\u00edfico da \u00e1gua \u00e9 1cal\/(g.<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C) e que 1 cal = 4,2 J. <\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>4,40.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> J <\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>8,40.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> J <\/b><\/span><\/span><\/span><\/p>\n

c) <\/b><\/span><\/span><\/span>44,0.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> J <\/b><\/span><\/span><\/span><\/p>\n

d) <\/b><\/span><\/span><\/span>48,0.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> J <\/b><\/span><\/span><\/span><\/p>\n

e) <\/b><\/span><\/span><\/span>84,0.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> J<\/b><\/span><\/span><\/span><\/p>\n

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica da <\/span><\/span><\/span>Faculdade de Medicina de Taubat\u00e9 \u2013 FMT \u2013 019<\/b><\/span><\/span><\/span><\/p>\n

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

Colocando as for\u00e7as que agem sobre cada fio (figura 1) e decompondo-as (figura 2).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe <\/b><\/span><\/span><\/span>na figura 3 que as proje\u00e7\u00f5es de T na horizontal de anulam <\/b><\/span><\/span><\/span>(equil\u00edbrio horizontal).<\/b><\/span><\/span><\/span><\/p>\n

Proje\u00e7\u00e3o de T na vertical T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> = T.sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> <\/b>= 2k.0,5 <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> = 1 kN.<\/b><\/span><\/span><\/span><\/p>\n

Equil\u00edbrio na vertical 2T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> = P <\/b><\/span><\/span><\/span> 2.1k = P <\/b><\/span><\/span><\/span>P <\/b><\/span><\/span><\/span>= 2 kN <\/b><\/span><\/span><\/span>= 2000 N.<\/b><\/span><\/span><\/span><\/p>\n

P = m.g <\/b><\/span><\/span><\/span> 2000 = m.10 m = 2000\/10 m = 200 kg.<\/b><\/span><\/span><\/span><\/p>\n

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

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

C\u00e1lculo da acelera\u00e7\u00e3o a do corpo <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>massa m = 90 kg <\/b><\/span><\/span><\/span>pela <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o de Torricelli <\/b><\/span><\/span><\/span>ap\u00f3s o <\/b><\/span><\/span><\/span>deslocamento de d = 100 m <\/b><\/span><\/span><\/span> <\/b>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> + 2.a.d<\/b><\/span><\/span><\/span> 30<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.a.100 900 = 200.a a = 900\/200 <\/b><\/span><\/span><\/span>a = 4,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Como o <\/b><\/span><\/span><\/span>movimento \u00e9 na horizontal as for\u00e7as verticais normal e peso se anulam <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>n\u00e3o existindo atrito a for\u00e7a resultante \u00e9 F tal que,<\/b><\/span><\/span><\/span> pela <\/b><\/span><\/span><\/span>segunda lei de Newton F = m.a <\/b><\/span><\/span><\/span>= 90.4,5 <\/b><\/span><\/span><\/span>= 405 N.<\/b><\/span><\/span><\/span><\/p>\n

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

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

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

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

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

Como <\/b><\/span><\/span><\/span>os atritos s\u00e3o desprezados <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>sistema \u00e9 conservativo <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>a energia mec\u00e2nica em B que \u00e9 a cin\u00e9tica E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span> = 0,1 J <\/b><\/span><\/span><\/span>(E<\/b><\/span><\/span><\/span>pB<\/b><\/span><\/span><\/sub><\/span> = 0, pois em B h = 0) <\/b><\/span><\/span><\/span>\u00e9 a mesma que em A que \u00e9 a potencial gravitacional E<\/b><\/span><\/span><\/span>MA<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>pA<\/b><\/span><\/span><\/sub><\/span> = 0,1 <\/b><\/span><\/span><\/span>j (E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span> = 0, pois a velocidade em A \u00e9 nula).<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = m.g.h <\/b><\/span><\/span><\/span> 0,1 = 0,01.10.h h = 0,1\/0,1 <\/b><\/span><\/span><\/span>h = 1m <\/b><\/span><\/span><\/span>(veja figura abaixo)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

No <\/b><\/span><\/span><\/span>tri\u00e2ngulo hachurado cos\u03b8 <\/b><\/span><\/span><\/span>= cateto adjacente\/hipotenusa <\/b><\/span><\/span><\/span>= 1\/2 <\/b><\/span><\/span><\/span> <\/b>cos\u03b8 = 0,50.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Corpo abandonado de certa altura h do solo<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

C\u00e1lculo do tempo que demora para chegar ao solo em queda livre percorrendo h = 80 m h = g.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/<\/b><\/span><\/span><\/span>2 80 =10t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 16 <\/b><\/span><\/span><\/span>t = 4 s.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da velocidade m\u00e9dia da pessoa para chegar exatamente no ponto de colis\u00e3o do objeto com o solo que demora t = 4 s percorrendo os d = 40 m V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = d\/t V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = 40\/4 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = 10 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Analisando cada afirmativa voc\u00ea comprova que a (A) \u00e9 a correta<\/b><\/span><\/span><\/span>, pois se <\/b><\/span><\/span><\/span>V = 5 m\/s em t = 4s a pessoa percorrer\u00e1 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = d\/t <\/b><\/span><\/span><\/span> 5 = d\/4 <\/b><\/span><\/span><\/span>d = 20 m e estar\u00e1 a 20 m do ponto de colis\u00e3o quando o objeto atingir o solo que ocorre quando ela percorre 40 m.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Veja abaixo um resumo da teoria:<\/b><\/span><\/span><\/span><\/p>\n

Caracter\u00edsticas do Vetor Campo El\u00e9trico<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Carga el\u00e9trica\u00a0q\u00a0lan\u00e7ada com velocidade\u00a0\u00a0lan\u00e7ada perpendicularmente \u00e0s linhas de indu\u00e7\u00e3o de um campo magn\u00e9tico uniforme<\/b><\/span><\/span><\/span><\/p>\n

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

For\u00e7a sobre q devido ao campo el\u00e9trico <\/b><\/span><\/span><\/span> E = F\/q <\/b><\/span><\/span><\/span>F = q.E (I)<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a sobre q devido ao campo magn\u00e9tico F = q.V.B (II)<\/b><\/span><\/span><\/span><\/p>\n

(I) = (II) <\/b><\/span><\/span><\/span> q.E = q.V.B <\/b><\/span><\/span><\/span>V = E\/B.<\/b><\/span><\/span><\/span><\/p>\n

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

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

C\u00e1lculo da <\/b><\/span><\/span><\/span>resist\u00eancia do resistor equivalente onde R = 100 \u2126 e r = 20 \u2126 est\u00e3o em s\u00e9rie <\/b><\/span><\/span><\/span> <\/b>R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = R + r<\/b><\/span><\/span><\/span> = 100 + 20 <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = 120 \u2126<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da <\/b><\/span><\/span><\/span>corrente el\u00e9trica i no circuito R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = (U = E)\/i <\/b><\/span><\/span><\/span> 120 = 120\/i i 120\/120 <\/b><\/span><\/span><\/span>i = 1 A.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>pot\u00eancia dissipada no gerador \u00e9 devida a sua resist\u00eancia interna r = 20 \u2126<\/b><\/span><\/span><\/span> e fornecida por <\/b><\/span><\/span><\/span>P = r.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>= 20.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 20.1 <\/b><\/span><\/span><\/span>P = 20 W<\/b><\/span><\/span><\/span><\/p>\n

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

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

W = Q <\/b><\/span><\/span><\/span>= 4,2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> j\/4,2 <\/b><\/span><\/span><\/span>= 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> cal.<\/b><\/span><\/span><\/span><\/p>\n

Q = m.c \u2206t <\/b><\/span><\/span><\/span> 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> = 400.1.\u2206t \u2206t = 10000\/400 <\/b><\/span><\/span><\/span>\u2206t = 25 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> C.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Q = m.c.\u2206t <\/b><\/span><\/span><\/span>= 200.1.100 <\/b><\/span><\/span><\/span>W <\/b><\/span><\/span><\/span>= 2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> calx4,2 <\/b><\/span><\/span><\/span>= 8,4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> J.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/a><\/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":"

UNITAU \u2013 MEDICINA \u2013 2019 Faculdade de Medicina de Taubat\u00e9 (FMT), ligada hoje \u00e0\u00a0Universidade de Taubat\u00e9, \u00e9 uma\u00a0escola\u00a0de\u00a0medicinado\u00a0Interior de S\u00e3o Paulo, localizada no Campus do Bom Conselho, em\u00a0Taubat\u00e9. \u00c9 a \u00fanica escola de medicina do\u00a0Vale do Para\u00edba\u00a0e uma das mais antigas do interior do\u00a0Estado de S\u00e3o Paulo. Por pertencer a uma\u00a0universidade\u00a0de car\u00e1ter\u00a0municipal\u00a0e filantr\u00f3pica, a faculdade possui diversos programas de bolsas<\/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-4741","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4741","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=4741"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4741\/revisions"}],"predecessor-version":[{"id":10272,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4741\/revisions\/10272"}],"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=4741"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}