{"id":1946,"date":"2016-02-11T23:39:51","date_gmt":"2016-02-11T23:39:51","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1946"},"modified":"2024-09-03T13:58:20","modified_gmt":"2024-09-03T13:58:20","slug":"ufsc-2016-resolucao-comentada","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2016\/ufsc-2016\/ufsc-2016-resolucao-comentada\/","title":{"rendered":"UFSC 2016 – Resolu\u00e7\u00e3o Comentada"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es do vestibular da UFSC \u2013 2016<\/span><\/span><\/span><\/p>\n

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

01 \"\" Verdadeiro \"\" Galileu <\/b><\/span><\/span><\/span>foi um dos que se <\/b><\/span><\/span><\/span>influenciaram nas ideias de Francis Bacon<\/b><\/span><\/span><\/span>, que defendia a <\/b><\/span><\/span><\/span>observa\u00e7\u00e3o da natureza em conjunto da coleta de fatos.<\/b><\/span><\/span><\/span> <\/b>Investigava a natureza<\/b><\/span><\/span><\/span> com base em <\/b><\/span><\/span><\/span>observa\u00e7\u00f5es e experi\u00eancia emp\u00edrica,<\/b><\/span><\/span><\/span> reduzindo os <\/b><\/span><\/span><\/span>problemas a um simples conjunto de termos e resolvendo-os usando matem\u00e1tica.<\/b><\/span><\/span><\/span><\/p>\n

02<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Falso \"\" <\/b><\/span><\/span><\/span> Descoberta por Galileu, a <\/b><\/span><\/span><\/span>Lei da queda dos corpos diz que todos os corpos caem com acelera\u00e7\u00e3o constante, n\u00e3o dependendo da massa.<\/b><\/span><\/span><\/span><\/p>\n

04<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Falso \"\" <\/b><\/span><\/span><\/span>Galileu era um dos <\/b><\/span><\/span><\/span>defensores da teoria Helioc\u00eantrica<\/b><\/span><\/span><\/span>, aonde os <\/b><\/span><\/span><\/span>planetas giram em torno do Sol.<\/b><\/span><\/span><\/span><\/p>\n

08 \"\"<\/b><\/span><\/span><\/span> Falso \"\" <\/b><\/span><\/span><\/span> O <\/b><\/span><\/span><\/span>inventor<\/b><\/span><\/span><\/span> do telesc<\/b><\/span><\/span><\/span>\u00f3pio \u00e9 <\/b><\/span><\/span><\/span>Hans Lippershey,<\/b><\/span><\/span><\/span> Galileu apenas <\/b><\/span><\/span><\/span>aperfei\u00e7oou esse equipamento.<\/b><\/span><\/span><\/span><\/p>\n

16 \"\"<\/b><\/span><\/span><\/span> <\/b>Verdadeiro \"\" <\/b><\/span><\/span><\/span> Galileu foi respons<\/b><\/span><\/span><\/span>\u00e1vel pelos <\/b><\/span><\/span><\/span>primeiros estudos consistentes do movimento<\/b><\/span><\/span><\/span> <\/b>uniformemente acelerado, tamb\u00e9m enunciou a lei dos corpos e o princ\u00edpio da in\u00e9rcia.<\/b><\/span><\/span><\/span><\/p>\n

32<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span> <\/b>Falso \"\" <\/b><\/span><\/span><\/span> <\/b>Emp<\/b><\/span><\/span><\/span>\u00e9docles<\/b><\/span><\/span><\/span> foi o <\/b><\/span><\/span><\/span>primeiro a declarar essa teoria<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span>Galileu derrubou-a<\/b><\/span><\/span><\/span> ao afirmar que assim como na <\/b><\/span><\/span><\/span>Terra os astros tamb\u00e9m passavam por transforma\u00e7\u00f5es, portanto deveriam ser compostos de uma mesma mat\u00e9ria.<\/b><\/span><\/span><\/span><\/p>\n

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

01<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>Verdadeiro \"\"<\/b><\/span><\/span><\/span> pelo fato de o <\/b><\/span><\/span><\/span>sistema ser constitu<\/b><\/span><\/span><\/span>\u00eddo por duas roldanas <\/b><\/span><\/span><\/span>acontece o <\/b><\/span><\/span><\/span>seguinte equil\u00edbrio de for\u00e7as:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde as <\/b><\/span><\/span><\/span>setas de mesma cor t\u00eam a mesma intensidade.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>seta azul representa a for\u00e7a exercida pelo professor (350N). <\/b><\/span><\/span><\/span>Por ser uma <\/b><\/span><\/span><\/span>roldana e pelas propriedades de a\u00e7\u00e3o e rea\u00e7\u00e3o uma outra for\u00e7a igual e de mesma dire\u00e7\u00e3o \u00e9 aplicada no outro fio.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>soma dessas duas for\u00e7as <\/b><\/span><\/span><\/span>resulta na for\u00e7a representada pela <\/b><\/span><\/span><\/span>seta vermelha, <\/b><\/span><\/span><\/span>que ter\u00e1 <\/b><\/span><\/span><\/span>intensidade de 350 + 350 = 700N. <\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>mesmo racioc\u00ednio \u00e9 v\u00e1lido para se calcular a for\u00e7a representada pela seta amarela, <\/b><\/span><\/span><\/span>s\u00f3 que nesse caso <\/b><\/span><\/span><\/span>com for\u00e7as de 700N, ou seja, 700 + 700 = 1400N.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A <\/b><\/span><\/span><\/span>seta vermelha \u00e9 a for\u00e7a normal, <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>nesse caso tem mesma intensidade da for\u00e7a peso, j\u00e1 que possuem sentidos opostos e n\u00e3o existem outras for\u00e7as agindo verticalmente no bloco<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

N = P \"\" <\/b><\/span><\/span><\/span> <\/b>N = 4000N.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>seta amarela \u00e9 a tra\u00e7\u00e3o, calculada no (01), logo tem intensidade de F=1400N.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>seta verde representa a for\u00e7a de atrito<\/b><\/span><\/span><\/span>, nesse caso estudaremos o movimento, portanto <\/b><\/span><\/span><\/span>deveremos considerar o coeficiente de atrito cin\u00e9tico (<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 0,25). <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>for\u00e7a de atrito cin\u00e9tico (F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>pode ser <\/b><\/span><\/span><\/span>calculada por \"\"<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>c <\/b><\/span><\/span><\/sub><\/span>= N.<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/sub><\/span>Fc = 4000.0,25<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/sub><\/span>Fc = 1000N.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Essa <\/b><\/span><\/span><\/span>for\u00e7a ter\u00e1 sentido para a direita, <\/b><\/span><\/span><\/span>pois <\/b><\/span><\/span><\/span>a soma das for\u00e7as exercidas horizontalmente no lado direito do corpo ter\u00e3o maior intensidade que as do esquerdo. <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>intensidade dessa for\u00e7a <\/b><\/span><\/span><\/span>pode ser calculada por \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span> = F \u2013 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span> = 1400 – 1000<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span> = 400 N.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

04 <\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>Falso <\/b><\/span><\/span><\/span> \"\" Como estudado no item <\/b><\/span><\/span><\/span>anterior (02), <\/b><\/span><\/span><\/span>existe uma <\/b><\/span><\/span><\/span>outra for<\/b><\/span><\/span><\/span>\u00e7a al\u00e9m das citadas, a for\u00e7a normal, que surge do contato entre o corpo e o ch\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

08 <\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>Verdadeiro \"\" Tamb<\/b><\/span><\/span><\/span>\u00e9m demonstrado no item (02), <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>for\u00e7a resultante \u00e9 de 400N.<\/b><\/span><\/span><\/span><\/p>\n

16<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Falso<\/b><\/span><\/span><\/span> \"\" Nesse caso <\/b><\/span><\/span><\/span>vamos estudar o objeto antes do movimento<\/b><\/span><\/span><\/span>, com isso utilizaremos o <\/b><\/span><\/span><\/span>coeficiente de atrito est<\/b><\/span><\/span><\/span>\u00e1tico (<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>),<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>for\u00e7a exercida para que o bloco se movimente deve ser maior que a for\u00e7a de atrito est\u00e1tico (F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>), calculada a seguir \"\" <\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/sub><\/span>= N.<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> <\/b> \"\" Fe = 4000.0.3<\/b><\/span><\/span><\/span> <\/b> \"\" <\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 1200 N.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

1 \u2013 <\/b><\/span><\/span><\/span>\u00c1rea do tri\u00e2ngulo<\/b><\/span><\/span><\/span> \"\" (b.h)\/2 \"\" (20.10)\/2 \"\"100m<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 <\/b><\/span><\/span><\/span>\u00c1rea do ret\u00e2ngulo<\/b><\/span><\/span><\/span> \"\" b.h \"\" 8.20 \"\" 160m<\/b><\/span><\/span><\/span><\/p>\n

3 \u2013 <\/b><\/span><\/span><\/span>\u00c1rea do paralelogramo<\/b><\/span><\/span><\/span> \"\" ((B+b).h)\/2 \"\" ((40+20).8)\/2 \"\" 240m<\/b><\/span><\/span><\/span><\/p>\n

4 \u2013 <\/b><\/span><\/span><\/span>\u00c1rea do ret\u00e2ngulo <\/b><\/span><\/span><\/span> \"\" b.h \"\" 40.3 \"\" 120m<\/b><\/span><\/span><\/span><\/p>\n

5 \u2013 <\/b><\/span><\/span><\/span>\u00c1rea do paralelogramo <\/b><\/span><\/span><\/span> <\/b> \"\"<\/b><\/span><\/span><\/span> ((B+b).h)\/2 \"\" ((40+20).5)\/2 \"\" 150m<\/b><\/span><\/span><\/span><\/p>\n

6 \u2013 <\/b><\/span><\/span><\/span>\u00c1rea do ret\u00e2ngulo<\/b><\/span><\/span><\/span> \"\" b.h \"\" 10.20 \"\" 200m<\/b><\/span><\/span><\/span><\/p>\n

7 \u2013 <\/b><\/span><\/span><\/span>\u00c1rea do paralelogramo <\/b><\/span><\/span><\/span> \"\" ((B+b).h)\/2 \"\" ((40+20).8)\/2 \"\" <\/b><\/span><\/span><\/span>240m<\/b><\/span><\/span><\/span><\/p>\n

8 \u2013 <\/b><\/span><\/span><\/span>\u00c1rea do ret\u00e2ngulo <\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>b.h \"\" 40.5 \"\" <\/b><\/span><\/span><\/span>200m<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>deslocamento total<\/b><\/span><\/span><\/span> (D) \u00e9 a <\/b><\/span><\/span><\/span>soma<\/b><\/span><\/span><\/span> de todas esses deslocamentos:<\/b><\/span><\/span><\/span><\/p>\n

D=100+160+240+120+150+200+240+200 \"\" D=1410m<\/b><\/span><\/span><\/span><\/p>\n

02 \"\" Falso. O carro realmente <\/b><\/span><\/span><\/span>est\u00e1<\/b><\/span><\/span><\/span> em <\/b><\/span><\/span><\/span>movimento uniforme<\/b><\/span><\/span><\/span>, devido \u00e0 sua <\/b><\/span><\/span><\/span>velocidade constante<\/b><\/span><\/span><\/span>, por\u00e9m o gr\u00e1fico <\/b><\/span><\/span><\/span>n\u00e3o pode garantir<\/b><\/span><\/span><\/span> que esse movimento era necessariamente <\/b><\/span><\/span><\/span>retil\u00edneo<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

04 \"\" Verdadeiro. Podemos calcular a <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span> do carro nos dois instantes <\/b><\/span><\/span><\/span>como<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

a1=\u0394V1\/\u0394t1<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>a2=\u0394V2\/\u0394t2<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>a1=(40-20)\/(28-20) a2=(40-20)\/(52-44) \"\" a1=20\/8 a2=20\/8<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span>a1=a2=2.5m\/s\u00b2<\/b><\/span><\/span><\/span><\/p>\n

08 \"\" Falso \"\" Como provado no <\/b><\/span><\/span><\/span>caso (01)<\/b><\/span><\/span><\/span> os dados s\u00e3o dos carros <\/b><\/span><\/span><\/span>5 ou 6<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

16 \"\" Falso \"\" Para o carro estar em <\/b><\/span><\/span><\/span>repouso<\/b><\/span><\/span><\/span> sua velocidade tem que ser igual a <\/b><\/span><\/span><\/span>zero<\/b><\/span><\/span><\/span>, portanto ele s\u00f3 est\u00e1 em repouso no <\/b><\/span><\/span><\/span>t=0<\/b><\/span><\/span><\/span>. Quando o gr\u00e1fico est\u00e1 na horizontal <\/b><\/span><\/span><\/span>n\u00e3o<\/b><\/span><\/span><\/span> quer dizer que o carro est\u00e1 em <\/b><\/span><\/span><\/span>repouso<\/b><\/span><\/span><\/span>, e sim que o carro mant\u00e9m <\/b><\/span><\/span><\/span>velocidade constante<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

01 <\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> Falso \"\" A esta\u00e7\u00e3o espacial deve ter <\/b><\/span><\/span><\/span>velocidade angular igual \u00e0 velocidade angular da Terra<\/b><\/span><\/span><\/span>, pois no caso da <\/b><\/span><\/span><\/span>velocidade linear (\"\")<\/b><\/span><\/span><\/span>, ela <\/b><\/span><\/span><\/span>varia<\/b><\/span><\/span><\/span> com a <\/b><\/span><\/span><\/span>dist\u00e2ncia<\/b><\/span><\/span><\/span>, como a esta\u00e7\u00e3o espacial est\u00e1 em uma <\/b><\/span><\/span><\/span>altura maior<\/b><\/span><\/span><\/span> que a da Terra, sua <\/b><\/span><\/span><\/span>velocidade<\/b><\/span><\/span><\/span> seria maior, com isso a esta\u00e7\u00e3o <\/b><\/span><\/span><\/span>perderia<\/b><\/span><\/span><\/span> sua posi\u00e7\u00e3o geoestacion\u00e1ria, por terem <\/b><\/span><\/span><\/span>velocidades diferentes<\/b><\/span><\/span><\/span>. Diferente da <\/b><\/span><\/span><\/span>velocidade<\/b><\/span><\/span><\/span> <\/b>angular (\"\")<\/b><\/span><\/span><\/span>, que por <\/b><\/span><\/span><\/span>n\u00e3o<\/b><\/span><\/span><\/span> depender da <\/b><\/span><\/span><\/span>dist\u00e2ncia, manteria<\/b><\/span><\/span><\/span> essa <\/b><\/span><\/span><\/span>posi\u00e7\u00e3o geoestacion\u00e1ria<\/b><\/span><\/span><\/span>, por <\/b><\/span><\/span><\/span>manter<\/b><\/span><\/span><\/span> sempre a <\/b><\/span><\/span><\/span>posi\u00e7\u00e3o relativa<\/b><\/span><\/span><\/span> com a Terra.<\/b><\/span><\/span><\/span><\/p>\n

02 \"\" Falso \"\"Ainda <\/b><\/span><\/span><\/span>haveria atra\u00e7\u00e3o gravitacional<\/b><\/span><\/span><\/span>, que <\/b><\/span><\/span><\/span>perderia for\u00e7as<\/b><\/span><\/span><\/span> conforme a esta\u00e7\u00e3o ganhasse <\/b><\/span><\/span><\/span>altitude<\/b><\/span><\/span><\/span>. Por\u00e9m, devido \u00e0 <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o de subida<\/b><\/span><\/span><\/span> da esta\u00e7\u00e3o, uma <\/b><\/span><\/span><\/span>for\u00e7a contr\u00e1ria<\/b><\/span><\/span><\/span> pressionaria essas pessoas contra o ch\u00e3o,<\/b><\/span><\/span><\/span> similar<\/b><\/span><\/span><\/span> ao que acontece nos elevadores comuns.<\/b><\/span><\/span><\/span><\/p>\n

04 \"\" Verdadeiro \"\" Como j\u00e1 estudado no caso (01).<\/b><\/span><\/span><\/span><\/p>\n

08 <\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> Falsa \"\" varia intensidade, dire\u00e7\u00e3o e sentido.<\/b><\/span><\/span><\/span><\/p>\n

16 \"\" Verdadeiro \"\" Pois no caso da <\/b><\/span><\/span><\/span>velocidade<\/b><\/span><\/span><\/span> <\/b>angular (\"\")<\/b><\/span><\/span><\/span>, para <\/b><\/span><\/span><\/span>manter<\/b><\/span><\/span><\/span> a velocidade da Terra e da esta\u00e7\u00e3o igual, o <\/b><\/span><\/span><\/span>per\u00edodo<\/b><\/span><\/span><\/span> deve ser o <\/b><\/span><\/span><\/span>mesmo<\/b><\/span><\/span><\/span>, j\u00e1 que o <\/b><\/span><\/span><\/span>\u00e2ngulo percorrido<\/b><\/span><\/span><\/span> entre os dois \u00e9 sempre <\/b><\/span><\/span><\/span>igual<\/b><\/span><\/span><\/span> e de valor 360\u00b0.<\/b><\/span><\/span><\/span><\/p>\n

32 \"\" Falso \"\" A for\u00e7a de <\/b><\/span><\/span><\/span>atra\u00e7\u00e3o gravitacional da Terra<\/b><\/span><\/span><\/span> n\u00e3o ser\u00e1 a centr\u00edfuga, que exerce uma for\u00e7a <\/b><\/span><\/span><\/span>para fora<\/b><\/span><\/span><\/span> do movimento circular, e sim a <\/b><\/span><\/span><\/span>centr\u00edpeta<\/b><\/span><\/span><\/span>, que exerce uma for\u00e7a <\/b><\/span><\/span><\/span>para dentro<\/b><\/span><\/span><\/span> do movimento circular, ou seja, em dire\u00e7\u00e3o a Terra, sendo <\/b><\/span><\/span><\/span>similar<\/b><\/span><\/span><\/span> \u00e0 for\u00e7a gravitacional.<\/b><\/span><\/span><\/span><\/p>\n

64 \"\" Verdadeiro \"\" <\/b><\/span><\/span><\/span>A for\u00e7a gravitacional (\"\")<\/b><\/span><\/span><\/span> varia conforme a <\/b><\/span><\/span><\/span>dist\u00e2ncia<\/b><\/span><\/span><\/span> entre as duas massas, com isso, ao <\/b><\/span><\/span><\/span>aumentar a altitude<\/b><\/span><\/span><\/span>, a dist\u00e2ncia <\/b><\/span><\/span><\/span>aumenta<\/b><\/span><\/span><\/span> e a for\u00e7a gravitacional <\/b><\/span><\/span><\/span>diminui<\/b><\/span><\/span><\/span>,<\/b><\/span><\/span><\/span> <\/b>por serem inversamente proporcionais.<\/b><\/span><\/span><\/span><\/p>\n

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

01 \"\" Falso \"\" As luzes prim\u00e1rias s\u00e3o: vermelha, verde e azul. <\/b><\/span><\/span><\/span><\/p>\n

02 \"\" correta \"\" como a l\u00e2mpada 2 emite luz verde (figura 3), na figura 1 onde tamb\u00e9m se enxerga o verde, imp\u00f5e que as demais incidentes s\u00e3o absorvidas. <\/b><\/span><\/span><\/span><\/p>\n

04 \"\" Correta \"\" <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>l\u00e2mpada 2<\/b><\/span><\/span><\/span> emite luz de cor <\/b><\/span><\/span><\/span>verde<\/b><\/span><\/span><\/span>, por isso a palavra F\u00cdSICA, na <\/b><\/span><\/span><\/span>figura 3<\/b><\/span><\/span><\/span>, aparece na <\/b><\/span><\/span><\/span>cor verde j\u00e1 que essa palavra <\/b><\/span><\/span><\/span>reflete <\/b><\/span><\/span><\/span>difusamente a cor verde.<\/b><\/span><\/span><\/span><\/p>\n

08 \"\" Falso <\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> No <\/b><\/span><\/span><\/span>espectro eletromagn\u00e9tico<\/b><\/span><\/span><\/span> a cor <\/b><\/span><\/span><\/span>vermelha<\/b><\/span><\/span><\/span> possui <\/b><\/span><\/span><\/span>menor frequ\u00eancia<\/b><\/span><\/span><\/span>, o <\/b><\/span><\/span><\/span>verde<\/b><\/span><\/span><\/span> tem uma <\/b><\/span><\/span><\/span>frequ\u00eancia intermedi\u00e1ria<\/b><\/span><\/span><\/span> e o <\/b><\/span><\/span><\/span>azul<\/b><\/span><\/span><\/span> uma <\/b><\/span><\/span><\/span>frequ\u00eancia maior<\/b><\/span><\/span><\/span>, sendo assim, a ordem das l\u00e2mpadas deveria ser <\/b><\/span><\/span><\/span>azul, verde e vermelho<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

16 \"\" Falso \"\" Para se formar a <\/b><\/span><\/span><\/span>cor preta<\/b><\/span><\/span><\/span> \u00e9 necess\u00e1rio, no m\u00ednimo, 3 cores, <\/b><\/span><\/span><\/span>vermelho, amarelo e azul.<\/b><\/span><\/span><\/span><\/p>\n

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

01 \"\" Falso \"\" O <\/b><\/span><\/span><\/span>per\u00edodo<\/b><\/span><\/span><\/span> e a <\/b><\/span><\/span><\/span>frequ\u00eancia<\/b><\/span><\/span><\/span> dependem apenas do <\/b><\/span><\/span><\/span>n\u00famero de ciclos <\/b><\/span><\/span><\/span>e do <\/b><\/span><\/span><\/span>intervalo de tempo<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

02 \"\" Verdadeiro \"\" A <\/b><\/span><\/span><\/span>frequ\u00eancia<\/b><\/span><\/span><\/span> no MHS pode ser calculada como:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde T \u00e9 o <\/b><\/span><\/span><\/span>per\u00edodo<\/b><\/span><\/span><\/span>, m a <\/b><\/span><\/span><\/span>massa desejada<\/b><\/span><\/span><\/span> e k a <\/b><\/span><\/span><\/span>constante el\u00e1stica<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span>Para calcularmos a frequ\u00eancia, precisamos elevar toda a equa\u00e7\u00e3o por -1, pois a frequ\u00eancia \u00e9 o inverso do per\u00edodo, nesse caso apenas vamos inverter as fra\u00e7\u00f5es.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Colocando os valores:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Transformando 25.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> em 2500.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Saindo da raiz:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Simplificando:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

04 \"\" Verdadeiro. <\/b><\/span><\/span><\/span>Antes<\/b><\/span><\/span><\/span> de o carro entrar em <\/b><\/span><\/span><\/span>movimento<\/b><\/span><\/span><\/span> h\u00e1 o <\/b><\/span><\/span><\/span>m\u00e1ximo de energia potencial gravitacional<\/b><\/span><\/span><\/span>, com o carro <\/b><\/span><\/span><\/span>j\u00e1 em movimento <\/b><\/span><\/span><\/span>essa energia se <\/b><\/span><\/span><\/span>transforma em energia cin\u00e9tica<\/b><\/span><\/span><\/span>, at\u00e9 o ponto de <\/b><\/span><\/span><\/span>equil\u00edbrio da mola<\/b><\/span><\/span><\/span>, aonde o carro <\/b><\/span><\/span><\/span>para<\/b><\/span><\/span><\/span> de se movimentar, mas com a <\/b><\/span><\/span><\/span>energia cin\u00e9tica<\/b><\/span><\/span><\/span> em seu <\/b><\/span><\/span><\/span>m\u00e1ximo<\/b><\/span><\/span><\/span>, pois toda a energia que antes era <\/b><\/span><\/span><\/span>potencial<\/b><\/span><\/span><\/span>, transformou-se em <\/b><\/span><\/span><\/span>cin\u00e9tica<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

08 \"\" Verdadeiro Observe a imagem abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Utilizando a massa <\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/span> como a <\/b><\/span><\/span><\/span>massa do carro e dos 4 amigos<\/b><\/span><\/span><\/span> a mola se <\/b><\/span><\/span><\/span>deforma<\/b><\/span><\/span><\/span>, o <\/b><\/span><\/span><\/span>tamanho da deforma\u00e7\u00e3o<\/b><\/span><\/span><\/span> \u00e9 indicado por <\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/span>, no nosso caso sendo igual a <\/b><\/span><\/span><\/span>5 cm<\/b><\/span><\/span><\/span>. Para calcular a <\/b><\/span><\/span><\/span>constante el\u00e1stica<\/b><\/span><\/span><\/span> da mola vamos utilizar <\/b><\/span><\/span><\/span>Fel=K*x<\/b><\/span><\/span><\/span>, aonde Fel \u00e9 a <\/b><\/span><\/span><\/span>for\u00e7a realizada pela mola<\/b><\/span><\/span><\/span>, K \u00e9 a <\/b><\/span><\/span><\/span>constante el\u00e1stica<\/b><\/span><\/span><\/span> e x a <\/b><\/span><\/span><\/span>deforma\u00e7\u00e3o<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

A seta vermelha representa a <\/b><\/span><\/span><\/span>for\u00e7a peso <\/b><\/span><\/span><\/span>exercida pela massa M \"\"P=M.g \"\" P=1250.10 \"\"P=12500N<\/b><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>equil\u00edbrio<\/b><\/span><\/span><\/span> as duas for\u00e7as, peso e el\u00e1stica(seta amarela), t\u00eam <\/b><\/span><\/span><\/span>mesma intensidade<\/b><\/span><\/span><\/span>, com isso <\/b><\/span><\/span><\/span>Fel=P<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span> <\/b>Passando para a f\u00f3rmula<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span>12500=K.5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>(transforma\u00e7\u00e3o de cent\u00edmetros para metros)<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>K=12500\/5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>K=25.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N\/m<\/b><\/span><\/span><\/span><\/p>\n

16 \"\" Falso. O per\u00edodo pode ser calculado como:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Dessa vez a <\/b><\/span><\/span><\/span>massa \u00e9 s\u00f3 do carro<\/b><\/span><\/span><\/span>, 1000kg.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

Tirando da raiz<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

01 \"\" Falso \"\" Essa quest\u00e3o pode ser respondida <\/b><\/span><\/span><\/span>sem utilizarmos contas<\/b><\/span><\/span><\/span>, basta observarmos que na <\/b><\/span><\/span><\/span>grande maioria da dist\u00e2ncia percorrida pelo lateral ele se encontra abaixo da velocidade de 16,75 m\/s<\/b><\/span><\/span><\/span>, portanto ele n\u00e3o pode manter essa m\u00e9dia.<\/b><\/span><\/span><\/span><\/p>\n

02 \"\" Verdadeiro \"\" Para calcularmos a <\/b><\/span><\/span><\/span>energia cin\u00e9tica m\u00ednima<\/b><\/span><\/span><\/span>, utilizaremos a <\/b><\/span><\/span><\/span>menor velocidade<\/b><\/span><\/span><\/span>, que no caso \u00e9 de <\/b><\/span><\/span><\/span>23,4 km\/h<\/b><\/span><\/span><\/span> \"\"Transformando a velocidade de <\/b><\/span><\/span><\/span>Km\/h <\/b><\/span><\/span><\/span>para <\/b><\/span><\/span><\/span>m\/s<\/b><\/span><\/span><\/span>, basta dividirmos por <\/b><\/span><\/span><\/span>3,6<\/b><\/span><\/span><\/span> \"\" 23,4\/3,6 \"\" 6.5 m\/s \"\" Colocando na <\/b><\/span><\/span><\/span>f\u00f3rmula da energia cin\u00e9tica \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=(m.V\u00b2)\/2 \"\" E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=(80*6.5\u00b2)\/2 \"\" E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=1690J.<\/b><\/span><\/span><\/span><\/p>\n

04 \"\" Falso \"\" Mesmo com o jogador em <\/b><\/span><\/span><\/span>repouso<\/b><\/span><\/span><\/span>, h\u00e1 o <\/b><\/span><\/span><\/span>custo energ\u00e9tico para se manter em p\u00e9 e para as atividades fisiol\u00f3gicas<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

08 \"\" Correta. Primeiramente, vamos verificar a pot\u00eancia \u00fatil para o movimento do zagueiro:<\/b><\/span><\/span><\/span><\/p>\n

24 kcal _____ 1 min _____ 100% <\/b><\/span><\/span><\/span><\/p>\n

6 kcal _____ 1 min _____ 25%<\/b><\/span><\/span><\/span><\/p>\n

No Sistema Internacional de Unidades, temos: E = 6 kcal x 4 = 24 kJ.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o, calculamos a pot\u00eancia: <\/b><\/span><\/span><\/span><\/p>\n

P = E\/\u2206t = 24000\/60 = 400 W<\/b><\/span><\/span><\/span><\/p>\n

Finalmente, determinando a pot\u00eancia \u00fatil para que o mesmo indiv\u00edduo suba com velocidade constante de 0,5 m\/s:<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span> = F . vm <\/b><\/span><\/span><\/span>\u21d2<\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span> = 800 . 0,5 <\/b><\/span><\/span><\/span>\u21d2<\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>u<\/b><\/span><\/span><\/sub><\/span> = 400 W<\/b><\/span><\/span><\/span><\/p>\n

<\/a> 16 \"\" Verdadeiro \"\" Primeiramente vamos <\/b><\/span><\/span><\/span>passar<\/b><\/span><\/span><\/span> todas as velocidades para <\/b><\/span><\/span><\/span>m\/s \"\"<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>= 12,6\/3,6 V<\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/sub><\/span>= 16,2\/3,6 \"\" V<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=19,8\/3,6<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3,5 m\/s V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=4,5 m\/s V<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=5,5 m\/s \"\" Agora vamos calcular o <\/b><\/span><\/span><\/span>tempo, em s, <\/b><\/span><\/span><\/span>que eles correram em cada velocidade<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>=\u0394<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> t<\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/sub><\/span>= \u0394<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> t<\/b><\/span><\/span><\/span>4 <\/b><\/span><\/span><\/sub><\/span>= \u0394<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>= 1470\/3,5 t<\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/sub><\/span>= 1647\/4,5 t<\/b><\/span><\/span><\/span>4 <\/b><\/span><\/span><\/sub><\/span>= 693\/5,5 \"\" T<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>= 420 s t<\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/sub><\/span>= 366 s t<\/b><\/span><\/span><\/span>4 <\/b><\/span><\/span><\/sub><\/span>= 126 s<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span>Passando para <\/b><\/span><\/span><\/span>minutos<\/b><\/span><\/span><\/span>, basta <\/b><\/span><\/span><\/span>dividir por 60 \"\" <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=7 min t<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=6,1 min t<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=2,1 min\"\"Podemos calcular o <\/b><\/span><\/span><\/span>gasto cal\u00f3rico <\/b><\/span><\/span><\/span>multiplicando o <\/b><\/span><\/span><\/span>tempo<\/b><\/span><\/span><\/span> pela <\/b><\/span><\/span><\/span>quantidade de Kcal<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=7.18 Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=6,1.24 Q<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=2,1.27 \"\" Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=126 Kcal Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=146,4 Kcal Q<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=56,7 Kcal \"\" Somando tudo, teremos o <\/b><\/span><\/span><\/span>gasto cal\u00f3rico total \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=126+146,4+56,7 \"\" Q<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=329,1Kcal<\/b><\/span><\/span><\/span><\/p>\n

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

01 \"\" Verdadeiro \"\" Einstein teorizou que as <\/b><\/span><\/span><\/span>ondas eletromagn\u00e9ticas<\/b><\/span><\/span><\/span> (modelo ondulat\u00f3rio) s\u00f3 fariam com que os el\u00e9trons fossem ejetados <\/b><\/span><\/span><\/span>instantaneamente<\/b><\/span><\/span><\/span> se elas se comportassem como <\/b><\/span><\/span><\/span>part\u00edculas<\/b><\/span><\/span><\/span> (modelo corpuscular).<\/b><\/span><\/span><\/span><\/p>\n

02 \"\" Falso \"\" A <\/b><\/span><\/span><\/span>pot\u00eancia<\/b><\/span><\/span><\/span> altera sim a quantidade de <\/b><\/span><\/span><\/span>el\u00e9trons arrancados<\/b><\/span><\/span><\/span>, quanto <\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/span> a pot\u00eancia, <\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/span> o n\u00famero de el\u00e9trons ejetados. No caso o que altera a <\/b><\/span><\/span><\/span>energia cin\u00e9tica <\/b><\/span><\/span><\/span>\u00e9 a <\/b><\/span><\/span><\/span>frequ\u00eancia<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

04 \"\" Verdadeiro \"\" De acordo com o primeiro postulado da teoria da relatividade (princ\u00edpio da relatividade), as leis que governam as mudan\u00e7as de estado em <\/b><\/span><\/span><\/span>quaisquer<\/b><\/span><\/span><\/span> sistemas f\u00edsicos tomam a <\/b><\/span><\/span><\/span>mesma forma <\/b><\/span><\/span><\/span>em <\/b><\/span><\/span><\/span>quaisquer<\/b><\/span><\/span><\/span> sistemas de <\/b><\/span><\/span><\/span>coordenadas inerciais<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

08 \"\" Verdadeiro \"\" De acordo com o segundo postulado da teoria da relatividade (invari\u00e2ncia da velocidade da luz), a luz tem velocidade <\/b><\/span><\/span><\/span>invariante<\/b><\/span><\/span><\/span> igual a c em rela\u00e7\u00e3o a <\/b><\/span><\/span><\/span>qualquer<\/b><\/span><\/span><\/span> sistema de coordenadas inercial.<\/b><\/span><\/span><\/span><\/p>\n

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

01 \"\" Verdadeiro \"\" Uma <\/b><\/span><\/span><\/span>for\u00e7a<\/b><\/span><\/span><\/span> \u00e9 exercida na bola e essa bola sofre um <\/b><\/span><\/span><\/span>deslocamento<\/b><\/span><\/span><\/span>, logo foi realizado um <\/b><\/span><\/span><\/span>trabalho<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

02 \"\"<\/b><\/span><\/span><\/span><\/p>\n

04 \"\" Falsa \"\" Parte da energia recebida \u00e9 <\/b><\/span><\/span><\/span>dissipada<\/b><\/span><\/span><\/span> de v\u00e1rias formas, seja pelo <\/b><\/span><\/span><\/span>barulho<\/b><\/span><\/span><\/span> ap\u00f3s a bola ser chutada, pelo <\/b><\/span><\/span><\/span>atrito<\/b><\/span><\/span><\/span> com ch\u00e3o e vento, entre outros.<\/b><\/span><\/span><\/span><\/p>\n

08 \"\" Correta. \u00c9 o trabalho da for\u00e7a el\u00e9trica que vai desencadear a corrente el\u00e9trica que poder\u00e1 carregar ou permitir o uso de aparelhos el\u00e9tricos conectados \u00e0 bola.<\/b><\/span><\/span><\/span><\/p>\n

16 \"\" Verdadeiro \"\" A <\/b><\/span><\/span><\/span>lei de Faraday<\/b><\/span><\/span><\/span> explica a <\/b><\/span><\/span><\/span>indu\u00e7\u00e3o<\/b><\/span><\/span><\/span> de uma <\/b><\/span><\/span><\/span>for\u00e7a eletromotriz <\/b><\/span><\/span><\/span>a partir de uma <\/b><\/span><\/span><\/span>varia\u00e7\u00e3o do fluxo magn\u00e9tico<\/b><\/span><\/span><\/span>, com isso qualquer <\/b><\/span><\/span><\/span>movimento<\/b><\/span><\/span><\/span> da bola altera o <\/b><\/span><\/span><\/span>fluxo magn\u00e9tico <\/b><\/span><\/span><\/span>do gerador, gerando a energia el\u00e9trica.<\/b><\/span><\/span><\/span><\/p>\n

32 \"\" Falso \"\" A fun\u00e7\u00e3o da <\/b><\/span><\/span><\/span>bateria<\/b><\/span><\/span><\/span> \u00e9 de apenas <\/b><\/span><\/span><\/span>armazenar<\/b><\/span><\/span><\/span> essa energia el\u00e9trica, o <\/b><\/span><\/span><\/span>gerador<\/b><\/span><\/span><\/span> que tem essa fun\u00e7\u00e3o de <\/b><\/span><\/span><\/span>aumentar<\/b><\/span><\/span><\/span> a energia.<\/b><\/span><\/span><\/span><\/p>\n

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

01 \"\" Verdadeiro \"\" O professor <\/b><\/span><\/span><\/span>Wilhelm Conrad Roentgen <\/b><\/span><\/span><\/span>foi o respons\u00e1vel pela <\/b><\/span><\/span><\/span>descoberta dos raios-X no ano de 1895<\/b><\/span><\/span><\/span>, enquanto trabalhava com um <\/b><\/span><\/span><\/span>tubo cat\u00f3dico <\/b><\/span><\/span><\/span>no seu laborat\u00f3rio. <\/b><\/span><\/span><\/span><\/p>\n

02 \"\" Falso \"\" Os raios-X foram assim nomeados pela comunidade cient\u00edfica <\/b><\/span><\/span><\/span>devido ao desconhecimento, na \u00e9poca, a respeito da natureza dessa radia\u00e7\u00e3o<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

04 \"\" Falso \"\" Apesar do raio-X ter realmente um <\/b><\/span><\/span><\/span>baixo poder de penetra\u00e7\u00e3o<\/b><\/span><\/span><\/span>, pelo fato de ser uma <\/b><\/span><\/span><\/span>radia\u00e7\u00e3o alfa<\/b><\/span><\/span><\/span>, os profissionais que trabalham com ele devem seguir r\u00edgidas regras de seguran\u00e7a, pois <\/b><\/span><\/span><\/span>a exposi\u00e7\u00e3o prolongada <\/b><\/span><\/span><\/span>pode ter efeitos <\/b><\/span><\/span><\/span>colaterais permanentes<\/b><\/span><\/span><\/span>. Uma placa de madeira \u00e9 <\/b><\/span><\/span><\/span>facilmente atravessada <\/b><\/span><\/span><\/span>por essa radia\u00e7\u00e3o, tendo o profissional ficar dentro de um <\/b><\/span><\/span><\/span>biombo<\/b><\/span><\/span><\/span> ou com <\/b><\/span><\/span><\/span>aventais de chumbo<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

08 \"\" Verdadeiro \"\" A <\/b><\/span><\/span><\/span>exposi\u00e7\u00e3o prolongada <\/b><\/span><\/span><\/span>ao raio-X pode provocar vermelhid\u00e3o na pele e queimaduras, assim como <\/b><\/span><\/span><\/span>morte celular <\/b><\/span><\/span><\/span>e muta\u00e7\u00f5es g\u00eanicas.<\/b><\/span><\/span><\/span><\/p>\n

16 \"\" Falso \"\" A radia\u00e7\u00e3o <\/b><\/span><\/span><\/span>n\u00e3o permanece no corpo humano<\/b><\/span><\/span><\/span>, como qualquer radia\u00e7\u00e3o ela apenas se propaga, <\/b><\/span><\/span><\/span>n\u00e3o sendo armazenada <\/b><\/span><\/span><\/span>pelo nosso organismo. O <\/b><\/span><\/span><\/span>efeito cumulativo <\/b><\/span><\/span><\/span>nas c\u00e9lulas se trata de uma <\/b><\/span><\/span><\/span>exposi\u00e7\u00e3o prolongada <\/b><\/span><\/span><\/span>e\/ou <\/b><\/span><\/span><\/span>acima dos n\u00edveis tolerados<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es do vestibular da UFSC \u2013 2016 01- 01 Verdadeiro Galileu foi um dos que se influenciaram nas ideias de Francis Bacon, que defendia a observa\u00e7\u00e3o da natureza em conjunto da coleta de fatos. Investigava a natureza com base em observa\u00e7\u00f5es e experi\u00eancia emp\u00edrica, reduzindo os problemas a um simples conjunto de termos e resolvendo-os usando matem\u00e1tica.<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1944,"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-1946","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1946","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=1946"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1946\/revisions"}],"predecessor-version":[{"id":1948,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1946\/revisions\/1948"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1944"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1946"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}