{"id":4772,"date":"2019-11-07T02:51:20","date_gmt":"2019-11-07T02:51:20","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4772"},"modified":"2024-07-11T03:05:05","modified_gmt":"2024-07-11T03:05:05","slug":"eear-2019-20","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2019\/eear-2019-20\/","title":{"rendered":"EEAR 2019\/20"},"content":{"rendered":"

<\/p>\n

Curso de Forma\u00e7\u00e3o de Sargentos (CFS)<\/b><\/span><\/span><\/span><\/p>\n

Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0(<\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span>) \u00e9 o maior complexo de ensino t\u00e9cnico militar da Am\u00e9rica do Sul sendo uma organiza\u00e7\u00e3o do\u00a0Comando da Aeron\u00e1utica<\/a>, diretamente subordinada ao diretor-geral do Departamento de Ensino da Aeron\u00e1utica (DEPENS), que tem por finalidade a forma\u00e7\u00e3o e o aperfei\u00e7oamento de graduados da Aeron\u00e1utica.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

Est\u00e1 localizada no munic\u00edpio de\u00a0Guaratinguet\u00e1<\/a>, entre as escarpas da\u00a0Serra da Mantiqueira<\/a>\u00a0e das bordas da\u00a0Serra da Quebra-Cangalha<\/a>\u00a0e\u00a0Serra do Mar<\/a>.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>Curso de Forma\u00e7\u00e3o de Sargentos (CFS)<\/b><\/span><\/span><\/span> tem a dura\u00e7\u00e3o de dois anos e exige que os candidatos tenham conclu\u00eddo o ensino m\u00e9dio. J\u00e1 o Est\u00e1gio de Adapta\u00e7\u00e3o \u00e0 Gradua\u00e7\u00e3o de Sargento (EAGS) tem dura\u00e7\u00e3o de aproximadamente um ano, e o candidato deve ter conclu\u00eddo o curso t\u00e9cnico relativo \u00e0 especialidade de interesse da For\u00e7a A\u00e9rea. Para realizar inscri\u00e7\u00e3o para ambos \u00e9 necess\u00e1rio ser brasileiro(a), ter conclu\u00eddo o ensino m\u00e9dio, n\u00e3o ter menos de dezessete anos ou mais de vinte e quatro anos at\u00e9 a data da matr\u00edcula.<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Dois vetores V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> e V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> formam entre si um \u00e2ngulo <\/b><\/span><\/span><\/span>\uf071<\/b><\/span><\/span><\/span> <\/b>e <\/b><\/span><\/span><\/span>possuem m\u00f3dulos iguais a 5 unidades<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>12 unidades, respectivamente.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Se a resultante entre eles tem m\u00f3dulo igual a 13 unidades,<\/b><\/span><\/span><\/span> <\/b>podemos afirmar corretamente que o \u00e2ngulo <\/b><\/span><\/span><\/span>\uf071<\/b><\/span><\/span><\/span> entre os vetores V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> e V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>vale: <\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>0\u00ba <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span>45\u00ba <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>90\u00ba <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>180\u00ba <\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Uma <\/b><\/span><\/span><\/span>mola est\u00e1 suspensa verticalmente pr\u00f3xima \u00e0 superf\u00edcie terrestre, <\/b><\/span><\/span><\/span>onde <\/b><\/span><\/span><\/span>a acelera\u00e7\u00e3o da gravidade pode ser adotada como 10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span><\/span>extremidade livre da mola \u00e9 colocada uma cestinha de massa desprez\u00edvel, <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>ser\u00e1 preenchida com bolinhas de gude, de 15g cada.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

<\/p>\n

Ao acrescentar bolinhas \u00e0 cesta, verifica-se que a mola sofre uma elonga\u00e7\u00e3o proporcional ao peso aplicado.<\/b><\/span><\/span><\/span> Sabendo-se que <\/b><\/span><\/span><\/span>a mola tem uma constante el\u00e1stica k = 9,0N\/m, quantas bolinhas \u00e9 preciso acrescentar \u00e0 cesta para que a mola estique exatamente 5 cm?<\/b><\/span><\/span><\/span> <\/b><\/p>\n

a) <\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>5 <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>10 <\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Um atleta pratica salto ornamental,<\/b><\/span><\/span><\/span> fazendo uso de <\/b><\/span><\/span><\/span>uma plataforma situada a 5m do n\u00edvel da \u00e1gua da piscina.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

<\/p>\n

Se o atleta saltar desta plataforma, a partir do repouso, com que velocidade se chocar\u00e1 com a \u00e1gua? Obs.: despreze a resist\u00eancia do ar <\/b><\/span><\/span><\/span>e considere o <\/b><\/span><\/span><\/span>m\u00f3dulo da acelera\u00e7\u00e3o da gravidade g = 10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

a) <\/b><\/span><\/span><\/span>10 m\/s. <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span>20 m\/s. <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>30 m\/s. <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>50 m\/s. <\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Uma <\/b><\/span><\/span><\/span>part\u00edcula com carga el\u00e9trica igual a 3,2 <\/b><\/span><\/span><\/span>\uf06d<\/b><\/span><\/span><\/span>C e velocidade de 2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> m\/s \u00e9 lan\u00e7ada perpendicularmente a um campo magn\u00e9tico uniforme <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>sofre a a\u00e7\u00e3o de uma for\u00e7a magn\u00e9tica de intensidade igual a 1,6 .10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> N.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Determine a intensidade do campo magn\u00e9tico <\/b><\/span><\/span><\/span>(em Tesla) <\/b><\/span><\/span><\/span>no qual a part\u00edcula foi lan\u00e7ada.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

a) <\/b><\/span><\/span><\/span>0,25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> <\/b>b) <\/b><\/span><\/span><\/span>2,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> <\/b>c) <\/b><\/span><\/span><\/span>2,5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> <\/b>d) <\/b><\/span><\/span><\/span>0,25.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span> <\/b><\/p>\n

05-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Um <\/b><\/span><\/span><\/span>adolescente de 12 anos,<\/b><\/span><\/span><\/span> percebendo <\/b><\/span><\/span><\/span>altera\u00e7\u00f5es em sua voz, comunicou \u00e0 sua m\u00e3e a situa\u00e7\u00e3o observada com certa regularidade. <\/b><\/span><\/span><\/span>Em <\/b><\/span><\/span><\/span>determinados momentos apresentava tom de voz fina em outros momentos tom de voz grossa<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>quest\u00e3o relatada pelo adolescente refere-se a uma qualidade do som denominada:<\/b><\/span><\/span><\/span> <\/b><\/p>\n

a) <\/b><\/span><\/span><\/span>altura. <\/b><\/span><\/span><\/span> b) <\/b><\/span><\/span><\/span>timbre. <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>velocidade. <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>intensidade. <\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Considere <\/b><\/span><\/span><\/span>quatro esferas met\u00e1licas id\u00eanticas, A, B, C e D, inicialmente separadas entre si<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

Duas delas, B e D, est\u00e3o inicialmente neutras, <\/b><\/span><\/span><\/span>enquanto <\/b><\/span><\/span><\/span>as esferas A e C possuem cargas el\u00e9tricas iniciais, respectivamente, iguais a 3Q e -Q.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

<\/p>\n

Determine a carga el\u00e9trica final da esfera C ap\u00f3s contatos sucessivos com as esferas A, B e D, nessa ordem, considerando que ap\u00f3s cada contato, as esferas s\u00e3o novamente separadas. <\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>Q\/4 <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span> Q\/2 <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>2Q <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>4Q <\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Um astronauta de massa m e peso P foi levado da superf\u00edcie da Terra para a superf\u00edcie de um planeta cuja acelera\u00e7\u00e3o da gravidade, em m\u00f3dulo, \u00e9 igual a um ter\u00e7o da acelera\u00e7\u00e3o da gravidade registrada na superf\u00edcie terrestre.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

No <\/b><\/span><\/span><\/span>novo planeta, os valores da massa e do peso desse astronauta, em fun\u00e7\u00e3o de suas intensidades na Terra, ser\u00e3o respectivamente:<\/b><\/span><\/span><\/span> <\/b><\/p>\n

a) <\/b><\/span><\/span><\/span>m\/3 , P <\/b><\/span><\/span><\/span>b) <\/b><\/span><\/span><\/span>m, P <\/b><\/span><\/span><\/span>c) <\/b><\/span><\/span><\/span>m, P\/3 <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>m\/3 , P\/3 <\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Roberto, <\/b><\/span><\/span><\/span>empolgado com as aulas de F\u00edsica, decide construir um term\u00f4metro que trabalhe com uma escala escolhida por ele, a qual chamou de escala R. <\/b><\/span><\/span><\/span>Para tanto, <\/b><\/span><\/span><\/span>definiu – 20\u00b0R como ponto de fus\u00e3o do gelo e 80\u00b0R como temperatura de ebuli\u00e7\u00e3o da \u00e1gua, sendo estes os pontos fixos desta escala. Sendo R a temperatura na escala criada por Roberto e C a temperatura na escala Celsius, e considerando que o experimento seja realizado ao n\u00edvel do mar, a express\u00e3o que relaciona corretamente as duas escalas <\/b><\/span><\/span><\/span>ser\u00e1: <\/b><\/span><\/span><\/span><\/p>\n

a) <\/b><\/span><\/span><\/span>C = R \u2013 20 <\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>C = R + 20 <\/b><\/span><\/span><\/span><\/p>\n

c) <\/b><\/span><\/span><\/span>C = (R <\/b><\/span><\/span><\/span>\uf02b<\/b><\/span><\/span><\/span> 20)\/2 <\/b><\/span><\/span><\/span><\/p>\n

d) <\/b><\/span><\/span><\/span>C = (R <\/b><\/span><\/span><\/span>\uf02d<\/b><\/span><\/span><\/span> 20)\/2 <\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span> (Escola de Especialistas de Aeron\u00e1utica<\/b><\/span><\/span><\/span>\u00a0\u2013 <\/b><\/span><\/span><\/span>EEAR<\/b><\/span><\/span><\/span> – 019\/020)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

O <\/b><\/span><\/span><\/span>gr\u00e1fico a seguir corresponde ao comportamento da corrente el\u00e9trica <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>percorre um condutor, em fun\u00e7\u00e3o da diferen\u00e7a de potencial a ele aplicada.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

<\/p>\n

Sabendo-s<\/b><\/span><\/span><\/span>e que este condutor \u00e9 <\/b><\/span><\/span><\/span>constitu\u00eddo de um fio de 2m de comprimento e de um material cuja resistividade, a 20\u00b0C, vale 1,75. 10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span> <\/b>\uf057<\/b><\/span><\/span><\/span>.m,<\/b><\/span><\/span><\/span> <\/b>determine a \u00e1rea da se\u00e7\u00e3o transversal do fio e o valor da resist\u00eancia el\u00e9trica desse condutor na referida temperatura.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

a) <\/b><\/span><\/span><\/span>0,7.10<\/b><\/span><\/span><\/span>-4 <\/b><\/span><\/span><\/sup><\/span>cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> e 0,5<\/b><\/span><\/span><\/span>\uf057<\/b><\/span><\/span><\/span> <\/b><\/p>\n

b) <\/b><\/span><\/span><\/span>0,7.10<\/b><\/span><\/span><\/span>-4 <\/b><\/span><\/span><\/sup><\/span>cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> e 500<\/b><\/span><\/span><\/span>\uf057<\/b><\/span><\/span><\/span> <\/b><\/p>\n

c) <\/b><\/span><\/span><\/span>0,83.10<\/b><\/span><\/span><\/span>-4 <\/b><\/span><\/span><\/sup><\/span>cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> e 12,5<\/b><\/span><\/span><\/span>\uf057<\/b><\/span><\/span><\/span> <\/b><\/p>\n

d) <\/b><\/span><\/span><\/span>0,83.10<\/b><\/span><\/span><\/span>-4 <\/b><\/span><\/span><\/sup><\/span>cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> e 500<\/b><\/span><\/span><\/span>\uf057<\/b><\/span><\/span><\/span><\/p>\n

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica da <\/span><\/span><\/span>Escola de Especialistas de Aeron\u00e1utica<\/span><\/span><\/span>\u00a0\u2013 <\/span><\/span><\/span>EEAR<\/span><\/span><\/span> – 019\/020<\/span><\/span><\/span><\/p>\n

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

Soma de dois vetores pelo M\u00e9todo do paralelogramo<\/b><\/span><\/span><\/span><\/p>\n

Coloque a\u00a0origem dos dois vetores em um mesmo ponto\u00a0e, em seguida,\u00a0trace pelas extremidades de cada um deles, uma paralela ao outro, com linha pontilhada.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Unindo os pontos indicados<\/b><\/span><\/span>\u00a0e colocando a\u00a0<\/b><\/span><\/span>seta conforme a figura acima<\/b><\/span><\/span>, voc\u00ea obter\u00e1 o\u00a0<\/b><\/span><\/span>vetor\u00a0que \u00e9 o vetor\u00a0soma ou o vetor resultante.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Sendo\u00a0<\/b><\/span><\/span>\u03b2<\/b><\/span><\/span>\u00a0o \u00e2ngulo\u00a0<\/b><\/span><\/span>entre os dois vetores<\/b><\/span><\/span>, pode-se determinar o\u00a0<\/b><\/span><\/span>m\u00f3dulo ou intensidade do vetor\u00a0\u00a0pela lei dos cossenos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

No caso do exerc\u00edcio V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 5 u, V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 12 u, a o vetor resultante tem V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = 13 u e o \u00e2ngulo entre eles \u00e9 \u03b8.<\/b><\/span><\/span><\/span><\/p>\n

Aplicando a <\/b><\/span><\/span><\/span>lei dos cossenos V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> + V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.cos\u03b8 <\/b><\/span><\/span><\/span> 13<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 12<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.5.12.cos\u03b8 <\/b><\/span><\/span><\/span><\/p>\n

169 = 25 + 144 + 120.cos\u03b8 169 \u2013 169 = 120.cos\u03b8 cos\u03b8 = 0\/120 <\/b><\/span><\/span><\/span> cos\u03b8 = 0 se cos\u03b8 = 0 <\/b><\/span><\/span><\/span> <\/b><\/p>\n

<\/p>\n

\u03b8 = 90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> e V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> e V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> s\u00e3o perpendiculares entre si.<\/b><\/span><\/span><\/span><\/p>\n

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

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

For\u00e7a el\u00e1stica\u00a0\u2013 Lei de Hooke <\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

C\u00e1lculo do m\u00f3dulo da for\u00e7a el\u00e1stica<\/b><\/span><\/span><\/span> (vertical e para cima) <\/b><\/span><\/span><\/span>quando <\/b><\/span><\/span><\/span>(com as bolinhas) a <\/b><\/span><\/span><\/span>mola est\u00e1 deformada de x = 5 cm = 0,05 m e com K = 9 N\/m<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = K.x <\/b><\/span><\/span><\/span>= 9.0,05 <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 0,45 N. <\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Com a <\/b><\/span><\/span><\/span>cesta em equil\u00edbrio o peso P das n bolinhas<\/b><\/span><\/span><\/span> (vertical e para baixo) <\/b><\/span><\/span><\/span>\u00e9 equilibrado pela for\u00e7a el\u00e1stica, <\/b><\/span><\/span><\/span>ou seja <\/b><\/span><\/span><\/span>P =n.m.g <\/b><\/span><\/span><\/span>= n.0,015.10 <\/b><\/span><\/span><\/span>P = n.0,15 = F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> <\/b><\/span><\/span><\/span> 0,15n = 0,45 n = 0,45\/0,15 <\/b><\/span><\/span><\/span>n = 3 bolinhas.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Queda livre vertical<\/b><\/span><\/span><\/span><\/p>\n

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

<\/p>\n

Aplicando a equa\u00e7\u00e3o de Torricelli V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 2.g.h <\/b><\/span><\/span><\/span>= 2.10.5 =100 <\/b><\/span><\/span><\/span>V = 10 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

04-<\/span><\/span><\/span><\/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

<\/p>\n

S\u00e3o dados: q <\/b><\/span><\/span><\/span>= 3,2\u00b5C <\/b><\/span><\/span><\/span>= 3,2.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span> C<\/b><\/span><\/span><\/span>; <\/b><\/span><\/span><\/span>V = 2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> m\/s<\/b><\/span><\/span><\/span>; <\/b><\/span><\/span><\/span>F = 1,6.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> N <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>B = ?<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = q.V.B <\/b><\/span><\/span><\/span> 1,6.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 3,2.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.B 1,6.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 6,4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.B B = 1,6.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/6,4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> <\/b><\/span><\/span><\/span>B <\/b><\/span><\/span><\/span>= 0,25.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> = <\/b><\/span><\/span><\/span>2,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> T.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Altura do som<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>velocidade<\/b><\/span><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span><\/span>vibra\u00e7\u00e3o da\u00a0fonte sonora (l\u00e2mina, corda, membrana, etc.)<\/b><\/span><\/span><\/span>\u00a0\u00e9 que vai\u00a0<\/b><\/span><\/span><\/span>definir sua altura.<\/b><\/span><\/span><\/span><\/p>\n

As\u00a0<\/b><\/span><\/span><\/span>vibra\u00e7\u00f5es lentas produzem sons graves (baixos e grossos)\u00a0<\/b><\/span><\/span><\/span>e as\u00a0<\/b><\/span><\/span><\/span>vibra\u00e7\u00f5es r\u00e1pidas produzem sons agudos (altos e finos).<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>altura dos sons\u00a0<\/b><\/span><\/span><\/span>depende tamb\u00e9m do\u00a0<\/b><\/span><\/span><\/span>tamanho dos corpos que vibram<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span>Uma corda fina e curta produz sons mais agudos\u00a0<\/b><\/span><\/span><\/span>que os de uma\u00a0<\/b><\/span><\/span><\/span>corda longa e grossa.<\/b><\/span><\/span><\/span><\/p>\n

Uma\u00a0<\/b><\/span><\/span><\/span>flauta pequenina de tubo bem fino\u00a0<\/b><\/span><\/span><\/span>tamb\u00e9m produz\u00a0<\/b><\/span><\/span><\/span>sons mais agudos\u00a0<\/b><\/span><\/span><\/span>do que um\u00a0<\/b><\/span><\/span><\/span>instrumento de sopro com um tubo longo e grosso como a tuba.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>altura do som est\u00e1 relacionada com sua frequ\u00eancia,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>ou seja, a\u00a0<\/b><\/span><\/span><\/span>altura (tom) \u00e9 a qualidade do som que permite ao ouvido\u00a0distinguir um som grave, de baixa frequ\u00eancia, de um som agudo, de alta frequ\u00eancia.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

Eletriza\u00e7\u00e3o por contato<\/b><\/span><\/span><\/span><\/p>\n

Pode ocorrer<\/b><\/span><\/span><\/span>\u00a0entre\u00a0<\/b><\/span><\/span><\/span>dois condutores<\/b><\/span><\/span><\/span>\u00a0(cargas el\u00e9tricas se distribuem em suas superf\u00edcie externas) ou entre um\u00a0<\/b><\/span><\/span><\/span>condutor e um isolante<\/b><\/span><\/span><\/span>\u00a0(no isolante as cargas el\u00e9tricas ficam somente no local do contato).<\/b><\/span><\/span><\/span><\/p>\n

Considere\u00a0<\/b><\/span><\/span><\/span>um condutor A eletrizado com carga positiva (Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= + Q)<\/b><\/span><\/span><\/span>\u00a0e outro\u00a0<\/b><\/span><\/span><\/span>condutor B, eletricamente
\nneutro<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>(Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0)<\/b><\/span><\/span><\/span>\u00a0presos a\u00a0<\/b><\/span><\/span><\/span>suportes isolantes<\/b><\/span><\/span><\/span>\u00a0(para que n\u00e3o descarreguem, inicialmente separados.<\/b><\/span><\/span><\/span><\/p>\n

Quando s\u00e3o\u00a0<\/b><\/span><\/span><\/span>colocados em contato<\/b><\/span><\/span><\/span>, os dois se comportam como se fossem\u00a0<\/b><\/span><\/span><\/span>um \u00fanico corpo e haver\u00e1 uma distribui\u00e7\u00e3o de cargas nas superf\u00edcies de A e de B.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Ap\u00f3s essa distribui\u00e7\u00e3o de cargas\u00a0<\/b><\/span><\/span><\/span>s\u00e3o separados e A ficar\u00e1 com carga Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0e B com carga Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>, tal que\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0+ Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019 = + Q (princ\u00edpio da conserva\u00e7\u00e3o das cargas el\u00e9tricas).<\/b><\/span><\/span><\/span><\/p>\n

Observe que, se\u00a0<\/b><\/span><\/span><\/span>os\u00a0condutores forem id\u00eanticos<\/b><\/span><\/span><\/span>\u00a0e com as\u00a0<\/b><\/span><\/span><\/span>mesmas dimens\u00f5es<\/b><\/span><\/span><\/span>, voc\u00ea teria\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u2019 = Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u2019= Q\u2019\u00a0<\/b><\/span><\/span><\/span>\u00a0Q\u2019 + Q\u2019 = + Q\u00a0\u00a0<\/b><\/span><\/span><\/span>Q\u2019=Q\/2<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Se os dois corpos\u00a0<\/b><\/span><\/span><\/span>forem\u00a0id\u00eanticos<\/b><\/span><\/span><\/span>\u00a0e estiverem\u00a0<\/b><\/span><\/span><\/span>inicialmente eletrizados com cargas\u00a0Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0e Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>ap\u00f3s\u00a0<\/b><\/span><\/span><\/span>o contato eles ter\u00e3o cargas id\u00eanticas Q\u2019 tal que\u00a0<\/b><\/span><\/span><\/span>Q\u2019= (Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)\/2.<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

A\u00a0<\/b><\/span><\/span><\/span>massa uma\u00a0grandeza escalar, positiva\u00a0e\u00a0invari\u00e1vel para cada corpo n\u00e3o dependendo do lugar onde ele se\u00a0encontra.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

A <\/b><\/span><\/span><\/span>massa \u00e9 a mesma<\/b><\/span><\/span><\/span> (Veja teoria acima).<\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span><\/span>express\u00e3o P = m.g, como m \u00e9 constante, o peso P \u00e9 diretamente proporcional \u00e0 acelera\u00e7\u00e3o da gravidade g.<\/b><\/span><\/span><\/span> Assim, se <\/b><\/span><\/span><\/span>g do planeta \u00e9 3 vezes menor a que a da Terra, o peso P do astronauta no planeta tamb\u00e9m ser\u00e1 3 vezes menor.<\/b><\/span><\/span><\/span><\/p>\n

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

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

<\/p>\n

Relacionando as escala Celsius e R:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

Primeira lei de Ohm<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Segunda lei de Ohm (Resistividade)<\/b><\/span><\/span><\/span><\/p>\n

\"EEAR\"<\/p>\n

O <\/b><\/span><\/span><\/span>gr\u00e1fico Vxi fornecido refere-se \u00e0 primeira lei de Ohm <\/b><\/span><\/span><\/span>(teoria acima) onde <\/b><\/span><\/span><\/span>R \u00e9 constante para quaisquer valores de V e i <\/b><\/span><\/span><\/span>e, por exemplo, escolhendo <\/b><\/span><\/span><\/span>quando V = 50 V, i <\/b><\/span><\/span><\/span>= 100 mA = 100.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> A = <\/b><\/span><\/span><\/span>0,1 A <\/b><\/span><\/span><\/span> <\/b>R = v\/i <\/b><\/span><\/span><\/span> R = 50\/0,1 <\/b><\/span><\/span><\/span>R = 500 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

Aplicando agora a <\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span> lei de O<\/b><\/span><\/span><\/span>hm (veja teoria acima) <\/b><\/span><\/span><\/span>sendo dados: R = <\/b><\/span><\/span><\/span>500 \u2126;<\/b><\/span><\/span><\/span> L =<\/b><\/span><\/span><\/span> 2 m e 1,75. 10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span> <\/b>\uf057<\/b><\/span><\/span><\/span>.m <\/b><\/span><\/span><\/span> R = <\/b><\/span><\/span><\/span>\/<\/b><\/span><\/span><\/span>S 500 = 1,75.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.2\/S 500S = 3,5.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span> S = 3,5.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\/5.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 0,7.10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span> m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 0,7.<\/b><\/span><\/span><\/span><\/p>\n

10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> <\/b><\/span><\/span><\/span>S = 0,7.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span> cm<\/b><\/span><\/span><\/span>2.<\/b><\/span><\/span><\/sup><\/span><\/p>\n

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

<\/a><\/p>\n

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

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

Curso de Forma\u00e7\u00e3o de Sargentos (CFS) Escola de Especialistas de Aeron\u00e1utica\u00a0(EEAR) \u00e9 o maior complexo de ensino t\u00e9cnico militar da Am\u00e9rica do Sul sendo uma organiza\u00e7\u00e3o do\u00a0Comando da Aeron\u00e1utica, diretamente subordinada ao diretor-geral do Departamento de Ensino da Aeron\u00e1utica (DEPENS), que tem por finalidade a forma\u00e7\u00e3o e o aperfei\u00e7oamento de graduados da Aeron\u00e1utica. Est\u00e1 localizada no munic\u00edpio de\u00a0Guaratinguet\u00e1, entre as<\/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-4772","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4772","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=4772"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4772\/revisions"}],"predecessor-version":[{"id":10274,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4772\/revisions\/10274"}],"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=4772"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}