{"id":1742,"date":"2015-12-31T03:28:39","date_gmt":"2015-12-31T03:28:39","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1742"},"modified":"2024-09-03T14:08:24","modified_gmt":"2024-09-03T14:08:24","slug":"resolucao-comentada-da-unifesp-2016","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2016\/unifesp-2016\/resolucao-comentada-da-unifesp-2016\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da UNIFESP 2016"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios do vestibular da UNIFESP – 2016<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>m\u00f3dulo das velocidades escalares m\u00e9dias de A e de B, em m\/s, durante os 120 s<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Em <\/b><\/span><\/span><\/span>todo gr\u00e1fico da velocidade escalar em fun\u00e7\u00e3o do tempo,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>\u00e1rea A <\/b><\/span><\/span><\/span>entre a <\/b><\/span><\/span><\/span>reta representativa<\/b><\/span><\/span><\/span> (linha cheia) e o <\/b><\/span><\/span><\/span>eixo dos tempos<\/b><\/span><\/span><\/span> \u00e9 <\/b><\/span><\/span><\/span>numericamente igual ao deslocamento \u2206S <\/b><\/span><\/span><\/span>efetuado pelo m\u00f3vel em <\/b><\/span><\/span><\/span>determinado intervalo de tempo \u2206t.<\/b><\/span><\/span><\/span><\/p>\n

Calculando o <\/b><\/span><\/span><\/span>deslocamento do m\u00f3vel A<\/b><\/span><\/span><\/span> entre <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 e t = 120 s pela \u00e1rea A<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Calculando o <\/b><\/span><\/span><\/span>deslocamento do m\u00f3vel B<\/b><\/span><\/span><\/span> entre <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 e t = 120 s pela \u00e1rea A<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>dist\u00e2ncia entre os ve\u00edculos, em metros<\/b><\/span><\/span><\/span>, no <\/b><\/span><\/span><\/span>instante t = 60 s.<\/b><\/span><\/span><\/span><\/p>\n

Observe que, no <\/b><\/span><\/span><\/span>instante t = 60 s eles possuem a mesma velocidade,<\/b><\/span><\/span><\/span> que \u00e9 o <\/b><\/span><\/span><\/span>ponto (P)<\/b><\/span><\/span><\/span> onde as <\/b><\/span><\/span><\/span>retas<\/b><\/span><\/span><\/span><\/b><\/p>\n

\"\"<\/p>\n

Acontece que <\/b><\/span><\/span><\/span>no instante t = 60s, V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = V\u2019<\/b><\/span><\/span><\/span> <\/b>= 12 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Deslocamentos de A e de B entre 0 e 60 s pela \u00e1rea:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

02- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>m\u00f3dulo da for\u00e7a de atrito<\/b><\/span><\/span><\/span>, em <\/b><\/span><\/span><\/span>N,<\/b><\/span><\/span><\/span> entre a <\/b><\/span><\/span><\/span>caixa e a rampa no ponto B<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Observe na figura a for\u00e7a peso decomposta em suas duas parcelas com as respectivas fun\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Decompondo a for\u00e7a peso,<\/b><\/span><\/span><\/span> obtemos a <\/b><\/span><\/span><\/span>express\u00e3o matem\u00e1tica de cada uma dessas parcelas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

No <\/b><\/span><\/span><\/span>caso do exerc\u00edcio,<\/b><\/span><\/span><\/span> na rampa, a <\/b><\/span><\/span><\/span>for\u00e7a de atrito pedida<\/b><\/span><\/span><\/span> tem intensidade F<\/b><\/span><\/span><\/span>atAC<\/b><\/span><\/span><\/sub><\/span> = \u00b5Pcos\u03b8 = \u00b5mgcos\u03b8 =<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

0,25.40.10.0,8 \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>atAC<\/b><\/span><\/span><\/sub><\/span> = 80 N.<\/b><\/span><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>dist\u00e2ncia percorrida<\/b><\/span><\/span><\/span> pelo garoto, <\/b><\/span><\/span><\/span>em metros,<\/b><\/span><\/span><\/span> desde o <\/b><\/span><\/span><\/span>ponto A at\u00e9 o ponto D.<\/b><\/span><\/span><\/span><\/p>\n

\"\" For\u00e7as <\/b><\/span><\/span><\/span>que agem sobre o corpo na rampa no <\/b><\/span><\/span><\/span>trecho AC:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Teorema da energia cin\u00e9tica<\/b><\/span><\/span><\/span> \"\" o <\/b><\/span><\/span><\/span>trabalho de todas as for\u00e7as<\/b><\/span><\/span><\/span> que agem sobre o corpo no <\/b><\/span><\/span><\/span>trecho total AD \u00e9 igual \u00e0 vara\u00e7\u00e3o da energia cin\u00e9tica entre os pontos A de velocidade V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 1m\/s e D, de velocidade zero (fornecidos pelo enunciado).<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>FatAC<\/b><\/span><\/span><\/sub><\/span> + W<\/b><\/span><\/span><\/span>Pp <\/b><\/span><\/span><\/sub><\/span>+ W<\/b><\/span><\/span><\/span>FatCD<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cD<\/b><\/span><\/span><\/sub><\/span> \u2013 E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>– 800 + 2400 – 100d<\/b><\/span><\/span><\/span>CD<\/b><\/span><\/span><\/sub><\/span> = 0 \u2013 mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>1600 – 100d<\/b><\/span><\/span><\/span>CD<\/b><\/span><\/span><\/sub><\/span> = – 40.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>100d<\/b><\/span><\/span><\/span>CD<\/b><\/span><\/span><\/sub><\/span> = 1620<\/b><\/span><\/span><\/span> \"\" d<\/b><\/span><\/span><\/span>CD<\/b><\/span><\/span><\/sub><\/span> = 16,2 m.<\/b><\/span><\/span><\/span><\/p>\n

\u00c9 pedido d<\/b><\/span><\/span><\/span>AD<\/b><\/span><\/span><\/sub><\/span> = 10 + 16,2 \"\" <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>AD<\/b><\/span><\/span><\/sub><\/span> = 26,2 m.<\/b><\/span><\/span><\/span><\/p>\n

03- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>quantidade de calor,<\/b><\/span><\/span><\/span> em <\/b><\/span><\/span><\/span>cal,<\/b><\/span><\/span><\/span> necess\u00e1ria para <\/b><\/span><\/span><\/span>elevar a temperatura do copo com \u00e1gua l\u00edquida de<\/b><\/span><\/span><\/span> <\/b>20 \u00baC para 50 \u00baC.<\/b><\/span><\/span><\/span><\/b><\/p>\n

Copo \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> para a <\/b><\/span><\/span><\/span>temperatura<\/b><\/span><\/span><\/span> do copo subir de <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 20<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C para t = 50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C, sendo o calor espec\u00edfico do copo c<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 0,2 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C e m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 200 g \"\" Q<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) =<\/b><\/span><\/span><\/span>100.0,2.(50 \u2013 20<\/b><\/span><\/span><\/span>) \"\" Q<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 600 cal.<\/b><\/span><\/span><\/span><\/p>\n

\u00c1gua \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> para a <\/b><\/span><\/span><\/span>temperatura<\/b><\/span><\/span><\/span> da \u00e1gua subir de <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 20<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C para t = 50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C, sendo o calor espec\u00edfico da c<\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/sub><\/span>= 1 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C e m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = 100 g \"\" Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) = <\/b><\/span><\/span><\/span>200.1.(50 \u2013 20<\/b><\/span><\/span><\/span>) \"\" Q<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 6000 cal.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de calor pedida<\/b><\/span><\/span><\/span> \"\" Q<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span> = 600 + 6000 \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span> = 6600 cal.<\/b><\/span><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>massa de vapor de \u00e1gua<\/b><\/span><\/span><\/span>, em <\/b><\/span><\/span><\/span>gramas,<\/b><\/span><\/span><\/span> necess\u00e1ria para <\/b><\/span><\/span><\/span>elevar a temperatura do copo com \u00e1gua<\/b><\/span><\/span><\/span> <\/b>l\u00edquida at\u00e9 atingir o equil\u00edbrio t\u00e9rmico a 50 \u00baC.<\/b><\/span><\/span><\/span><\/p>\n

Pelo enunciado a <\/b><\/span><\/span><\/span>temperatura inicial do vapor de \u00e1gua era t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 120 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C,<\/b><\/span><\/span><\/span> que deve se <\/b><\/span><\/span><\/span>transformar<\/b><\/span><\/span><\/span> em <\/b><\/span><\/span><\/span>\u00e1gua l\u00edquida a 50 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C. Etapas:<\/b><\/span><\/span><\/span><\/p>\n

1<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span> \"\" Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>quantidade de calor cedida<\/b><\/span><\/span><\/span> pelo <\/b><\/span><\/span><\/span>vapor de \u00e1gua<\/b><\/span><\/span><\/span> para passar de <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 120<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C para t =<\/b><\/span><\/span><\/span> <\/b>100<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C, sem mudan\u00e7a de estado<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> = m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.0,5.(100 \u2013 120) \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = – 10m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

2\u00aa<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/sup><\/span>Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>quantidade de calor cedida pelo vapor de \u00e1gua<\/b><\/span><\/span><\/span> para passar de <\/b><\/span><\/span><\/span>vapor de \u00e1gua a 100<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>\u00e1gua l\u00edquida a 100<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C (condensa\u00e7\u00e3o, liquefa\u00e7\u00e3o)<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.L<\/b><\/span><\/span><\/span>= m<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>.(-540) \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = – 540m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

3<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span> \"\" Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>quantidade de calor cedida pela \u00e1gua l\u00edquida<\/b><\/span><\/span><\/span> para passar de <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 100<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C para t = 50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C, sem mudan\u00e7a de estado<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> = m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.1.(50 \u2013 100)\"\" Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> = – 50m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de calor <\/b><\/span><\/span><\/span>cedida pelo vapor de \u00e1gua<\/b><\/span><\/span><\/span> para se transformar <\/b><\/span><\/span><\/span>em \u00e1gua l\u00edquida a 50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

Q\u2019 = Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> + Q<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> – 10 m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> \u2013 540m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> \u2013 50m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>Q\u2019<\/b><\/span><\/span><\/span> <\/b>= – 600m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Essa quantidade de calor <\/b><\/span><\/span><\/span>Q\u2019 cedida pelo vapor de \u00e1gua para ir de 120<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a \u00e1gua a 50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span> \u00e9 a <\/b><\/span><\/span><\/span>mesma<\/b><\/span><\/span><\/span> <\/b>recebida pelo sistema (copo + \u00e1gua) para ir de 20<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a 50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C (achada no item Q<\/b><\/span><\/span><\/span>t <\/b><\/span><\/span><\/sub><\/span>= 6600 cal)<\/b><\/span><\/span><\/span>, ou seja, no <\/b><\/span><\/span><\/span>equil\u00edbrio t\u00e9rmico<\/b><\/span><\/span><\/span> a soma dessas quantidades de calor <\/b><\/span><\/span><\/span>trocadas entre o vapor e o copo com \u00e1gua deve ser nula.<\/b><\/span><\/span><\/span><\/p>\n

– 600m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> + 6600 = 0 \"\" m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> = 6600\/600 \"\" <\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> = 11 g.<\/b><\/span><\/span><\/span><\/p>\n

04- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>dist\u00e2ncia, em metros, da imagem dos ve\u00edculos ao espelho.<\/b><\/span><\/span><\/span><\/p>\n

Existe uma <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o matem\u00e1tica entre a posi\u00e7\u00e3o (localiza\u00e7\u00e3o) do objeto<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>,<\/b><\/span><\/span><\/span>\u00a0a da <\/b><\/span><\/span><\/span>imagem<\/b><\/span><\/span><\/span>\u00a0P\u2019<\/b><\/span><\/span><\/span>\u00a0e a <\/b><\/span><\/span><\/span>dist\u00e2ncia focal<\/b><\/span><\/span><\/span>\u00a0f\u00a0<\/b><\/span><\/span><\/span>para os espelhos esf\u00e9ricos,<\/b><\/span><\/span><\/span> denominada <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o dos pontos conjugados<\/b><\/span><\/span><\/span>, no referencial de Gauss, que \u00e9 fornecida pela <\/b><\/span><\/span><\/span>express\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o<\/b><\/span><\/span><\/span> entre o <\/b><\/span><\/span><\/span>comprimento do di\u00e2metro da imagem do pneu de um dos carros, indicada por d na figura, e o comprimento real do di\u00e2metro desse pneu.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o do aumento linear transversal<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

05- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>resist\u00eancia equivalente<\/b><\/span><\/span><\/span>, em <\/b><\/span><\/span><\/span>\u03a9<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Observe a <\/b><\/span><\/span><\/span>sequ\u00eancia abaixo<\/b><\/span><\/span><\/span> para o <\/b><\/span><\/span><\/span>c\u00e1lculo da resist\u00eancia equivalente R<\/b><\/span><\/span><\/span>eq <\/b><\/span><\/span><\/sub><\/span>no circuito da figura 2, onde os tr\u00eas resistores est\u00e3o em paralelo<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>pot\u00eancia total dissipada<\/b><\/span><\/span><\/span>, em <\/b><\/span><\/span><\/span>W.<\/b><\/span><\/span><\/span><\/p>\n

Voc\u00ea pode calcular a <\/b><\/span><\/span><\/span>pot\u00eancia total dissipada<\/b><\/span><\/span><\/span> utilizando a <\/b><\/span><\/span><\/span>resist\u00eancia equivalente<\/b><\/span><\/span><\/span> que est\u00e1 submetida \u00e0 uma <\/b><\/span><\/span><\/span>tens\u00e3o de U = 12 V.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios do vestibular da UNIFESP – 2016 01- a) o m\u00f3dulo das velocidades escalares m\u00e9dias de A e de B, em m\/s, durante os 120 s. Em todo gr\u00e1fico da velocidade escalar em fun\u00e7\u00e3o do tempo, a \u00e1rea A entre a reta representativa (linha cheia) e o eixo dos tempos \u00e9 numericamente igual ao deslocamento \u2206S efetuado<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1740,"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-1742","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1742","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=1742"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1742\/revisions"}],"predecessor-version":[{"id":1746,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1742\/revisions\/1746"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1740"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1742"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}