{"id":3459,"date":"2018-01-19T00:56:42","date_gmt":"2018-01-19T00:56:42","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3459"},"modified":"2024-09-04T12:35:09","modified_gmt":"2024-09-04T12:35:09","slug":"resolucao-comentada-das-questoes-de-fisica-da-fuvest-sp-2018","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2018\/fuvest-2018\/resolucao-comentada-das-questoes-de-fisica-da-fuvest-sp-2018\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da FUVEST-SP \u2013 2018"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de <\/b><\/span><\/span><\/span><\/p>\n

F\u00cdSICA da FUVEST-SP \u2013 2018<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

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

Veja a express\u00e3o matem\u00e1tica da equa\u00e7\u00e3o das lentes esf\u00e9ricas:<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de Gauss<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S\u00e3o dados: <\/b><\/span><\/span><\/span>P <\/b><\/span><\/span><\/span>= 2 m <\/b><\/span><\/span><\/span>= 2 000 mm \"\"<\/b><\/span><\/span><\/span> <\/b>f = 20 mm.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

Observe o fen\u00f4meno representado na figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do <\/b><\/span><\/span><\/span>volume V de \u00e1gua <\/b><\/span><\/span><\/span>de chuva coletado na <\/b><\/span><\/span><\/span>precipita\u00e7\u00e3o de h = 1,5 cm = 1,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> m de chuva por dia (\u2206t = 8,6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> s) <\/b><\/span><\/span><\/span>em uma regi\u00e3o plana de raio<\/b><\/span><\/span><\/span> R = 660 km = 66.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>m.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

Caracter\u00edsticas da queda livre vertical<\/b><\/span><\/span><\/span><\/p>\n

Qualquer<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>corpo abandonado a\u00a0<\/b><\/span><\/span><\/span>partir do repouso (V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0),<\/b><\/span><\/span><\/span>\u00a0ou lan\u00e7ado verticalmente para baixo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>com velocidade inicial V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2260 0<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>descreve um movimento\u00a0uniformemente acelerado, com acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a=g.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

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

Observe<\/b><\/span><\/span><\/span> no <\/b><\/span><\/span><\/span>gr\u00e1fico <\/b><\/span><\/span><\/span>que \u00e0 <\/b><\/span><\/span><\/span>medida que a profundidade h <\/b><\/span><\/span><\/span>da \u00e1gua <\/b><\/span><\/span><\/span>vai aumentando, <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>velocidade de propaga\u00e7\u00e3o da onda tamb\u00e9m aumenta.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

C\u00e1lculo da <\/b><\/span><\/span><\/span>intensidade da velocidade inicial V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> <\/b>do <\/b><\/span><\/span><\/span>sistema <\/b><\/span><\/span><\/span>sabendo que ele est\u00e1 <\/b><\/span><\/span><\/span>freando <\/b><\/span><\/span><\/span>com acelera\u00e7\u00e3o <\/b><\/span><\/span><\/span>a = – 5 m\/s<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>at\u00e9 parar V = 0 <\/b><\/span><\/span><\/span>ap\u00f3s se <\/b><\/span><\/span><\/span>deslocar de \u2206S = 10 m.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equa\u00e7\u00e3o de Torricelli \"\" V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.a. \u2206S \"\" <\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2.5.10 \"\"V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 100 \"\"<\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 10 m\/s <\/b><\/span><\/span><\/span>(velocidade dos dois ve\u00edculos juntos ap\u00f3s o impacto).<\/b><\/span><\/span><\/span><\/p>\n

Agora voc\u00ea deve <\/b><\/span><\/span><\/span>aplicar<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da quantidade de moviment<\/b><\/span><\/span><\/span>o imediatamente <\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/span> e imediatamente <\/b><\/span><\/span><\/span>ap\u00f3s o impacto<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Antes<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> + m<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span> <\/b>= 2000.V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> + 1000.0 = 2000V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> + 0 \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = 2000V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Depois \"\" Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = M.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> <\/b>= 3000.10 \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = 30000 kgm\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Como <\/b><\/span><\/span><\/span>qualquer tipo de atrito <\/b><\/span><\/span><\/span>ou <\/b><\/span><\/span><\/span>resist\u00eancia ao movimento \u00e9 desprezada <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>sistema<\/b><\/span><\/span><\/span> pode ser considerado c<\/b><\/span><\/span><\/span>onservativo<\/b><\/span><\/span><\/span> e voc\u00ea pode aplicar o <\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da energia mec\u00e2nica <\/b><\/span><\/span><\/span>entre os pontos <\/b><\/span><\/span><\/span>P e Q.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na <\/b><\/span><\/span><\/span>posi\u00e7\u00e3o mais baixa Q <\/b><\/span><\/span><\/span>as <\/b><\/span><\/span><\/span>for\u00e7as<\/b><\/span><\/span><\/span> que agem <\/b><\/span><\/span><\/span>sobre<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>pessoa<\/b><\/span><\/span><\/span> s\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

Seu <\/b><\/span><\/span><\/span>peso P, vertical<\/b><\/span><\/span><\/span> e para <\/b><\/span><\/span><\/span>baixo<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span><\/span>intensidade P =m.g<\/b><\/span><\/span><\/span> = 60.10 \"\" <\/b><\/span><\/span><\/span>P = 600 N.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

Como, pelo <\/b><\/span><\/span><\/span>enunciado o el\u00e9tron se desloca de A para B,<\/b><\/span><\/span><\/span> ent\u00e3o a <\/b><\/span><\/span><\/span>placa A \u00e9 negativa <\/b><\/span><\/span><\/span>(menor potencial), pois deve <\/b><\/span><\/span><\/span>repelir o el\u00e9tron (carga negativa) <\/b><\/span><\/span><\/span>e a <\/b><\/span><\/span><\/span>placa B \u00e9 positiva <\/b><\/span><\/span><\/span>(maior potencial) <\/b><\/span><\/span><\/span>devendo atrair o el\u00e9tron.<\/b><\/span><\/span><\/span><\/p>\n

I.<\/b><\/span><\/span><\/span> Correta \"\"<\/b><\/span><\/span><\/span> Considerando a <\/b><\/span><\/span><\/span>for\u00e7a el\u00e9trica <\/b><\/span><\/span><\/span>como a <\/b><\/span><\/span><\/span>for\u00e7a resultante sobre o el\u00e9tron <\/b><\/span><\/span><\/span>(\u00fanica for\u00e7a que age sobre ele) e lembrando que o <\/b><\/span><\/span><\/span>campo el\u00e9trico entre as placas \u00e9 uniforme,<\/b><\/span><\/span><\/span> vamos calcular o <\/b><\/span><\/span><\/span>trabalho da for\u00e7a el\u00e9trica entre A e B (W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>), <\/b><\/span><\/span><\/span>em <\/b><\/span><\/span><\/span>fun\u00e7\u00e3o da diferen\u00e7a de potencial entre A e B sabendo que o el\u00e9tron possui carga negativa e que V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> < V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

\"\"W<\/b><\/span><\/span><\/span>BA<\/b><\/span><\/span><\/sub><\/span> = q.(V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>= (-1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>).(- 2,0.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>) <\/b><\/span><\/span><\/span>= 3,2.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span> J. <\/b><\/span><\/span><\/span>(Observe que esse <\/b><\/span><\/span><\/span>trabalho independe da trajet\u00f3ria dependendo apenas de q e de (V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>s\u00e3o os mesmos para as duas trajet\u00f3rias)<\/b><\/span><\/span><\/span><\/p>\n

Esse trabalho tamb\u00e9m <\/b><\/span><\/span><\/span>\u00e9 o mesmo <\/b><\/span><\/span><\/span>quando <\/b><\/span><\/span><\/span>calculado pela varia\u00e7\u00e3o da energia cin\u00e9tica <\/b><\/span><\/span><\/span>(Teorema da energia cin\u00e9tica) \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span> \u2013 E<\/b><\/span><\/span><\/span>cA<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> 3,2.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span> = E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span> \u2013 mV<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span> \u2013 m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span> \u2013 0 \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>cB<\/b><\/span><\/span><\/sub><\/span> = 3,2.10<\/b><\/span><\/span><\/span>-15<\/b><\/span><\/span><\/sup><\/span> J.(energia cin\u00e9tica com que o el\u00e9tron atinge a placa B).<\/b><\/span><\/span><\/span><\/p>\n

II.<\/b><\/span><\/span><\/span> Correta \"\" Como j\u00e1 vimos em I.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

III.<\/b><\/span><\/span><\/span> Falsa \"\" Veja I e II.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Os valores nominais do LED s\u00e3o P = 8 W e U = 4 V fornecidos pelo enunciado (<\/b><\/span><\/span><\/span>LED (L) de 8 W, <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>opera<\/b><\/span><\/span><\/span> com <\/b><\/span><\/span><\/span>4 V).<\/b><\/span><\/span><\/span><\/p>\n

Pelos valores nominais voc\u00ea pode calcular o valor da resist\u00eancia do LED atrav\u00e9s da express\u00e3o P = \"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\" 8 = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>R =<\/b><\/span><\/span><\/span> \"\" \"\" <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span> = 2 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

Para que o <\/b><\/span><\/span><\/span>LED opere segundo sua tens\u00e3o nominal (U = 4 V) <\/b><\/span><\/span><\/span>ele <\/b><\/span><\/span><\/span>dever\u00e1 ser percorrido <\/b><\/span><\/span><\/span>por uma <\/b><\/span><\/span><\/span>corrente el\u00e9trica de R = \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> 2 = \"\" \"\" <\/b><\/span><\/span><\/span>i = 2 A.<\/b><\/span><\/span><\/span><\/p>\n

Como <\/b><\/span><\/span><\/span>R e R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span> est\u00e3o em s\u00e9rie <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>corrente el\u00e9trica neles \u00e9 a mesma <\/b><\/span><\/span><\/span>e a resist\u00eancia do <\/b><\/span><\/span><\/span>resistor equivalente ser\u00e1 R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>R + R<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span> <\/b>= R + 2. <\/b><\/span><\/span><\/span><\/p>\n

\"\"Esse <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = R + 2 <\/b><\/span><\/span><\/span>est\u00e1 ligado na <\/b><\/span><\/span><\/span>fonte de U = 6 V \"\"<\/b><\/span><\/span><\/span> <\/b>lei de Ohm R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = \"\" \"\"<\/b><\/span><\/span><\/span> R + 2 = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>R = 3 – 2<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

R = 1 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

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

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

L\u00e2mpada fluorescente<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do <\/b><\/span><\/span><\/span>custo total de uma l\u00e2mpada fluorescente <\/b><\/span><\/span><\/span>de pot\u00eancia <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 28 W = 28.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> kW <\/b><\/span><\/span><\/span>durante sua <\/b><\/span><\/span><\/span>vida \u00fatil fornecida (\"\"<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 10000 h = 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> h).<\/b><\/span><\/span><\/span><\/p>\n

Energia el\u00e9trica (W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>) consumida <\/b><\/span><\/span><\/span>por essa l\u00e2mpada <\/b><\/span><\/span><\/span>durante sua vida \u00fatil \"\"<\/b><\/span><\/span><\/span> <\/b>W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> <\/b>= 28.10<\/b><\/span><\/span><\/span>-3 <\/b><\/span><\/span><\/sup><\/span>kW. 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>h \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 280 kWh.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Custo total de cada l\u00e2mpada fluorescente<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>C$ = 70,00 + pre\u00e7o de uma l\u00e2mpada <\/b><\/span><\/span><\/span>= 70,00 + 20,00 \"\" <\/b><\/span><\/span><\/span>C$ = 90,00.<\/b><\/span><\/span><\/span><\/p>\n

L\u00e2mpada incandescente<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do <\/b><\/span><\/span><\/span>custo total de uma l\u00e2mpada incandescente <\/b><\/span><\/span><\/span>de pot\u00eancia <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 100 W = 10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span> kW <\/b><\/span><\/span><\/span>durante a mesma vida \u00fatil que a da fluorescente<\/b><\/span><\/span><\/span> (\"\"<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 10 000 h = 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> h).<\/b><\/span><\/span><\/span><\/p>\n

Energia el\u00e9trica (W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>) consumida <\/b><\/span><\/span><\/span>por essa l\u00e2mpada <\/b><\/span><\/span><\/span>durante sua vida \u00fatil \"\"<\/b><\/span><\/span><\/span> <\/b>W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> <\/b>= 10<\/b><\/span><\/span><\/span>-1 <\/b><\/span><\/span><\/sup><\/span>kW. 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>h \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> kWh = 1 000 kWh.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que voc\u00ea <\/b><\/span><\/span><\/span>necessita de 10 l\u00e2mpadas incandescentes <\/b><\/span><\/span><\/span>(vida \u00fatil 10 000 h) para <\/b><\/span><\/span><\/span>operar na mesma vida \u00fatil (mesmo per\u00edodo de tempo)<\/b><\/span><\/span><\/span> que a <\/b><\/span><\/span><\/span>das fluorescentes <\/b><\/span><\/span><\/span>(vida \u00fatil 1 000 h).<\/b><\/span><\/span><\/span><\/p>\n

Assim, o <\/b><\/span><\/span><\/span>custo total de <\/b><\/span><\/span><\/span>cada incandescente<\/b><\/span><\/span><\/span> <\/b>dever\u00e1 ser <\/b><\/span><\/span><\/span>C$ = custo de uma incandescente + 10xpre\u00e7o de uma l\u00e2mpada incandescente = 250,00 + 10×4,00 \"\" <\/b><\/span><\/span><\/span>C$ = 290,00 <\/b><\/span><\/span><\/span>(custo total de cada incandescente para a mesma vida \u00fatil de cada fluorescente)<\/b><\/span><\/span><\/span><\/p>\n

Assim, o <\/b><\/span><\/span><\/span>valor poupado <\/b><\/span><\/span><\/span>ser\u00e1 <\/b><\/span><\/span><\/span>C$ = <\/b><\/span><\/span><\/span>290,00 \u2013 90,00 <\/b><\/span><\/span><\/span>= 200,00.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/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 de F\u00cdSICA da FUVEST-SP \u2013 2018 01- R- B 02- Veja a express\u00e3o matem\u00e1tica da equa\u00e7\u00e3o das lentes esf\u00e9ricas: Equa\u00e7\u00e3o de Gauss S\u00e3o dados: P = 2 m = 2 000 mm f = 20 mm. R – E 03- Observe o fen\u00f4meno representado na figura abaixo: C\u00e1lculo do volume V de \u00e1gua de chuva coletado<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3457,"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-3459","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3459","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=3459"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3459\/revisions"}],"predecessor-version":[{"id":3462,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3459\/revisions\/3462"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3457"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3459"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}