{"id":6480,"date":"2015-12-31T02:59:39","date_gmt":"2015-12-31T02:59:39","guid":{"rendered":"http:\/\/localhost\/fisicaevestibular\/?page_id=1716"},"modified":"2024-08-19T16:31:09","modified_gmt":"2024-08-19T16:31:09","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-potencia-eletrica-energia-eletrica-2","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/potencia-eletrica-energia-eletrica\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-potencia-eletrica-energia-eletrica-2\/","title":{"rendered":"Energia el\u00e9trica – Resolu\u00e7\u00e3o"},"content":{"rendered":"

\u00a0Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

Pot\u00eancia el\u00e9trica \u2013 Energia el\u00e9trica<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0a) A l\u00e2mpada de menor pot\u00eancia \u2013 40W<\/b><\/span><\/span><\/span>
\nb) O de maior pot\u00eancia \u2013 caminh\u00e3o<\/b><\/span><\/span><\/span>
\nc) L\u00e2mpada de 40W, l\u00e2mpada de 100W, ferro el\u00e9trico, fusca e caminh\u00e3o<\/b><\/span><\/span><\/span>
\n02-<\/b><\/span><\/span><\/span>\u00a0Quanto maior o n\u00famero de equipamentos el\u00e9tricos, maior o consumo de energia (III)\u00a0 —\u00a0 observe na express\u00e3o W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t que o consumo de energia W depende tamb\u00e9m da pot\u00eancia do equipamento P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0(I) e do tempo \u0394t de funcionamento (II)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span>
\n03-<\/b><\/span><\/span><\/span>\u00a0Observe que em cada uma das recomenda\u00e7\u00f5es voc\u00ea tentando diminuir o efeito Joule, ou seja, a transforma\u00e7\u00e3o de energia el\u00e9trica em t\u00e9rmica\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span>
\n04-<\/b><\/span><\/span><\/span>\u00a0A energia el\u00e9trica \u00e9 expressa por W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t, medida em megawatts.hora MW.h) e n\u00e3o megawatts por hora (MW\/h)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span>
\n05-<\/b><\/span><\/span><\/span>\u00a0a) O disjuntor desliga quando a pot\u00eancia total for superior a P<\/b><\/span><\/span><\/span>ot<\/b><\/span><\/span><\/sub><\/span>=i.U=15.110\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>ot<\/b><\/span><\/span><\/sub><\/span>=1.650W\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>o ferro el\u00e9trico e a l\u00e2mpada.<\/b><\/span><\/span><\/span>
\nb) n=1.650W\/100W\u00a0 —\u00a0 n=16,5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0n=16 l\u00e2mpadas<\/b><\/span><\/span><\/span>
\n06-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=i.U=0,2.10\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=2,0W<\/b><\/span><\/span><\/span>
\n07-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=i.U=2.600\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1.200W\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u0394t\u00a0 —\u00a0 1.200=W\/3.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 W=3,6J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span>
\n08-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.U<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.110\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.110\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 110.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=220.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a corrente que percorre o primeiro \u00e9 2 vezes maior que a corrente que percorre o segundo, mas o consumo de energia \u00e9 o mesmo, pois a pot\u00eancia \u00e9 a mesma\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/span>
\n09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span>
\n10-<\/b><\/span><\/span><\/span>\u00a0a) TENS\u00c3O DE ALIMENTA\u00c7\u00c3O;<\/b><\/span><\/span><\/span>\u00a024V<\/b><\/span><\/span><\/span>
\n\u00a0\u00a0\u00a0 POT\u00caNCIA CONSUMIDA:\u00a0<\/b><\/span><\/span><\/span>150W<\/b><\/span><\/span><\/span>
\nb) P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=i.U\u00a0 —\u00a0 150=i.24\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>i=6,25A<\/b><\/span><\/span><\/span>
\n11-<\/b><\/span><\/span><\/span>\u00a0W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=30(1,5.8 + 3,3.1\/3 + 0,2.10 + 0,36.10 + 0,1.6)=30.(12 + 1,1 + 2,0 + 3,5 + 0,6)=30.(19,4)\u00a0 —\u00a0 W=582Wh\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1kWh \u2013 0,40\u00a0 —\u00a0 1.362kWh \u2013 x\u00a0 —\u00a0 x=232,8\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span>
\n12-<\/b><\/span><\/span><\/span><\/span>\u00a0a) P<\/b><\/span><\/span><\/span><\/span>ot<\/b><\/span><\/span><\/span><\/sub><\/span>=4.000×60 + 600×200\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span><\/span>ot<\/b><\/span><\/span><\/span><\/sub><\/span>=240.000 + 120.000\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span><\/span>ot<\/b><\/span><\/span><\/span><\/sub><\/span>=360.000W\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span><\/span>ot<\/b><\/span><\/span><\/span><\/sub><\/span>=i.U\u00a0 —\u00a0 360.000=120.i\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>i=3.000A<\/b><\/span><\/span><\/span><\/span>
\nb) W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>ot<\/b><\/span><\/span><\/sub><\/span>.\u0394t=360.000×120=43.200.000Wh=43.200kWh\u00a0 —\u00a0 R$=43.200×0,40\u00a0 —<\/b><\/span><\/span><\/span>\u00a0R$=17.280,00<\/b><\/span><\/span><\/span>
\n13-<\/b><\/span><\/span><\/span>\u00a0W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=12.8800\u00a0 —\u00a0 W=105.600Wh\u00a0 —\u00a0 W=105,6kWh\u00a0 —\u00a0 R$=105,6.0,5=52,80\u00a0 —<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span>
\n14-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=i.U=2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.2,5.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=5,0.10<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0 W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=5.10<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 W=5.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>J\u00a0 — \u00a0regra de tr\u00eas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>1kWh – 3,6.10<\/b><\/span><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 xkWh \u2013 5,0.10<\/b><\/span><\/span><\/span><\/span>9<\/b><\/span><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 x=5.10<\/b><\/span><\/span><\/span><\/span>9<\/b><\/span><\/span><\/span><\/sup><\/span>\/3,6.10<\/b><\/span><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup><\/span>=1,39.10<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 W=1.390kWh\u00a0 —\u00a0 C$=0,38.1390\u00a0 —\u00a0 R$=530,00\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/span>
\n15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0a) C\u00e1lculo da carga el\u00e9trica total Q pela soma das \u00e1reas dos 4 ret\u00e2ngulos do gr\u00e1fico acima\u00a0 —\u00a0 Q=(10.100 + 20.100 + 30.100 + 10.100)\u00a0 —\u00a0 Q=7.000A.min=7.000.60\u00a0 —\u00a0 Q=420.000C\u00a0 —\u00a0 W=Q.U-420.000.120\u00a0 —\u00a0 W=50.400.000J\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>
\nregra de tr\u00eas\u00a0 —\u00a0 1kWh \u2013 3,6.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 xkWh \u2013 50,4.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 x=W=14kWh\u00a0 —\u00a0 custo=14.0,20\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0C$2,80<\/b><\/span><\/span><\/span>
\nb) Observe no gr\u00e1fico que nos primeiros 42 minutos a corrente \u00e9 de 10\u00aa\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=i.U=10.120\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1.200W\u00a0 —\u00a0 W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=1.200.(42.60)\u00a0 —\u00a0 W=3.024.000J\u00a0 —\u00a0 apenas 1\/3 dessa energia foi utilizada para aquecer a \u00e1gua\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>calor<\/b><\/span><\/span><\/sub><\/span>=3.024.000\/3\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>calor<\/b><\/span><\/span><\/sub><\/span>=1.008.000J\u00a0 —\u00a0 equa\u00e7\u00e3o fundamental da calorimetria\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>calor<\/b><\/span><\/span><\/sub><\/span>=m.c.\u0394\u03b8\u00a0 —\u00a0 1.008.000=10.4,2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>. \u0394\u03b8\u00a0 —\u00a0 \u0394\u03b8=1.008.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/42.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0\u0394\u03b8=24<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>
\n16-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo da corrente na l\u00e2mpada\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=i.U\u00a0 —\u00a0 6=i.12\u00a0 —\u00a0 i=0,5A (C\/s)\u00a0 —\u00a0 carga na bateria\u00a0 —\u00a0 Q=45\/2A.h=22,5.3.600\u00a0 —\u00a0 Q=81.000C\u00a0 —\u00a0 i=Q\/\u0394t\u00a0 —\u00a0 0,5=81.000\/\u0394t\u00a0 —\u00a0 \u0394t=162.000s=162.000\/3.600\u00a0 —\u00a0 \u0394t=45h\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span>
\n17-<\/b><\/span><\/span><\/span>\u00a0Campo Grande\u00a0 —\u00a0 anual\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>=700W\u00a0 —\u00a0 \u00e1rea 1m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u0394t=365diasx6h\u00a0 —\u00a0 \u0394t=2.190h\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u0394t\u00a0 —\u00a0 700W=W\/2.190h\u00a0 —\u00a0 W=2.190.700=1.533.000W.h\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>Campo Grande<\/b><\/span><\/span><\/sub><\/span>=1.533kWh\u00a0 —\u00a0 energia coletada em Campo Grande, por m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0de \u00e1rea durante um ano, 6h ao dia\u00a0 —\u00a0 equipamento de capta\u00e7\u00e3o\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=1.533kWh x 0,6(60%) x 4(\u00e1rea de capta\u00e7\u00e3o 4 vezes maior)\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=3.679,2kWh\u00a0 —\u00a0 economiza 3.679,2 x 0,60\u00a0 —\u00a0 C$=2.207,00\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span>
\n18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Televis\u00e3o de 29 polegadas\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u0394t\u00a0 —\u00a0 70=W\/6.30\u00a0 —\u00a0 W=12.600Wh=12,6kWh\u00a0 —\u00a0 televis\u00e3o de 42 polegadas\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u0394t\u00a0 —\u00a0 220=W\/180\u00a0 —\u00a0 W=39.600Wh=39,6kWh\u00a0 —\u00a0 diferen\u00e7a\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=39,6 \u2013 12,6\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=27kWh\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>
\n19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Decodificador\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>od<\/b><\/span><\/span><\/sub><\/span>.\u0394t=6.30,24\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=4.320W.h\u00a0 —\u00a0 l\u00e2mpada\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>ol<\/b><\/span><\/span><\/sub><\/span>.\u0394t\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=4.320=60.\u0394t\u00a0 —\u00a0<\/b><\/span><\/span><\/span>
\n\u0394t=4.320\/60\u00a0 —\u00a0 \u0394t=72h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span>
\n20-<\/b><\/span><\/span><\/span>\u00a0no Equador\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=792.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 aquecedor\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=i.U.\u0394t=20.220. \u0394<\/b><\/span><\/span><\/span>t=4.400.<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>t\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 792.10<\/b><\/span><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup><\/span>=4.400.<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>t=792.10<\/b><\/span><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup><\/span>\/44.10<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>t=180.000s\/3.300=50h\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/span><\/span>
\n21-<\/b><\/span><\/span><\/span><\/span>\u00a0a) V=<\/b><\/span><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span>f\u00a0 —\u00a0 3.10<\/b><\/span><\/span><\/span><\/span>8<\/b><\/span><\/span><\/span><\/sup><\/span>=<\/b><\/span><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span>830.10<\/b><\/span><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span>=0,36m<\/b><\/span><\/span><\/span><\/span>
\nb) Q=0,8Ah=0,8C\/s.1h=0,8C\/s.3600s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Q=2.880C<\/b><\/span><\/span><\/span>
\nc) i=Q\/<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>t=2.880\/110.60\u00a0 —<\/b><\/span><\/span><\/span><\/span>\u00a0 i=0,44 A<\/b><\/span><\/span><\/span><\/span>\u00a0\u00a0—\u00a0 W=Q.U=2.880.6\u00a0 —\u00a0 W=17.280J\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>=W\/\u0394t=17.280\/110.60=17.280\/6.600\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>=2,62W<\/b><\/span><\/span><\/span><\/span>
\n22-<\/b><\/span><\/span><\/span>\u00a0Comparando, em termos de porcentagem, 60lm\/W com 15lm\/W, voc\u00ea verifica que, ao substituir uma l\u00e2mpada de 15lm\/W por outra de 60lm\/W, o rendimento ser\u00e1 de 75%\u00a0 —\u00a0 como cada l\u00e2mpada tem 100W e s\u00e3o 10 l\u00e2mpadas, \u00a0a economia de pot\u00eancia ser\u00e1 P =10.0,75.100= 750 W, que fornece, em um m\u00eas (30 dias), uma economia de energia (W) que,funcionando 6h por dia\u00a0 —\u00a0 W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=750.30.6\u00a0 —\u00a0 W=135.000Wh\u00a0 —\u00a0 W=135kWh\u00a0 —\u00a0 economia=135.0,30\u00a0 —\u00a0<\/b><\/span><\/span><\/span>
\nC$=40,50\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>
\n23-<\/b><\/span><\/span><\/span>\u00a0W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=200.60\u00a0 —\u00a0 W=12.000J\u00a0 —\u00a0 1cal \u2013 4J\u00a0 —\u00a0 Qcal \u2013 12.000J\u00a0 —\u00a0 4Q=12.000\u00a0 —\u00a0 Q=3.000cal\u00a0 —\u00a0 Q=m.c.\u0394\u03b8\u00a0 —\u00a0<\/b><\/span><\/span><\/span>
\n3.000=100.1. .\u0394\u03b8\u00a0 —\u00a0 .\u0394\u03b8=30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span>
\n24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Decodificador\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>od<\/b><\/span><\/span><\/sub><\/span>.\u0394t=6.30,24\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=4.320W.h\u00a0 —\u00a0 l\u00e2mpada\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>ol<\/b><\/span><\/span><\/sub><\/span>.\u0394t\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=4.320=60.\u0394t\u00a0 —\u00a0<\/b><\/span><\/span><\/span>
\n\u0394t=4.320\/60\u00a0 —\u00a0 \u0394t=72h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span>
\n25-<\/b><\/span><\/span><\/span>\u00a0a)Energia de Itaipu\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=87.600GWh=87.600.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>Wh\u00a0 —\u00a0 energia captada pela \u00e1rea S em 1 ano\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=0,2.P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=0,2.170.W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.S.360.24=293.760S\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 293.760S=87.600.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 S=87.600.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\/293.760\u00a0 —\u00a0\u00a0 S=0,3.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 S=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 S=3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0S=3.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>km<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>
\nb) Porcentual m\u00e9dio\u00a0 —\u00a0 P=87.600.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>Wh\/14.000.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>.24.360=87.600.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\/14.24.360.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 P=0,724\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P=72,4%<\/b><\/span><\/span><\/span>
\n26-<\/b><\/span><\/span><\/span>\u00a0Comparando, em termos de porcentagem, 60lm\/W com 15lm\/W, voc\u00ea verifica que, ao substituir uma l\u00e2mpada de 15lm\/W por outra de 60lm\/W, o rendimento ser\u00e1 de 75%\u00a0 —\u00a0 como cada l\u00e2mpada tem 100W e s\u00e3o 10 l\u00e2mpadas, \u00a0a economia de pot\u00eancia ser\u00e1 P =10.0,75.100= 750 W, que fornece, em um m\u00eas (30 dias), uma economia de energia (W) que,funcionando 6h por dia\u00a0 —\u00a0 W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=750.30.6\u00a0 —\u00a0 W=135.000Wh\u00a0 —\u00a0 W=135kWh\u00a0 —\u00a0 economia=135.0,30\u00a0 —\u00a0<\/b><\/span><\/span><\/span>
\nC$=40,50\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>
\n27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=i.U=400.12\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=4.800W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span>
\n28-<\/b><\/span><\/span><\/span>\u00a0Gr\u00e1fico que fornece a quantidade de energia na forma de luz vis\u00edvel pode ser calculado como sendo aproximadamente a<\/b><\/span><\/span><\/span><\/p>\n

\"Energia<\/p>\n

\u00a0\u00e1rea do trap\u00e9zio da figura\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=(B + b).h\/2=(3,8 + 0,5).0,4\/2\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,86\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 \u00e1rea do gr\u00e1fico que<\/b><\/span><\/span><\/span><\/p>\n

\"Energia<\/p>\n

\u00a0fornece quantidade total de energia gasta para o funcionamento da l\u00e2mpada \u2013 \u00e1rea do tri\u00e2ngulo\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=b.h\/2=1,7.4\/2\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3,4\u00a0 —\u00a0 efici\u00eancia\u00a0 —\u00a0 e=S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,86\/3,4=0,252 x 100=25,2%\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/span>
\n29-<\/b><\/span><\/span><\/span>\u00a0Excluindo as l\u00e2mpadas, em cada c\u00f4modo da casa existe um aparelho el\u00e9trico citado na primeira tabela\u00a0 —\u00a0 soma de suas pot\u00eancias\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=120 + 3.000 + 500 + 200 + 200 + 50=4.070\u00a0 — \u00a0P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4.070W\u00a0 —\u00a0 cada l\u00e2mpada\u00a0 —\u00a0 na sala \u2013 \u00e1rea=3.3=9m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 100W\u00a0 —\u00a0 no banheiro \u2013 \u00e1rea=1,5 x 2,1=3,1m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 60W\u00a0 —\u00a0 no corredor \u2013 \u00e1rea=0,9 x 1,5=1,3m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 60W\u00a0 —\u00a0 no quarto \u2013 \u00e1rea=2,8 x 3=8,4m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 100W\u00a0 —\u00a0 pot\u00eancia das l\u00e2mpadas \u2013 P<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=100 + 60 + 60 + 100=320W\u00a0 —\u00a0 pot\u00eancia total instalada na casa\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=4.070 + 320\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=4.390W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span>
\n30-<\/b><\/span><\/span><\/span>\u00a0Energia el\u00e9trica consumida em um m\u00eas funcionando 10h por dia\u00a0 —\u00a0 W=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u0394t=180.30.10=54.000Wh=54kWh\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 54kWh \u2013 50%\u00a0 —\u00a0 x \u2013 100%\u00a0 —\u00a0 x=54.100\/25=216kWh\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span>
\n31-<\/b><\/span><\/span><\/span>\u00a0As l\u00e2mpadas fluorescentes produzem maior luminosidade utilizando a mesma pot\u00eancia e materiais com melhores caracter\u00edsticas condutoras reduzem a perda de energia\u00a0 por calor (efeito Joule)\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- E<\/b><\/span><\/span><\/span>
\n32-<\/b><\/span><\/span><\/span>\u00a0U=1,1V\u00a0 — \u00a0m=63,5\u03bcg=63,5.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>g\u00a0 —\u00a0 \u0394t=193s<\/b><\/span><\/span><\/span>
\n\"Energia\u00a0<\/span>
\nP<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=i.U=1,0.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.1,1=1,1.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>W=1,1mW\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span>
\n33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da corrente el\u00e9trica fornecida pela bateria\u00a0 —\u00a0 P=iU\u00a0 —\u00a0 60=i.12 \u00a0—\u00a0 i=5 A\u00a0 —\u00a0 a carga Q fornecida permanecer\u00e1 a mesma, fornecendo 5 A de corrente durante\u00a0 —\u00a0 i=Q\/\u0394t\u00a0 —\u00a0 5= 60\/\u0394t\u00a0 —\u00a0 \u0394t=60\/5\u00a0 —\u00a0 \u0394t=12h\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span>
\n34-<\/b><\/span><\/span><\/span> R- A\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span>
\n35-<\/b><\/span><\/span><\/span>
\n\"Energia
\n36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) P=iU\u00a0 —\u00a0 7.200=i.120\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>i=60 A<\/b><\/span><\/span><\/span>
\nb) W<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=4kW.2h + 6kW.2h + 2kW.2h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=24kWh<\/b><\/span><\/span><\/span>
\nc) W=24.30=720kWh\u00a0 —\u00a0 pre\u00e7o\u00a0 —\u00a0 p=720.0,5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0p=R$ 360,00<\/b><\/span><\/span><\/span>
\n37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Aplicando a 1\u00aa Lei de Ohm a esse resistor, temos\u00a0 —\u00a0 U = R I\u00a0 —\u00a0 120 = R (8)\u00a0 —\u00a0 R = 15 \u03a9\u00a0 —\u00a0 P=U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=150<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/15\u00a0 P=1.500W<\/b><\/span><\/span><\/span>
\nR- B<\/b><\/span><\/span><\/span>
\n38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: P = 4.500 W\u00a0 —\u00a0 \u00a0U = 110 V.<\/b><\/span><\/span><\/span>
\nP = i U\u00a0 —\u00a0 \u00a0i = P\/U=4.500\/110=40,9 A\u00a0 —\u00a0 portanto o disjuntor escolhido dever\u00e1 ser o de 45 A, que \u00e9 o valor mais pr\u00f3ximo do acima do calculado\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span>
\n39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo N a quantidade de dispositivos, P a pot\u00eancia el\u00e9trica e \u0394t, o tempo efetivo de uso di\u00e1rio, e C (R$ 0,50) o custo do kWh, o gasto di\u00e1rio dos dispositivos \u00e9\u00a0 —\u00a0 G = N x P x \u0394t x C\u00a0 —\u00a0 aplicando essa express\u00e3o, voc\u00ea obt\u00e9m a tabela<\/b><\/span><\/span><\/span><\/p>\n

\"Energia<\/p>\n

\u00a0 —\u00a0 gasto mensal (30 dias)\u00a0 —\u00a0 G<\/b><\/span><\/span><\/span>mensal<\/b><\/span><\/span><\/sub><\/span>\u00a0= 30.( 12,00 + 2,10 + 3,00) = 30 (17,10)\u00a0 —\u00a0 G<\/b><\/span><\/span><\/span>mensal<\/b><\/span><\/span><\/sub><\/span>\u00a0= R$ 513,00\u00a0 —<\/b><\/span><\/span><\/span>
\nR- B<\/b><\/span><\/span><\/span>
\n40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 — \u00a0V = 120 L\u00a0 —\u00a0 \u00a0m = 120 kg\u00a0 —\u00a0 \u0394T = 30\u00b0C\u00a0 —\u00a0 c = 1 cal.g<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.\u00b0C<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0= 4.200 J.kg<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.\u00b0C<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 quantidade de calor gasta diariamente\u00a0 —\u00a0 Q = m \u00adc \u0394T = 120 x 4.200 x 30 = 15,12.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0 equival\u00eancia entre quilowatt hora e joule\u00a0 —\u00a0<\/b><\/span><\/span><\/span>
\n\u00a01 kWh = (10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0W) x (3.600 s) = 3,6 x 106<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>W.s =\u00a0 3,6 x 10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0J\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 3,6.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J \u2013 1 kWh\u00a0 —\u00a0 15,12.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>J \u2013 Q\u00a0 —\u00a0 Q=15,12.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\/3,6.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Q=4,2kWh\u00a0 —\u00a0 gasto total com esse dispositivo em 30 dias\u00a0 — \u00a0G<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>\u00a0= 30 x 4,2 x 0,50\u00a0 —\u00a0\u00a0\u00a0 G<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>\u00a0= R$ 63,00\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>
\n41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=i.U=2.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.5.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=10.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0,1W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span>
\n42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394t=45\/60hx30\u00a0 —\u00a0 \u0394t=22,5h\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u0394t\u00a0 —\u00a0 5.000=W\/22,5\u00a0 —\u00a0 W=11.250Wh=11,25kWh\u00a0 —\u00a0 gasto\u00a0 —\u00a0 g=11,25×0,20=22,40<\/b><\/span><\/span><\/span>
\nR- C<\/b><\/span><\/span><\/span>
\n43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Recebe – P<\/b><\/span><\/span><\/span>or<\/b><\/span><\/span><\/sub><\/span>=50J\/s\u00a0 —\u00a0 libera \u2013 P<\/b><\/span><\/span><\/span>ol<\/b><\/span><\/span><\/sub><\/span>=20J\/s\u00a0 —\u00a0 ganha continuamente P<\/b><\/span><\/span><\/span>og<\/b><\/span><\/span><\/sub><\/span>=(50 \u2013 20)=30J\/s\u00a0 —\u00a0 ao atingira 1.000J ele derrete e para que isso ocorra ele deve receber W=(1.000 \u2013 550)=450J de energia\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/t\u00a0 —\u00a0 30=450\/t\u00a0 —\u00a0 t=15s\u00a0<\/b><\/span><\/span><\/span>
\nR- D<\/b><\/span><\/span><\/span>
\n44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\/\u0394t\u00a0 —\u00a0 5.000=W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u0394t\u00a0 —\u00a0 \u0394t=W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/5.000 (I)\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u0394t\u00a0 —\u00a0 4.000=W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u0394t\u00a0 —\u00a0 \u0394t=W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/4.000 (II)\u00a0 —\u00a0 igualando I com II\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/5.000=W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/4.000\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,8W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 o novo chuveiro utiliza 80% da pot\u00eancia do anterior e economiza 20%\u00a0 —\u00a0 economizar\u00e1\u00a0 —\u00a0 0,2×40=8,00\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>
\n45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O rel\u00f3gio indicou um consumo de 2.563 kwh no m\u00eas anterior e de 2.783 kwh na \u00faltima leitura\u00a0 —\u00a0 a diferen\u00e7a entre essas duas medidas determina um consumo de 220 kwh no per\u00edodo de um m\u00eas\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 kwh – R$ 0,20\u00a0 —\u00a0 220 kwh \u2013 C\u00a0 —\u00a0 C=0,20×220\u00a0 —\u00a0 C= R$ 44,00\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span>
\n46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A tens\u00e3o alternada produzida pelos geradores na usina hidrel\u00e9trica \u00e9 relativamente baixa. Assim, para que se possa abastecer diferentes centros utilizando linhas de transmiss\u00e3o, essa tens\u00e3o \u00e9 aumentada at\u00e9 centenas ou milhares de kV por meio de transformadores. Ao atingir os centros de consumo, a tens\u00e3o \u00e9 reduzida, por exemplo, a algumas dezenas de kV, pelos transformadores das subesta\u00e7\u00f5es e distribu\u00edda para o p\u00fablico.\u00a0 A tens\u00e3o de transmiss\u00e3o \u00e9 elevada e a corrente diminu\u00edda porque assim se podem utilizar fios condutores mais finos (mais leves, mais baratos) j\u00e1 que, para que a pot\u00eancia seja a mesma (P=Ui) se voc\u00ea aumentar a tens\u00e3o (U) voc\u00ea deve diminuir a corrente i.<\/b><\/span><\/span><\/span>
\nR- E<\/b><\/span><\/span><\/span>
\n47<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=U.i=\u03b5.i=12.100=1.300W\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u0394t\u00a0 —\u00a0 1.200=W\/60\u00a0 —\u00a0 W=1.200×60=7.200W\u00a0 —\u00a0 W=7,2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>W\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span>
\n48<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>
\n\"Energia
\nR- A<\/b><\/span><\/span><\/span>
\n\u00a0<\/span>
\n49-(UFPA-PA)<\/b><\/span><\/span><\/span>
\nPo=W\/\u2206t\u00a0 —\u00a0 W=Po.\u2206t\u00a0 —\u00a0 l\u00e2mpadas\u00a0 —\u00a0 W1=0,1.10=1kWh\u00a0 —\u00a0 televis\u00e3o\u00a0 —\u00a0 W2=0,1.8=0,8kWh\u00a0 —\u00a0 geladeira\u00a0 —\u00a0 W3=0,3.24=7,2kWh\u00a0 —\u00a0 ventilador\u00a0 —\u00a0 W4=0,125.8=1,0kWh\u00a0 —\u00a0 Wtotal=1,0 + 0,8 + 7,2 + 1,0=10,0kWh\u00a0 —\u00a0 1 painel \u2013 0,5kWh\u00a0 —\u00a0 n pain\u00e9is \u2013 10,0kWh\u00a0 —\u00a0 N=20 pain\u00e9is\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span>
\n50-(<\/b><\/span><\/span><\/span>ENEM-MEC)\u00a0\u00a0<\/b><\/span><\/span><\/span>
\n\u00a0<\/span>–\u00a0<\/b><\/span><\/span><\/span>Os biocombustiveis de primeira gera\u00e7\u00e3o s\u00e3o produzidos a partir da fermenta\u00e7\u00e3o alco\u00f3lica dos a\u00e7\u00facares vegetais da biomassa que contenha amido ou sacarose, como por exemplo o milho, o trigo, o sorgo, o amendoim, a soja, a beterraba, a batata doce, o girassol e a cana-de-a\u00e7\u00facar, n\u00e3o utilizando a celulose —\u00a0\u00a0 por\u00e9m, para evitar que se atinja o limite da oferta ou venha a ocorrer a competi\u00e7\u00e3o pelo uso da terra para a produ\u00e7\u00e3o de biocombust\u00edveis e de alimentos, \u00e9 necess\u00e1rio investir no desenvolvimento de tecnologias de segunda gera\u00e7\u00e3o para produ\u00e7\u00e3o de etanol com o aproveitamento eficiente da celulose\u00a0 —\u00a0 \u00a0estimativa \u00e9 de que o aproveitamento do baga\u00e7o e parte das palhas e pontas da cana-de-a\u00e7\u00facar eleve a produ\u00e7\u00e3o de \u00e1lcool em 30% a 40%, para uma mesma \u00e1rea plantada\u00a0 — \u00a0demais mat\u00e9rias-primas para as quais se buscam tecnologias de processamento da celulose, tais como capim-elefante, braqui\u00e1rias, panicuns e \u00e1rvores de crescimento r\u00e1pido podem representar alternativas competitivas e eficientes para locais onde n\u00e3o se cultiva ou cultivar\u00e1 cana-de-a\u00e7\u00facar, podendo gerar novos empregos\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A.<\/b><\/span><\/span><\/span>
\n\u00a0<\/span><\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a>\u00a0<\/span><\/h3>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

\u00a0Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Pot\u00eancia el\u00e9trica \u2013 Energia el\u00e9trica 01-\u00a0a) A l\u00e2mpada de menor pot\u00eancia \u2013 40W b) O de maior pot\u00eancia \u2013 caminh\u00e3o c) L\u00e2mpada de 40W, l\u00e2mpada de 100W, ferro el\u00e9trico, fusca e caminh\u00e3o 02-\u00a0Quanto maior o n\u00famero de equipamentos el\u00e9tricos, maior o consumo de energia (III)\u00a0 —\u00a0 observe na express\u00e3o W=Po.\u0394t que o consumo<\/p>\n","protected":false},"author":2,"featured_media":0,"parent":1711,"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-6480","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/6480","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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=6480"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/6480\/revisions"}],"predecessor-version":[{"id":10544,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/6480\/revisions\/10544"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1711"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=6480"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}