{"id":3929,"date":"2018-04-30T18:56:30","date_gmt":"2018-04-30T18:56:30","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3929"},"modified":"2024-09-04T12:33:55","modified_gmt":"2024-09-04T12:33:55","slug":"resolucao-comentada-das-questoes-de-fisica-da-faculdade-pernambucana-de-saude-fps-pe-2018","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2018\/faculdade-pernambucana-de-saude-fps-2018\/resolucao-comentada-das-questoes-de-fisica-da-faculdade-pernambucana-de-saude-fps-pe-2018\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da Faculdade Pernambucana de Sa\u00fade \u2013 FPS \u2013 PE \u2013 2018"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da<\/span><\/span><\/span><\/p>\n

Faculdade Pernambucana de Sa\u00fade \u2013 FPS \u2013 PE \u2013 2018<\/b><\/span><\/span><\/span><\/p>\n

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

Colocando a <\/b><\/span><\/span><\/span>origem da trajet\u00f3ria no posto rodovi\u00e1rio, orientando-a para a direita <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>deduzindo<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o hor\u00e1ria <\/b><\/span><\/span><\/span>do <\/b><\/span><\/span><\/span>autom\u00f3vel <\/b><\/span><\/span><\/span>e da <\/b><\/span><\/span><\/span>moto.<\/b><\/span><\/span><\/span><\/p>\n

Autom\u00f3vel \"\"<\/b><\/span><\/span><\/span> est\u00e1 em <\/b><\/span><\/span><\/span>movimento uniforme <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>velocidade constante de V<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> <\/b>= 108 km\/h\/3,6 <\/b><\/span><\/span><\/span>= 30 m\/s, passando por S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 \"\"<\/b><\/span><\/span><\/span> equa\u00e7\u00e3o do MU \"\" <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + V.t <\/b><\/span><\/span><\/span>= 0 + 30t \"\" <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = 30t.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

Fun\u00e7\u00e3o hor\u00e1ria do MCU<\/b><\/span><\/span><\/span><\/p>\n

Num\u00a0<\/b><\/span><\/span><\/span>MCU\u00a0a\u00a0velocidade angular\u00a0(W)\u00a0e a escalar\u00a0(V)\u00a0<\/b><\/span><\/span><\/span>s\u00e3o\u00a0tem sempre o <\/b><\/span><\/span><\/span>mesmo m\u00f3dulo\u00a0em\u00a0qualquer intervalo de\u00a0tempo.<\/b><\/span><\/span><\/span><\/p>\n

Dividindo\u00a0todos\u00a0os termos da\u00a0equa\u00e7\u00e3o hor\u00e1ria do movimento uniforme\u00a0(S = So\u00a0+ V.t)\u00a0pelo\u00a0raio R\u00a0da trajet\u00f3ria\u00a0\"\"\u00a0S\/R= S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/R + Vt\/R\u00a0\u00a0\"\"\u00a0 S\/R =<\/b><\/span><\/span><\/span> \u03c6<\/b><\/span><\/span><\/span>\u00a0\u00a0\"\"S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/R = <\/b><\/span><\/span><\/span>\u03c6<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0\"\"\u00a0 V\/R=W\u00a0\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u00a0\u03c6 = \u03c6<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W.t.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

Energia el\u00e9trica W<\/b><\/span><\/span><\/span> consumida por <\/b><\/span><\/span><\/span>cada aparelho, por dia:<\/b><\/span><\/span><\/span><\/p>\n

Geladeira \"\" W<\/b><\/span><\/span><\/span>g<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>g<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>W<\/b><\/span><\/span><\/span>g<\/b><\/span><\/span><\/sub><\/span> = 0,25.\"\"<\/b><\/span><\/span><\/span> <\/b>= 6,0 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Computador \"\" W<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 0,50.\"\"<\/b><\/span><\/span><\/span> = <\/b><\/span><\/span><\/span>1,0 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Televis\u00e3o \"\" W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span> = 0,10.\"\"<\/b><\/span><\/span><\/span> <\/b>= 0,5 Wh .<\/b><\/span><\/span><\/span><\/p>\n

Ar condicionado \"\" W<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>W<\/b><\/span><\/span><\/span>g<\/b><\/span><\/span><\/sub><\/span> = 3,00.\"\"<\/b><\/span><\/span><\/span> <\/b>= 15 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Chuveiro \"\" W<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>W<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 4,00.\"\"<\/b><\/span><\/span><\/span> <\/b>= 2,0 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Forno \"\" 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> \"\"<\/b><\/span><\/span><\/span> <\/b>W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 1,50.1\/3<\/b><\/span><\/span><\/span> <\/b>= 0,5 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Energia total consumida por dia \"\" W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span> = 25 kWh<\/b><\/span><\/span><\/span><\/p>\n

Como <\/b><\/span><\/span><\/span>cada painel <\/b><\/span><\/span><\/span>de 1,0 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> de \u00e1rea <\/b><\/span><\/span><\/span>gera, em m\u00e9dia, 0,5 kWh por dia, <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>n\u00famero m\u00ednimo <\/b><\/span><\/span><\/span>de pain\u00e9is ser\u00e1 25\/0,5 <\/b><\/span><\/span><\/span>= 50 pain\u00e9is.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

No trecho de <\/b><\/span><\/span><\/span>A \"\" B a transforma\u00e7\u00e3o \u00e9 isob\u00e1rica (press\u00e3o constante) de P = 2,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> com o volume diminuindo de \"\"<\/b><\/span><\/span><\/span> <\/b>= V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> <\/b>= (2,0 \u2013 5,0) <\/b><\/span><\/span><\/span>= – 3,0 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> com o trabalho W<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = P.\"\"<\/b><\/span><\/span><\/span> = <\/b><\/span><\/span><\/span>2,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.(- 3,0) \"\"<\/b><\/span><\/span><\/span> W<\/b><\/span><\/span><\/span>AB <\/b><\/span><\/span><\/sub><\/span>= – 6,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> J.<\/b><\/span><\/span><\/span><\/p>\n

No trecho de <\/b><\/span><\/span><\/span>B \"\" C a transforma\u00e7\u00e3o \u00e9 uma isoc\u00f3rica (isom\u00e9trica ou isovolum\u00e9trica) que ocorre a volume constante e, como o volume n\u00e3o varia, \"\"<\/b><\/span><\/span><\/span> <\/b>= 0 e W<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/sub><\/span> = P.\"\"<\/b><\/span><\/span><\/span> = P.0 = 0.<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>trabalho total pedido, <\/b><\/span><\/span><\/span>nos dois trechos, \u00e9 a <\/b><\/span><\/span><\/span>soma dos trabalhos parciais,<\/b><\/span><\/span><\/span> ou seja, <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span> <\/b>= 0 + (- 6,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> J) <\/b><\/span><\/span><\/span>= – 6,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> J.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

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

Lei de Snell-Descartes\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

Veja na <\/b><\/span><\/span><\/span>figura abaixo <\/b><\/span><\/span><\/span>a sequ\u00eancia para o <\/b><\/span><\/span><\/span>c\u00e1lculo do resistor equivalente:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 6 = \"\" \"\" i = \"\" \"\"<\/b><\/span><\/span><\/span> i = 2 A.<\/b><\/span><\/span><\/span><\/p>\n

Como os <\/b><\/span><\/span><\/span>resistores est\u00e3o em s\u00e9rie <\/b><\/span><\/span><\/span>a corrente el\u00e9trica <\/b><\/span><\/span><\/span>i = 2 A \u00e9 a mesma em cada um deles.<\/b><\/span><\/span><\/span><\/p>\n

Pot\u00eancia dissipada pelo resistor R = 5,0 \u2126 <\/b><\/span><\/span><\/span>percorrido por <\/b><\/span><\/span><\/span>i = 2 A<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>P = R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>= 5.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>P = 20 W. <\/b><\/span><\/span><\/span><\/p>\n

R- A<\/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 Faculdade Pernambucana de Sa\u00fade \u2013 FPS \u2013 PE \u2013 2018 01- Colocando a origem da trajet\u00f3ria no posto rodovi\u00e1rio, orientando-a para a direita e deduzindo a equa\u00e7\u00e3o hor\u00e1ria do autom\u00f3vel e da moto. Autom\u00f3vel est\u00e1 em movimento uniforme com velocidade constante de Va = 108 km\/h\/3,6 = 30 m\/s, passando por So =<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3927,"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-3929","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3929","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=3929"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3929\/revisions"}],"predecessor-version":[{"id":3930,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3929\/revisions\/3930"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3927"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3929"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}