{"id":3708,"date":"2018-02-13T17:01:15","date_gmt":"2018-02-13T17:01:15","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3708"},"modified":"2024-08-20T23:28:36","modified_gmt":"2024-08-20T23:28:36","slug":"resolucao-comentada-hidrostatica-experiencia-de-torricelli-principio-de-pascal","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/hidrostatica-experiencia-de-torricelli-principio-de-pascal\/resolucao-comentada-hidrostatica-experiencia-de-torricelli-principio-de-pascal\/","title":{"rendered":"Experi\u00eancia de Torricelli \u2013 Princ\u00edpio de Pascal – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada <\/b><\/span><\/span><\/span><\/p>\n

Hidrost\u00e1tica<\/b><\/span><\/span><\/span><\/p>\n

Experi\u00eancia de Torricelli \u2013 Princ\u00edpio de Pascal<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\u00a0<\/span>01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe atentamente a figura que representa a experi\u00eancia de Torricelli realizada ao n\u00edvel do mar e leia com aten\u00e7\u00e3o o texto que a segue:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

–\u00a0\u00a0 Torricelli concluiu que a atmosfera exerce press\u00e3o sobre a superf\u00edcie do merc\u00fario dentro do recipiente e como o merc\u00fario est\u00e1 em equil\u00edbrio, a press\u00e3o no tubo ao n\u00edvel P, (figura acima), \u00e9 igual \u00e0 press\u00e3o atmosf\u00e9rica no n\u00edvel Q da superf\u00edcie do l\u00edquido contido no recipiente.. Como na parte superior do tubo n\u00e3o h\u00e1 ar, a press\u00e3o no ponto P (press\u00e3o devida \u00e0 coluna de merc\u00fario dentro do tubo) \u00e9 igual \u00e0 press\u00e3o no ponto Q (press\u00e3o atmosf\u00e9rica ao n\u00edvel do mar). Portanto, a press\u00e3o atmosf\u00e9rica ao n\u00edvel do ar corresponde \u00e0 press\u00e3o que mant\u00e9m uma coluna de merc\u00fario a uma altura de 76cm, ou 760mm ou ainda 0,76m. Lembrando que g=9,8m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0e que d<\/b><\/span><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub><\/span>=13.600kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>, tem-se\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub><\/span>.g.h\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>=13.600.9,8.0,76\u00a0 —\u00a0\u00a0P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>=1,01N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(Pa)<\/b><\/span><\/span><\/span><\/p>\n

I. Falsa\u00a0 —\u00a0 como a press\u00e3o atmosf\u00e9rica \u00e9 devido ao peso da coluna de ar, quanto maior a altitude <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

menor ser\u00e1 esse peso e,consequentemente, menor ser\u00e1 a altura da coluna que o equilibrar\u00e1\u00a0 —\u00a0 a press\u00e3o atmosf\u00e9rica diminui com a altitude.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta\u00a0 — \u00a0Se a experi\u00eancia de Torricelli for realizada ao n\u00edvel do mar, mas com \u00e1gua, cuja densidade \u00e9 13,6 vezes menor que a do merc\u00fario, a altura da coluna de \u00e1gua ser\u00e1 aproximadamente igual a 10,3m, pois\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.g .h<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub><\/span>= P<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0 —\u00a0 13,6d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.g .h<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 13,6.h<\/b><\/span><\/span><\/span>merc\u00fario<\/b><\/span><\/span><\/sub><\/span>=h<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 Torricelli fixou uma escala vertical graduada no tubo de merc\u00fario utilizado nesta experi\u00eancia, obtendo desta forma o<\/b><\/span><\/span><\/span><\/p>\n

aparelho de medi\u00e7\u00e3o da press\u00e3o atmosf\u00e9rica, o bar\u00f3metro de merc\u00fario\u00a0 —\u00a0 assim, bar\u00f4metro \u00e9 um <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

aparelho que mede press\u00e3o atmosf\u00e9rica,mas tamb\u00e9m serve para medir a altitude de um lugar, pois P=d.g.h \u00e9 uma fun\u00e7\u00e3o tamb\u00e9m da altura h.<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00c0 medida que os sacos sobem sobe seu volume vai aumentando porque a diferen\u00e7a entre a press\u00e3o do ar dentro dos sacos e a press\u00e3o cada vez mais reduzida fora dela gera uma for\u00e7a resultante que empurra o pl\u00e1stico do saco de salgadinhos para fora\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quanto maior a altitude,\u00a0menor\u00a0\u00e9 a press\u00e3o e quanto menor a press\u00e3o\u00a0menor\u00a0\u00e9 a temperatura de ebuli\u00e7\u00e3o da \u00e1gua\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>a \u00e1gua, em press\u00f5es muito baixas, ferve \u00e0 temperaturas bem inferiores \u00e0 100\u00a0\u00b0C\u00a0 —\u00a0 no princ\u00edpio do funcionamento de uma panela de press\u00e3o, h\u00e1 o aumento da press\u00e3o atmosf\u00e9rica dentro da panela e isso faz com que o seu ponto de ebuli\u00e7\u00e3o seja maior , podendo atingir temperaturas de at\u00e9 120\u00b0 da \u00e1gua e haja um cozimento mais r\u00e1pido do alimento.<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Qualquer que seja a natureza ou densidade do l\u00edquido, a superf\u00edcie livre do mesmo fica a uma mesma altura h, nos dois compartimentos do bule (corpo principal e bico), em rela\u00e7\u00e3o \u00e0 linha de refer\u00eancia do sistema\u00a0 —\u00a0 isso ocorre porque a press\u00e3o exercida na\u00a0<\/b><\/span><\/span><\/span>superf\u00edcie<\/b><\/span><\/span><\/span>\u00a0dos dois ramos do sistema de vasos comunicantes \u00e9 a\u00a0<\/b><\/span><\/span><\/span>press\u00e3o atmosf\u00e9rica\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que os pontos M e N da figura abaixo est\u00e3o numa mesma horizontal, suportando assim <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a mesma press\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+\u03c1<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g.h = P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+\u00a0\u03c1<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g.0,75h\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=0,75\u03c1<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/\u03c1<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=0,75\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 h=10cm=0,1m\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=0,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 sendo a se\u00e7\u00e3o reta circular (S=1cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=1,0.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) no diafragma a mesma nos dois lados, e como a densidade do l\u00edquido A \u00e9 maior que a do l\u00edquido B, a intensidade da for\u00e7a resultante sobre o diafragma vale\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/S\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.S\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.S\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.S\u00a0\u2013 P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.S=S(P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=S.(d<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g.h \u2013 d<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g.h)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=1,0.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>.(5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.0,1 \u2013 0,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.0,1)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=1.0.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>.4,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=4,5.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=4.500.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

07-\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo que os pontos M e N na mesma horizontal suportam a mesma press\u00e3o\u00a0 –<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span>\u00a0+d.g.h\u00a0 —P<\/b><\/span><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub><\/span>=76cmHg + 30cmHg\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>g\u00e1s<\/b><\/span><\/span><\/sub><\/span>=106cmHg\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Princ\u00edpio enunciado por Pascal, f\u00edsico e matem\u00e1tico franc\u00eas (1623 \u2013 1663), conhecido como\u00a0princ\u00edpio de Pascal: \u201c O acr\u00e9scimo de press\u00e3o exercido num ponto de um l\u00edquido ideal em equil\u00edbrio \u00e9 transmitida integralmente a todos os pontos desse l\u00edquido e tamb\u00e9m \u00e0s paredes do recipiente onde est\u00e1 contido\u201d\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Princ\u00edpio de Pascal: \u201c O acr\u00e9scimo de press\u00e3o exercido num ponto de um l\u00edquido ideal em equil\u00edbrio \u00e9 transmitida integralmente a todos os pontos desse l\u00edquido e tamb\u00e9m \u00e0s paredes do recipiente onde est\u00e1 contido\u201d\u00a0 —\u00a0 ent\u00e3o, a press\u00e3o em cada \u00eambolo a mesma\u00a0 —\u00a0F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Princ\u00edpio de Pascal\u00a0 — \u00a0F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u03c0R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u03c0R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 2\/\u03c0(5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u03c0(1.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\/25.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,08.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=8N \u00a0—\u00a0 essa for\u00e7a equilibra uma massa\u00a0 —\u00a0 p=Mg\u00a0 —\u00a0 8=M.10\u00a0 —\u00a0 M=0,8kg=800g\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pelo princ\u00edpio de Pascal\u00a0 — P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 — \u00a0F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/\u03c0D<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/\u03c0(4D)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/1=F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/16\u00a0 —\u00a0\u00a0F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=16\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 4.000 N\u00a0 —\u00a0 A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10\u03c0 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0,05\u03c0 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u00a0L = 1 m\u00a0 —\u00a0 \u00a0m = 1 g = 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

quando voc\u00ea aplica a for\u00e7a\u00a0\"\"\u00a0sobre o pist\u00e3o de \u00e1rea A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, atrav\u00e9s do ar comprimido, transmite-se sobre o chumbinho de \u00e1rea A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, a for\u00e7a\u00a0\"\"\u00a0\u00a0\u00a0—\u00a0 a rela\u00e7\u00e3o entre essas grandezas \u00e9 dada pelo princ\u00edpio de Pascal\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 4.000\/10=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/0,05\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=20N\u00a0 —\u00a0 essa for\u00e7a\u00a0\"\"\u00a0, tamb\u00e9m suposta constante, realiza trabalho sobre o chumbinho ao longo do cano da espingarda, aumentando sua velocidade a partir do repouso (v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0)\u00a0 —\u00a0 pelo teorema da energia cin\u00e9tica, considerando\u00a0\"\"\u00a0\u00a0como a for\u00e7a resultante sobre o chumbinho\u00a0 —\u00a0 WF<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=\u0394E<\/b><\/span><\/span><\/span>cin\u00e9tica<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.d.cos\u03b1= E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.L.1=mV<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 – 0\u00a0 —\u00a0 20.1=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=\u221a(40 000)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=200m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span>Dados: V = 1.200 L; h = 30 m; L = 200 m\u00a0 —\u00a0 seguindo as instru\u00e7\u00f5es do fabricante, entremos com os dados na tabela para obtermos o valor de H.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como mostrado, obtemos H = 45 m\u00a0 —\u00a0 analisando o gr\u00e1fico dado, temos os valores mostrados: H\u00a0 = 45 m \u00a0—\u00a0 \u00a0Q = 900 L\/h.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Calculando o tempo para encher o reservat\u00f3rio\u00a0 —\u00a0 Q=V\/t\u00a0 —\u00a0 900=1.200\/t\u00a0 —\u00a0 t=1.200\/900\u00a0 —\u00a0 t=4\/3h\u00a0 —\u00a0 t=80min\u00a0 —\u00a0 t=1h e 20min\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R-\u00a0<\/b><\/span><\/span><\/span>E<\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o Comentada Hidrost\u00e1tica Experi\u00eancia de Torricelli \u2013 Princ\u00edpio de Pascal \u00a0 \u00a001-\u00a0Observe atentamente a figura que representa a experi\u00eancia de Torricelli realizada ao n\u00edvel do mar e leia com aten\u00e7\u00e3o o texto que a segue: –\u00a0\u00a0 Torricelli concluiu que a atmosfera exerce press\u00e3o sobre a superf\u00edcie do merc\u00fario dentro do recipiente e como o merc\u00fario est\u00e1 em equil\u00edbrio, a press\u00e3o<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3706,"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-3708","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3708","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=3708"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3708\/revisions"}],"predecessor-version":[{"id":10675,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3708\/revisions\/10675"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3706"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3708"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}