{"id":1544,"date":"2015-11-16T15:38:08","date_gmt":"2015-11-16T15:38:08","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1544"},"modified":"2024-08-23T14:35:08","modified_gmt":"2024-08-23T14:35:08","slug":"resolucao-comentada-das-questoes-de-vestibulares-sobre-teorema-de-stevin","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/hidrostatica\/teorema-de-stevin-pressao-hidrostatica-vasos-comunicantes\/resolucao-comentada-das-questoes-de-vestibulares-sobre-teorema-de-stevin\/","title":{"rendered":"Teorema de Stevin – Resolu\u00e7\u00e3o"},"content":{"rendered":"

\u00a0Resolu\u00e7\u00e3o comentada das quest\u00f5es de vestibulares sobre<\/b><\/span><\/span><\/span><\/b><\/p>\n

Teorema de Stevin \u2013 Press\u00e3o Hidrost\u00e1tica \u2013 Vasos Comunicantes<\/span><\/span><\/span><\/p>\n

01<\/b><\/span><\/span>– A press\u00e3o hidrost\u00e1tica \u00e9 fornecida por P<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>caf\u00e9.<\/b><\/span><\/span><\/sub>g.h e observe que ela n\u00e3o depende das dimens\u00f5es do tubo e da cafeteira e, como os pontos A e B est\u00e3o abertos, a press\u00e3o nos dois \u00e9 a mesma e ficam na mesma altura\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

02<\/b><\/span><\/span>– \u0394P=P<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=1,02 \u2013 1,00\u00a0 —\u00a0 \u0394P=0,2atm=0,2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394P=dgh\u00a0 —\u00a0 2.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h\u00a0 —\u00a0 h=2m\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 5m \u2013 0,5atm\u00a0 —\u00a0 2m – xatm\u00a0 —\u00a0 5x=2.0,5\u00a0 —\u00a0 x=1\/5=0,2m=20cm\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span>\u00a0P=d.g.h\u00a0 —\u00a0 0,05.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.h\u00a0 —\u00a0 h=0,5m\u00a0 —<\/b><\/span><\/span>\u00a0h=50cm<\/b><\/span><\/span><\/p>\n

04<\/b><\/span><\/span>–\u00a0<\/b><\/span><\/span>R- D\u00a0<\/b><\/span><\/span>\u00a0(veja teoria)<\/b><\/span><\/span><\/p>\n

05<\/b><\/span><\/span>– A press\u00e3o no fundo do recipiente varia linearmente com a altura e independe da \u00e1rea de se\u00e7\u00e3o reta de cada cilindro\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0Como as bases tem a mesma \u00e1rea (S), o l\u00edquido \u00e9 o mesmo (\u00e1gua) e a as alturas das colunas l\u00edquidas s\u00e3o as mesmas, a press\u00e3o (P) no fundo dos recipientes \u00e9 a mesma\u00a0 —\u00a0 P=F\/S\u00a0 —\u00a0 portanto os valores das for\u00e7as exercidas pela \u00e1gua nas bases dos recipientes \u00e9 a mesma (\"\")\u00a0 —\u00a0 as for\u00e7as exercidas pelas balan\u00e7as sobre as bases correspondem ao peso dos volumes dos l\u00edquidos contidos em cada recipiente\u00a0 —\u00a0 maior volume, maior peso (\"\")\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0Quando voc\u00ea abre a v\u00e1lvula,a \u00e1gua flui da caixa da esquerda para a da direita at\u00e9 que elas tenham a mesma altura\u00a0 —\u00a0 a altura da caixa da esquerda diminui e consequentemente a press\u00e3o na base da mesma tamb\u00e9m diminui\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

08<\/b><\/span><\/span>–\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span>\u00a0 (veja teoria)<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span>\u00a0P=dgh\u00a0 —\u00a0 P=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.4\u00a0 —\u00a0 P=4.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=F\/S\u00a0 —\u00a0 4.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=F\/12\u00a0 —\u00a0 F=4.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\/12.10<\/b><\/span><\/span>-4\u00a0\u00a0<\/b><\/span><\/span><\/sup>—<\/b><\/span><\/span>–\u00a0 F=48N\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0a) Com a torneira aberta, o volume de \u00e1gua que preencheu o recipiente II \u00e9 o mesmo que escoou do recipiente I\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c0R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>.h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= \u03c0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>.0,5= (2R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/sup>\u00a0\u00a0—\u00a0<\/b><\/span><\/span>\u00a0h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,125m<\/b><\/span><\/span><\/p>\n

b) P=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d.g.h\u00a0 —\u00a0 P= 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P=1,1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

11-<\/b><\/span><\/span>\u00a0A cada 10m de profundidade que voc\u00ea desce sob a superf\u00edcie da \u00e1gua sua press\u00e3o devido \u00e0 coluna l\u00edquida aumenta de 1 atm\u00a0 — P<\/b><\/span><\/span>Valdirene<\/b><\/span><\/span><\/sub>= 1.000atm – P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>(1atm)\u00a0 —\u00a0 P<\/b><\/span><\/span>Valdirene<\/b><\/span><\/span><\/sub>=999atm\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1atm \u2013 10m\u00a0 —\u00a0 999atm – x\u00a0 —<\/b><\/span><\/span><\/p>\n

x=999 x 10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x=h=9.990m<\/b><\/span><\/span><\/p>\n

12<\/b><\/span><\/span>– a) P= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ dgh\u00a0 —\u00a0 4.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0= 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h\u00a0 —\u00a0 h=3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0h=30m<\/b><\/span><\/span><\/p>\n

b) Para sofrer na subida vertical uma varia\u00e7\u00e3o de press\u00e3o \u0394P=10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, em \u0394t=1s, ele deve se deslocar\u00a0 —\u00a0 \u0394P=d.g.h\u00a0 —\u00a0 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h\u00a0 —\u00a0 h=1m\u00a0 —\u00a0 V= \u0394h\/\u0394t\u00a0 —\u00a0 V=1\/1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=1m\/s<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span>\u00a0Press\u00e3o exercida pelo \u00eambolo na profundidade h\u00a0 —\u00a0 P=P<\/b><\/span><\/span>atm\u00a0<\/b><\/span><\/span><\/sub>+ F\/S \u00a0—\u00a0 P=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 6,28\/\u03c0.(R)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>atm +\u00a0<\/b><\/span><\/span><\/sub>6,28\/3,14.2\/2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+2.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Press\u00e3o exercida pelo l\u00edquido na profundidade h\u00a0 —\u00a0 P<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ dgh=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 1,25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h\u00a0 —\u00a0 P<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 1,25.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.h\u00a0 —\u00a0 P=P<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 2.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ 1,25.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.h\u00a0 —\u00a0 h=2\/1,25\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>h=1,6m<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>sangu\u00ednea<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>hidrost\u00e1tica<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>.g.h<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.0,5=13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.h<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 h<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>=0,0367mHg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0h<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>=36,7mmHg<\/b><\/span><\/span><\/p>\n

15<\/b><\/span><\/span>– P=d.g.h\u00a0 —\u00a0 9.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h\u00a0 —\u00a0 h=0,9m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

16-\u00a0<\/b><\/span><\/span>A\u00a0\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>press\u00e3o do ar aprisionado no barco \u00e0 profundidade h=(2,20 \u2013 1,70)=0,5m \u00e9 dada por P= P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d.g.h\u00a0 —\u00a0 P -P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=d.g.h=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.0,5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P \u2013 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=5,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

17-<\/b><\/span><\/span>\u00a0P=1atm=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=10m\u00a0 —\u00a0 P=dgh\u00a0 —\u00a0 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=d.10.10\u00a0 —\u00a0 d=1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>k\/m<\/b><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 d\u2019=0,8d\u00a0 —\u00a0 P\u2019=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P\u2019=d\u2019.<\/b><\/span><\/span>g.h\u2019\u00a0 —\u00a0 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=0,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h\u2019\u00a0 —\u00a0 h\u2019=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h\u2019=12,5m\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

18- P=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ dgh=1,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 1,00.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.(1,0 + 2,0)\u00a0 —\u00a0 P=1,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 3,00.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=1,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 0,30.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=1,31.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

19- P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03c1.g.h\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1,2\u03c1.g.h\u2019\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u03c1.g.h\/1,2\u03c1.g.h\u2019\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 1= \u03c1.g.h\/1,2\u03c1.g.h\u2019\u00a0 —\u00a0 h\u2019=1\/1,2h10\/12h=5\/6h\u00a0 —\u00a0 h \u2013 5\/6h=(6h \u2013 5h)\/6=h\/6\u00a0 —\u00a0 h\u2019=h\/6\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>\u00a0 (veja teoria)<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span>\u00a0Do gr\u00e1fico\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>=0,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=5,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 quando h=15m \u2013 P=5,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 P=d.g.h\u00a0 —\u00a0 5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=d.10.15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d=0,033.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=3,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

 <\/p>\n

22-\u00a0<\/b><\/span><\/span>Como haver\u00e1 redu\u00e7\u00e3o na altura do n\u00edvel de l\u00edquido na caixa da esquerda, pela lei de Stevin, p =<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0d.g.h, haver\u00e1 menor press\u00e3o no fundo da caixa\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

23-\u00a0<\/b><\/span><\/span>Aumentando-se a altitude, diminui a espessura da camada atmosf\u00e9rica sobre a superf\u00edcie, diminuindo a press\u00e3o\u00a0 —\u00a0 Curitiba encontra-se na altitude de 945m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

24-\u00a0<\/b><\/span><\/span>Dados\u00a0 — \u00a0A = 2,4 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0h = 1,8 m\u00a0 — \u00a0d = 1,2 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 1.200 kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u00a0—\u00a0 \u00a0g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a for\u00e7a m\u00e1xima suportada correspode ao peso do l\u00edquido\u00a0 —\u00a0 F = P = m.g = d.V.g = d.A.h.g = 1.200.(2,4).(1,8).(10)\u00a0 —\u00a0 F = 51.840 N\u00a0 —\u00a0 50.100 N < F < 52.000 N\u00a0 —\u00a0<\/b><\/span><\/span>R- D\u00a0\u00a0<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

25-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 Z = 100 mL\/s = 0,1 L\/s\u00a0 —\u00a0 \u00a0d = 1 kg\/L = 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>col.de\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>Pa\u00a0 —\u00a0\u00a0 g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0pelo teorema de Stevin a press\u00e3o da coluna l\u00edquida \u00e9\u00a0 —\u00a0 P<\/b><\/span><\/span>col.de\u00e1gua<\/b><\/span><\/span><\/sub>=dgh\u00a0 —\u00a0 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h\u00a0 —\u00a0 h=0,1m\u00a0 —\u00a0 caixa c\u00fabica de volume 1m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0tem \u00e1rea da base A<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=1m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 c\u00e1lculo do volume derramado\u00a0 —\u00a0 V=A<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>.h=1.(0,1)\u00a0 —\u00a0 V=0,1m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=100L\u00a0 —\u00a0 a vaz\u00e3o (Z) \u00e9 a raz\u00e3o entre o volume derramado e o tempo\u00a0 —\u00a0 Z=V\/\u0394t=100\/0,1\u00a0 —\u00a0 \u0394t=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

26-\u00a0<\/b><\/span><\/span>De acordo com Teorema de Stevin, pontos de mesmo l\u00edquido em repouso que est\u00e3o na mesma horizontal est\u00e3o sob mesma press\u00e3o — ent\u00e3o, no ponto\u00a0<\/b><\/span><\/span>A<\/b><\/i><\/span><\/span>\u00a0<\/b><\/span><\/span>da horizontal que passa pela interface entre a \u00e1gua e o l\u00edquido mais denso, a press\u00e3o deve ser a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

mesma que no ponto\u00a0<\/b><\/span><\/span>B<\/b><\/i><\/span><\/span>\u00a0situado na \u00e1gua, na mesma horizontal\u00a0 —\u00a0 p<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0= p<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\"\"\u00a0\u00a0—\u00a0 como d<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>\u00a0> d<\/b><\/span><\/span>\u00e1g\u00a0 —\u00a0<\/b><\/span><\/span><\/sub>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

h<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>\u00a0< h<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

27-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 d<\/b><\/span><\/span>ag<\/b><\/span><\/span><\/sub>\u00a0= 1 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>;\u00a0 —\u00a0 d<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>\u00a0= 1,3 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 1,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 h<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>= 10 cm = 10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0m; h<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>\u00a0= 1 m\u00a0 —\u00a0\u00a0 g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 p<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0= 1,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0Pa\u00a0 —\u00a0 press\u00e3o da coluna de \u00e1gua\u00a0 —\u00a0 P<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>gh<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>=(10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>).10.(10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 P<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>=1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>Pa=0,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa\u00a0 —\u00a0 press\u00e3o da coluna de glicerina\u00a0 —\u00a0 P<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>gh<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>=(1,3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>).(10).(1)\u00a0 —\u00a0 P<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>=1,3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>Pa=0,13.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa\u00a0 —\u00a0 na superf\u00edcie que separa \u00e1gua-glicerina a press\u00e3o a press\u00e3o \u00e9 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>=1,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 0,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1,02.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa\u00a0 —\u00a0 no fundo do recipiente, a press\u00e3o (p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) \u00e9 vale\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm + P<\/b><\/span><\/span><\/sub>\u00e1gua + P<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 0,01.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 0,13.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1,15.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0Pa.\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

Obs: segundo o Sistema Internacional de Unidades o plural das unidades \u00e9 feito apenas acrescentando a letra s no final, quando n\u00e3o terminada em s. Caso seja terminada em s, n\u00e3o sofre flex\u00e3o, e quando grafadas por extenso, deve ser em letras min\u00fasculas. Exemplos: pascal – pascals; decibel – decibels; newton – newtons.<\/b><\/span><\/span><\/p>\n

28-\u00a0<\/b><\/span><\/span>Dados: m = 900 kg; A<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 2.500 cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 25.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0; A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= 25 cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 25.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>; d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>\u00a0= 900 kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>; h = 4m\u00a0 —\u00a0 de acordo com o teorema de Stevin, pontos de um mesmo l\u00edquido em repouso que est\u00e3o na mesma horizontal suportam a mesma press\u00e3o\u00a0 — a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

press\u00e3o no ponto (1) provocada pelo peso do carro \u00e9 igual \u00e0 press\u00e3o no ponto (2) provocada pela for\u00e7a\u00a0\"\", somada \u00e0 da coluna l\u00edquida\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mg\/A<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F\/A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>gh\u00a0 —\u00a0 9000\/25.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=F\/25.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0+ 900.10.4\u00a0 —\u00a0 36.000 \u2013 36.000=F\/25.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=0\u00a0 —<\/b><\/span><\/span><\/p>\n

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

29-\u00a0<\/b><\/span><\/span>\u00a076cm Hg ————– 10m de H2O—————–1,0atm—————– 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

X Hg——————-\u00a0 5 m de H2O ————–Yatm——————–Z N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

X=76×5\/10\u00a0 —\u00a0 X=38cmHg\u00a0 —\u00a0 Y=5×1\/10\u00a0 —\u00a0 Y=0,5atm\u00a0 —\u00a0 Z=0,5×10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/1\u00a0 —\u00a0 Z=0,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 apress\u00e3o no fundo \u00e9 a soma da press\u00e3o atmosf\u00e9rica com a press\u00e3o da coluna de \u00e1gua\u00a0 —\u00a0 P=1,0 + 0,5\u00a0 —\u00a0 1,5atm\u00a0 —\u00a0 P=76 + 38\u00a0 —\u00a0 P=114cmHg\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ 0,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=1,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Nm<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

30-\u00a0<\/b><\/span><\/span>A diferen\u00e7a de press\u00e3o hidrost\u00e1tica (\u0394P) entre dois pontos de desn\u00edvel h, para um l\u00edquido de densidade d<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>, \u00e9 dada pelo teorema de Stevin\u00a0 —\u00a0 \u0394P = d<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>.g.h\u00a0 —\u00a0 assim,, essa diferen\u00e7a s\u00f3 depende da densidade do l\u00edquido, do desn\u00edvel e da gravidade local\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

31- h<\/b><\/span><\/span>\u00a0= 2.000 m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>g<\/b><\/span><\/span>\u00a0= 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>;\u00a0<\/b><\/span><\/span>r<\/b><\/span><\/span>\u00a0= 0,9 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 9<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b410<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 teorema de Stevin\u00a0 —\u00a0 DP = r<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>g<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>h = 9<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b410<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 180<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0DP = 1,8<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00b4<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>\u00a0Pa\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

32- P<\/b><\/span><\/span>torneira<\/b><\/span><\/span><\/sub>\u00a0= P<\/b><\/span><\/span>\u00e1gua\u00a0<\/b><\/span><\/span><\/sub>+ P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F\/S = dgh + P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 80\/4.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0= 1.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10.h + 1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>h + 1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=10m<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

33- Observe na figura abaixo que a press\u00e3o no ponto B no interior do l\u00edquido \u00e9 fornecida por P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0=<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g.h\u00a0 —\u00a0\u00a0 essa express\u00e3o \u00e9 chamada de Teorema Fundamental da Hidrost\u00e1tica ou de Teorema de Stevin\u00a0 —\u00a0 uma das conseq\u00fc\u00eancias do teorema de Stevin \u00e9 de que todos os pontos de uma superf\u00edcie horizontal (a uma mesma altura h) suportam a mesma press\u00e3o, desde que o l\u00edquido seja o mesmo.<\/b><\/span><\/span><\/p>\n

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

34- Teorema de Stevin\u00a0 — observe a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0Teorema de Stevin\u00a0 —\u00a0 A press\u00e3o no ponto B devida apenas \u00e0 coluna l\u00edquida P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g.h \u00e9 chamada depress\u00e3o hidrost\u00e1tica\u00a0e P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g.h \u00e9 chamada de press\u00e3o total, press\u00e3o absoluta ou simplesmente press\u00e3o.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0A press\u00e3o exercida por uma coluna l\u00edquida n\u00e3o depende das dimens\u00f5es do recipiente que a cont\u00e9m, mas apenas da natureza do l\u00edquido, fornecida pela sua densidade (d), do local (g) e da altura da coluna (h).<\/b><\/span><\/span><\/p>\n

\"\"\u00a0O gr\u00e1fico da press\u00e3o total P em fun\u00e7\u00e3o da altura \u201cprofundidade\u201d h, (P=Patm+ d.g.h, que \u00e9 uma fun\u00e7\u00e3o do primeiro<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

grau)), \u00e9 uma reta inclinada.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0A press\u00e3o devido \u00e1 coluna l\u00edquida \u00e9 maior quanto maior for a profundidade, ou seja, a maior a altura da coluna l\u00edquida,<\/b><\/span><\/span><\/p>\n

Observe na teoria acima, que a press\u00e3o que a \u00e1gua deve exercer para acionar o sif\u00e3o depende apenas da altura da coluna l\u00edquida, n\u00e3o dependendo das dimens\u00f5es do recipiente, ou seja, do volume de \u00e1gua nele armazenada\u00a0 —\u00a0 assim, mesmo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

um pequeno volume de \u00e1gua armazenado, mas com altura suficiente pode acionar o sif\u00e3o\u00a0 —\u00a0 portanto, a economia deve-se ao volume de \u00e1gua armazenada no tanque\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span>.35-(EsPCEx-012)<\/b><\/span><\/span><\/p>\n

Teorema de Stevin\u00a0 —\u00a0 P = P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ d.g.h\u00a0 —\u00a0 2,2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0= 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ d.10.5\u00a0 —\u00a0 d=1,2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/50\u00a0 —\u00a0 d=0,024.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=2,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

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

36-<\/b><\/span><\/span> A) Press\u00e3o sobre o mergulhador que se encontra a uma profundidade de 30 m \u00e9 determinada a partir da express\u00e3o\u00a0 —\u00a0 P = P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ \u03bcgh\u00a0 – (Teorema Fundamental da Hidrost\u00e1tica ou de Teorema de Stevin), onde P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u00e9 a press\u00e3o atmosf\u00e9rica ao n\u00edvel do mar, \u03bc \u00e9 a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

densidade da \u00e1gua do mar 1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>, h \u00e9 a profundidade em que se encontra o mergulhador e g \u00e9 a acelera\u00e7\u00e3o da gravidade 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0—\u00a0 P=1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>+10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>x10x30\u00a0 —\u00a0 P=1.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>+3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=4.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0ou P=4 atm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a press\u00e3o hidrost\u00e1tica (devida somente \u00e0 coluna de \u00e1gua) \u00e9 P=3,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(Pa) ou P=3,0 atm e a press\u00e3o (absoluta, total) \u00e9 P=4,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>(Pa)ou P=4,0 atm.\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Observa\u00e7\u00e3o: \u00a0\u00e0 cada 10m de profundidade na \u00e1gua, a press\u00e3o devido \u00e0 coluna l\u00edquida aumenta de 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa ou de 1 atm\u00a0 —\u00a0 Exemplo: Se um peixe est\u00e1 a 20m de profundidade da superf\u00edcie l\u00edquida da \u00e1gua, ele suporta uma press\u00e3o \u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

hidrost\u00e1tica de P<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=2,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa ou 2atm. A press\u00e3o total deve ser acrescida da press\u00e3o atmosf\u00e9rica na superf\u00edcie\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa + 2.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa=3.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa ou de 3 atm.<\/b><\/span><\/span><\/p>\n

B) Trata-se de uma transforma\u00e7\u00e3o isot\u00e9rmica (temperatura constante)\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= P.V\/T\u00a0 —\u00a0 T= T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= P.V=constante (Lei de Boyle)\u00a0 —\u00a0 press\u00e3o na superf\u00edcie\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=1 atm\u00a0 —\u00a0 volume na superf\u00edcie\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00e0 uma profundidade onde a press\u00e3o \u00e9 P, o volume \u00e9 V=25% de V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0,25V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= P.V\u00a0 —\u00a0 1.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0,25V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.P\u00a0 —\u00a0 P=4 atm\u00a0 —\u00a0 observe que essa press\u00e3o \u00e9 exatamente igual \u00e0 press\u00e3o sobre o mergulhador a 30m de profundidade, ent\u00e3o ele n\u00e3o poder\u00e1 ultrapassar essa profundidade sem que o seu pulm\u00e3o possa vir a sofrer danos<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

\u00a0Resolu\u00e7\u00e3o comentada das quest\u00f5es de vestibulares sobre Teorema de Stevin \u2013 Press\u00e3o Hidrost\u00e1tica \u2013 Vasos Comunicantes 01– A press\u00e3o hidrost\u00e1tica \u00e9 fornecida por Ph=dcaf\u00e9.g.h e observe que ela n\u00e3o depende das dimens\u00f5es do tubo e da cafeteira e, como os pontos A e B est\u00e3o abertos, a press\u00e3o nos dois \u00e9 a mesma e ficam na mesma altura\u00a0 —\u00a0\u00a0R- A<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1541,"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-1544","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1544","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=1544"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1544\/revisions"}],"predecessor-version":[{"id":10894,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1544\/revisions\/10894"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1541"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1544"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}