{"id":4895,"date":"2020-01-10T03:16:51","date_gmt":"2020-01-10T03:16:51","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4895"},"modified":"2024-09-04T14:47:41","modified_gmt":"2024-09-04T14:47:41","slug":"resolucao-hidrostatica-2018-2019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/hidrostatica-2018-2019\/resolucao-hidrostatica-2018-2019\/","title":{"rendered":"Resolu\u00e7\u00e3o – Hidrost\u00e1tica \u2013 2018\/2019"},"content":{"rendered":"

Hidrost\u00e1tica \u2013 2018\/2019<\/span><\/span><\/span><\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

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

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

<\/p>\n

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

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

Em primeiro lugar, deve-se encontrar a rela\u00e7\u00e3o entre as\u00a0<\/b><\/span><\/span>densidades do objeto e da \u00e1gua\u00a0<\/b><\/span><\/span>utilizando a f\u00f3rmula para\u00a0<\/b><\/span><\/span>a for\u00e7a de empuxo E\u00a0<\/b><\/span><\/span>realizada em um corpo com\u00a0<\/b><\/span><\/span>um volume submerso Vs<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>no flu\u00eddo de\u00a0<\/b><\/span><\/span>densidade \u03c1<\/b><\/span><\/span>, sendo esse empuxo calculado como<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como o corpo de\u00a0<\/b><\/span><\/span>densidade \u03c1o<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>volume total Vo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>est\u00e1 em\u00a0<\/b><\/span><\/span>equil\u00edbrio<\/b><\/span><\/span>, a for\u00e7a de empuxo sobre o corpo\u00a0<\/b><\/span><\/span>ser\u00e1 igual a for\u00e7a peso<\/b><\/span><\/span>, utilizando\u00a0<\/b><\/span><\/span>\u03c1a\u00a0para denotar a densidade da \u00e1gua\u00a0<\/b><\/span><\/span>e usando que\u00a0<\/b><\/span><\/span>o volume de \u00e1gua deslocado pelo flu\u00eddo Vs\u00a0\u00e9 3\/5 do volume do objeto Vo\u00a0(60%),<\/b><\/span><\/span>\u00a0temos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

Empuxo<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

<\/p>\n

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

a) A partir da\u00a0<\/b><\/span><\/span>defini\u00e7\u00e3o de press\u00e3o<\/b><\/span><\/span>, podemos escrever:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Sendo que a\u00a0<\/b><\/span><\/span>\u00e1rea A \u00e9 a que est\u00e1 sujeita a for\u00e7a aplicada F<\/b><\/span><\/span>. Repare que a for\u00e7a a qual a superf\u00edcie est\u00e1 sujeita \u00e9 a do peso do bloquinho:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

No entanto, a massa do bloco n\u00e3o varia, logo, a\u00a0<\/b><\/span><\/span>for\u00e7a \u00e9 constante para qualquer posicionamento do bloquinho<\/b><\/span><\/span>, portanto:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Sendo\u00a0<\/b><\/span><\/span>e as \u00e1reas de contato da primeira e da segunda configura\u00e7\u00e3o<\/b><\/span><\/span>, respectivamente. As \u00e1reas de contato t\u00eam a forma de ret\u00e2ngulos, fazendo com que seus valores sejam o\u00a0<\/b><\/span><\/span>produto simples de uma dimens\u00e3o pela outra<\/b><\/span><\/span>. Substituindo os valores, obtemos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

b) Nesta situa\u00e7\u00e3o, teremos\u00a0<\/b><\/span><\/span>tr\u00eas for\u00e7as atuando no bloquinho<\/b><\/span><\/span>, a for\u00e7a\u00a0<\/b><\/span><\/span>peso<\/b><\/span><\/span>, a for\u00e7a\u00a0<\/b><\/span><\/span>normal<\/b><\/span><\/span>\u00a0\u00e0 superf\u00edcie e a for\u00e7a de\u00a0<\/b><\/span><\/span>empuxo<\/b><\/span><\/span>\u00a0exercida pelo flu\u00eddo e contr\u00e1ria \u00e0 for\u00e7a peso. O esbo\u00e7o abaixo retrata as tr\u00eas for\u00e7as atuantes.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Aplicando a\u00a0<\/b><\/span><\/span>segunda lei de Newton\u00a0<\/b><\/span><\/span>e considerando que as for\u00e7as que apontam para cima possuem sinal negativo, teremos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Por\u00e9m, o\u00a0<\/b><\/span><\/span>bloquinho encontra-se parado<\/b><\/span><\/span>, logo possui\u00a0<\/b><\/span><\/span>acelera\u00e7\u00e3o nula<\/b><\/span><\/span>, logo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

A for\u00e7a peso \u00e9 o produto da massa do objeto pela acelera\u00e7\u00e3o gravitacional:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

A for\u00e7a de empuxo \u00e9 dada por:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Em que V \u00e9 o volume de l\u00edquido deslocado, \u00e9 a densidade do l\u00edquido e g \u00e9 a acelera\u00e7\u00e3o gravitacional. Pelo\u00a0<\/b><\/span><\/span>princ\u00edpio de Arquimedes<\/b><\/span><\/span>, o\u00a0<\/b><\/span><\/span>volume de l\u00edquido deslocado \u00e9 igual ao volume do objeto submerso no l\u00edquido<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Usando as equa\u00e7\u00f5es e o princ\u00edpio de Arquimedes, obtemos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

A massa do bloquinho pode ser obtida utilizando o conceito de press\u00e3o apresentado no item a, utilizando a press\u00e3o e a \u00e1rea da primeira configura\u00e7\u00e3o, tal que:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Substituindo na equa\u00e7\u00e3o da for\u00e7a normal:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

Para este exerc\u00edcio, vamos utilizar o\u00a0<\/b><\/span><\/span>Teorema Fundamental da Hidrost\u00e1tica<\/b><\/span><\/span>, que se encontra resumido abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

A situa\u00e7\u00e3o descrita no enunciado encontra-se esbo\u00e7ada abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

<\/p>\n

Vcasca\u00a0= Vcaro\u00e7o\u00a0= V<\/b><\/span><\/span><\/span><\/span><\/p>\n

dcasca\u00a0= mcasca\/V\u00a0<\/b><\/span><\/span>\u00a00,4 = mcasca\/V\u00a0\u00a0<\/b><\/span><\/span>mcasca\u00a0= 0,4V<\/b><\/span><\/span><\/span><\/span><\/p>\n

dcaro\u00e7o\u00a0= mcaro\u00e7o\/V\u00a0<\/b><\/span><\/span>\u00a0dcaro\u00e7o\u00a0= mcaro\u00e7o\/V\u00a0\u00a0<\/b><\/span><\/span>mcaro\u00e7o\u00a0= dcaro\u00e7o.V<\/b><\/span><\/span><\/span><\/span><\/p>\n

dcorpo\u00a0= mcorpo\/Vcorpo\u00a0<\/b><\/span><\/span>\u00a0dcorpo\u00a0= (mcasca\u00a0+ mcaro\u00e7o)\/2V\u00a0\u00a0dcorpo\u00a0= (0,4V + dcaro\u00e7o.V)\/2V<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 a seguir:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Enunciado do princ\u00edpio de Arquimedes<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Express\u00e3o matem\u00e1tica do Empuxo<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Express\u00e3o matem\u00e1tica do peso (for\u00e7a vertical e para baixo)<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Veja figura abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

<\/p>\n

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

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

Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 a seguir.<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

<\/p>\n

No\u00a0<\/b><\/span><\/span>caso do exerc\u00edcio voc\u00ea ter\u00e1 Pfundo<\/b><\/span><\/span>\u00a0= 11 000\/10 + 1 atm\u00a0<\/b><\/span><\/span>= 1 101 atm.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

Teorema Fundamental da Hidrost\u00e1tica ou Teorema de Stevin<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe\u00a0<\/b><\/span><\/span><\/span>pelo teorema de Stevin que quanto maior a altura h da coluna l\u00edquida\u00a0<\/b><\/span><\/span><\/span>(no caso do fundo do aqu\u00e1rio),\u00a0<\/b><\/span><\/span><\/span>maior ser\u00e1 a press\u00e3o devido a \u00e1gua nesse ponto.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Assim,\u00a0<\/b><\/span><\/span>h2\u00a0> h1\u00a0> h (veja figuras acima)<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

Resumo te\u00f3rico<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Colocando as for\u00e7as que agem sobre o bal\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

<\/p>\n

<\/p>\n

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

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

01. Falso \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>como\u00a0<\/b><\/span><\/span><\/span>todas as melancias tem densidade menor que a da \u00e1gua todas devem flutuar\u00a0<\/b><\/span><\/span><\/span>(mais denso, mais pesado em baixo e menos denso menos pesado, em cima).<\/b><\/span><\/span><\/span><\/p>\n

02. Falso \u2013\u00a0dependendo da altura que s\u00e3o soltas algumas podem ficar submersas<\/b><\/span><\/span><\/span>, mas,\u00a0<\/b><\/span><\/span><\/span>no equil\u00edbrio final todas ficam flutuando (veja 01).<\/b><\/span><\/span><\/span><\/p>\n

04. Falso \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/span>for\u00e7a peso, vertical e para baixo \u00e9 constante,\u00a0<\/b><\/span><\/span><\/span>mas a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a de empuxo vertical e para cim<\/b><\/span><\/span><\/span>a (peso do volume de l\u00edquido deslocado)\u00a0<\/b><\/span><\/span><\/span>vai aumentando \u00e0 medida que a melancia vai entrando na \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

Assim,\u00a0<\/b><\/span><\/span><\/span>durante a entrada na \u00e1gua a for\u00e7a resultante (FR=\u00a0P- E) vai diminuindo at\u00e9 que quando P = E ela fica nula<\/b><\/span><\/span><\/span>. Portanto a\u00a0<\/b><\/span><\/span><\/span>velocidade da melancia durante a penetra\u00e7\u00e3o na \u00e1gua vai diminuindo.<\/b><\/span><\/span><\/span><\/p>\n

08- Correto \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>veja 04.<\/b><\/span><\/span><\/span><\/p>\n

16- Falso\u00a0\u2013\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante s\u00f3 \u00e9 nula quando a melancia retorna\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>fica flutuando o que s\u00f3 ocorre quando P = E.<\/b><\/span><\/span><\/span><\/p>\n

R \u2013 08<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

Enunciado do princ\u00edpio de Arquimedes<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Se o corpo estiver parcialmente imerso<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

No\u00a0<\/b><\/span><\/span><\/span>caso do exerc\u00edcio s\u00e3o dados:<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>densidade do bloco de madeira dcorpo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 680 kg\/m3;\u00a0<\/b><\/span><\/span><\/span>volume da parte imersa Vimersa<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 2\u00d70,6\u00d70,25\u00a0<\/b><\/span><\/span><\/span>= 0,3 m3<\/b><\/span><\/span><\/span>;\u00a0<\/b><\/span><\/span><\/span>densidade da \u00e1gua d\u00e1gua\u00a0<\/b><\/span><\/span><\/span>= 1,0.103\u00a0kg\/m3\u00a0e\u00a0<\/b><\/span><\/span><\/span>volume total = volume do corpo (imerso) + volume da cabe\u00e7a (emerso<\/b><\/span><\/span><\/span>) = 0,3 m3\u00a0+ Vcabe\u00e7a.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

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

Veja a teoria a seguir:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Sem alternativa \u2013 quest\u00e3o anulada.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

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

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

Trata-se do Empuxo de Arquimedes \u2013 corpo parcialmente imerso<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

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

<\/p>\n

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

Enquanto o\u00a0<\/b><\/span><\/span><\/span>gelo n\u00e3o se transformar totalmente em \u00e1gua<\/b><\/span><\/span><\/span>\u00a0(fus\u00e3o do gelo) a\u00a0<\/b><\/span><\/span><\/span>temperatura do sistema permanece constante (0oC).<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>massa de gelo que se transforma em \u00e1gua \u00e9 a mesma\u00a0<\/b><\/span><\/span><\/span>e, como a\u00a0<\/b><\/span><\/span><\/span>densidade da \u00e1gua \u00e9 maior\u00a0<\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>a do gelo\u00a0<\/b><\/span><\/span><\/span>(gelo menos denso \u201cmais leve\u201d flutua na \u00e1gua)\u00a0<\/b><\/span><\/span><\/span>um volume de mesma massa de \u00e1gua \u00e9 menor que o de mesma massa de gelo,\u00a0<\/b><\/span><\/span><\/span>pois a\u00a0<\/b><\/span><\/span><\/span>express\u00e3o d,\u00a0<\/b><\/span><\/span><\/span>mostra que sendo\u00a0<\/b><\/span><\/span><\/span>m constante, densidade e volume s\u00e3o inversamente proporcionais (V\u00e1gua\u00a0< Vgelo).<\/b><\/span><\/span><\/span><\/p>\n

\u00c0 medida que o\u00a0<\/b><\/span><\/span><\/span>gelo vai derretendo o volume de \u00e1gua l\u00edquida V\u00e1gua\u00a0do sistema vai aumentando\u00a0<\/b><\/span><\/span><\/span>e o respectivo\u00a0<\/b><\/span><\/span><\/span>volume de gelo que era essa \u00e1gua (Vgelo) diminuindo,\u00a0<\/b><\/span><\/span><\/span>o que provoca uma\u00a0<\/b><\/span><\/span><\/span>diminui\u00e7\u00e3o do volume total do sistema, pois V\u00e0gua\u00a0< Vgelo).<\/b><\/span><\/span><\/span><\/p>\n

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

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

a)<\/b><\/span><\/span><\/span><\/p>\n

Defini\u00e7\u00e3o do Princ\u00edpio Fundamental da Din\u00e2mica ou Segunda lei de Newton<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Aplicando a\u00a0<\/b><\/span><\/span><\/span>segunda lei de Newton ( FR\u00a0= m.a),\u00a0<\/b><\/span><\/span><\/span>vamos\u00a0<\/b><\/span><\/span><\/span>calcular a acelera\u00e7\u00e3o a\u00a0<\/b><\/span><\/span><\/span>de subida da esfera sendo:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

b)<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7as que agem sobre a esfera:<\/b><\/span><\/span><\/span><\/p>\n

Peso, vertical e para baixo\u00a0<\/b><\/span><\/span><\/span>de\u00a0<\/b><\/span><\/span><\/span>intensidade<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>P = m.g\u00a0<\/b><\/span><\/span><\/span>= 0,05.10\u00a0\u00a0<\/b><\/span><\/span><\/span>P = 0,5 N.<\/b><\/span><\/span><\/span><\/p>\n

Empuxo de Arquimedes<\/b><\/span><\/span><\/span>\u00a0\u00a0for\u00e7a vertical e para cima fornecida por:<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

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

Pelo gr\u00e1fico,\u00a0<\/b><\/span><\/span><\/span>quando o cilindro\u00a0<\/b><\/span><\/span><\/span>est\u00e1 totalmente emerso (no ar)\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a de tra\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>\u00e9 constante e vale\u00a0<\/b><\/span><\/span><\/span>T = 120 N, que \u00e9 o pr\u00f3prio peso do cilindro\u00a0\u00a0P = 120 N.<\/b><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>gr\u00e1fico<\/b><\/span><\/span><\/span>\u00a0tamb\u00e9m\u00a0<\/b><\/span><\/span><\/span>informa<\/b><\/span><\/span><\/span>\u00a0que, quando o\u00a0<\/b><\/span><\/span><\/span>cilindro est\u00e1 totalmente imerso na \u00e1gua\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a de tra\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>vale\u00a0<\/b><\/span><\/span><\/span>T = 40 N\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>nesse caso, al\u00e9m do peso P = 120 N\u00a0<\/b><\/span><\/span><\/span>(vertical e para baixo) voc\u00ea tem o\u00a0<\/b><\/span><\/span><\/span>empuxo E\u00a0<\/b><\/span><\/span><\/span>(vertical e para cima).<\/b><\/span><\/span><\/span><\/p>\n

Intensidade do empuxo<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

<\/p>\n

<\/p>\n

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

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

Hidrost\u00e1tica \u2013 2018\/2019 01- R- E 02- R- A 03- Em primeiro lugar, deve-se encontrar a rela\u00e7\u00e3o entre as\u00a0densidades do objeto e da \u00e1gua\u00a0utilizando a f\u00f3rmula para\u00a0a for\u00e7a de empuxo E\u00a0realizada em um corpo com\u00a0um volume submerso Vs\u00a0no flu\u00eddo de\u00a0densidade \u03c1, sendo esse empuxo calculado como Como o corpo de\u00a0densidade \u03c1o\u00a0e\u00a0volume total Vo\u00a0est\u00e1 em\u00a0equil\u00edbrio, a for\u00e7a de empuxo sobre o<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4893,"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-4895","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4895","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=4895"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4895\/revisions"}],"predecessor-version":[{"id":4896,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4895\/revisions\/4896"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4893"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4895"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}