{"id":3271,"date":"2017-09-30T23:51:05","date_gmt":"2017-09-30T23:51:05","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3271"},"modified":"2024-09-04T14:47:11","modified_gmt":"2024-09-04T14:47:11","slug":"resolucao-comentada-hidrostatica-2017-2016-e-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/hidrostatica-2017-2016-e-2015\/resolucao-comentada-hidrostatica-2017-2016-e-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada HIDROST\u00c1TICA \u2013 2017 \u2013 2016 e 2015"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada <\/span><\/span><\/span><\/p>\n

HIDROST\u00c1TICA \u2013 2017 \u2013 2016 e 2015<\/span><\/span><\/span><\/p>\n

Densidade e Press\u00e3o<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>Para essa quest\u00e3o apenas\u00a0<\/b><\/span><\/span>utilizaremos a f\u00f3rmula da press\u00e3o<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde:<\/b><\/span><\/span><\/p>\n

P \u00e9 a press\u00e3o<\/b><\/span><\/span><\/p>\n

F \u00e9 a for\u00e7a exercida<\/b><\/span><\/span><\/p>\n

A \u00e9 a \u00e1rea aonde essa for\u00e7a foi exercida<\/b><\/span><\/span><\/p>\n

Bom, como foi dito no pr\u00f3prio exerc\u00edcio\u00a0<\/b><\/span><\/span>essa press\u00e3o se deve ao peso da coluna de ar<\/b><\/span><\/span>, ent\u00e3o\u00a0<\/b><\/span><\/span>podemos substituir essa for\u00e7a por m.g<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"Isolando a massa:<\/b><\/span><\/span><\/p>\n

\"\"Colocando os valores:<\/b><\/span><\/span><\/p>\n

\"\"m = 2.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>\u00a0Kg<\/b><\/span><\/span><\/p>\n

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

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

01)<\/b><\/span><\/span>\u00a0Press\u00e3o \u00e9 uma grandeza vetorial, portanto tem m\u00f3dulo, dire\u00e7\u00e3o e sentido definidos<\/b><\/span><\/span><\/p>\n

01. Falsa<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Se a for\u00e7a\u00a0\"\"\u00a0estiver inclinada de um\u00a0<\/b><\/span><\/span>\u00e2ngulo \u03b2<\/b><\/span><\/span>\u00a0em rela\u00e7\u00e3o \u00e0 horizontal, a parcela dessa<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

for\u00e7a\u00a0<\/b><\/span><\/span>que exerce a press\u00e3o \u00e9 a\u00a0<\/b><\/span><\/span>componente vertical de\u00a0\"\", ou seja, \u00e9 Fsen\u03b2, conforme a figura acima.<\/b><\/span><\/span><\/p>\n

A press\u00e3o sempre vai estar relacionada ao\u00a0<\/b><\/span><\/span>m\u00f3dulo da for\u00e7a que age perpendicularmente<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00e0 superf\u00edcie<\/b><\/span><\/span>. Ent\u00e3o, <\/b><\/span><\/span>n\u00e3o importa a dire\u00e7\u00e3o<\/b><\/span><\/span>\u00a0que a for\u00e7a \u00e9 aplicada, ela\u00a0<\/b><\/span><\/span>n\u00e3o muda a dire\u00e7\u00e3o da press\u00e3o<\/b><\/span><\/span>, que \u00e9 sempre <\/b><\/span><\/span>perpendicular a superf\u00edcie<\/b><\/span><\/span>.
\n<\/b><\/span><\/span>S\u00f3\u00a0<\/b><\/span><\/span>o que leva em conta na press\u00e3o, portanto, \u00e9 a\u00a0<\/b><\/span><\/span>intensidade.<\/b><\/span><\/span><\/p>\n

02)<\/b><\/span><\/span>\u00a0A diferen\u00e7a de press\u00e3o entre dois pontos no interior de um l\u00edquido homog\u00eaneo e em equil\u00edbrio \u00e9<\/b><\/span><\/span><\/p>\n

proporcional \u00e0 diferen\u00e7a de profundidade entre esses dois pontos.<\/b><\/span><\/span><\/p>\n

02. Correta\u00a0\"\"\u00a0Teorema de Stevin<\/b><\/span><\/span>\u00a0\"\"\u00a0Observe a figura abaixo onde, quem varia \u00e9\u00a0<\/b><\/span><\/span>apenas a press\u00e3o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>vertical<\/b><\/span><\/span>, onde o l\u00edquido que \u00e9 homog\u00eaneo e incompress\u00edvel est\u00e1 em equil\u00edbrio.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na superf\u00edcie livre superior (A) desse l\u00edquido age a press\u00e3o atmosf\u00e9rica (Patm),\u00a0exercida pela coluna de ar que est\u00e1 sobre ele. Num ponto qualquer\u00a0<\/b><\/span><\/span>B do interior do l\u00edquido<\/b><\/span><\/span>, a\u00a0<\/b><\/span><\/span>press\u00e3o (PB)<\/b><\/span><\/span>\u00a0que age \u00e9 a\u00a0<\/b><\/span><\/span>soma da press\u00e3o atmosf\u00e9rica com a press\u00e3o exercida pela coluna l\u00edquida acima dele, devido \u00e0 seu peso.<\/b><\/span><\/span><\/p>\n

PB=Patm\u00a0+ Pl\u00edquido\u00a0 \u2014\u00a0 PB=Patm\u00a0+ F\/S\u00a0 \u2014\u00a0 PB= Patm\u00a0+ peso do l\u00edquido\/S\u00a0 \u2014\u00a0 PB=Patm\u00a0+ (ml\u00edquido.g)\/S\u00a0 \u2014\u00a0 dl\u00edquido=ml\u00edquido\/Vl\u00edquido\u00a0 \u2014\u00a0 dl\u00edquido=ml\u00edquido\/S.h\u00a0\u00a0\u00a0\u00a0\u00a0\u2014\u00a0 ml\u00edquido=dl\u00edquido.S.h\u00a0 \u2014 \u00a0PB=Patm\u00a0+ (dl\u00edquido.S.h)\/S\u00a0 \u2014\u00a0\u00a0<\/b><\/span><\/span>PB=Patm+ dl\u00edquido.g.h\u00a0 \u2014\u00a0 esta express\u00e3o \u00e9 chamada de Teorema Fundamental da Hidrost\u00e1tica ou de Teorema de Stevin<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

04)<\/b><\/span><\/span>\u00a0No sistema internacional de medidas (SI) a unidade de press\u00e3o \u00e9 o Pascal (Pa).<\/b><\/span><\/span><\/p>\n

04. Correta\u00a0\"\"\u00a0<\/b><\/span><\/span>Pa = N\/m2.<\/b><\/span><\/span><\/p>\n

08)<\/b><\/span><\/span>\u00a0Quando um dado ponto de um l\u00edquido em equil\u00edbrio no interior de um recipiente sofre uma varia\u00e7\u00e3o de press\u00e3o, essa varia\u00e7\u00e3o de press\u00e3o \u00e9 transmitida a todos os pontos do l\u00edquido e das paredes do recipiente.<\/b><\/span><\/span><\/p>\n

08. Correta<\/b><\/span><\/span>\u00a0\"\"\u00a0Considere um\u00a0<\/b><\/span><\/span>fluido<\/b><\/span><\/span>\u00a0(l\u00edquidos e gases que fluem para as regi\u00f5es inferiores de um recipiente at\u00e9 preench\u00ea-los totalmente) em<\/b><\/span><\/span>equil\u00edbrio (vertical e horizontal)<\/b><\/span><\/span>\u00a0no interior de um recipiente.<\/b><\/span><\/span><\/p>\n

Esse l\u00edquido exerce sobre as paredes do recipiente que o contem for\u00e7as que se tornam de maior intensidade \u00e0 medida que a profundidade aumenta.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que as\u00a0<\/b><\/span><\/span>for\u00e7as, na mesma horizontal, em ambos os extremos, tem a mesma intensidade, pois o l\u00edquido est\u00e1 em equil\u00edbrio horizontal, caso contr\u00e1rio, ele se moveria nessa dire\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

16)<\/b><\/span><\/span>\u00a0Estando um l\u00edquido em um recipiente aberto, a press\u00e3o na superf\u00edcie do l\u00edquido \u00e9 nula.<\/b><\/span><\/span><\/p>\n

16. Falsa<\/b><\/span><\/span>\u00a0\"\"\u00a0veja 02.<\/b><\/span><\/span><\/p>\n

R- (02 + 04 + 08) = 14<\/b><\/span><\/span><\/p>\n

03- <\/b><\/span><\/span><\/span>F = ? \u00a0\"\"\u00a0S =<\/b><\/span><\/span>\u00a01,0 mm<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span>= 1,0.10-6\u00a0m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0a\u00a0<\/b><\/span><\/span>press\u00e3o sist\u00f3lica\u00a0<\/b><\/span><\/span>(sangue) de valor\u00a0<\/b><\/span><\/span>Ps\u00a0= 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.160 = 160.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0Pa (N\/m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>) que \u00e9 a press\u00e3o exercida por uma coluna de h = 160 mm = 0,16 m de merc\u00fario,\u00a0<\/b><\/span><\/span>de valor\u00a0\"\"\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>Hg<\/b><\/span><\/sub><\/span>\u00a0= d.g.h\u00a0<\/b><\/span><\/span>= 13,6.103.10.16.10-2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 217,6.10<\/b><\/span><\/span>2 <\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0P<\/b><\/span><\/span>Hg<\/b><\/span><\/sub><\/span>\u00a0= 21760 Pa (N\/m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>).<\/b><\/span><\/span><\/p>\n

\u00c9 essa press\u00e3o\u00a0<\/b><\/span><\/span>(21760 Pa) devido \u00e0 coluna de Hg (mec\u00e2nica) que age sobre a \u00e1rea S = 1,0.<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>-6<\/b><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>,\u00a0<\/b><\/span><\/span>devido \u00e0\u00a0<\/b><\/span><\/span>for\u00e7a\u00a0<\/b><\/span><\/span>de intensidade\u00a0<\/b><\/span><\/span>F tal que P =\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a021760 =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>F = 2,176.<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>-2<\/b><\/span><\/sup><\/span>\u00a0N.<\/b><\/span><\/span><\/p>\n

Teorema de Stevin \u2013 press\u00e3o hidrost\u00e1tica – vasos comunicantes<\/span><\/span><\/span><\/p>\n

04- <\/b><\/span><\/span><\/span>Devido ao\u00a0<\/b><\/span><\/span>teorema de Stevin, P=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span>, para um mesmo l\u00edquido, <\/b><\/span><\/span>quanto maior a profundidade, maior ser\u00e1 a press\u00e3o<\/b><\/span><\/span>. Observe na figura abaixo, onde a esfera est\u00e1 imersa, que <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

as\u00a0<\/b><\/span><\/span>for\u00e7as exercidas pelo l\u00edquido sobre ela s\u00e3o maiores nas \u00e1reas inferiores (onde h \u00e9 maior) que nas superiores (onde h\u00e9 menor).<\/b><\/span><\/span>\u00a0A\u00a0<\/b><\/span><\/span>for\u00e7a resultante de todas essas for\u00e7as est\u00e1 dirigida para cima<\/b><\/span><\/span>, e corresponde ao\u00a0<\/b><\/span><\/span>empuxo\u00a0<\/b><\/span><\/span>que age sobre essa esfera. Nas\u00a0<\/b><\/span><\/span>laterais essas for\u00e7as se anulam,<\/b><\/span><\/span><\/p>\n

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

05- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span>\u00a0Se o processo ocorre \u00e0\u00a0<\/b><\/span><\/span>temperatura constante<\/b><\/span><\/span>\u00a0trata-se de uma transforma\u00e7\u00e3o\u00a0<\/b><\/span><\/span>isot\u00e9rmica\u00a0<\/b><\/span><\/span>e, aplicando a equa\u00e7\u00e3o dos gases ideais\u00a0\"\"<\/b><\/span><\/span>Po.Vo\/To\u00a0= P.V\/T\u00a0\"\"Po.Vo\u00a0= P.V<\/b><\/span><\/span>\u00a0\"\"\u00a0volume inicial de ar no tubo cil\u00edndrico de \u00e1rea da base S\u00a0\"\"\u00a0<\/b><\/span><\/span>Vo<\/b><\/span><\/span>\u00a0\"\"\u00a0volume final de ar no tubo cil\u00edndrico\u00a0\"\"\u00a0<\/b><\/span><\/span>V =(4\/5)Vo<\/b><\/span><\/span>\u00a0\"\"\u00a0Po\u00a0= 1 atm\u00a0\"\"\u00a0Po.Vo\u00a0= P.V\u00a0\"\"\u00a01.Vo\u00a0= P.(4\/5)Vo\u00a0\"\"\u00a0P = 5\/4\u00a0\"\"\u00a0<\/b><\/span><\/span>P = 1,25 atm.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Mas, uma das consequ\u00eancias do princ\u00edpio de Stevin \u00e9 que\u00a0<\/b><\/span><\/span>todos os pontos do mesmo n\u00edvel<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>horizontal suportam a mesma press\u00e3o <\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>M<\/b><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span>N <\/b><\/span><\/sub><\/span>\u00a0\"\"<\/b><\/span><\/span>1,35.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0+ 10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span> H \u00a0\"\"\u00a01,35.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0\u2013 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span> H \u00a0\"\"\u00a0H = 0,35.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span> \/10<\/b><\/span><\/span>4 <\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0H = 0,35.10\u00a0\"\"\u00a0<\/b><\/span><\/span>H = 3,5m<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

\"\"<\/p>\n

07- <\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span>press\u00e3o total\u00a0<\/b><\/span><\/span>no fundo do lago devido apenas\u00a0<\/b><\/span><\/span>\u00e0 agua vale P =\u00a0<\/b><\/span><\/span>5 atm \u2013 1 atm\u00a0<\/b><\/span><\/span>= 4 atm.<\/b><\/span><\/span><\/p>\n

Como para\u00a0<\/b><\/span><\/span>cada 1 atm de press\u00e3o a altura equivale a 10m,\u00a0<\/b><\/span><\/span>o lago ter\u00e1\u00a0<\/b><\/span><\/span>h = 4\u00d710 = 40 m\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>profundidade.<\/b><\/span><\/span><\/p>\n

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

Experi\u00eancia de Torricelli<\/span><\/span><\/span><\/p>\n

08- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span>\u00a0Se o processo ocorre \u00e0\u00a0<\/b><\/span><\/span>temperatura constante<\/b><\/span><\/span>\u00a0trata-se de uma transforma\u00e7\u00e3o\u00a0<\/b><\/span><\/span>isot\u00e9rmica\u00a0<\/b><\/span><\/span>e, aplicando a equa\u00e7\u00e3o dos gases ideais\u00a0\"\"<\/b><\/span><\/span>Po.Vo\/To\u00a0= P.V\/T\u00a0\"\"Po.Vo\u00a0= P.V<\/b><\/span><\/span>\u00a0\"\"\u00a0volume inicial de ar no tubo cil\u00edndrico de \u00e1rea da base S\u00a0\"\"\u00a0<\/b><\/span><\/span>Vo<\/b><\/span><\/span>\u00a0\"\"\u00a0volume final de ar no tubo cil\u00edndrico\u00a0\"\"\u00a0<\/b><\/span><\/span>V =(4\/5)Vo<\/b><\/span><\/span>\u00a0\"\"\u00a0Po\u00a0= 1 atm\u00a0\"\"\u00a0Po.Vo\u00a0= P.V\u00a0\"\"\u00a01.Vo\u00a0= P.(4\/5)Vo\u00a0\"\"\u00a0P = 5\/4\u00a0\"\"\u00a0<\/b><\/span><\/span>P = 1,25 atm.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Mas, uma das consequ\u00eancias do princ\u00edpio de Stevin \u00e9 que\u00a0<\/b><\/span><\/span>todos os pontos do mesmo n\u00edvel<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>horizontal suportam a mesma press\u00e3o <\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>M<\/b><\/span><\/sub><\/span>\u00a0= P<\/b><\/span><\/span>N <\/b><\/span><\/sub><\/span>\u00a0\"\"<\/b><\/span><\/span>1,35.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0+ 10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span> H \u00a0\"\"\u00a01,35.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0\u2013 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span> H \u00a0\"\"\u00a0H = 0,35.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span> \/10<\/b><\/span><\/span>4 <\/b><\/span><\/sup><\/span>\u00a0\"\"\u00a0H = 0,35.10\u00a0\"\"\u00a0<\/b><\/span><\/span>H = 3,5m<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Princ\u00edpio de Pascal<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Teorema de Arquimedes \u2013 Empuxo<\/b><\/span><\/span><\/span><\/p>\n

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

\u201c<\/span>Todo corpo total ou parcialmente<\/b><\/span><\/span><\/span>\u00a0mergulhado\u00a0<\/b><\/span><\/span><\/span>num l\u00edquido em equil\u00edbrio<\/b><\/span><\/span><\/span>, recebe\u00a0<\/b><\/span><\/span><\/span>uma for\u00e7a de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>dire\u00e7\u00e3o vertical e sentido\u00a0 para cima<\/b><\/span><\/span><\/span>\u00a0denominada\u00a0<\/b><\/span><\/span><\/span>de Empuxo, cuja intensidade\u00a0 \u00e9 igual ao peso do volume de l\u00edquido deslocado\u201c<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

\u201c<\/span>Todo corpo\u00a0<\/b><\/span><\/span><\/span>total ou parcialmente mergulhado num l\u00edquido em equil\u00edbrio,<\/b><\/span><\/span><\/span>\u00a0recebe uma\u00a0<\/b><\/span><\/span><\/span>for\u00e7a dedire\u00e7\u00e3o vertical e sentido\u00a0 para cima<\/b><\/span><\/span><\/span>\u00a0denominada de\u00a0<\/b><\/span><\/span><\/span>Empuxo,<\/b><\/span><\/span><\/span>\u00a0cuja\u00a0<\/b><\/span><\/span><\/span>intensidade\u00a0 \u00e9 igual ao peso do volume de l\u00edquido deslocado\u201c<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o navio est\u00e1\u00a0<\/b><\/span><\/span>em equil\u00edbrio vertical\u00a0<\/b><\/span><\/span>(flutuando) o\u00a0<\/b><\/span><\/span>empuxo<\/b><\/span><\/span>\u00a0(peso do volume do l\u00edquido deslocado), vertical e para cima deve ser\u00a0<\/b><\/span><\/span>igual ao peso do navio,<\/b><\/span><\/span>\u00a0vertical e para baixo.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

15- <\/span><\/span><\/span>Para esse exerc\u00edcio precisamos analisar a\u00a0<\/b><\/span><\/span>f\u00f3rmula do empuxo:<\/b><\/span><\/span><\/p>\n

E = d<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>Fd<\/b><\/span><\/span><\/sub><\/span>.g<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde:<\/b><\/span><\/span><\/p>\n

E \u00e9 o empuxo<\/b><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u00e9 a densidade do flu\u00eddo<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>Fd<\/b><\/span><\/sub><\/span>\u00a0\u00e9 o volume de flu\u00eddo deslocado<\/b><\/span><\/span><\/p>\n

g \u00e9 a acelera\u00e7\u00e3o da gravidade<\/b><\/span><\/span><\/p>\n

Com a\u00a0<\/b><\/span><\/span>adi\u00e7\u00e3o de sal a densidade da \u00e1gua aumenta<\/b><\/span><\/span>, por\u00e9m quando se <\/b><\/span><\/span>aumenta a densidade parte da esfera sai da \u00e1gua, e o volume deslocado diminui<\/b><\/span><\/span>, sendo assim\u00a0<\/b><\/span><\/span>a alternativa D \u00e9 a correta<\/b><\/span><\/span>, pois o\u00a0<\/b><\/span><\/span>empuxo n\u00e3o se altera,<\/b><\/span><\/span>\u00a0j\u00e1\u00a0<\/b><\/span><\/span>que o peso da esfera \u00e9 o mesmo em ambos os casos.<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

Massa do corpo s\u00f3lido\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>ms<\/b><\/span><\/span>\u00a0= 600g\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Vs\u00f3lido<\/b><\/span><\/span>\u00a0= 1 L\u00a0<\/b><\/span><\/span>= 10<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>\u00a0cm<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>\u00a0\"\" <\/b><\/span><\/span>densidade do corpo s\u00f3lid<\/b><\/span><\/span>o\u00a0\"\"\u00a0<\/b><\/span><\/span>ds\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>ds\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>ds\u00a0= 0,6 g\/cm3.<\/b><\/span><\/span><\/p>\n

Volume total =<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>1,0.10\u00a03\u00a0cm3<\/b><\/span><\/span><\/p>\n

Volume imerso\u00a0<\/b><\/span><\/span>= 80% do Vtotal\u00a0= 0,8.103\u00a0cm3.<\/b><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>=<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>l\u00b4quido<\/b><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 0,75 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/p>\n

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

17- <\/span><\/span><\/span>Para que a\u00a0<\/b><\/span><\/span>pessoa + peso\u00a0<\/b><\/span><\/span>fique em\u00a0<\/b><\/span><\/span>equil\u00edbrio \u00e9 necess\u00e1rio que o empuxo\u00a0<\/b><\/span><\/span>(for\u00e7a vertical e para cima)\u00a0<\/b><\/span><\/span>equilibre os pesos da pessoa (Pp) somado ao peso do objetos (Po), verticais e para baixo.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

E = Pp\u00a0+ Po<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= mp.g + Po\u00a0= 70.10 + Po\u00a0\"\"\u00a0<\/b><\/span><\/span>E = 700 + Po.<\/b><\/span><\/span><\/p>\n

Pessoa<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>dpessoa\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a00,93.103\u00a0=\u00a0\"\"\u00a0\"\"\u00a0Vpessoa\u00a0=\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>Vpessoa<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>=\u00a0\"\"\u00a0<\/b><\/span><\/span>= 75,27.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/p>\n

Empuxo sobre a pessoa totalmente imersa\u00a0\"\"\u00a0E = d\u00e1gua.Vpessoa.g\u00a0<\/b><\/span><\/span>= 10<\/b><\/span><\/span>3\"\"<\/b><\/span><\/sup><\/span>. 75,27.10<\/b><\/span><\/span>-3<\/b><\/span><\/sup><\/span>.10\u00a0\"\"\u00a0<\/b><\/span><\/span>E = 752,7 N.<\/b><\/span><\/span><\/p>\n

E = 700 + Po\u00a0\"\"\u00a0752,7<\/b><\/span><\/span>\u00a0= 700 + Po\u00a0\"\"\u00a0Po\u00a0= 752,7 \u2013 700\u00a0\"\"\u00a0<\/b><\/span><\/span>Po\u00a0= 52,7 N<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

Teorema de Arquimedes \u2013 Empuxo<\/b><\/span><\/span><\/span><\/p>\n

\u201c<\/span>Todo corpo<\/b><\/span><\/span><\/span>total ou parcialmente mergulhado num l\u00edquido em equil\u00edbrio,<\/b><\/span><\/span><\/span>\u00a0recebe uma\u00a0<\/b><\/span><\/span><\/span>for\u00e7a dedire\u00e7\u00e3o vertical e sentido\u00a0para cima<\/b><\/span><\/span><\/span>\u00a0denominada de\u00a0<\/b><\/span><\/span><\/span>Empuxo,<\/b><\/span><\/span><\/span>\u00a0cuja\u00a0<\/b><\/span><\/span><\/span>intensidade\u00a0 \u00e9 igual ao peso do volume de l\u00edquido deslocado\u201c<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio\u00a0<\/b><\/span><\/span>peso total<\/b><\/span><\/span>\u00a0= peso da embarca\u00e7\u00e3o = mg=10.000\u00d710\u00a0<\/b><\/span><\/span>= 100.000 N\u00a0<\/b><\/span><\/span>+ peso dos dois caminh\u00f5es = 2.mg = 2.10.000\u00d710\u00a0<\/b><\/span><\/span>= 200.000 N\u00a0<\/b><\/span><\/span>+ peso da carga dos dois caminh\u00f5es = 2.m.10\u00a0<\/b><\/span><\/span>=20m.<\/b><\/span><\/span><\/p>\n

Ptotal\u00a0=300.000 + 20m<\/b><\/span><\/span>\u00a0sendo\u00a0<\/b><\/span><\/span>m\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>massa de cada caminh\u00e3o.<\/b><\/span><\/span><\/p>\n

Volume imerso = Vi\u00a0=\u00a0<\/b><\/span><\/span>\u00e1rea da base x altura = 10.10.0,4\u00a0\"\"<\/b><\/span><\/span>Vi\u00a0=40m<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span>\"\"<\/p>\n

Densidade da \u00e1gua\u00a0\"\"\u00a0da\u00a0= 1g\/c3\u00a0= 10<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span> kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/p>\n

C\u00e1lculo do empuxo\u00a0\"\"\u00a0E = da.Vi.g\u00a0<\/b><\/span><\/span>= 10<\/b><\/span><\/span>3<\/b><\/span><\/sup><\/span>.40.10\"\"<\/b><\/span><\/span>\u00a0E = 400.000 N<\/b><\/span><\/span><\/p>\n

Como flutua parcialmente imerso est\u00e1 em equil\u00edbrio vertical e a for\u00e7a resultante sobre ele \u00e9 nula, assim o peso total deve anular o empuxo\"\"\u00a0Pt\u00a0= E\u00a0\"\"<\/b><\/span><\/span>300.000 + 20m = 400.000\u00a0\"\"\u00a0m = 100.000\/20\"\" <\/b><\/span><\/span>m = 5.000 kg.<\/b><\/span><\/span><\/p>\n

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

19- <\/span><\/span><\/span>De acordo com o\u00a0<\/b><\/span><\/span>Princ\u00edpio de Arquimedes\u00a0<\/b><\/span><\/span>todo corpo imerso sofre a\u00e7\u00e3o de uma for\u00e7a, denominada empuxo, que \u00e9 vertical para cima, sua intensidade \u00e9 igual ao peso do\u00a0<\/b><\/span><\/span>fluido deslocado pelo corpo<\/b><\/span><\/span>. No caso do estudante, o\u00a0<\/b><\/span><\/span>volume deslocado\u00a0<\/b><\/span><\/span>foi utilizado, visto que a garrafa sempre estaria\u00a0<\/b><\/span><\/span>ocupando<\/b><\/span><\/span>\u00a0esse volume, com isso ele economizou, pois o vaso\u00a0<\/b><\/span><\/span>n\u00e3o enche tanto<\/b><\/span><\/span>, n\u00e3o\u00a0<\/b><\/span><\/span>gasta tanta \u00e1gua<\/b><\/span><\/span>\u00a0e ainda por cima\u00a0<\/b><\/span><\/span>n\u00e3o interfere no rendimento\u00a0<\/b><\/span><\/span>da descarga.<\/b><\/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 HIDROST\u00c1TICA \u2013 2017 \u2013 2016 e 2015 Densidade e Press\u00e3o 01- Para essa quest\u00e3o apenas\u00a0utilizaremos a f\u00f3rmula da press\u00e3o: Onde: P \u00e9 a press\u00e3o F \u00e9 a for\u00e7a exercida A \u00e9 a \u00e1rea aonde essa for\u00e7a foi exercida Bom, como foi dito no pr\u00f3prio exerc\u00edcio\u00a0essa press\u00e3o se deve ao peso da coluna de ar, ent\u00e3o\u00a0podemos substituir essa for\u00e7a<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3269,"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-3271","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3271","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=3271"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3271\/revisions"}],"predecessor-version":[{"id":3276,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3271\/revisions\/3276"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3269"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3271"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}