{"id":1567,"date":"2015-11-28T03:34:02","date_gmt":"2015-11-28T03:34:02","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1567"},"modified":"2024-08-23T14:38:31","modified_gmt":"2024-08-23T14:38:31","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-teorema-de-arquimedes-empuxo","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/hidrostatica\/teorema-de-arquimedes-empuxo\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-teorema-de-arquimedes-empuxo\/","title":{"rendered":"Teorema de Arquimedes – Empuxo – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- C\u00a0<\/b><\/span><\/span><\/span>\u2013 Veja teoria<\/b><\/span><\/span><\/span><\/p>\n

02<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/span>\u00a0\u2013 (Veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span><\/span>\u00a0(O1)- Correta \u2013 em seu interior existe ar, que faz diminuir sua densidade m\u00e9dia, ficando menor do que a da \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

(02)- Correta \u2013 ele est\u00e1 flutuando e em equil\u00edbrio, ent\u00e3o a for\u00e7a resultante sobre ele \u00e9 nula.<\/b><\/span><\/span><\/span><\/p>\n

(04)- Correta \u2013 o empuxo \u00e9 igual ao peso do volume de l\u00edquido deslocado e, no caso \u00e9 igual ao pr\u00f3prio peso, pois ele est\u00e1 em equil\u00edbrio.<\/b><\/span><\/span><\/span><\/p>\n

(08)- Falsa \u2013 Veja (04)<\/b><\/span><\/span><\/span><\/p>\n

(16)- Correta \u2013 nesse caso, sua densidade m\u00e9dia ser\u00e1 a da \u00e1gua mais a das chapas de a\u00e7o<\/b><\/span><\/span><\/span><\/p>\n

(32)- Falsa \u2013 ele desloca apenas a parte de \u00e1gua e o volume do navio \u00e9 diferente do volume de \u00e1gua deslocada e, se o volume \u00e9 diferente as densidades devem ser diferentes para manter a igualdade \u2013 d<\/b><\/span><\/span><\/span>navio<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>navio<\/b><\/span><\/span><\/span><\/sub>.g=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>\u00e1gua deslocada<\/b><\/span><\/span><\/span><\/sub>.g<\/b><\/span><\/span><\/span><\/p>\n

(01+02+04+16)=23<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span><\/span>\u00a0Enquanto ele est\u00e1 no ar a tens\u00e3o no cabo \u00e9 constante e tem valor m\u00e1ximo\u00a0 —\u00a0 \u00e0 medida que o bloco vai penetrando na \u00e1gua, ele vai deslocando mais l\u00edquido, o empuxo vai aumentando e a tens\u00e3o no cabo diminuindo\u00a0 —\u00a0 quando ele est\u00e1 totalmente imerso e descendo, o empuxo \u00e9 m\u00e1ximo (independente da profundidade) e a tens\u00e3o \u00e9 constante e m\u00ednima.<\/b><\/span><\/span><\/span><\/p>\n

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

05-<\/b><\/span><\/span><\/span><\/span>\u00a0Sendo o l\u00edquido que envolve o recipiente a \u00e1gua e o l\u00edquido que o est\u00e1 preenchendo tamb\u00e9m a \u00e1gua (mesma densidade), para cada unidade preenchida com \u00e1gua, o recipiente desce tamb\u00e9m uma unidade\u00a0 — Observe atentamente a sequ\u00eancia de figuras abaixo e verifique que a resposta \u00e9 a C\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

06-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A diferen\u00e7a na leitura da balan\u00e7a corresponde a ao empuxo sofrido pela m\u00e3o ao ser mergulhada, (veja teoria)\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0 —\u00a0 Volume de l\u00edquido deslocado=volume da m\u00e3o =V\u00a0 —\u00a0 E=d\u00ad<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V.g\u00a0 —\u00a0 4,5=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.V.10\u00a0 —\u00a0 V=4,5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=4,5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=4,5.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0ou V=450 cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0A maioria dos peixes \u00f3sseos apresenta bexiga natat\u00f3ria\u00a0<\/b><\/span><\/span>(atualmente denominada ves\u00edcula gasosa)<\/b><\/span><\/span><\/span>, uma bolsa cheia de gases acima do est\u00f4mago cujo volume \u00e9 regulado por meio de trocas de gases com o sangue e, pela sua dila\u00e7\u00e3o ou contra\u00e7\u00e3o, determina a posi\u00e7\u00e3o do peixe na \u00e1gua.\u00a0<\/b><\/span><\/span>Para aumentar a profundidade, os peixes contraem a bexiga natat\u00f3ria e, com isso, aumentam a sua densidade tornando-se mais pesado que a \u00e1gua e descendo. Ao subir, fazem o contr\u00e1rio.<\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span>\u00a0\u2013 Veja teoria<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/span>\u00a0\u2013 Veja teoria<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Estando o contrapeso em equil\u00edbrio, em cada l\u00edquido, o peso \u00e9 igual ao empuxo\u00a0 —\u00a0 como o contrapeso \u00e9 o mesmo, o peso \u00e9 o mesmo e, portato o empuxo \u00e9 o mesmo\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span><\/span>\u00a0Sua densidade \u00e9 menor que a de B, ent\u00e3o ele flutua em B, e maior que a de A, ent\u00e3o ele afunda em A\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na figura I, N<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\u00a0\u00e9 a indica\u00e7\u00e3o da balan\u00e7a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=P<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na figura II, se o corpo imerso recebe do l\u00edquido uma for\u00e7a vertical e para cima (Empuxo), pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>o corpo reage sobre o l\u00edquido\u00a0<\/b><\/span><\/span><\/span>com for\u00e7a de mesma intensidade<\/b><\/span><\/span><\/span>(Empuxo)<\/b><\/span><\/span><\/span>, mesma dire\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>(vertical<\/b><\/span><\/span><\/span>) e sentido sentido contr\u00e1rio (<\/b><\/span><\/span><\/span>para baixo<\/b><\/span><\/span><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A balan\u00e7a indica apenas as for\u00e7as que agem no l\u00edquido, indicadas na figura da direita acima, que s\u00e3o: peso P do sistema (recipiente mais l\u00edquido), empuxo sobre o l\u00edquido (E) e a rea\u00e7\u00e3o normal da balan\u00e7a (indica\u00e7\u00e3o da balan\u00e7a N)\u00a0 —\u00a0 N=P + E\u00a0 —\u00a0 da figura I\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=P\u00a0 —\u00a0 N=N<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\u00a0+ E\u00a0 —<\/b><\/span><\/span><\/span>\u00a0N \u2013 N<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=E\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 assim,<\/b><\/span><\/span><\/span>\u00a0o empuxo \u00e9 fornecido pela diferen\u00e7a entre as indica\u00e7\u00f5es da balan\u00e7a antes e depois de imergir a esfera.<\/b><\/span><\/span><\/span><\/p>\n

R- (4 + 8)=12<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span><\/span> O peso do bloco \u00e9 constante\u00a0 —\u00a0 \u00e0 medida que o cilindro vai imergindo na \u00e1gua, o empuxo vai<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

aumentando e consequentemente a tra\u00e7\u00e3o no fio tamb\u00e9m vai aumentando, mas a diferen\u00e7a entre eles, que \u00e9 o peso permanece constante\u00a0 —\u00a0 T + P=E\u00a0 —\u00a0 T \u2013 E=P (constante)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a001- Falsa \u2013 era gra\u00e7as ao empuxo que ele recebia do ar, vertical e para cima, maior que seu peso.
\n02- Falsa \u2013 o princ\u00edpio de Arquimedes \u00e9 v\u00e1lido para qualquer corpo imerso em qualquer fluido (l\u00edquidos e gases).<\/b><\/span><\/span><\/p>\n

04- Correta \u2013 ele \u00e9 causado pela varia\u00e7\u00e3o da press\u00e3o com a profundidade ou altitude, sendo que nos pontos inferiores do corpo, a for\u00e7a que causa o empuxo \u00e9 maior que nos pontos superiores.<\/b><\/span><\/span><\/p>\n

08- Falsa \u2013 o empuxo \u00e9 igual ao peso do volume de ar deslocado.<\/b><\/span><\/span><\/p>\n

16- Correta \u2013 E=d.V.g=1,3×2.000×10\u00a0 —\u00a0 E=2,6.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N.<\/b><\/span><\/span><\/p>\n

32- Falsa \u2013 Veja 01.<\/b><\/span><\/span><\/p>\n

64- Correta \u2013 diminuindo parte do g\u00e1s, diminu\u00eda o volume dos bal\u00f5es, diminuindo assim o volume de ar deslocado, o que implica em diminuir o empuxo.<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>01- Correta \u2013 P=mg=4.10\u00a0 —\u00a0 P=40N<\/b><\/span><\/span><\/span><\/p>\n

02- d=m\/V=4\/5=0,8kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/span><\/p>\n

04- E=dVg=1,3.5.10\u00a0 —\u00a0 E=65,0N\u00a0 —\u00a0 Correta<\/b><\/span><\/span><\/span><\/p>\n

08- Falsa \u2013 para mant\u00ea-lo em equil\u00edbrio, o empuxo \u00e9 igual ao peso.<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span> Verdadeira \u2013 n\u00e3o haveria fluido para ser deslocado.<\/b><\/span><\/span><\/span><\/p>\n

(01 + 04 + 16)=21<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Se as caixas tem a mesma massa, t\u00eam o mesmo peso (P=mg) e, se est\u00e3o em equil\u00edbrio, o peso de cada uma \u00e9 igual ao respectivo empuxo\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span> a) Sim ,ele flutua quando nas c\u00e2maras a \u00e1gua \u00e9 expulsa e substitu\u00edda por ar, tornando o peso do submarino menor que o empuxo.<\/b><\/span><\/span><\/p>\n

b) N\u00e3o depende, pois o volume de l\u00edquido deslocado (empuxo) \u00e9 o mesmo em qualquer profundidade.<\/b><\/span><\/span><\/p>\n

c) Ponto C, devido ao teorema de Stevin\u00a0 —\u00a0 P=d.g<\/b><\/span><\/span>.h<\/b><\/span><\/span>\u00a0 —\u00a0 d e g s\u00e3o os mesmos\u00a0 —\u00a0 maior h, maior press\u00e3o \u00a0<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>For\u00e7as que agem sobre o bloco\u00a0 —\u00a0 peso P (vertical e para baixo)\u00a0 —\u00a0 N for\u00e7a que o bloco troca<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

com a tampa (vertical e para baixo)\u00a0 —\u00a0 empuxo E (vertical e para cima)\u00a0 —\u00a0 P + N=E\u00a0 —\u00a0\u00a0 mg + N=d.V.g\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>bloco<\/b><\/span><\/span><\/span><\/sub>.V.<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/sub>g + N=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V.g\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

N=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V.g \u2013 d<\/b><\/span><\/span><\/span>bloco<\/b><\/span><\/span><\/span><\/sub>.V.g\u00a0 —\u00a0 N=(1,0.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0\u2013 0,25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>).V.g\u00a0 — \u00a0N=0,75.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.V.g<\/b><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P= 0,25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.V.<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/sub>g\u00a0 —\u00a0 N\/P=(0,75.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.V.g)\/(0,25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.V.g)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0N\/P=3<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O empuxo, com o submarino totalmente submerso, \u00e9 sempre o mesmo (peso do volume de l\u00edquido deslocado), expulsando a \u00e1gua,<\/b><\/span><\/span>\u00a0o peso torna-se menor que o empuxo.<\/b><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0a) As for\u00e7as que atuam sobre o submarino s\u00e3o o peso e o empuxo e, como ele se encontra em repouso (equil\u00edbrio est\u00e1tico), os m\u00f3dulos destas for\u00e7as s\u00e3o iguais. Portanto E = P.
\nb) O empuxo diminui pois o volume de l\u00edquido deslocado \u00e9 menor.
\n<\/b><\/span><\/span>21-\u00a0<\/b><\/span><\/span>Ambos aumentaram o empuxo sobre a massa. Arquimedes aumentou a densidade do l\u00edquido e Ulisses aumentou o volume de l\u00edquido deslocado.<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Com \u00e1gua\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>tubo<\/b><\/span><\/span><\/span><\/sub>=E\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>tubo.<\/b><\/span><\/span><\/span><\/sub>V<\/b><\/span><\/span><\/span>tubo<\/b><\/span><\/span><\/span><\/sub>.g=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>tubo.<\/b><\/span><\/span><\/span><\/sub>S.h.=1.S.10\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>tubo<\/b><\/span><\/span><\/span><\/sub>=10\/h\u00a0 —\u00a0 com l\u00edquido\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>tubo.<\/b><\/span><\/span><\/span><\/sub>V<\/b><\/span><\/span><\/span>tubo<\/b><\/span><\/span><\/span><\/sub>.g=d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>tubo.<\/b><\/span><\/span><\/span><\/sub>S.h.g=d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>tubo.<\/b><\/span><\/span><\/span><\/sub>S.h=d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/span><\/sub>.S.8\u00a0 —\u00a0 (10\/h).S.h=d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/span><\/sub>.S.8\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/span><\/sub>=1,25g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Colocando as for\u00e7as que agem sobre o cubo\u00a0 —\u00a0 peso (P) vertical e para baixo\u00a0 —\u00a0 indica\u00e7\u00e3o do dinam\u00f4metro (for\u00e7a de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

tra\u00e7\u00e3o no fio) \u2013 T=18N\u00a0 —\u00a0 empuxo (E) vertical e para cima\u00a0 —\u00a0 equil\u00edbrio\u00a0 —\u00a0 T + E=P\u00a0 —\u00a0 18 + d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.g = m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —\u00a0 18 + 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.S.h<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.10=m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>.10\u00a0 —\u00a0 18 + 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.(0,1).(0,1).0,02 = 10.m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>=2010\u00a0 — m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>=2kg\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span><\/span>\u00a0Come\u00e7ando pela letra b:<\/b><\/span><\/span><\/span><\/p>\n

b) bloco flutuando com 3\/4 de seu volume submerso\u00a0 —\u00a0 equil\u00edbrio\u00a0 —\u00a0 P=E\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/span><\/sub>.g=d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

d<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/span><\/sub>.60.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/span><\/sup>=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.(60.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/4)\u00a0 —\u00a0 d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=180.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/240\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=0,75.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0ou d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=7,5.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0ou d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=0,75g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

a) corpo totalmente suberso e atado pelo fio ao fundo do recipiente\u00a0 —\u00a0 T = E – P —\u00a0 T = d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>.V.g – d<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/span><\/sub>.V.g\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0T=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.60.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/span><\/sup>.10 \u2013 0,7510<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.60.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/span><\/sup>.10\u00a0 —\u00a0 T= 0,6 \u2013 0,45\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0T=0,15N ou T=1,5.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O m\u00f3dulo do empuxo sobre o corpo imerso \u00e9 igual ao m\u00f3dulo do peso do fluido deslocado. Tanto a esfera quanto o barquinho deslocaram a mesma quantidade de \u00e1gua, pois os n\u00edveis de \u00e1gua nos dois recipientes subiram a mesma altura.<\/b><\/span><\/span><\/span><\/p>\n

Desse modo, os m\u00f3dulos dos dois empuxos s\u00e3o iguais ao m\u00f3dulo do peso dessa mesma quantidade de \u00e1gua, ou seja,\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub>\u00a0= E<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/span><\/sub>\u00a0.<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span><\/span>\u00a0a) volume de \u00e1gua deslocada\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>=S.H=8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>.5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/span><\/sup>m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 estando o recipiente em equil\u00edbrio\u00a0 —\u00a0 P=E\u00a0 —\u00a0 mg=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —\u00a0 m=1.000.4.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>m=0,4kg<\/b><\/span><\/span><\/span><\/p>\n

b) nesse caso, o recipiente est\u00e1 na imin\u00eancia de afundar, e sua massa ser\u00e1 m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>\u00a0(massa dos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0chumbinhos) + m (massa do recipiente)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>= S.h=8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>.8.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>=64.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/span><\/sup>m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 peso do recipiente + peso dos chumbinhos=P<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>=(m + m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>).g\u00a0 —\u00a0 equil\u00edbrio\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>=E\u00a0 —\u00a0 (m + m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>).g=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —\u00a0 (0,4 + m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>)=1.000.64.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>=0,64 \u2013 0,4\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>=0,24kg=240g\u00a0 —\u00a0 n<\/b><\/span><\/span><\/span>chumbinhos<\/b><\/span><\/span><\/span><\/sub>=240\/12\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>chumbinhos<\/b><\/span><\/span><\/span><\/sub>=20<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0<\/b><\/span><\/span><\/span>n\u00e3o mudariam<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 observe na express\u00e3o seguinte que a acelera\u00e7\u00e3o da gravidade se cancela\u00a0 —\u00a0 P=E\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>.g<\/b><\/span><\/span><\/span>marte<\/b><\/span><\/span><\/span><\/sub>=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.g<\/b><\/span><\/span><\/span>marte<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 assim , g n\u00e3o interfere na rwesolu\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span><\/span>\u00a0Colocando as for\u00ebas que agem sobre o cilindro —\u00a0 peso (P) vertical e para baixo\u00a0 —\u00a0 empuxo (E)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0vertical e para cima —\u00a0 como a mola n\u00e3o est\u00e1 deformada ela n\u00e3o influi na resolu\u00e7\u00e3o\u00a0 —\u00a0 P=E\u00a0 —\u00a0\u00a0 d<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub>Vg=d<\/b><\/span><\/span><\/span>a.<\/b><\/span><\/span><\/span><\/sub>V.g\u00a0 —\u00a0 0,5.1.16.g=0,8.1.h.g\u00a0 — 8= 0,8h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>h=10cm<\/b><\/span><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Rompendo o fio as fo\u00e7as que agem sobre a bolinha durante sua subida no trecho (2) s\u00e3o\u00a0 —\u00a0 peso (P) e empuxo (E)\u00a0 —\u00a0<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 E \u2013 P=ma\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>Vg \u2013 mg=ma\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=5d<\/b><\/span><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/span><\/sub>\u00a0—\u00a0 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=m\/V\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=0,01\/V\u00a0 —\u00a0 V=0.01\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 5d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>.0,01\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>.10 \u2013 0,01.10=0,01.a\u00a0 —\u00a0 0,5 \u2013 0,4=0,01.a\u00a0 —\u00a0 a=40m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 aplicando Torricelli ainda no meio (2)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 2.a (h \u2013 0,2)\u00a0 —\u00a0 8<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0+2.40.(h \u2013 0,2)\u00a0 —\u00a0 h=0,8 + 0,2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0h=1,0m<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span><\/span>\u00a0Estando o iceberg flutuando, ele est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 P=E\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>gelo<\/b><\/span><\/span><\/span><\/sub>.g=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>gelo<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>gelo<\/b><\/span><\/span><\/span><\/sub>=d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>gelo<\/b><\/span><\/span><\/span><\/sub>\/d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>=V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>\/V<\/b><\/span><\/span><\/span>gelo<\/b><\/span><\/span><\/span><\/sub>\u00a0 — 0,92\/1,03= V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/span><\/sub>\/V<\/b><\/span><\/span><\/span>gelo<\/b><\/span><\/span><\/span><\/sub>\u00a0\u00a0 —\u00a0 0,893=<\/b><\/span><\/span><\/span>fra\u00e7\u00e3o submersa\/fra\u00e7\u00e3o total=89,3%<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span><\/span>\u00a0a) A massa do recipiente, da \u00e1gua e do barquinho sobre a balan\u00e7a \u00e9 a mesma, quer o barquinho esteja flutuando, quer esteja submerso. Portanto,\u00a0<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0= M<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\u00a0.<\/b><\/span><\/span><\/span><\/p>\n

b) Quando o barquinho est\u00e1 flutuando, o empuxo sobre ele \u00e9 igual a seu peso e, portanto, maior do que o empuxo quando submerso. Sendo maior o empuxo no barquinho flutuando, o volume da \u00e1gua por ele deslocado nesse caso \u00e9 maior do que o volume da \u00e1gua por ele deslocado quando est\u00e1 submerso. Uma vez que o volume dentro do recipiente sob o n\u00edvel da superf\u00edcie livre da \u00e1gua \u00e9 o volume da \u00e1gua acrescido do volume de \u00e1gua deslocado, conclu\u00edmos que o volume dentro do recipiente sob o n\u00edvel da superf\u00edcie livre \u00e9 maior com o barquinho flutuando do que com o barquinho submerso. Assim,a altura da superf\u00edcie livre com o barquinho flutuando \u00e9 maior do que a\u00a0 altura da superf\u00edcie livre com o barquinho submerso, ou seja,<\/b><\/span><\/span><\/span>\u00a0h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0> h<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\u00a0.<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span><\/span>\u00a0O acr\u00e9scimo de peso na balan\u00e7a corresponde ao empuxo, que \u00e9 igual ao peso do volume de l\u00edquido deslocado e, observe, que ele \u00e9 maior em 2 do que em 3\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Sem carga\u00a0 —\u00a0 P=E=d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span><\/sub>.g=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.S.d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>.10=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>(20.5).d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>.10\u00a0 —\u00a0 P=10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup>.d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 com carga de 10 autom\u00f3veis\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P\u2019=10×1.200×10=12.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>N\u00a0 —\u00a0 E\u2019=P + 12.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>.S.d=P + 12.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.(20.5).d=10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup>.d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\u00a0+ 12.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup>d \u2013 10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup>d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=12.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup>(d \u2013 d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>)=12.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 (d \u2013 d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>)=12.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>\/10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 (d \u2013 d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>)=12.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup>m=12cm\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span><\/span>\u00a0a)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) A redu\u00e7\u00e3o da indica\u00e7\u00e3o do dinam\u00f4metro representa o empuxo E=0,075N=75.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>N\u00a0 —\u00a0 E=d.V.g\u00a0 —\u00a0 75.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>=<\/b><\/span><\/span><\/span><\/p>\n

=d.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/span><\/sup>.10\u00a0 —\u00a0 d=75.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>\/10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>d=7,5.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup><\/p>\n

34-<\/b><\/span><\/span><\/span><\/span>\u00a0C\u00e1lculo do empuxo que age sobre cada bexiga\u00a0 —\u00a0 E=d.V.g=1,2.2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>.10\u00a0 —\u00a0 E=2,4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup>N\u00a0 —\u00a0 peso da menina\u00a0 —\u00a0 P=mg=24.10\u00a0 —\u00a0 P=240N\u00a0 —\u00a0 n\u00famero de bexigas\u00a0 —\u00a0 n=240\/2,4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 n=240\/0,024\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 n=10.000 bexigas<\/b><\/span><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em ambos os casos o peso do sistema \u00e9 equilibrado pelo empuxo aplicado pela \u00e1gua.<\/b><\/span><\/span><\/p>\n

P = E = d<\/b><\/span><\/span>ll<\/b><\/span><\/span><\/sub>\u00a0V g \u00a0em ambos os casos o volume imerso \u00e9 o mesmo e a altura h n\u00e3o se altera.<\/b><\/span><\/span><\/p>\n

V (1\u00ba caso) = V (2\u00ba caso)\u00a0 —\u00a0 3\/5.5V=x\u00a0 —\u00a0 x=3V\u00a0 —\u00a0\u00a0 como o bloco menor tem volume V, ent\u00e3o, um volume\u00a0 2V do bloco maior ficar\u00e1 imerso, o que corresponde a uma fra\u00e7\u00e3o y do volume total (5V) dada por\u00a0 —\u00a0 y=2V\/5V\u00a0 —\u00a0 y=2\/5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Colocando as for\u00e7as que agem sobre o cilindro\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=P\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>.g + \u03c1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>.g = \u03c1(V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>).g\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>.S.h\/3 + \u03c1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>.S.2h\/3= \u03c1.S.h\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\/3 + \u03c1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>.(2\/3)= \u03c1<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/sub>\u00a0—\u00a0 \u03c1= (\u03c1<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0+ 2\u03c1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>)\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span><\/span>\u00a0Como a vela se mant\u00e9m sempre em equil\u00edbrio \u00e0 medida que vai queimando\u00a0 —\u00a0 E=P\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>\u00e1gua.<\/b><\/span><\/span><\/span><\/sub>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span><\/sub>.g = d<\/b><\/span><\/span><\/span>chumbo<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>chumbo<\/b><\/span><\/span><\/span><\/sub>.g + d<\/b><\/span><\/span><\/span>vela<\/b><\/span><\/span><\/span><\/sub>.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span><\/sub>.g\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>.S.(e + y).g=d<\/b><\/span><\/span><\/span>chumbo<\/b><\/span><\/span><\/span><\/sub>.S.e.g + d<\/b><\/span><\/span><\/span>vela<\/b><\/span><\/span><\/span><\/sub>.S.(x + y).g\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>e + d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>y=d<\/b><\/span><\/span><\/span>chumbo<\/b><\/span><\/span><\/span><\/sub>e + d<\/b><\/span><\/span><\/span>vela<\/b><\/span><\/span><\/span><\/sub>x + d<\/b><\/span><\/span><\/span>vela<\/b><\/span><\/span><\/span><\/sub>y\u00a0 —\u00a0<\/b><\/span><\/span><\/span>(d<\/b><\/span><\/span><\/span>chumbo\u00a0<\/b><\/span><\/span><\/span><\/sub>\u2013d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>)e + d<\/b><\/span><\/span><\/span>vela<\/b><\/span><\/span><\/span><\/sub>x = (d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/span><\/sub>\u00a0\u2013 d<\/b><\/span><\/span><\/span>vela\u00a0<\/b><\/span><\/span><\/span><\/sub>)y\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>observe nessa express\u00e3o que, se x diminui, y tamb\u00e9m diminui ( \u00e0 medida que a vela queima,em rela\u00e7\u00e3o \u00e0 superf\u00edcie da \u00e1gua, a altura da chama (x) diminui e a parte imersa (Y) tamb\u00e9m diminui e sobe)\u00a0 —\u00a0 quando quando a chama chega \u00e0 superf\u00edcie da \u00e1gua (x=0), ainda existe parte imersa, pois, y\u22600\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span><\/span>\u00a0Se o peso do conjunto (boia + flutuador) \u00e9 desprez\u00edvel, sobre ele, com volume imerso V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span><\/sub>=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

age apenas o empuxo (E), vertical e para cima\u00a0 —\u00a0 sobre a alavanca, no ponto C, reage a for\u00e7a da v\u00e1lvula (F), horizontal e para a esquerda\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

E=d.V.g=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>.10\u00a0 —\u00a0 E=10N\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>BC<\/b><\/span><\/span><\/span><\/sub>=1\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/span><\/sub>=5\u00a0 —\u00a0 a soma dos momentos das for\u00e7as E e F em rela\u00e7\u00e3o ao polo B deve ser nula\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span><\/sub>\u00a0+ M<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span><\/sub>=0\u00a0 —\u00a0 E.d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/span><\/sub>=F.d<\/b><\/span><\/span><\/span>BC\u00a0<\/b><\/span><\/span><\/span><\/sub>\u00a0—\u00a0 10.5=F.1\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F=50N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span><\/span>\u00a0For\u00e7as que agem sobre o submarino:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(01)- d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>=3d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 c\u00e1lculoda for\u00e7a de tra\u00e7\u00e3o no fio\u00a0 —\u00a0 equil\u00edbrio\u00a0 —\u00a0 E= P + T\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/sub>.Vg= d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>Vg + T\u00a0 —\u00a0\u00a0 3d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>Vg –\u00a0 d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>Vg = T\u00a0 —\u00a0 T=2d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>Vg (tra\u00e7\u00e3o \u00e0 que o fio est\u00e1 submetido)\u00a0 —\u00a0 o fio suporta\u00a0 —\u00a0 T=3P\u00a0 —\u00a0 T=3d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>Vg\u00a0 —\u00a0 a tra\u00e7\u00e3o que ele suporta \u00e9 maior que a traa\u00e7\u00e3o \u00e0 que ele est\u00e1 submetido\u00a0 —\u00a0 o fio n\u00e3o n\u00e3o rompe\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/span><\/p>\n

(02)- Verdadeira- Veja (01)<\/b><\/span><\/span><\/span><\/p>\n

(04)- E=3d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>Vg\u00a0 —\u00a0 P=d<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/span><\/sub>Vg\u00a0 —\u00a0 o m\u00f3dulo do empuxo \u00e9 3 vezes maior que o m\u00f3dulo da for\u00e7a peso\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/span><\/p>\n

(08)- Verdadeira \u2013 veja (01)<\/b><\/span><\/span><\/span><\/p>\n

(16) Falsa \u2013 massas diferentes, pois as densidades s\u00e3o diferentes<\/b><\/span><\/span><\/span><\/p>\n

(32) Falsa \u2013 empuxo \u00e9 igual ao peso do volume de \u00e1gua deslocada e P=mg<\/b><\/span><\/span><\/span><\/p>\n

(02) + (08) = 20<\/b><\/span><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span><\/span>\u00a0a) P=P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/span><\/sub>\u00a0+ d.g.h=1.0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>.10.2\u00a0 —\u00a0 P=1,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 0,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P=1,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/span><\/sup>Nm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/p>\n

b)For\u00e7as que agem sobre o homem:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=T=40N\u00a0 —\u00a0 T + E=P\u00a0 —\u00a0 40 + d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>Vg=mg\u00a0 —\u00a0 40 + 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.V.10=80.10\u00a0 —\u00a0 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>V=800 \u2013 40\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0V=760.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=76L<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como as bolas possuem densidades diferentes, sofrem empuxos diferentes e a mais densa fica em baixo e a menos densa em cima, do \u00e1lcool l\u00edquido\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Densidade da crian\u00e7a\u00a0 —\u00a0 d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=m\/V\u00a0 —\u00a0 quando inspira a crian\u00e7a aumenta seu volume diminuindo sua densidade\u00a0 —\u00a0 sendo o peso do ar inspirado praticamente desprez\u00edvel, o peso da crian\u00e7a n\u00e3o se altera\u00a0 —\u00a0 E=P\u00a0 —\u00a0 d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>.g=d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>corpo\u00ad<\/b><\/span><\/span><\/sub>\/dagua\u00a0 —\u00a0 observe nessa express\u00e3o que, se a densidade do corpo diminui, a raz\u00e3o entre o volume imerso (V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>) e o volume total (V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>) tamb\u00e9m diminui\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

43-<\/b><\/span><\/span><\/span>\u00a0Observe na figura a for\u00e7as que agem sobre a esfera totalmente imersa\u00a0 —\u00a0 d<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 5 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>e d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0= 1<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 como a esfera est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 N + E = P\u00a0 —\u00a0 N = P \u2013 E\u00a0 —\u00a0 N = d<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0V g \u2013 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0V g \u00a0—\u00a0 N = (d<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>)V g\u00a0 —\u00a0 N\/P=(d<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0\u2013 d\u00ad<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>)Vg\/d<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>Vg\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

N\/P=(5 \u2013 1)\/5\u00a0 —\u00a0 N\/P=0,8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

44-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A figura mostra as for\u00e7as agindo nos objetos A e B\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0m\u00f3dulos dos pesos de A e B,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>\u00a0 — \u00a0m\u00f3dulo do peso da \u00e1gua em cada recipiente\u00a0 —\u00a0 E<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e E<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>: m\u00f3dulos dos empuxos aplicados pela \u00e1gua nos objetos A e B, respectivamente\u00a0 —\u00a0 N<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e N<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>: m\u00f3dulos das rea\u00e7\u00f5es aos empuxos em A e B, respectivamente\u00a0 —\u00a0 T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>: m\u00f3dulos das for\u00e7as que as hastes exercem em A e B, respectivamente\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0s\u00e3o os volumes de A e B, respectivamente\u00a0 —\u00a0 como esses volumes s\u00e3o iguais\u00a0 —\u00a0 V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0= V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0= V\u00a0 —\u00a0 d<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>\u00ad, d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 s\u00e3o as densidades da \u00e1gua, e dos objetos A e B, respectivamente\u00a0 —\u00a0 m<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>\u00a0\u00e9 a massa de \u00e1gua contida em cada recipiente, ambas iguais\u00a0 —\u00a0 g \u00e9 a acelera\u00e7\u00e3o da gravidade local\u00a0 —\u00a0 analisando cada uma das afirmativas:<\/b><\/span><\/span><\/p>\n

(01) Errada\u00a0 —\u00a0 a indica\u00e7\u00e3o da \u201cbalan\u00e7a\u201d \u00e9 a intensidade resultante da for\u00e7a normal aplicada no prato\u00a0 —\u00a0 cada um dos pratos recebe uma normal devida ao peso da \u00e1gua e outra normal devida ao empuxo\u00a0 —\u00a0 lembre-se que: {E = d<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>\u00a0V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\u00a0g} e {P = d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\u00a0V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\u00a0g}\u00a0 —\u00a0 como os corpos est\u00e3o totalmente imersos, V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\u00a0= V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\u00a0 — \u00a0indica\u00e7\u00e3o da \u201cbalan\u00e7a\u201d da esquerda\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>, as duas balan\u00e7as fornecem a mesma indica\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

(02) Correta\u00a0 —\u00a0 como os objetos est\u00e3o em equil\u00edbrio, as for\u00e7as atuantes em cada um deles est\u00e3o equilibradas. Note que o objeto A, de corti\u00e7a, tem densidade menor que a da \u00e1gua. Por isso a tend\u00eancia dele \u00e9 flutuar. Logo a haste exerce nele for\u00e7a de compress\u00e3o para baixo\u00a0\"\". J\u00e1, o objeto B, de chumbo, mais denso que a \u00e1gua, tende a afundar. Assim, a haste exerce nele for\u00e7a de tra\u00e7\u00e3o para cima\u00a0\"\"\u00a0\u00a0—\u00a0 o chumbo tem densidade bem superior \u00e0 da \u00e1gua, certamente mais que o dobro. Assim, a diferen\u00e7a entre as densidades do chumbo e da \u00e1gua \u00e9 maior que 1\u00a0 —\u00a0 (d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 d<\/b><\/span><\/span>\u00e1g<\/b><\/span><\/span><\/sub>\u00ad) > 1\u00a0 —\u00a0 T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0< T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.
\n(04) Errada\u00a0 —\u00a0 como j\u00e1 destacado em (01), os volumes deslocados de \u00e1gua s\u00e3o iguais, portanto a \u00e1gua exerce for\u00e7as de mesma intensidade nos dois objetos.<\/b><\/span><\/span><\/p>\n

(08) Errada\u00a0 —\u00a0 antes, a indica\u00e7\u00e3o de cada \u201cbalan\u00e7a\u201d era apenas a de massa de \u00e1gua, a mesma nas duas \u201cbalan\u00e7as\u201d. Fazendo-se a imers\u00e3o dos corpos, como os empuxos s\u00e3o iguais, os acr\u00e9scimos nas duas balan\u00e7as tamb\u00e9m ser\u00e3o iguais.<\/b><\/span><\/span><\/p>\n

(16) Correta. E empuxo \u00e9 igual ao peso de l\u00edquido deslocado. Assim, o acr\u00e9scimo de massa corresponde a massa de \u00e1gua deslocada.\u00a0<\/b><\/span><\/span><\/p>\n

R- (02 + 16) = 18<\/b><\/span><\/span><\/p>\n

45-\u00a0<\/b><\/span><\/span><\/span>(01) Errada\u00a0 —\u00a0 o l\u00edquido menos denso fica em cima e sua superf\u00edcie livre fica num n\u00edvel mais alto\u00a0 —\u00a0 assim: d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0< d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

(02) Correta\u00a0 —\u00a0 note na figura que o desn\u00edvel entre os pontos 1 e 3 (h<\/b><\/span><\/span>1,3<\/b><\/span><\/span><\/sub>) \u00e9 igual ao desn\u00edvel entre os pontos 2 e 4 (h<\/b><\/span><\/span>2,4<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 h<\/b><\/span><\/span>1,3<\/b><\/span><\/span><\/sub>\u00a0= h<\/b><\/span><\/span>2,4<\/b><\/span><\/span><\/sub>\u00a0= h\u00a0 —\u00a0 pelo teorema de Stevin, as press\u00f5es nos pontos 3 e 4 s\u00e3o iguais\u00a0 —\u00a0 p<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0= p<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 p<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0g h = p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0g h\u00a0 —\u00a0 subtraindo membro a membro\u00a0 —\u00a0 p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 p<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= (d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>) gh\u00a0 —\u00a0 como d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0> d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 p<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0> 0\u00a0 —\u00a0 p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0> p<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

(04) Errada\u00a0 —\u00a0 os pontos 5 e 6 est\u00e3o no mesmo l\u00edquido e pertencem \u00e0 mesma horizontal\u00a0 —\u00a0 logo, est\u00e3o sob mesma press\u00e3o.<\/b><\/span><\/span><\/p>\n

(08) Correta\u00a0 —\u00a0 embre-se que\u00a0 —\u00a0 E = d<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>\u00a0V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\u00a0g\u00a0 —\u00a0 como d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0< d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0, num mesmo corpo totalmente imerso, o l\u00edquido B exerce menor empuxo.<\/b><\/span><\/span><\/p>\n

(16) Correta\u00a0 —\u00a0 vide (02).\u00a0<\/b><\/span><\/span><\/p>\n

R- (02 + 08 + 16) = 26<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 — \u00a0\u03c1<\/b><\/span><\/span>iceberg<\/b><\/span><\/span><\/sub>\/\u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=0,90\u00a0 — \u00a0A = 30 km<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h<\/b><\/span><\/span>emersa<\/b><\/span><\/span><\/sub>\u00a0= 100 m\u00a0 —\u00a0 como o iceberg est\u00e1 em equil\u00edbrio, a resultante de for\u00e7as nele agindo (peso e empuxo) \u00e9 nula.<\/b><\/span><\/span><\/p>\n

E = P\u00a0 — \u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>submerso<\/b><\/span><\/span><\/sub>\u00a0g = \u03c1<\/b><\/span><\/span>iceberg<\/b><\/span><\/span><\/sub>V g \u00a0—\u00a0 V<\/b><\/span><\/span>submerso<\/b><\/span><\/span><\/sub>\/V=\u03c1<\/b><\/span><\/span>iceberg<\/b><\/span><\/span><\/sub>\/\u03c1<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=0,9\u00a0 —\u00a0 \u00a0A.h<\/b><\/span><\/span>submersa<\/b><\/span><\/span><\/sub>\/A(h<\/b><\/span><\/span>submersa<\/b><\/span><\/span><\/sub>\u00a0+ h<\/b><\/span><\/span>emersa<\/b><\/span><\/span><\/sub>)=0,9\u00a0 —\u00a0 h<\/b><\/span><\/span>submersa<\/b><\/span><\/span><\/sub>\/(h<\/b><\/span><\/span>submersa\u00a0<\/b><\/span><\/span><\/sub>+ 100)=0,9\u00a0 —\u00a0 h<\/b><\/span><\/span>submersa<\/b><\/span><\/span><\/sub>\u00a0\u00a0 =\u00a0\u00a0 0,9 h<\/b><\/span><\/span>submersa<\/b><\/span><\/span><\/sub>\u00a0+ 90\u00a0 —\u00a0 0,1 h<\/b><\/span><\/span>submersa<\/b><\/span><\/span><\/sub>\u00a0= 90\u00a0 —\u00a0 h<\/b><\/span><\/span>submersa<\/b><\/span><\/span><\/sub>= 900 m = 0,9 km\u00a0 —\u00a0 volume submerso\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>submerso<\/b><\/span><\/span><\/sub>\u00a0= Axh<\/b><\/span><\/span>submersa<\/b><\/span><\/span><\/sub>\u00a0= 30×0,9\u00a0 —\u00a0 V<\/b><\/span><\/span>submerso<\/b><\/span><\/span><\/sub>\u00a0= 27 km<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.\u00a0<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>I. Errada\u00a0 —\u00a0 se colocado em \u00e1gua doce o ovo vai para o fundo, \u00e9 porque ele \u00e9 mais denso que ela\u00a0 —\u00a0 d<\/b><\/span><\/span>ovo<\/b><\/span><\/span><\/sub>\u00a0> d<\/b><\/span><\/span>\u00e1guadoce<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 d<\/b><\/span><\/span>\u00e1guadoce<\/b><\/span><\/span><\/sub>\u00a0< d<\/b><\/span><\/span>ovo<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0se na \u00e1gua salgada o ovo flutua\u00a0 —\u00a0 d<\/b><\/span><\/span>ovo<\/b><\/span><\/span><\/sub>\u00a0< d<\/b><\/span><\/span>\u00e1guasalgada<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 comparando as duas conclus\u00f5es\u00a0 —\u00a0 \u00a0d<\/b><\/span><\/span>doce<\/b><\/span><\/span><\/sub>\u00a0< d<\/b><\/span><\/span>ovo<\/b><\/span><\/span><\/sub>\u00a0< d<\/b><\/span><\/span>\u00e1guasalgada<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 o empuxo \u00e9 igual ao peso de l\u00edquido deslocado.<\/b><\/span><\/span><\/p>\n

III. Errada\u00a0 —\u00a0 a press\u00e3o n\u00e3o afeta, pois qualquer varia\u00e7\u00e3o \u00e9 transmitida integralmente em todas as dire\u00e7\u00f5es a todos os pontos do l\u00edquido.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

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

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe as figuras abaixo\u00a0 — \u00a0d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>, a densidade da \u00e1gua\u00a0 — \u00a0d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>, a densidade da bola\u00a0 —\u00a0 \u00a0V, o volume da bola\u00a0 —\u00a0 \u00a0V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, o volume imerso da bola\u00a0 —\u00a0 na Fig. 1 o sistema est\u00e1 em repouso\u00a0 —\u00a0 o empuxo \u00e9 igual ao peso da bola\u00a0 —\u00a0 E=P\u00a0 —\u00a0 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>gV<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=mg\u00a0 —\u00a0 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>V\u00a0 —\u00a0 d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/V (I)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

considere, agora, a Fig 2 para o caso do movimento ter acelera\u00e7\u00e3o vertical\u00a0 —\u00a0 para tal, vamos considerar a \u00e1gua um fluido incompress\u00edvel e desprezar eventuais for\u00e7as de viscosidade\u00a0 —\u00a0 considere, por exemplo, que a acelera\u00e7\u00e3o do elevador seja a para cima\u00a0 —\u00a0 \u00a0camada de \u00e1gua em contato com o fundo do recipiente recebe desse fundo uma for\u00e7a normal\u00a0\"\"\u00a0de intensidade\u00a0 maior<\/b><\/span><\/span><\/p>\n

que o peso, para haver acelera\u00e7\u00e3o\u00a0 —\u00a0 N \u2013 P = m a\u00a0 —\u00a0 N \u2013 m g = ma\u00a0 —\u00a0 N = m(a + g)\u00a0 —\u00a0 portanto, a press\u00e3o hidrost\u00e1tica no fundo do recipiente tamb\u00e9m aumenta, assim como em todos os pontos do l\u00edquido\u00a0 —\u00a0 ou seja, numa profundidade h a press\u00e3o hidrost\u00e1tica passa a ser\u00a0 —\u00a0 p = d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0(g + a) h\u00a0 —\u00a0 o empuxo ocorre pela diferen\u00e7a de press\u00e3o entre as faces superior e inferior do corpo, que \u00e9 a pr\u00f3pria press\u00e3o hidrost\u00e1tica\u00a0 —\u00a0 como press\u00e3o \u00e9 o produto da for\u00e7a pela \u00e1rea\u00a0 —\u00a0 E\u2019 = \u0394p.A\u00a0 —\u00a0 E\u2019 = d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>(g + a).(h<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>.A)\u00a0 —\u00a0 nessa express\u00e3o, h<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0\u00e9 a altura imersa do corpo e A \u00e9 a \u00e1rea de atua\u00e7\u00e3o do empuxo\u00a0 —\u00a0 mas o produto (h<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>.A) = ao volume imerso (V\u2019)\u00a0 —\u00a0 dessa forma, para o corpo acelerando para cima, o empuxo \u00e9\u00a0 —\u00a0 E\u2019 = d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>V\u2019(g + a)\u00a0 —\u00a0 \u00a0princ\u00edpio fundamental da din\u00e2mica\u00a0 —\u00a0 E\u2019 \u2013 P = m a\u00a0 —\u00a0 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>V\u2019\u00ad(g + a ) \u2013 m g = m a\u00a0 —\u00a0 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>V\u2019\u00ad(g + a ) = d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>V(g + a)\u00a0 —\u00a0 d<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=V\u2019\/V (II)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

comparando (I) e (II), voc\u00ea conclui que V\u2019 = V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 se o elevador acelerasse para baixo, voc\u00ea obteria o mesmo resultado\u00a0 —\u00a0 assim, o volume imerso (V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) independe da acelera\u00e7\u00e3o do elevador\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Nos recipientes II e III, os volumes deslocados de \u00e1gua s\u00e3o iguais aos volumes das por\u00e7\u00f5es imersas de gelo e de bolas\u00a0 —\u00a0 \u00a0\u00a0volume imerso de gelo = volume de \u00e1gua deslocado pelo gelo\u00a0 —\u00a0 volume imerso de bolas = volume de \u00e1gua deslocado pelas bolas\u00a0 —\u00a0 VII\u00a0 —\u00a0 volume de \u00e1gua no recipiente II\u00a0 —\u00a0 VIII\u00a0 —\u00a0 volume de \u00e1gua no recipiente III\u00a0 —\u00a0 como nos tr\u00eas recipientes a \u00e1gua est\u00e1 no mesmo n\u00edvel, nos recipientes II e III os volumes de \u00e1gua somados aos volumes imersos d\u00e3o o mesmo volume de \u00e1gua (VI) contida no recipiente I\u00a0 —\u00a0 VII + Vgelo = VI\u00a0 —\u00a0 \u00a0VIII + Vbolas = VI\u00a0 —\u00a0 o empuxo \u00e9 igual ao peso do volume de \u00e1gua deslocado (E = d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>desl<\/b><\/span><\/span><\/sub>.g)\u00a0 —\u00a0 recipiente I\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u00a0PI<\/b><\/span><\/span>\u00a0= P<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>= d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.VI.g\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 recipiente II\u00a0 —\u00a0 PII = P<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 PII = d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.VII.g\u00a0 + d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 PII = d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.(VII + V<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>).g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>PII = d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.VI.g<\/b><\/span><\/span>\u00a0 —\u00a0 recipiente III\u00a0 —\u00a0 PIII = P<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>bola\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 PIII = d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.VIII.g + d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>bolas<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 PIII = d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.(VIII + V<\/b><\/span><\/span>bolas<\/b><\/span><\/span><\/sub>).g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>PIII = d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.VI.g<\/b><\/span><\/span>\u00a0 —\u00a0 \u00a0PI = PII = PIII\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Dados: d<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>\u00a0= 0,920 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>; d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0= 1,025 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 for\u00e7as que agem sobre o cone de gelo\u00a0 —\u00a0 peso<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0e empuxo\u00a0 —\u00a0 a fra\u00e7\u00e3o submersa\u00a0 de volume do cone vale\u00a0 —\u00a0 f=V<\/b><\/span><\/span>subm<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 como ele est\u00e1 em equil\u00edbrio E=P\u00a0 —\u00a0 d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>.g = d<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\u00a0= f=d<\/b><\/span><\/span>gelo<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 f=0,920\/1,025\u00a0 —\u00a0 f=0,898\u00a0 —\u00a0 f=89,8%\u00a0 —\u00a0 se o cone fosse invertido, essa fra\u00e7\u00e3o continuaria a mesma, pois o empuxo seria o mesmo, resultando na mesma equa\u00e7\u00e3o do item anterior que mostra que a fra\u00e7\u00e3o imersa depende apenas das densidades do gelo e da \u00e1gua, que s\u00e3o constantes.<\/b><\/span><\/span><\/p>\n

b) Os fatores mencionados (varia\u00e7\u00f5es da acelera\u00e7\u00e3o da gravidade e da press\u00e3o atmosf\u00e9rica) em nada afetam o experimento\u00a0 —\u00a0 a justificativa est\u00e1 na pr\u00f3pria express\u00e3o encontrada no item anterior que mostra que a fra\u00e7\u00e3o imersa depende apenas das densidades do gelo e da \u00e1gua, que independem das varia\u00e7\u00f5es da press\u00e3o e da acelera\u00e7\u00e3o da gravidade.<\/b><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) A densidade \u00e9 inversamente proporcional ao volume\u00a0 —\u00a0 o gr\u00e1fico \u00a0mostra que quando a temperatura diminui, at\u00e9 4 \u00b0C ocorre diminui\u00e7\u00e3o do volume, portanto um aumento da densidade\u00a0 —\u00a0 ou seja, quando a temperatura diminui correntes \u00a0de convec\u00e7\u00e3o\u00a0 formam-se na \u00e1gua\u00a0 —\u00a0 \u00a0a \u00e1gua da superf\u00edcie se resfria, aumenta a densidade e desce, subindo a \u00e1gua que est\u00e1 no fundo\u00a0 —\u00a0 por\u00e9m, quando a temperatura baixa de 4\u00b0C o volume come\u00e7a a aumentar diminuindo a densidade, cessando o processo de convec\u00e7\u00e3o\u00a0 —\u00a0 assim, a \u00e1gua da superf\u00edcie congela, e o gelo tamb\u00e9m \u00e9 menos denso que a \u00e1gua, ficando na superf\u00edcie\u00a0 —\u00a0 essa camada de gelo, que \u00e9 um isolante t\u00e9rmico, impede a perda de calor da \u00e1gua que est\u00e1 abaixo para o meio ambiente.<\/b><\/span><\/span><\/p>\n

b) T = 0,2 N\u00a0 —\u00a0 V = 1.000 cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 d = 998 kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 d\u2019 = 1.000 kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 na figura, a 20<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C \u2013<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

equil\u00edbrio\u00a0 —\u00a0P + T = E\u00a0 —\u00a0 P = d V g \u2013 T\u00a0 —\u00a0 P\u00a0 = 998.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10\u00a0 \u2013 0,2\u00a0 —\u00a0 P = 9,98 \u2013 0,2\u00a0 —\u00a0 P = 9,78 N\u00a0 —\u00a0 a 4\u00b0C\u00a0 —\u00a0 P + T\u2019 = E\u2019\u00a0 —\u00a0 T\u2019 = E\u2019 \u2013 P\u00a0 —\u00a0 T\u2019 = d\u2019 V g \u2013 P\u00a0 —\u00a0 T\u2019 = 1.000.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10 \u2013 9,78\u00a0 —\u00a0 T\u2019= 10 \u2013 9,78\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T\u201d = 0,22 N.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>V = 26 cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 26.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 d = 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 g = 9,8 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 — E = d<\/b><\/span><\/span>liq<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 E = (10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>).(26.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>).(9,8)\u00a0 —\u00a0 E = 0,2548 N\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O empuxo \u00e9 uma for\u00e7a vertical, para cima que o l\u00edquido exerce sobre um corpo nele imerso e \u00e9 avaliado pelo princ\u00edpio de Arquimedes: o empuxo tem a mesma intensidade do peso de l\u00edquido deslocado.\u00a0<\/b><\/span><\/span><\/p>\n

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

54-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Se o volume emerso \u00e9 1\/8 do volume do corpo, o volume imerso \u00e9 7\/8 desse volume\u00a0 —\u00a0 como o corpo est\u00e1 em equil\u00edbrio, o peso e o empuxo t\u00eam a mesma intensidade\u00a0 —\u00a0 P=E\u00a0 —\u00a0 d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>.g=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\/1=(7\/8)V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>=7\/8\u00a0 —\u00a0 d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>=0,875 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

55-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>P\/E=12,5\u00a0 —\u00a0 E=P\/12,5\u00a0 —\u00a0 princ\u00edpio fundamental da din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P \u2013 E = ma\u00a0 —\u00a0 mg \u2013 (mg\/12,5) = ma\u00a0 —\u00a0 10 \u2013 10\/12,5=a\u00a0 —\u00a0 a=10 \u2013 0,8\u00a0 —\u00a0 a=9,2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

56-\u00a0<\/b><\/span><\/span>Veja figuras\u00a0 —\u00a0 a = 4 cm\u00a0 —\u00a0 \u00a0d\u00e1gua = 1 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 A<\/b><\/span><\/span>imersa<\/b><\/span><\/span><\/sub>\u00a0= 0,7.A<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u0394h = 0,50 cm\u00a0 —\u00a0 a \u00e1rea<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

imersa \u00e9 a \u00e1rea do fundo mais uma parte da \u00e1rea das 4 paredes laterais, de altura h\u00a0 —\u00a0 de acordo com o enunciado\u00a0 —\u00a0 A<\/b><\/span><\/span>imersa<\/b><\/span><\/span><\/sub>=0,7\u00aa<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 4ah=0,7.6.a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

4 h = 3,2 a\u00a0 —\u00a0 h = 0,8 a\u00a0 —\u00a0 como o cubo \u00e9 um s\u00f3lido reto e est\u00e1 em equil\u00edbrio em \u00e1gua, seu peso \u00e9 equilibrado pelo empuxo:<\/b><\/span><\/span><\/p>\n

P = E\u00a0 —\u00a0 d<\/b><\/span><\/span>cubo<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>.g = d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 d<\/b><\/span><\/span>cubo<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>cubo<\/b><\/span><\/span><\/sub>\/1=a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>h\/a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>cubo<\/b><\/span><\/span><\/sub>=a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.0,8\u00aa\/a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>cubo<\/b><\/span><\/span><\/sub>=0,80 g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>\u00a0\u00a0— o aumento do empuxo equilibra\u00a0 o peso da r\u00e3\u00a0 —\u00a0 P<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=\u0394E\u00a0 —\u00a0 mg=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.\u0394V.g\u00a0 —\u00a0 m=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.A.\u0394h\u00a0 —\u00a0 m=1,0.16.0,5\u00a0 —\u00a0 m=8,0g\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

57-<\/b><\/span><\/span><\/span> I. Falsa\u00a0 —\u00a0 enunciado do princ\u00edpio de Arquimedes: \u201cTodo corpo total ou parcialmente mergulhado num l\u00edquido em equil\u00edbrio, recebe uma for\u00e7a de dire\u00e7\u00e3o vertical e sentido\u00a0 para cima denominada de Empuxo, cuja intensidade\u00a0 \u00e9 igual ao peso do volume de l\u00edquido deslocado\u201c<\/b><\/span><\/span><\/p>\n

Caracter\u00edsticas do empuxo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na express\u00e3o acima que a intensidade do empuxo (for\u00e7a para cima) \u00e9 diretamente proporcional \u00e0 densidade do l\u00edquido no qual ele est\u00e1 total ou parcialmente imerso\u00a0 —\u00a0 como a densidade da \u00e1gua do Mar Morto \u00e9 maior que a do Mar Vermelho, a embarca\u00e7\u00e3o sofrer\u00e1 maior empuxo no Mar Morto.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 quanto maior a profundidade maior ser\u00e1 a press\u00e3o e, quanto maior a press\u00e3o maior ser\u00e1 a temperatura de ebuli\u00e7\u00e3o da \u00e1gua\u00a0 —\u00a0 assim, o ponto de ebuli\u00e7\u00e3o da \u00e1gua ao n\u00edvel do Mar Morto \u00e9 maior que o ponto de ebuli\u00e7\u00e3o da \u00e1gua ao n\u00edvel do mar que \u00e9 de 100<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C=292<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>F<\/b><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 pelo texto fornecido a diferen\u00e7a de altura entre o Mar Morto e o Mar Vermelho \u00e9 de h=420m\u00a0 —\u00a0 W<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=<\/b><\/span><\/span><\/p>\n

P.h=m.g.h=1000x10x420=4 200 000J.<\/b><\/span><\/span><\/p>\n

IV. Falsa\u00a0 —\u00a0 a press\u00e3o hidrost\u00e1tica (devida somente \u00e0 altura da coluna l\u00edquida) \u00e9 fornecida por P<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.g.h\u00a0 —\u00a0 observe nesta express\u00e3o que, com g e h s\u00e3o os mesmos, ela depende apenas da densidade do l\u00edquido\u00a0 —\u00a0 ser\u00e1 maior no Mar Morto porque a\u00ed a densidade da \u00e1gua \u00e9 maior.<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

58-<\/b><\/span><\/span><\/span> Volume do cubo\u00a0 —\u00a0 V=0,1.0,1.0,1=10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 quando o cubo est\u00e1 no ar o dinam\u00f4metro no qual ele est\u00e1 suspenso pelo fio, a tra\u00e7\u00e3o no mesmo \u00e9 seu peso\u00a0 — T= P=40N (figura 1)\u00a0 —\u00a0 quando ele se encontra com metade de seu volume imerso num l\u00edquido, ele recebe um empuxo, vertical e para cima, de intensidade\u00a0 —\u00a0 E=d<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>.g .V\/2= d<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>.10.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/2\u00a0 —<\/b><\/span><\/span><\/p>\n

E= d<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>.5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 nesse caso, a for\u00e7a de tra\u00e7\u00e3o no fio \u00e9 a indica\u00e7\u00e3o do dinam\u00f4metro de valor\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 T=32N\u00a0 —\u00a0 como o bloco est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 P = T + E (figura 2)\u00a0 —\u00a0 40 = 32 + E\u00a0 —\u00a0 E=8N\u00a0 —\u00a0 8= d<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>.5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>=1,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>= 1,6g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

59- <\/b><\/span><\/span><\/span>Com o dinam\u00f4metro no ar, a indica\u00e7\u00e3o da for\u00e7a de tra\u00e7\u00e3o no fio (indica\u00e7\u00e3o do dinam\u00f4metro) \u00e9 igual ao peso do bloco\u00a0 —\u00a0 T=P=m.g=3.10\u00a0 —\u00a0 P=30N\u00a0 —\u00a0 quando o bloco \u00e9 imerso na \u00e1gua, surge sobre ele um empuxo para cima, tornando-o mais leve (peso aparente) e, nesse caso, a for\u00e7a de tra\u00e7\u00e3o T \u00e9 a indica\u00e7\u00e3o do dinam\u00f4metro, ou seja, T=24N\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 como o empuxo \u00e9 o peso do volume de l\u00edquido deslocado (V\u2019), que \u00e9 igual ao peso da metade do volume (V) do corpo\u00a0 —\u00a0 volume do corpo V=\u2113<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=(10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 metade do volume\u00a0 —\u00a0 V\u2019=(10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/2)m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.V\u2019<\/b><\/span><\/span>l\u00edquidodeslocado<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 E=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.(10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/2).10\u00a0 —\u00a0 E=5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 como o bloco est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 P=T + E\u00a0 —\u00a0 30 = 24 + 5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=(6\/5).10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=1,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=1,2kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=1,2g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

60-<\/b><\/span><\/span><\/span> Com o peixe totalmente imerso na \u00e1gua o m\u00f3dulo do empuxo (for\u00e7a vertical e para cima) \u00e9 igual ao peso do volume do l\u00edquido deslocado — E= P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>\/V\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0 m<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 onde\u00a0 — Analisando as alternativas:<\/b><\/span><\/span><\/p>\n

a) Falsa\u00a0 —\u00a0 observe na express\u00e3o E=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.g que o empuxo E \u00e9 diretamente proporcional ao volume de l\u00edquido deslocado que \u00e9 igual ao volume do peixe.<\/b><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 como a massa n\u00e3o varia o peso tamb\u00e9m n\u00e3o variar\u00e1, pois, P=m.g (m e g s\u00e3o constantes)<\/b><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 a densidade da \u00e1gua \u00e9 a mesma, pr\u00f3ximo ou n\u00e3o do peixe.<\/b><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 quando enche a bexiga natat\u00f3ria de gases, o volume do peixe aumenta sem variar sua massa m\u00a0 —\u00a0 assim, a densidade do peixe diminui, pois d=m\/V (observe nesta express\u00e3o que d \u00e9 inversamente proporcional a V, pois m \u00e9 constante)<\/b><\/span><\/span><\/p>\n

e) Correta\u00a0 —\u00a0 o m\u00f3dulo da for\u00e7a peso da quantidade de \u00e1gua deslocada pelo corpo do peixe aumenta, pois esse m\u00f3dulo \u00e9 igual ao empuxo E.<\/b><\/span><\/span><\/p>\n

61-<\/b><\/span><\/span><\/span> Considere que o ar se comporta como um g\u00e1s ideal e note que o n\u00famero de moles de ar no interior do bal\u00e3o \u00e9 proporcional \u00e0 sua densidade.<\/b><\/span><\/span><\/p>\n

a) Volume do bal\u00e3o\u00a0 —\u00a0 1 L=1 dm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=3.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 c\u00e1lculo do empuxo (for\u00e7a vertical, para cima e de intensidade E= \u03c1<\/b><\/span><\/span>amb<\/b><\/span><\/span><\/sub>.V.g) do bal\u00e3o\u00a0 —\u00a0 E= \u03c1<\/b><\/span><\/span>amb<\/b><\/span><\/span><\/sub>.V.g=1,26.3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10 \u00a0—\u00a0\u00a0<\/b><\/span><\/span>E=3,78.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0N<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

b) Equa\u00e7\u00e3o geral dos gases perfeitos\u00a0 —\u00a0 P.V=n.R.T\u00a0 —\u00a0 P.V=(m\/M).R.T\u00a0 —\u00a0 o enunciado afirma que durante a transforma\u00e7\u00e3o a press\u00e3o P e o volume V permanecem constantes\u00a0 —\u00a0 P.V=constante, ent\u00e3o (m\/M).R.T=constante\u00a0 —\u00a0 como M e R j\u00e1 s\u00e3o constantes, ent\u00e3o o produto da massa m pela temperatura T\u00a0 tamb\u00e9m ser\u00e1 constante\u00a0 —\u00a0 m.T=constante\u00a0 —\u00a0 observe que o enunciado afirma que n\u00famero de moles de ar no interior do bal\u00e3o \u00e9 proporcional \u00e0 sua densidade, o que implica que a massa m tamb\u00e9m ser\u00e1 proporcional \u00e0 densidade \u03c1.T=constante\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>amb<\/b><\/span><\/span><\/sub>.T<\/b><\/span><\/span>amb<\/b><\/span><\/span><\/sub>\u00a0= \u03c1<\/b><\/span><\/span>quente<\/b><\/span><\/span><\/sub>.T<\/b><\/span><\/span>quente<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1.26.300 = 1,05.T<\/b><\/span><\/span>quente<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T<\/b><\/span><\/span>quente<\/b><\/span><\/span><\/sub>=360K<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

62-<\/b><\/span><\/span><\/span> <\/b>Se<\/b><\/span><\/span><\/span>\u00a0um corpo estiver flutuando parcialmente imerso num l\u00edquido, a for\u00e7a resultante sobre ele \u00e9<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

nula (equil\u00edbrio vertical) —\u00a0 P = E\u00a0\u00a0 — P=d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 E=d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>.g\u00a0 — P = E\u00a0 — \u00a0d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>.g=d.V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0\u00a0d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 c\u00e1lculo do raio do tronco\u00a0 —\u00a0 per\u00edmetro=2.\u03c0.R\u00a0 —\u00a0 1,2=2.3.R\u00a0 —\u00a0 R=0,2m\u00a0 —\u00a0 volume total do tronco\u00a0 —\u00a0 V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=\u03c0.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=3.(0,2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=0,36m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 densidade do tronco\u00a0 —\u00a0 d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>=0,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>\u00a0 — 0,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=0,36\/V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>imerso<\/b><\/span><\/span><\/sub>=0,288m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

63-<\/b><\/span><\/span><\/span> Considerando a carga totalmente imersa na \u00e1gua ela sofrer\u00e1 um empuxo (for\u00e7a vertical e para cima) de intensidade\u00a0 —\u00a0 E=d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>carga<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 E=(1,0.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>).(20.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>).(10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0E=200km\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=200N\u00a0 —\u00a0 peso da carga (vertical e para baixo de intensidade\u00a0 — \u00a0 P=mg=50.10\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0 P=500N\u00a0 —\u00a0 sobre a carga agem para cima duas for\u00e7as de tra\u00e7\u00e3o (2T) aplicadas pela corda\u00a0 —\u00a0 sendo a ascens\u00e3o com velocidade constante a for\u00e7a resultante sobre a carga \u00e9 nula\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 P=E + 2T\u00a0 —\u00a0 500=200 + 2T\u00a0 —\u00a0 T=150N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

64-<\/b><\/span><\/span><\/span> a) A for\u00e7a que o ar exerce sobre o bal\u00e3o corresponde ao empuxo (for\u00e7a vertical e para cima) cuja intensidade \u00e9 o peso do volume de ar deslocado pelo bal\u00e3o (que \u00e9 o pr\u00f3prio volume do bal\u00e3o) e fornecido por\u00a0 —\u00a0 E=\u03c1<\/b><\/span><\/span>ar<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>bal\u00e3o<\/b><\/span><\/span><\/sub>.g=1,3.0,5.10\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0E=6,5N.<\/b><\/span><\/span><\/p>\n

b) Dado do exerc\u00edcio\u00a0 —\u00a0 volume m\u00e1ximo do bal\u00e3o\u00a0 —\u00a0 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=0,5m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 o bal\u00e3o cont\u00e9m h\u00e9lio\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>He<\/b><\/span><\/span><\/sub>=0,18kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03c1<\/b><\/span><\/span>He<\/b><\/span><\/span><\/span><\/sub>=m<\/b><\/span><\/span><\/span>He<\/b><\/span><\/span><\/span><\/sub>\/V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 0,18=m<\/b><\/span><\/span><\/span>He<\/b><\/span><\/span><\/span><\/sub>\/0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>He<\/b><\/span><\/span><\/span><\/sub>=0,09kg.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

d) Estando o sistema em equil\u00edbrio, a for\u00e7a resultante sobre o bal\u00e3o \u00e9 nula\u00a0 —\u00a0\u00a0 T + P<\/b><\/span><\/span>bal\u00e3o<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>He<\/b><\/span><\/span><\/sub>\u00a0= E\u00a0 —\u00a0 T + m<\/b><\/span><\/span>bal\u00e3o<\/b><\/span><\/span><\/sub>.g +<\/b><\/span><\/span><\/p>\n

\u03c1<\/b><\/span><\/span>He<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>bal\u00e3o<\/b><\/span><\/span><\/sub>.g = E\u00a0 —\u00a0 T + 0,1.10 + 0,18.0,5.10 = 6,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=4,6N<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

65-<\/b><\/span><\/span><\/span> Empuxo\u00a0 —\u00a0 \u201cTodo corpo imerso num l\u00edquido recebe uma for\u00e7a para cima que \u00e9 igual ao peso do volume do l\u00edquido deslocado\u201d\u00a0 —\u00a0 o volume do l\u00edquido deslocado corresponde a 3\/4 do volume do cilindro (parte imersa)\u00a0 —\u00a0 V<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=(3\/4).S.h=(3\/4).(400.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>).\u00a0 (12.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 V<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=36.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E= \u03c1<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>.g=(0,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>).36.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 E=28,8 N\u00a0 —\u00a0 peso do cilindro\u00a0 —\u00a0 P<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>= mg\u00a0\u00a0 —\u00a0 P<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>.g=\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>.(S.h).g= d<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>.(400.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>).12.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 P<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>= d<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>.48.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 como o cilindro encontra-se em equil\u00edbrio\u00a0 —\u00a0 E = P\u00a0 —\u00a0 28,8 = 48.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>. d<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>=28,8\/48.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>=0,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=0,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0—\u00a0<\/b><\/span><\/span><\/p>\n

d<\/b><\/span><\/span>cilindro<\/b><\/span><\/span><\/sub>=0,6g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

66-<\/b><\/span><\/span><\/span>Todo corpo imerso em um l\u00edquido recebe uma for\u00e7a vertical e para cima denominada Empuxo que obedece \u00e0 seguinte equa\u00e7\u00e3o\u00a0 —E=densidadedol\u00edquidoxvolumedel\u00edquidodeslocadoxacelera\u00e7\u00e3odagravidade\u00a0 — observe que, quanto maior a densidade do l\u00edquido, no caso do mar morto (alta concentra\u00e7\u00e3o salina), maior ser\u00e1 a for\u00e7a do empuxo para cima e com mais facilidade ele flutua\u00a0 —\u00a0\u00a0R- B.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Teorema de Arquimedes – Empuxo 01-\u00a0R- C\u00a0\u2013 Veja teoria 02–\u00a0R- C\u00a0\u2013 (Veja teoria) 03-\u00a0(O1)- Correta \u2013 em seu interior existe ar, que faz diminuir sua densidade m\u00e9dia, ficando menor do que a da \u00e1gua. (02)- Correta \u2013 ele est\u00e1 flutuando e em equil\u00edbrio, ent\u00e3o a for\u00e7a resultante sobre ele \u00e9 nula. (04)- Correta<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1563,"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-1567","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1567","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=1567"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1567\/revisions"}],"predecessor-version":[{"id":10897,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1567\/revisions\/10897"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1563"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1567"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}