{"id":1535,"date":"2015-11-16T15:33:23","date_gmt":"2015-11-16T15:33:23","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1535"},"modified":"2024-08-23T14:30:12","modified_gmt":"2024-08-23T14:30:12","slug":"resolucao-comentada-das-questoes-de-vestibulares-sobre-densidade-e-pressao","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/hidrostatica\/densidade-e-pressao\/resolucao-comentada-das-questoes-de-vestibulares-sobre-densidade-e-pressao\/","title":{"rendered":"Densidade e Press\u00e3o – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

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

01-<\/b><\/span><\/span><\/span>\u00a0d=m\/V=(190 \u2013 50)\/(200\/2)\u00a0 —\u00a0 d=1,5g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a01L=1dm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>-3\u00a0<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 d=m\/V\u00a0 —\u00a0 13,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=m\/10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 m=13,6 kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0a) a densidade da esfera leva em conta o volume total\u00a0 —\u00a0 d=m\/v=50\/30\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=1,7g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

b) a massa espec\u00edfica leva em conta apenas a parte de volume que cont\u00e9m alum\u00ednio\u00a0 —\u00a0 \u03c1=m\/v=\/50(30 \u2013 10)\u00a0 —\u00a0 <\/b><\/span><\/span>\u03c1=2,5g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=2V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 d<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2,80=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/2V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=5,60V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1,60=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1,60V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=(m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)\/(V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 d<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=(5,60V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ 1,60V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)\/3V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=2,4g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0I- Na hora mais quente, o volume aumenta e, num mesmo volume haveria menos massa\u00a0 —\u00a0\u00a0 voc\u00ea levaria prejuiso\u00a0 — Falsa<\/b><\/span><\/span><\/p>\n

II- Com temperatura mais baixa, o volume diminui e, num mesmo volume haveria mais massa\u00a0 —\u00a0 voc\u00ea levaria vantagem\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/p>\n

III- Correta\u00a0 —\u00a0 voc\u00ea estaria comprando o que realmente interessa, que \u00e9 a massa.<\/b><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0figura 1\u00a0 —\u00a0\u00a0 sendo o mesmo volume, o corpo de maior massa ter\u00e1 maior densidade\u00a0 —\u00a0 d<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>>d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>>d<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>>d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 figura 2\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 como a massa \u00e9 a mesma para cada bloco, aquele que tiver maior densidade ter\u00e1 menor volume\u00a0 —\u00a0 V<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub><V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub><V<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub><V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 4(A); 1(B); 3(C) e 2(D)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0volume da mala\u00a0 —\u00a0 dm\u00b3\u00a0 = 20 000 cm\u00b3\u00a0 —\u00a0\u00a0\u00a0 com esse volume, achar a massa de cada metal\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>alum\u00ednio<\/b><\/span><\/span>\u00a0\u2013 d=m\/V\u00a0 —\u00a0 2,7=m\/20.000\u00a0 —\u00a0 m=54.000g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=54kg<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>zinco<\/b><\/span><\/span>\u00a0 —\u00a0 7,1=m\/20.000\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0m=141kg<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>prata\u00a0 —\u00a0 m=210kg<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>chumbo\u00a0 —\u00a0 m=228kg<\/b><\/span><\/span>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>0uro\u00a0 —\u00a0 m=388kg<\/b><\/span><\/span>\u00a0 —\u00a0 como ele s\u00f3 pode correr com uma massa igual \u00e0 sua pr\u00f3pria massa (70kg), a alternativa que mais satisfaz \u00e9 o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>alum\u00ednio (54kg)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0Em cada litro de \u00e1lcool hidratado t\u00eam-se\u00a0 —\u00a0 0,96L de \u00e1lcool (96%) e 0,04L de \u00e1gua (4%)\u00a0 —\u00a0 massa de \u00e1lcool\u00a0 —\u00a0 d<\/b><\/span><\/span>\u00e1lcool<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>\u00e1lcool<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>\u00e1lcool<\/b><\/span><\/span><\/sub>\u00a0 — 800g\/L=m<\/b><\/span><\/span>\u00e1lcool<\/b><\/span><\/span><\/sub>\/0,96L\u00a0 —\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>\u00e1lcool<\/b><\/span><\/span><\/sub>=768g<\/b><\/span><\/span>\u00a0 —\u00a0 massa de \u00e1gua\u00a0 —\u00a0 d<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1.000g\/L=m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>\/0,04L\u00a0 —\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>=40g<\/b><\/span><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=(m<\/b><\/span><\/span>\u00e1lcool<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>)\/V<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=(768 + 40)\/1\u00a0 —\u00a0 d<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=808g\/L\u00a0 —\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0a) Volume de \u00e1gua que caiu numa \u00e1rea de 100km<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=100.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, ocupando uma altura de h=10mm=<\/b><\/span><\/span><\/p>\n

10.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 V=\u00e1rea.altura=10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 densidade da \u00e1gua =1g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=m\/10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>kg<\/b><\/span><\/span><\/p>\n

b) \u00e1rea =1m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=100dm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 altura=10mm=10.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>dm\u00a0 —\u00a0 \u00e1gua contida numa \u00e1rea de 1m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0a uma altura de 10mm\u00a0 tem volume=\u00e1rea da base.altura=100.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>=10dm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=10L\u00a0 —\u00a0 10L em 1.200s\u00a0 —\u00a0 1L=1.000mL\u00a0 —\u00a0 10.000mL em 1200s\u00a0 —\u00a0 vaz\u00e3o da chuva\u00a0 —\u00a0 10.000mL\/1.200s=8,3mL\/s (caem 8,3mL por segundo de \u00e1gua)\u00a0 —\u00a0 1mL tem aproximadamente 20 gotas\u00a0 —\u00a0 8,3mL ter\u00e3o\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0n=8,3.20=166,6 gotas\u00a0<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0Quando o corpo estiver na imin\u00eancia de flutuar, a densidade do corpo \u00e9 igual \u00e0 do l\u00edquido (mistura)\u00a0 —\u00a0 d<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>=m\/V=90\/100\u00a0 —\u00a0 d=0,9g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A\u00a0<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=3m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2,7=3m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=3m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/2,7\u00a0 —\u00a0 d<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 7,5=m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/7,5\u00a0 —\u00a0 d<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=(m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>)\/(V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>)=4m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/(3m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/2,7 + m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/7,5)=(4m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/1) X (20,25\/25,2m<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 d<\/b><\/span><\/span>mistura<\/b><\/span><\/span><\/sub>=3,21g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

12<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> \u03c1<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=80\/10=8g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0,8g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\/\u03c1<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=8\/0,8=10\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Se a espessura de cada chapa \u00e9 a mesma, a massa de cada face \u00e9 diretamente proporcional \u00e1 \u00e1rea S de cada face\u00a0 —\u00a0 m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=kS<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=ka.a\u00a0 —\u00a0 m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=ka<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=k.2a.2a\u00a0 —\u00a0 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4ka<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=(2.a)<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=8.a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=d=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=k.a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4ka<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/8.a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/d=4ka<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/8.a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0X a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/k.a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/d=1\/2\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=d\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span><\/span>\u00a0V=30.20.50=30.000cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>=3.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/span><\/sup>=0,03kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0\u00a0 2,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>=m\/3.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>m=75kg<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0Se a massa \u00e9 a mesma, a massa espec\u00edfica \u00e9 inversamente proporcional ao volume (\u03c1=m\/V), assim, o cone circular reto \u00e9 o que tem menor volume e dever\u00e1 ter maior densidade\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0A for\u00e7a trocada entre cada dedo e a tachinha tem a mesma intensidade, mas a press\u00e3o \u00e9 maior no indicador, pois a\u00ed a \u00e1rea \u00e9 menor com a for\u00e7a se distribuindo com maior intensidade\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0Quanto menor a \u00e1rea de contato, maior ser\u00e1 a press\u00e3o, pois a for\u00e7a (peso) que a provoca \u00e9 o mesmo\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0Quanto maior a \u00e1rea, menor a press\u00e3o\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0Sendo o tijolo o mesmo, sua massa, seu volume e\u00a0 seu peso (que \u00e9 a for\u00e7a que ele exerce sobre o tampo) s\u00e3o os mesmos\u00a0 —\u00a0 P=F\/S=P\/S\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 menor \u00e1rea, maior press\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>>P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0Como a for\u00e7a sobre o plano liso \u00e9 a mesma (mesmo peso, pois a massa e o material \u201cmassa espec\u00edfica\u201d \u00e9 a mesma), a maior press\u00e3o \u00e9 daquele que exerce menor contato com o piso\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0a) O avi\u00e3o decola a partir do instante de<\/b><\/span><\/span>\u00a0t=10 s<\/b><\/span><\/span>, pois \u00e9 a partir deste instante que a for\u00e7a de sustenta\u00e7\u00e3o supera o peso da aeronave, que \u00e9 de 3000 N.<\/b><\/span><\/span><\/p>\n

b) Pelo gr\u00e1fico em t = 20s tem-se uma for\u00e7a de sustenta\u00e7\u00e3o F = 3000 N\u00a0 —\u00a0 essa for\u00e7a atua sobre uma \u00e1rea de S=50m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394p = F\/S = 3000\/50\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394P=60N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=60Pa<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0peso da v\u00e1lvula\u00a0 —\u00a0 p=mg=0,06.10\u00a0 —\u00a0 p=6.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 S=\u00e1rea do orif\u00edcio de raio=2,8.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 R=1,4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P=F\/S=P\/S=6.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\/\u03c0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=6.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\/3.(1,4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2\u00a0\u00a0<\/b><\/span><\/span><\/sup>—\u00a0 P=6.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\/3.1,96.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0\u00a0 —\u00a0 P=1,02.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>Pa\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0Mo\u00e7a\u00a0 —\u00a0 P=F\/S=peso\/S=mg\/S=60.10\/160\u00a0 —\u00a0 P=3,75kg\/cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 press\u00e3o da personagem=press\u00e3o da mo\u00e7a=P\u00a0 —\u00a0 P=F\/S\u00a0 —\u00a0 3,75=m.g\/S\u00a0 —\u00a0 3,75=m.10\/400\u00a0 —\u00a0 m=150kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0a) Estimativa do volume uma carga nova de uma caneta esferogr\u00e1fica (cilindro de raio R=1mm=10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m e altura h=10cm=10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>m)\u00a0 —\u00a0 V=\u00e1rea da base.altura=S.h=(\u03c0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>).10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>=3.(10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=3.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 esse volume \u00e9 o mesmo que o de um tra\u00e7o de comprimento c=3km=3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>m, largura l=0,5mm=0,5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m e espessura\u00a0<\/b><\/span><\/span>e<\/b><\/span><\/span>(pedida)\u00a0 —\u00a0 V<\/b><\/span><\/span>tra\u00e7o<\/b><\/span><\/span><\/sub>=c.l.e\u00a0 —\u00a0 3.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>=3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.e\u00a0 —\u00a0 e=3.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>\/15.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>e=0,2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>m ou e=0,2\u03bcm ou e=2,0.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span><\/p>\n

b) a largura da esfera da caneta \u00e9 a mesma que a do tra\u00e7o d=0,5mm=5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 raio da esfera\u00a0 —\u00a0 r=d\/2=5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 r=2,5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 \u00e1rea da esfera da caneta\u00a0 —\u00a0 S=\u03c0r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=3.(2,5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=18,75.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=F\/S=3\/18,75.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P=0,16.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=1,6.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>\u00a0Pa<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0C1\u00a0 —\u00a0 P1=F\/S1=P\/a.a=P\/a2\u00a0 —\u00a0 C2\u00a0 —\u00a0 P2=F\/S2=P\/2a.2a=P\/4a2\u00a0 —\u00a0 P2\/P1=P\/4a2 x a2\/P\u00a0 —\u00a0 P2=P1\/4\u00a0 —\u00a0 falsa, ficar\u00e1 4 vezes menor<\/b><\/span><\/span><\/p>\n

26- <\/b><\/span><\/span><\/span>As for\u00e7as respons\u00e1veis pela press\u00e3o na base do paralelep\u00edpedo s\u00e3o\u00a0 —\u00a0 parcela vertical de\u00a0\"\", de intensidade Fv=Fsen\u03b8=100.sen60o=100.0,87\u00a0 —\u00a0 Fv=87N\u00a0 —\u00a0 p=m.g=2,8.10\u00a0 —\u00a0 p=28N\u00a0 —\u00a0 FR=Fv + p=87 + 28\u00a0 —\u00a0 FR=115N\u00a0 —\u00a0 P=F\/S=115\/(10.5)\u00a0 —\u00a0 P=2,3kg\/cm2 \u00a0—\u00a0 P=2,3.104N\/m2\u00a0 —\u00a0 P=2,3.104Pa\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0\u0394P=1,0 \u2013 0,6=0,4atm=0,4.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394P=4.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394P=F\/S=F\/(0,5.0,25)\u00a0 —\u00a0 4.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=F\/0,125\u00a0 —\u00a0 F=0,5.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=5.000N\u00a0 —\u00a0 F=p=mg\u00a0 —\u00a0 5.000=m.10\u00a0 —\u00a0 m=500kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0A \u00e1rea de contato \u00e9 a do cubo inferior\u00a0 —\u00a0 S=10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0\u00a0—\u00a0 P=F\/S\u00a0 —\u00a0 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=F\/10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=10N=peso dos 4 cubos\u00a0 —\u00a0 cada cubo tem p=10\/4=2,5N\u00a0 —\u00a0 m=2,5\/10\u00a0 —\u00a0 m=0,25kg\u00a0 —\u00a0 d=m\/V =0,25\/10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=250kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/p>\n

29-<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>plat<\/b><\/span><\/span><\/span><\/sub>=90t=90.000kg=9.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>kg\u00a0 —\u00a0 terreno suporta\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>ter<\/b><\/span><\/span><\/span><\/sub>=F\/S<\/b><\/span><\/span><\/span>ter<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 0,15t\/cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=m<\/b><\/span><\/span><\/span>plat<\/b><\/span><\/span><\/span><\/sub>\/S<\/b><\/span><\/span><\/span>plat<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 1,5.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=9.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/span><\/sup>\/S<\/b><\/span><\/span><\/span>plat<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

S<\/b><\/span><\/span>plat<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\/1,5.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>plat<\/b><\/span><\/span><\/sub>=600cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 cada estaca tem \u00e1rea de base\u00a0 —\u00a0 S<\/b><\/span><\/span>est<\/b><\/span><\/span><\/sub>=10cm.10cm=100cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 n\u00famero de estacas\u00a0 —\u00a0 n=600100\u00a0 —\u00a0 n=6 estacas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) V=0,5m\/s\u00a0 —\u00a0 \u0394t=30dias.2h.3600s\u00a0 —\u00a0 \u0394t=216.000s\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0 0,5= \u0394S\/216.000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394S=108.000m=108km<\/b><\/span><\/span><\/p>\n

b) P=F\/A\u00a0 —\u00a0 2.10<\/b><\/span><\/span>7<\/b><\/span><\/span><\/sup>=F\/2,5.10<\/b><\/span><\/span>-11<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=5.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0(01)-\u00a0 Falsa \u2013 a for\u00e7a rem a mesma intensidade<\/b><\/span><\/span><\/p>\n

(02)- Verdadeira \u2013 menor \u00e1rea, maior press\u00e3o<\/b><\/span><\/span><\/p>\n

(04)- Falsa \u2013 veja (02)<\/b><\/span><\/span><\/p>\n

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

(16)- Falsa \u2013 depende tamb\u00e9m da \u00e1rea de contato<\/b><\/span><\/span><\/p>\n

(32)-<\/b><\/span><\/span><\/span> Verdadeira \u2013 a intensidade da for\u00e7a \u00e9 a mesma<\/b><\/span><\/span><\/p>\n

(2 + 32) = 34<\/b><\/span><\/span><\/p>\n

32-\u00a0<\/b><\/span><\/span>a) F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bc.N=0,6.3,0\u00a0 —\u00a0 Fat=1,8N\u00a0 —\u00a0 d=2cm=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 \u00e2ngulo entre o Fat e o deslocamento\u00a0 —\u00a0 \u03b2=180<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 cos180<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=-1 \u00a0—\u00a0 W=F.d.cos\u03b2=1,8.2.10<\/b><\/span><\/span>-2.<\/b><\/span><\/span><\/sup>.(-1)\u00a0 —\u00a0<\/b><\/span><\/span>W=-3,6.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>J<\/b><\/span><\/span><\/p>\n

b) Pelo gr\u00e1fico\u00a0 —\u00a0 quando U=20kV\u00a0 —\u00a0 P=2,0.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=0,25mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0,25.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P=F\/S\u00a0 —\u00a0 2.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>=F\/0,25.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=50N<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) jovem\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/A<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 1,2.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>=24\/A<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 A<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2,0.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 idoso\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2,0.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/0,8.2,0.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3,2N<\/b><\/span><\/span><\/p>\n

b) a cada ida e volta ele percorreu meia circunfer\u00eancia\u00a0 —\u00a0 \u0394S=2\u03c0R\/2=3.0,5=1,5m\u00a0 —\u00a0 em \u0394t=10s ele percorreu\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 0,6=\u0394S\/10\u00a0 —\u00a0 \u0394S=6m\u00a0 —\u00a0 n=6m\/1,5m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0n=4 oscila\u00e7\u00f5es<\/b><\/span><\/span><\/p>\n

 <\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>d =m\/V\u00a0 —\u00a0 como a se\u00e7\u00e3o transversal \u00e9 constante, o volume \u00e9 dado por V = A.L\u00a0 —\u00a0 \u00a0d = m\/A.L\u00a0 —\u00a0 na segunda parte do gr\u00e1fico, a linha se torna mais \u00edngreme, indicando que a densidade se torna maior\u00a0 —\u00a0 isso afirma que a primeira parte do gr\u00e1fico representa o alum\u00ednio e a segunda parte representa o cobre\u00a0 —\u00a0 as densidades do alum\u00ednio e do cobre s\u00e3o, respectivamente\u00a0 —\u00a0 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0=16\/40.A=2\/5.A\u00a0 e d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0=(90 \u2013 16)\/(100 \u2013 40).A\u00a0 —\u00a0 d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=4\/3.A \u00a0—\u00a0\u00a0 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=2\/(5.A)\/4\/(3.A)=(2\/5)x(3\/4)\u00a0 —\u00a0 d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=0,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0\u00a0 Mp\u00f3 = 8.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0kg\u00a0 —\u00a0\u00a0 d<\/b><\/span><\/span>biodie<\/b><\/span><\/span><\/sub>\u00a0= 900 kg\/m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 0,9 kg\/L\u00a0 —\u00a0\u00a0 M<\/b><\/span><\/span>biodie<\/b><\/span><\/span><\/sub>\u00a0= 15% M<\/b><\/span><\/span>p\u00f3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>biodie<\/b><\/span><\/span><\/sub>=M<\/b><\/span><\/span>biodie<\/b><\/span><\/span><\/sub>\/V\u00a0 —\u00a0 V=0,15M<\/b><\/span><\/span>p\u00f3<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>biodie<\/b><\/span><\/span><\/sub>\u00a0 —-\u00a0 V=1,33.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>L\u00a0 —\u00a0 V=1,33 bilh\u00e3o de litros\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 S = 6,25 cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0h = 20 cm\u00a0 —\u00a0 \u00a0m = 100 g\u00a0 —\u00a0 d=m\/V=m\/S.h=100\/6,25×20\u00a0 —\u00a0 d=0,8g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0O valor de M dever\u00e1 produzir uma tens\u00e3o T = M.g que dever\u00e1 equilibrar a for\u00e7a devida a press\u00e3o atmosf\u00e9rica sobre os dois vasos, dada por, P = F\/S\u00a0 —\u00a0 F = P.S\u00a0 —\u00a0 P.S = M.g\u00a0 —\u00a0 M = P.S\/g = 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.10\/10 = 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0kg = 10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0t \u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>M= 100 toneladas<\/b><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>int<\/b><\/span><\/span><\/sub>\u00a0= 0,95.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0= 1,00.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 A = 0,10 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P = 40 N\u00a0 — na placa agem as seguintes for\u00e7as\u00a0 —\u00a0 \u00a0\"\"– peso da placa\u00a0 —\u00a0\"\"– for\u00e7a devida \u00e0 diferen\u00e7a entre as press\u00f5es externa e interna na c\u00e2mara\u00a0 — \u00a0\"\"\u00a0– for\u00e7a vertical de contato entre as paredes da c\u00e2mara e a placa\u00a0 —\u00a0 essas for\u00e7as s\u00e3o mostradas na figura:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) A resultante das for\u00e7as que agem sobre a placa \u00e9 nula, pois ela est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 N + P = F\u00a0 —\u00a0 N=F \u2013 P=(P<\/b><\/span><\/span>atm<\/b><\/span><\/span><\/sub>\u00a0\u2013 P<\/b><\/span><\/span>int<\/b><\/span><\/span><\/sub>).A – P\u00a0 —\u00a0 N=(1,00 \u2013 0,95).10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.(0,10) \u2013 40=500 \u2013 40\u00a0 —\u00a0<\/b><\/span><\/span>N=460N<\/b><\/span><\/span><\/p>\n

b) O peso m\u00e1ximo com que a ventosa pode segurar a placa ocorre quando a for\u00e7a de contato entre elas \u00e9 nula, ou seja, quando N=0\u00a0 —\u00a0 P=F\u00a0 —\u00a0 P<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=(1,00 \u2013 0,95).10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.(0,10)\u00a0 —\u00a0<\/b><\/span><\/span>P=500N<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>O ato de sugar implica em aumentar o volume dos pulm\u00f5es e, consequentemente, diminuir a press\u00e3o interna da boca e do canudinho, tornando-a menor que a press\u00e3o atmosf\u00e9rica local na superf\u00edcie livre do l\u00edquido\u00a0 —\u00a0 essa diferen\u00e7a de press\u00e3o provoca uma for\u00e7a que empurra o l\u00edquido para cima, na tend\u00eancia de um novo equil\u00edbrio de press\u00f5es\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>P = 140 atm\u00a0 —\u00a0 P = 140 x 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0—\u00a0 P = 140 x 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0N\/(10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0—\u00a0 P= 14 N\/mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0—\u00a0 \u00e1rea=1,0mm.200mm\u00a0 —\u00a0 S=200mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 g=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 g=10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0mm\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 peso\u00a0 —\u00a0 P=mg\u00a0 —\u00a0 press\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=F\/S\u00a0 —\u00a0 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\/mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=mkgx10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>x<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m.s<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/200mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 m=280kg\u00a0 —\u00a0 \u00e9 muito improv\u00e1vel um patinador ter massa de 280kg ou seja, um peso de 2.800N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0Dados\u00a0 —\u00a0 \u00a0m = 62.823 kg\u00a0 — A = 105,4 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a for\u00e7a de sustenta\u00e7\u00e3o\u00a0\"\"\u00a0gerada nas asas, que equilibra o peso\u00a0\"\"\u00a0para que o avi\u00e3o voe horizontalmente, \u00e9 provocada pela diferen\u00e7a de press\u00e3o (\u0394P) acima e abaixo das asas\u00a0 —\u00a0 \u0394P=F<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\/S=mg\/S\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 \u0394P=628.230\/105,4\u00a0 —\u00a0 \u0394P=5.960,4 P<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Em grandes altitudes a press\u00e3o atmosf\u00e9rica diminui, o ar \u00e9 mais rarefeito, tornando-se mais dif\u00edcil a capta\u00e7\u00e3o de oxig\u00eanio pelo organismo\u00a0 —\u00a0 para se adaptar a essa baixa press\u00e3o atmosf\u00e9rica o corpo se autoregula, aumentando a frequ\u00eancia respirat\u00f3ria. Isso acarreta sintomas como dores de cabe\u00e7a, n\u00e1useas, lentid\u00e3o de racioc\u00ednio, dores musculares, fadiga e taquicardia.\u00a0\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>01. Correta\u00a0 — a massa de ar contida no bal\u00e3o varia com o aquecimento, pois o ar aquecido se expande, diminuindo sua densidade\u00a0 —\u00a0 \u00a0o empuxo torna-se maior que o peso e o bal\u00e3o se eleva.<\/b><\/span><\/span><\/p>\n

02. Falsa \u00a0—\u00a0 \u00a0como a \u00e1gua \u00e9 mais densa que o etanol, a densidade da mistura \u00e9 maior que a do etanol.<\/b><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 o volume pode ser determinado experimentalmente, por exemplo, por imers\u00e3o em \u00e1gua, revestindo-se o corpo com uma fina camada imperme\u00e1vel.<\/b><\/span><\/span><\/p>\n

08. Correta\u00a0<\/b><\/span><\/span><\/p>\n

R- (01 + 08)=09<\/b><\/span><\/span><\/p>\n

 <\/p>\n

44-<\/b><\/span><\/span><\/span> \u00a0Express\u00e3o da densidade\u00a0 —\u00a0 densidade = massa\/volume\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 como a massa de \u00e1gua \u00e9 a mesma, no caso, m=1g, a densidade \u00e9 inversamente proporcional ao volume, ou seja, a temperatura em que o volume \u00e9 m\u00ednimo (4<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C) a densidade \u00e9 m\u00e1xima<\/b><\/span><\/span><\/p>\n

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

45-<\/b><\/span><\/span><\/span> <\/b>Sendo a densidade da \u00e1gua d\u00e1gua=1,00g\/mL (quando voc\u00ea tem 0% de etanol) —\u00a0 1,00g \u2013 1mL\u00a0 —\u00a0 m\u00e1gua\u00a0g\u00a0 —\u00a0 50mL\u00a0 —\u00a0 m\u00e1gua\u00a0=50g\u00a0 —\u00a0 sendo a densidade do etanol detanol=0,79g\/mL(quando voc\u00ea tem 100% de etanol)\u00a0 —\u00a0 0,79g \u2013 1mL\u00a0 —\u00a0 metanol\u00a0g\u00a0 —\u00a0 50mL\u00a0 —\u00a0 metanol\u00a0=39,5g\u00a0 —\u00a0 massa total de \u00e1gua a 20oC\u00a0 —\u00a0 mtotal= 50g (\u00e1gua) + 39,5g (etanol)=89,5g (\u00e1gua + etanol)\u00a0 —\u00a0 se voc\u00ea observar no gr\u00e1fico ver\u00e1 que a densidade da mistura que cont\u00e9m 50% de etanol vale dmistura=0,93g\/mL\u00a0 —\u00a0 dmistura=mmistura\/Vmistura\u00a0 —\u00a0 0,93 = 89,5\/Vmistura\u00a0 —\u00a0 Vmistura\u2248 96 mL\u00a0 —\u00a0R- E<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span><\/span> <\/b>\u00a0C\u00e1lculo da intensidade da for\u00e7a exercida sobre o \u00eambolo que \u00e9 igual \u00e0 for\u00e7a el\u00e1stica ( dado do exerc\u00edcio:as for\u00e7as exercidas pela mola e pelo fluido, sobre o \u00eambolo, \u00a0s\u00e3o equilibradas)\u00a0 —\u00a0 F=kx=1000.0,05\u00a0 —\u00a0 F=50N\u00a0 —\u00a0 c\u00e1lculo da press\u00e3o total (Patm\u00a0+ Phidrost\u00e1tica) exercida sobre o \u00eambolo\u00a0 —\u00a0 Ptotal=F\/S=50\/3.10-4\u00a0 —\u00a0 Ptotal=(5\/3).105N\/m2\u00a0(Pa)\u00a0 —\u00a0 a press\u00e3o total na profundidade h sobre o \u00eambolo \u00e9 fornecida pelo teorema de Stevin\u00a0 —\u00a0 Ptotal=Patmosf\u00e9rica\u00a0+ d\u00e1gua.g.h\u00a0 —\u00a0 (5\/3).105\u00a0= 1,0.105\u00a0+ 1,0.103.10.h\u00a0 —\u00a0 (5\/3).105\u00a0– 1,0.105\u00a0= 1,0.103.10.h\u00a0 —\u00a0 (5.105\u00a0\u2013 3.105)\/3=104h\u00a0 —\u00a0 h=2.105\/3.104=0,666.101=6,66.100m\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada das quest\u00f5es de vestibulares sobre Densidade e Press\u00e3o 01-\u00a0d=m\/V=(190 \u2013 50)\/(200\/2)\u00a0 —\u00a0 d=1,5g\/cm3\u00a0 —\u00a0\u00a0R- B 02-\u00a01L=1dm3=10-3\u00a0m3\u00a0 —\u00a0 d=13,6.103kg\/m3\u00a0 —\u00a0\u00a0 d=m\/V\u00a0 —\u00a0 13,6.103=m\/10-3\u00a0 —\u00a0 m=13,6 kg\u00a0 —\u00a0\u00a0R- D 03-\u00a0a) a densidade da esfera leva em conta o volume total\u00a0 —\u00a0 d=m\/v=50\/30\u00a0 —\u00a0\u00a0d=1,7g\/cm3 b) a massa espec\u00edfica leva em conta apenas a parte de volume que cont\u00e9m alum\u00ednio\u00a0 —\u00a0 \u03c1=m\/v=\/50(30<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1532,"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-1535","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1535","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=1535"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1535\/revisions"}],"predecessor-version":[{"id":10888,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1535\/revisions\/10888"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1532"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1535"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}