{"id":1386,"date":"2015-09-29T20:19:24","date_gmt":"2015-09-29T20:19:24","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1386"},"modified":"2024-08-23T13:13:59","modified_gmt":"2024-08-23T13:13:59","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-impulso-e-quantidade-de-movimento","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/impulso-e-quantidade-de-movimento\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-impulso-e-quantidade-de-movimento\/","title":{"rendered":"Impulso e Quantidade de Movimento – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Impulso e Quantidade de Movimento<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0A for\u00e7a que age sobre cada bola \u00e9 a mesma (peso) e o impulso ser\u00e1 maior para aquela que tiver maior tempo de contato com a mesa<\/b><\/span><\/span><\/span>\u2013 R- B<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0R- A (veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0O impulso \u00e9 numericamente igual \u00e0 \u00e1rea do trap\u00e9zio\u00a0 —\u00a0 I=(B + b).h\/2=(25 + 10).30\/2\u00a0 —\u00a0 I=525N.m\u00a0 —\u00a0 I=F.\u2206t\u00a0 —\u00a0 525=F.25\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F=21N R- E<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0P=mg\u00a0 —\u00a0 360=m.10\u00a0 —\u00a0 m=36kg\u00a0 —\u00a0 Q=m.V=36.60\/3,6\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 Q=600kgm\/s R-C<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0t=0 \u2013 Q=12kgm\/s\u00a0 —\u00a0 Q=mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 12=3.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=4m\/s\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>co<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3.16\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>co<\/b><\/span><\/span><\/sub><\/span>=24J<\/b><\/span><\/span><\/span><\/p>\n

t=6.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>s\u00a0 – Q=0\u00a0 —\u00a0 Q=m.V\u00a0 —\u00a0 0=3V\u00a0 —\u00a0 V=0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=3.0\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0\u00a0\u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 E<\/b><\/span><\/span><\/span>co<\/b><\/span><\/span><\/sub><\/span>=0 \u2013 24\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>= -24J<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0Q=mV (fun\u00e7\u00e3o linear \u2013 1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0grau)\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 (fun\u00e7\u00e3o quadr\u00e1tica \u2013 do 2<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0grau)\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0Baixo \u2013 E<\/b><\/span><\/span><\/span>mb\u00a0<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 topo \u2013 E<\/b><\/span><\/span><\/span>mt<\/b><\/span><\/span><\/sub><\/span>=mgh\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=mgh\u00a0 —\u00a0\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10.0,8\u00a0 —\u00a0 V=4m\/s\u00a0 —\u00a0 Q=mV=60.4\u00a0 Q=240=2,4.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>kg.m\/s\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0Supondo que, ap\u00f3s o impulso contr\u00e1rio, ele continue se deslocando para a direita\u00a0 —\u00a0 I=Q \u2013 Q<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 40=mV \u2013 mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=2.15 \u2013 2V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=-5m\/s (ele volta com velocidade de 5m\/s\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

I= Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>= Q<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>= mV<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=0,01.3\u00a0 —\u00a0 I=3,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>N.s e para cima \u2013 veja figuras acima\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0a) 1<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span>\u00a0trecho<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u2206S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 0,5=0.1 + a.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 a=1m\/s<\/b><\/span><\/span><\/span>2\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>—\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0 + 1.1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1m\/s<\/b><\/span><\/span><\/span><\/p>\n

2<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0trecho<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 — V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1 + 1.1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2m\/s<\/b><\/span><\/span><\/span><\/p>\n

3<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span>\u00a0trecho<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ a.t=2 + 1.1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub><\/span>=3m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub><\/span>=5.3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub><\/span>=15kgm\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

b)\u00a0<\/b><\/span><\/span><\/span><\/span>\u2206<\/b><\/span><\/span><\/span>S=V<\/b><\/span><\/span><\/span><\/span>0<\/b><\/span><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2=2.1 + 1.12\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>D<\/b><\/span><\/span><\/span>S=2,5m<\/b><\/span><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0A energia mec\u00e2nica da pedra na altura de 72m \u00e9 s\u00f3 a energia potencial gravitacional – E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mgh=10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.10.72\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=72.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J\u00a0 —\u00a0 como 90% foram dissipadas, sobraram\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=10% de 72.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>=0,1.72.10<\/b><\/span><\/span><\/span>5\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=72.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>J que \u00e9 o valor da energia cin\u00e9tica da pedra na parte inferior\u00a0 —\u00a0 72.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 72.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span><\/span>4.<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V=12m\/s\u00a0 —\u00a0 Q=m.V=10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.12\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Q=1,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>kg.m\/s R- B<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0R \u2013B (veja teoria<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span> F.\u2206t=m\u2206V\u00a0 —\u00a0 F.0,01=0,12.(6)\u00a0 —\u00a0 F=0,72\/0,01\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F=72N\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0F.\u2206t=m\u2206V\u00a0 —\u00a0 F.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=0,1.30\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F=3.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>N\u00a0\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0F.\u2206t=m\u2206V\u00a0\u00a0 —\u00a0 F.\u2206t=m.V\/2\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 F=mV\/2\u2206t\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0F.\u2206t=m.\u00a0\u2206V\u00a0 —\u00a0 f\u00ad<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>. 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=m.4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 P=mg=10m\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\/P=4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m\/10m\u00a0\u00a0<\/b><\/span><\/span><\/span>— f<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\/P=400<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>= m A energia mec\u00e2nica da pedra na altura de 72m \u00e9 s\u00f3 a energia potencial gravitacional – E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=mgh=10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.10.72\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=72.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>J.<\/b><\/span><\/span><\/span><\/p>\n

Como 90% foram dissipadas, sobraram\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=10% de 72.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>=0,1.72.10<\/b><\/span><\/span><\/span>5\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>=72.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>J que \u00e9 o valor da energia cin\u00e9tica da pedra na parte inferior\u00a0 —\u00a0 72.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 72.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span><\/span>4.<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V=12m\/s\u00a0 —\u00a0 Q=m.V=10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.12\u00a0 —\u00a0<\/b><\/span><\/span><\/span>Q=1,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>kg.m\/s R- B<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 4\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2Vi<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\/2\u00a0 —\u00a0 V=2\u03c0m\/s\u00a0 —\u00a0 I=mV – mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 I=2.2\u03c0\u00a0– 0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>I=4\u03c0N.s<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0Velocidade com que ele chega \u00e0 cama a partir de uma altura de 3,2m\u00a0 —\u00a0 mgh=mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 10.3,2=V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=8m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Velocidade com que ele sai da cama atingindo h=2,45m\u00a0 —\u00a0 mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=mgh\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10.2,45\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=7m\/s<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F.\u2206t=m(V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 F.0,4=80(7 \u2013 (-8))\u00a0\u00a0 —\u00a0 F= 1.200\/0,4\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F=3,0.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>N<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>Veja o esquema abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

I = m.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 m.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 I = m.(+V) \u2013 m.(-V)\u00a0 —\u00a0 I = 0,3.5 \u2013 (- 0,3.5)\u00a0 —\u00a0 I= 1,5 + 1,5<\/b><\/span><\/span><\/span>—\u00a0 I=3,0kg.m\/s\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0P\u00e1ssaro- antes V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=0 e depois ele adquire a mesma velocidade que o avi\u00e3o V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=900\/3,6=250m\/s\u00a0 —\u00a0 F.\u2206t=m(V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0– V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>)\u00a0 — F.0,001=2.(250 \u2013 0)\u00a0 —\u00a0 F=500\/0,001\u00a0 —\u00a0<\/b><\/span><\/span><\/span>F=500.000N ou F=5,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>N<\/b><\/span><\/span><\/span><\/p>\n

21<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span>\u00a0a)Queda livre de uma altura de 80m\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.h\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.10.80\u00a0 —\u00a0 V=40m\/s\u00a0 —\u00a0\u00a0\u2206V=40m\/s\u00a0 —\u00a0 F.\u2206t=m.\u2206V\u00a0 —\u00a0 F.0,05=50.40\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F=40.000N<\/b><\/span><\/span><\/span><\/p>\n

b) a=\u2206V\/\u2206t=40\/0,05\u00a0 —\u00a0 a=800m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=8.10\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=80m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 a\/a<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=800\/80\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 a\/a<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P=mg=10m\u00a0 —\u00a0 F=60.p=60.10m\u00a0 —\u00a0 F=600m\u00a0 —\u00a0 F.Dt=m(V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 600m.0,2=m(+3V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) \u2013 (-V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 120=3V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=120\/4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=30m\/s\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0R- B (veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0a) Veja pela express\u00e3o F.\u2206t=m.\u2206V, que\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>\u2206V\u00a0<\/b><\/span><\/span><\/span>s\u00e3o constantes (mesma x\u00edcara e mesma varia\u00e7\u00e3o de velocidade) e assim, teremos:<\/b><\/span><\/span><\/span><\/p>\n

F.\u2206t=constante, portanto se voc\u00ea aumentar o intervalo de tempo do choque, estar\u00e1 diminuindo a intensidade da for\u00e7a m\u00e1xima que a x\u00edcara troca com o tapete. t<\/b><\/span><\/span><\/span><\/p>\n

b) F.\u2206t=m.\u2206V\u00a0 —\u00a0 F.0,5=0,1.2\u00a0 —\u00a0 F=0,2\/0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F=0,4N<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 F.0,01=0,1.2\u00a0 —\u00a0 F=0,2\/0,01\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F=20,0N<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>1- Verdadeira \u2013 quando ele deixa o degrau, est\u00e1 em queda livre e, nesse caso, a for\u00e7a resultante sobre ele \u00e9 seu peso e a varia\u00e7\u00e3o do seu momento linear (quantidade de movimento) \u00e9 I=P.\u2206t=m\u2206V=\u2206Q.<\/b><\/span><\/span><\/span><\/p>\n

2 -Verdadeira \u2013 c\u00e1lculo da velocidade com que ele chega ao solo \u2013 queda livre com Vo=0 \u2013 Torricelli \u2013 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=Vo<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.g.\u2206S\u00a0 — V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.10.0,5\u00a0 —\u00a0 V=\u221a10m\/s \u2013 velocidade com que chega ao solo\u00a0 —\u00a0 durante o choque com o solo sua velocidade varia de\u00a0\u221a10m\/s a zero e o m\u00f3dulo dessa varia\u00e7\u00e3o \u00e9\u2206V=\u221a10m\/s enquanto que sua altura diminui de\u00a0\u2206S=3cm=0,03m , numa desacelera\u00e7\u00e3o a de\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=Vo<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> +2.(-a).\u2206S\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(10)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> -2.a 0,03\u00a0 —\u00a0 a=500\/3m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 for\u00e7a que ele troca com o solo e que produz essa desacelera\u00e7\u00e3o\u00a0 —\u00a0 F=m.a=70.500\/3\u00a0 —\u00a0 F\u00bb11.667N\u00a0 —\u00a0 peso correspondente \u00e0 massa de 1ton=1.000kg \u2013 P=1.000X10 \u2013 P=10.000N\u00a0 —\u00a0 portanto – F>P<\/b><\/span><\/span><\/span><\/p>\n

3- Verdadeira \u2013 \u00e9 a for\u00e7a de contato, que os p\u00e9s trocam com o solo.<\/b><\/span><\/span><\/span><\/p>\n

4- Verdadeira \u2013 (veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0\u00a0 — \u00a0v = 162 km\/h = 45 m\/s\u00a0 —\u00a0 \u00a0m = 170 g = 0,17 g\u00a0 — \u0394t = 10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0s\u00a0 —\u00a0 considerando que for\u00e7a aplicada pelo stick \u00e9 a resultante, pelo teorema do impulso, vem:<\/b><\/span><\/span><\/span><\/p>\n

I =\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>Q\u00a0 —\u00a0 F.<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t = m.(v \u2013 v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 F.(10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>) = 0,17.(45)\u00a0 —\u00a0 F = 7,65×10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 \u00a0m = 0,8 kg\u00a0 — v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 93,6 km\/h = 26 m\/s\u00a0 —\u00a0 \u00a0v = 280,8 km\/h = 78 m\/s\u00a0 —\u00a0 o enunciado afirma que antes da colis\u00e3o a velocidade relativa entre a mola e o capacete \u00e9 V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=78 \u2013 26=52m\/s, e no instante ap\u00f3s a colis\u00e3o a velocidade relativa \u00e9 nula\u00a0 —\u00a0 assim, a quantidade de movimento da mola ap\u00f3s a colis\u00e3o em rela\u00e7\u00e3o ao capacete \u00e9 nula\u00a0 —\u00a0 I=\u0394Q\u00a0 —\u00a0 F.\u0394t=m\u0394V\u00a0 —\u00a0 F.\u0394t=m.(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 F.0,026=0,8.(52 \u2013 0)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F=1.600N<\/b><\/span><\/span><\/span><\/p>\n

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

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo as for\u00e7as que atuam sobre a polia\u00a0 —\u00a0 o dinam\u00f4metro indica a intensidade da for\u00e7a de tra\u00e7\u00e3o no fio<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0ao qual est\u00e1 ligado\u00a0 — T = 60 N\u00a0 —\u00a0 como o corpo C est\u00e1 em equil\u00edbrio\u00a0 — \u00a02T = P\u00a0 —\u00a0\u00a0 2(60) = m(10) \u00a0—\u00a0 \u00a0m = 12 kg\u00a0 —\u00a0 depois que o fio \u00e9 cortado o corpo de massa 12kg cai em queda livre com acelera\u00e7\u00e3o a=g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0e, ap\u00f3s 1s ter\u00e1 velocidade\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ gt\u00a0 —\u00a0 V=0 + 10.1\u00a0 —\u00a0 V=10m\/s\u00a0 —\u00a0 nesse instante ter\u00e1 quantidade de movimento\u00a0\u00a0 —\u00a0 Q=mV=12×10\u00a0 —\u00a0 Q=120kg.m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Em todo gr\u00e1fico F x t, o impulso \u00e9 numericamente igual a \u00e1rea compreendida entre a reta representativa e o eixo t, no caso a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e1rea do tri\u00e2ngulo\u00a0 —\u00a0 I=b.h\/2=0,6.8\/2\u00a0 —\u00a0 I=2,4N.s\u00a0 —\u00a0 pelo Teorema do Impulso\u00a0 —\u00a0 \u00a0o impulso da for\u00e7a resultante \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento (\u0394Q)\u00a0 —\u00a0 I = \u0394Q = m\u00a0\u0394v \u00a0— \u00a02,4 = 0,1 (v \u2013 0) \u00a0—\u00a0 \u00a0v = 24 m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe pelo gr\u00e1fico que, quando V=1m\/s, E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=1J\u00a0 —\u00a0 substituindo na equa\u00e7\u00e3o E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 1=m.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 m=2kg\u00a0 — quandoV=5m\/s, a quantidade de movimento Q ser\u00e1\u00a0 —\u00a0 Q=mV=2.5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>Q=10kg.m\/s
\n<\/b><\/span><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Considerando essa for\u00e7a como a resultante\u00a0 —\u00a0 I = \u0394Q\u00a0 —\u00a0 como h\u00e1 apenas mudan\u00e7a de sentido e n\u00e3o de dire\u00e7\u00e3o (movimento retil\u00edneo), o impulso pode ser obtido pela \u00e1rea entre a linha do gr\u00e1fico e o eixo t, como indicado na figura\u00a0 —\u00a0 como a for\u00e7a F tem<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

m\u00f3dulo constante em cada intervalo de tempo, o impulso tamb\u00e9m pode ser obtido pela express\u00e3o\u00a0 —\u00a0 I<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\u00a0= F.\u0394t\u00a0 — \u2502I<\/b><\/span><\/span><\/span>FR<\/b><\/span><\/span><\/sub><\/span>\u2502=\u2502\u0394Q\u2502=m\u2502V – V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u2502 —\u00a0 A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 A<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=m(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 1×20 + 1×10 \u2013 1×10=10(V \u2013 0)\u00a0 —\u00a0 10V=20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=2m\/s<\/b><\/span><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Justificando as falsas:<\/b><\/span><\/span><\/span><\/p>\n

01) Considerando que a for\u00e7a F \u00e0 que se refere a afirmativa seja a for\u00e7a resultante e que ela seja constante, temos, pelo teorema do impulso\u00a0 — I=\u0394Q = F\u0394t\u00a0 —\u00a0 F=\u0394Q\/\u0394t\u00a0 —\u00a0 nesta express\u00e3o, observe que o coeficiente angular dessa fun\u00e7\u00e3o \u00e9 F.<\/b><\/span><\/span><\/span><\/p>\n

02) A quantidade de movimento no instante t = 0 \u00e9 Q<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= m v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a quantidade de movimento somente ser\u00e1 nula se a velocidade inicial for nula.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- (04 + 08 + 16)=28<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe as figuras abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Teorema do impulso\u00a0 —\u00a0 I=\u0394Q\u00a0 —\u00a0 F.\u0394t=m\u0394V\u00a0 —\u00a0 F.0,04=0,75{10 \u2013 (-10)}\u00a0 —\u00a0 F=1,5\/0,004\u00a0 —\u00a0 F=375N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Correta\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

02- Falsa\u00a0 —\u00a0 n\u00e3o \u00e9 a varia\u00e7\u00e3o da quantidade de movimento que tem sempre a dire\u00e7\u00e3o e o sentido da velocidade mas sim, a quantidade de movimento.<\/b><\/span><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

08. Falsa\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

16- Falsa\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

R- (01 + 04)=05<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

35- <\/b><\/span><\/span><\/span>C\u00e1lculo da acelera\u00e7\u00e3o do atleta\u00a0 —\u00a0 quando t=2s, V=2,2m\/s\u00a0 —\u00a0 V=Vo\u00a0+ a.t\u00a0 —\u00a0 2,2=0 + a.2\u00a0 —\u00a0 a=1,1m\/s2\u00a0 —\u00a0 for\u00e7a resultante sobre o atleta\u00a0 —\u00a0 F=m.a=85.1,1\u00a0 —\u00a0 F=93,5N\u00a0 —\u00a0 impulso sobre o atleta nesses 2s\u00a0 —\u00a0 I=F.\u2206t=93,5.2\u00a0 —\u00a0 I=187N\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0Leopardo\u00a0 —\u00a0 Ql=ml.vl=120.60=Q1=7.200kg.m\/s\u00a0 —\u00a0 autom\u00f3vel\u00a0 —\u00a0 Qa=ma.va=1.100×70=Q2=77.000kg.m\/s\u00a0 —\u00a0 caminh\u00e3o\u00a0 —\u00a0 Qc=mc.vc=Q3=3.600×20=72.000kg.m\/s\u00a0 —\u00a0 cofre\u00a0 —\u00a0 velocidade com que ele chega ao solo\u00a0 —\u00a0 V2\u00a0= Vo2\u00a0+ 2.g.h = 0 + 2.10.5\u00a0 —\u00a0 V=10ms\u00a0 —\u00a0 Qco=mco.Vco=300.10\u00a0 —\u00a0 Qco=Q4=3.000kg.m\/s\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0Q=m.V=6.10-2.60\u00a0 —\u00a0 Q=3,6kg.m\/s\u00a0 —\u00a0 Ec=m.V2\/2=6.10-2.(60)2\/2\u00a0 —\u00a0 Ec=108J<\/b><\/span><\/span><\/p>\n

38- <\/b><\/span><\/span><\/span>Um\u00a0 \u00f4nibus\u00a0\u00a0 que\u00a0 trafegava\u00a0 inicialmente\u00a0 a\u00a0 54 km\/h foi freado bruscamente, como forma de impedir um impacto iminente com o ve\u00edculo<\/b><\/span><\/span><\/span><\/p>\n

\u00e0 sua frente. Um passageiro de massa igual a 60 kg, surpreendido pela manobra violenta, foi arr<\/b><\/span><\/span><\/span><\/p>\n

–<\/b><\/span><\/span><\/span>\u00a0I. Correta\u00a0 —\u00a0 F=ma\u00a0 —\u00a0 a=\u2206V\/\u2206t\u00a0 —\u00a0 F=m\u2206V\/\u2206t\u00a0 —\u00a0 F.\u2206t = m. \u2206V\u00a0 —\u00a0 F.0,3 = 60.15\u00a0 —\u00a0 F=900\/0,3\u00a0 —\u00a0 F=3.000N.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 \u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 m.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=60.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 60.15<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0 \u2013 13.500\/2\u00a0 —\u00a0 \u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>= – 6.750J.<\/b><\/span><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=W\/\u2206W\/\u2206t=\u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/\u2206t\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=6.750\/0,3\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=22.500W=22,5kW.<\/b><\/span><\/span><\/span><\/p>\n

IV. Falsa\u00a0 —\u00a0 \u2206Q=m. \u2206V=60.(0 \u2013 15)= – 900kg.m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

Voltar para os exerc\u00edcios<\/a>\u00a0<\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Impulso e Quantidade de Movimento 01-\u00a0A for\u00e7a que age sobre cada bola \u00e9 a mesma (peso) e o impulso ser\u00e1 maior para aquela que tiver maior tempo de contato com a mesa\u2013 R- B 02-\u00a0R- A (veja teoria) 03-\u00a0O impulso \u00e9 numericamente igual \u00e0 \u00e1rea do trap\u00e9zio\u00a0 —\u00a0 I=(B + b).h\/2=(25 + 10).30\/2\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1382,"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-1386","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1386","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=1386"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1386\/revisions"}],"predecessor-version":[{"id":10834,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1386\/revisions\/10834"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1382"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1386"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}