{"id":1079,"date":"2015-07-19T04:07:05","date_gmt":"2015-07-19T04:07:05","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1079"},"modified":"2024-08-23T12:51:40","modified_gmt":"2024-08-23T12:51:40","slug":"resolucao-dos-exercicios-sobre-vetores","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/vetores-diferencas-entre-grandezas-escalares-e-grandezas-vetoriais\/resolucao-dos-exercicios-sobre-vetores\/","title":{"rendered":"Resolu\u00e7\u00e3o dos Exerc\u00edcios sobre Vetores"},"content":{"rendered":"

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

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

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

\u00a0\"\"\u00a0 —\u00a0\u00a0\"\"\u00a0—\u00a0\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

b)\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

d)\u00a0\"\"\"\"<\/b><\/span><\/span><\/span><\/p>\n

02<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span><\/span>Dois<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 s\u00e3o perfeitamente definidas por um n\u00famero acrescido de uma unidade.<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a01) deslocamento\u00a0 —\u00a0 vetorial\u00a0 —\u00a0 2) \u00e1rea\u00a0 —\u00a0 escalar\u00a0 —\u00a0 3) for\u00e7a\u00a0 — \u00a0vetorial\u00a0 —\u00a0 4) velocidade\u00a0 —\u00a0 vetorial\u00a0 —\u00a0 5) tempo\u00a0 —\u00a0 escalar\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0A<\/b><\/span><\/span><\/span><\/p>\n

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

Soma vetorial\u00a0 —\u00a0\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

08-<\/b><\/span><\/span><\/span>\u00a0CD + DE + EA =\u00a0 CB + BA\u00a0 —\u00a0 EA \u2013 CB +DE = BA \u2013 CD\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0Somando os vetores pelo m\u00e9todo da linha poligonal<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

Lei dos cossenos\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.cos\u03b2\u00a0 —\u00a0 9=3,24 + 5,76 + 8,64cos\u03b2\u00a0 —\u00a0 9 \u2013 9=8,64 cos\u03b2\u00a0 —\u00a0 cos\u03b2=0\u00a0 —\u00a0 \u03b2=90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 6<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R=\u221a52\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

S<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ F<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 50<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ F<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 2.500=F<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ F<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0\u00a0 I<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

S=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 70=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=70 \u2013 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 II\u00a0 —\u00a0 II em I\u00a0 —\u00a0 2.500=(70 \u2013 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 2.500=4.900 \u2013 140F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>2=<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0— 2F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 140F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 + 2.400=0\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 70F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 1.200=0\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(70\u00b110)\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=30kgf\u00a0 e F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=40kgf\u00a0\u00a0 —\u00a0 ou\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=40kgf e F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=30kgf<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Pit\u00e1goras\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>– D<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0Na figura abaixo est\u00e1 a soma dos tr\u00eas vetores e, para calcular o m\u00f3dulo de\u00a0\u00a0\"\"\u00a0, aplica-se Pit\u00e1goras no tri\u00e2ngulo hachurado.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=14<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 6<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 S=15,2m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0Aplicando Pit\u00e1goras no tri\u00e2ngulo hachurado\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0\u00a0\"\"+\"\"=5\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0Dividindo o hex\u00e1gono conforme a figura abaixo e calculando o vetor soma (resultante) pelo m\u00e9todo da linha poligonal em cima e<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

em baixo, o vetor soma dos dois vetores de m\u00f3dulo 16u ser\u00e1 16u + 16u=32u\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0Dividindo o hex\u00e1gono e calculando o m\u00f3dulo de M<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Observe na figura acima que os vetores\u00a0\"\"\u00a0de cima e de baixo se anulam e o vetor soma ter\u00e1 intensidade S=2M=2.3.\u221aR\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>S=6\u221aR<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0Usando o m\u00e9todo da linha poligonal<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\u00a019-<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0 \u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

20-\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

22-<\/b><\/span><\/span><\/span>\u00a0Calculando S e aplicando Pit\u00e1goras no tri\u00e2ngulo hachurado<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 12<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 S=13m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

24-<\/b><\/span><\/span><\/span>\u00a0A dist\u00e2ncia percorrida \u0394S (indica\u00e7\u00e3o do od\u00f4metro do carro) \u00e9 fornecida pela soma alg\u00e9brica de todos os deslocamentos\u00a0 —\u00a0 \u0394S=5.20=100m\u00a0 —\u00a0 unindo A com B obt\u00e9m-se o vetor deslocamento que \u00e9 calculado aplicando-se Pit\u00e1goras no tri\u00e2ngulo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

hachurado\u00a0 —\u00a0 d=40<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 20<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 — d=\u221a2.000\u00a0 —\u00a0 d=20\u221a5m\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0a) Falsa\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b) Verdadeira\u00a0\u00a0\u00a0\u00a0\u00a0 c) verdadeira\u00a0\u00a0\u00a0\u00a0\u00a0 d) falsa\u00a0\u00a0\u00a0\u00a0\u00a0 e) falsa\u00a0\u00a0\u00a0\u00a0\u00a0 f) verdadeira\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0Somando todos os vetores pelo m\u00e9todo da linha poligonal<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S=4.0,50=2.0cm\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

R- 2 e 3<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

M\u00f3dulo\u00a0 —\u00a0 R=3N<\/b><\/span><\/span><\/span><\/p>\n

b) Ap\u00f3s a soma, a origem (P) tem que coincidir com a extremidade (P)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0Como n\u00e3o foi especificado a dire\u00e7\u00e3o e o sentido desses dois vetores, eles podem ser quaisquer\u00a0 —\u00a0 valor m\u00e1ximo de R=10 + 15=25N (mesma dire\u00e7\u00e3o e mesmo sentido)\u00a0 —\u00a0 valor m\u00ednimo de R=15 \u2013 10=5N (mesma dire\u00e7\u00e3o e sentidos contr\u00e1rios)\u00a0 —\u00a0 portanto R n\u00e3o pode ser superior a 25N, nem inferior a 5N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0Observe a figura abaixo onde os vetores foram decompostos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

34-<\/b><\/span><\/span><\/span>\u00a0Nas condi\u00e7\u00f5es do exerc\u00edcio\u00a0 —\u00a0 \u03b1=45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 lei dos cossenos\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 70<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 100<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.70.100.cos\u03b1\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 4.900 + 10.000 + 14.000.\u221a2\/2\u00a0 —\u00a0 R=157,2km\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

36–\u00a0<\/b><\/span><\/span><\/span>A figura mostra os deslocamentos citados e a dist\u00e2ncia procurada.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o tri\u00e2ngulo mostrado \u00e9 ret\u00e2ngulo \u00e9 s\u00f3 aplicarmos o teorema de Pit\u00e1goras\u00a0 —\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

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

38-\u00a0<\/b><\/span><\/span><\/span>a) O vetor A est\u00e1 orientado na mesma dire\u00e7\u00e3o e sentido do vetor B, ou seja, os vetores A e B s\u00e3o paralelos. Quando os vetores<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

se encontram na mesma dire\u00e7\u00e3o e sentido (neste exemplo, horizontal) e o m\u00f3dulo do vetor resultante (\"\") \u00e9 obtido somando-se os seus m\u00f3dulos, ou seja, C = A + B.<\/b><\/span><\/span><\/span><\/p>\n

b) O vetor\u00a0\"\"\u00a0est\u00e1 orientado em uma dire\u00e7\u00e3o perpendicular ao vetor\u00a0\"\". Quando os vetores s\u00e3o perpendiculares, a soma dos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

quadrados dos seus m\u00f3dulos \u00e9 igual ao quadrado do m\u00f3dulo do vetor resultante, ou seja, C<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0=\u00a0 A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

39-\u00a0<\/b><\/span><\/span><\/span>A escala gr\u00e1fica disp\u00f5e que cada cent\u00edmetro do mapa equivale a 250 quil\u00f4metros do terreno, o que facilita representar<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

vetorialmente o percurso feito pelo viajante e, inclusive representar seu deslocamento vetorial (em azul). Dessa forma ele caminhou 1000 km para o Sul (dire\u00e7\u00e3o f\u00e1cil de identificar, pois o Norte est\u00e1 indicado no mapa), saindo do Cear\u00e1 e passando por Pernambuco e Bahia. Nesse estado mudou de rumo e viajou 1.000 km para o Oeste, chegando a Goi\u00e1s, a partir de onde rumou mais 750 km para o Sul, chegando ao estado de S\u00e3o Paulo. Nesse trajeto o viajante avistou os ecossistemas da Caatinga, do Cerrado e da Mata Atl\u00e2ntica.<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

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

\u00a0Para calcular a intensidade da for\u00e7a resultante que age sobre a part\u00edcula c\u00f3smica voc\u00ea pode decompor as for\u00e7as nas dire\u00e7\u00f5es norte e leste\u00a0 —\u00a0 observe na sequ\u00eancia abaixo que a intensidade da for\u00e7a resultante \u00e9 de 1N no sentido leste\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como a velocidade inicial da part\u00edcula tem intensidade V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1200m\/s do norte para o sul e a for\u00e7a resultante sobre ela tem intensidade 1N do oeste para leste, o movimento da part\u00edcula tem as caracter\u00edsticas de composi\u00e7\u00e3o de dois movimentos, um no sentido leste e outro no sentido sul (veja figura)\u00a0 —\u00a0 no sentido leste, a proje\u00e7\u00e3o da velocidade inicial \u00e9 nula V<\/b><\/span><\/span><\/span>oL<\/b><\/span><\/span><\/sub><\/span>=0 e ela se desloca sob a\u00e7\u00e3o de uma for\u00e7a resultante de valor F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=1N e com acelera\u00e7\u00e3o\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.a\u00a0 —\u00a0 1=2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.a\u00a0 —\u00a0 a= 500m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 sua velocidade nessa dire\u00e7\u00e3o ap\u00f3s t=1s ter\u00e1 intensidade\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oL<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>.t=0 + 500.1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>=500m\/s\u00a0 —\u00a0 no sentido sul ela ser\u00e1 lan\u00e7ada para baixo com V<\/b><\/span><\/span><\/span>oS<\/b><\/span><\/span><\/sub><\/span>=1200m\/s, acelerando com acelera\u00e7\u00e3o da gravidade g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 ap\u00f3s t=1s, sua velocidade nessa dire\u00e7\u00e3o ser\u00e1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oS<\/b><\/span><\/span><\/sub><\/span>\u00a0+ g.t=1200 + 10.1=1210m\/s\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0observe na figura que essas duas velocidades s\u00e3o perpendiculares e, aplicando Pit\u00e1goras voc\u00ea obter\u00e1 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(500)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ (1210)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=\u221a(1714100)\u00a0 —\u00a0 V=1309m\/s=1,3km\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

41<\/b><\/span><\/span>–<\/span><\/span> \u00a0<\/span><\/span>Pelo princ\u00edpio da in\u00e9rcia, se as for\u00e7as deixarem de atura, a for\u00e7a resultante sobre ela ser\u00e1 nula e, ap\u00f3s esse innstante, por in\u00e9rcia, ela seguir\u00e1 em MRU com velocidade constante de 1,3km\/s\u00a0 —\u00a0 observe na resolu\u00e7\u00e3o do exerc\u00edcio anterior (08) que, antes de 1s a trajet\u00f3ria era parab\u00f3lica\u00a0 —\u00a0<\/span><\/span>\u00a0<\/span><\/span>R- D.<\/span><\/span><\/p>\n

42-<\/b><\/span><\/span><\/span>\u00a0A soma vetorial de\u00a0\"\"\u00a0com\u00a0\"\"\u00a0\u00a0fornece a velocidade\u00a0\"\"\u00a0do avi\u00e3o em rela\u00e7\u00e3o ao solo\u00a0\u00a0 —\u00a0 observe na figura que \u03b8 +<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u03b8\u2019=180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —cos\u03b8\u2019=cos(180 \u2013 \u03b8)= – cos \u03b8\u00a0 —\u00a0 – cos\u03b8=cateto adjacente\/hipotenusa=A\/B\u00a0 —\u00a0 cos\u03b8= – A\/B\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/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 sobre vetores \u00a0 01-\u00a0a) \u00a0\u00a0 —\u00a0\u00a0\u00a0—\u00a0\u00a0 b)\u00a0 c)\u00a0 d)\u00a0 02–\u00a0Dois\u00a0 03-\u00a0R- B\u00a0 —\u00a0 s\u00e3o perfeitamente definidas por um n\u00famero acrescido de uma unidade. 04-\u00a01) deslocamento\u00a0 —\u00a0 vetorial\u00a0 —\u00a0 2) \u00e1rea\u00a0 —\u00a0 escalar\u00a0 —\u00a0 3) for\u00e7a\u00a0 — \u00a0vetorial\u00a0 —\u00a0 4) velocidade\u00a0 —\u00a0 vetorial\u00a0 —\u00a0 5) tempo\u00a0 —\u00a0 escalar\u00a0 —\u00a0\u00a0R- E 05-\u00a0R- D 06-\u00a0A 07- Soma vetorial\u00a0 —\u00a0\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1075,"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-1079","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1079","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=1079"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1079\/revisions"}],"predecessor-version":[{"id":1081,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1079\/revisions\/1081"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1075"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1079"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}