{"id":845,"date":"2015-05-14T18:14:01","date_gmt":"2015-05-14T18:14:01","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=845"},"modified":"2024-09-03T13:14:33","modified_gmt":"2024-09-03T13:14:33","slug":"resolucao-comentada-da-upe-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-upe-2015\/resolucao-comentada-da-upe-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da UPE 2015"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>Colocando a <\/b><\/span><\/span>origem do referencial no ponto de partida da part\u00edcula 1<\/b><\/span><\/span><\/span> e orientando-a para a <\/b><\/span><\/span>direita,<\/b><\/span><\/span><\/span> vamos deduzir a <\/b><\/span><\/span>fun\u00e7\u00e3o hor\u00e1ria<\/b><\/span><\/span><\/span> de cada part\u00edcula, lembrando que elas est\u00e3o <\/b><\/span><\/span>freando ( a e V tem sentidos contr\u00e1rios)<\/b><\/span><\/span><\/span> com movimento <\/b><\/span><\/span>retardado <\/b><\/span><\/span><\/span>e, para cada uma, <\/b><\/span><\/span>a <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span>V <\/b><\/span><\/span><\/span>devem possuir <\/b><\/span><\/span>sinais contr\u00e1rios:<\/b><\/span><\/span><\/span><\/p>\n

Part\u00edcula 1\"\" <\/b><\/span><\/span>S<\/b><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/sub><\/span> = 0<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/sub><\/span> = + 10 m\/s<\/b><\/span><\/span><\/span> (a favor da orienta\u00e7\u00e3o da trajet\u00f3ria) \"\" <\/b><\/span><\/span>a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = – 1 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> (sinal contr\u00e1rio de V<\/b><\/span><\/span>o1<\/b><\/span><\/span><\/sub>).<\/b><\/span><\/span><\/p>\n

S<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span> = S<\/b><\/span><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/span><\/sub><\/span>t + a<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/span> = 0 + 10t \u2013 1t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\/2 \"\"<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span> = 10t \u2013 0,5t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Part\u00edcula 2\"\" <\/b><\/span><\/span>S<\/b><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/sub><\/span> = d<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/sub><\/span> = – 14 m\/s<\/b><\/span><\/span><\/span> (contra a orienta\u00e7\u00e3o da trajet\u00f3ria) \"\" <\/b><\/span><\/span>a<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = – +0,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> (sinal contr\u00e1rio de V<\/b><\/span><\/span>o2<\/b><\/span><\/span><\/sub>).<\/b><\/span><\/span><\/p>\n

S<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span> = S<\/b><\/span><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/span><\/sub><\/span>t + a<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/span> = d – 14t + 0,5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\/2 \"\"<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span> = d – 14t + 0,25t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No encontro S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub> = S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub> \"\" <\/b><\/span><\/span>10t \u2013 0,5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = d – 14t + 0,25t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" 24t = d + 0,75t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" 0,75t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 24t + d = 0.<\/b><\/span><\/span><\/span><\/p>\n

Para que o encontro, como pede o enunciado, ocorra <\/b><\/span><\/span><\/span>apenas uma vez<\/b><\/span><\/span><\/span>, a <\/b><\/span><\/span><\/span>raiz da equa\u00e7\u00e3o<\/b><\/span><\/span><\/span> deve ser <\/b><\/span><\/span><\/span>nula<\/b><\/span><\/span><\/span>, para que voc\u00ea obtenha <\/b><\/span><\/span><\/span>um \u00fanico tempo<\/b><\/span><\/span><\/span> para o encontro \"\" <\/b><\/span><\/span><\/span>\u2206 = 0.<\/b><\/span><\/span><\/span><\/p>\n

\u2206 = B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 4.A.C = 0 \"\" <\/b><\/span><\/span><\/span>(-24)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> \u2013 4.(0,75).d = 0<\/b><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>576 \u2013 3d<\/b><\/span><\/span><\/span> \"\" d = <\/b><\/span><\/span><\/span>576\/3<\/b><\/span><\/span><\/span> = 192 m.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>Pit\u00e1goras \"\" a<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup> = a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> + b<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> = A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> + B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> \"\" a<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub> = \u221a( A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> + B<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>) \"\" <\/b><\/span><\/span>a<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span> = \u221a( A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + B<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>1\/2<\/b><\/span><\/span><\/sup><\/span> kg.m\/s ou N.s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

03- <\/span><\/span><\/span>A <\/b><\/span><\/span>for\u00e7a m\u00e1xima F<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> ocorre quando o tempo \u00e9 <\/b><\/span><\/span>T\/2<\/b><\/span><\/span><\/span> e o <\/b><\/span><\/span>impulso I \u00e9 numericamente igual \u00e0 \u00e1rea<\/b><\/span><\/span><\/span> da figura entre <\/b><\/span><\/span>o e T\/2.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

I = (B + b).h\/2<\/b><\/span><\/span><\/span> = (F<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub> + F<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/2).(T\/2)\/2 \"\" <\/b><\/span><\/span>I =3F<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>T\/8,<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Observe que a <\/b><\/span><\/span>velocidade <\/b><\/span><\/span><\/span>da primeira esfera, durante <\/b><\/span><\/span>toda <\/b><\/span><\/span><\/span>a colis\u00e3o varia de <\/b><\/span><\/span>V a zero<\/b><\/span><\/span><\/span> (colis\u00e3o el\u00e1stica) e, quando a <\/b><\/span><\/span>for\u00e7a \u00e9 m\u00e1xima F<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> a velocidade <\/b><\/span><\/span>varia de v a v\/2 no intervalo de tempo T\/2.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o, a <\/b><\/span><\/span>for\u00e7a \u00e9 m\u00e1xima <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, quando <\/b><\/span><\/span>v<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = V\/2 e v<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = v<\/b><\/span><\/span><\/span> \"\" I = \u2206Q = m\u0399\u2206v\u0399 = m\u0399(v\/2 \u2013 v)\u0399 \"\" 3F<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>T\/8 = mv\/2 \"\" F<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub> = 8mv\/6T = 4mv\/3T.<\/b><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>aI. Falsa<\/b><\/span><\/span><\/span> \"\" trata-se do <\/b><\/span><\/span>princ\u00edpio de Pascal<\/b><\/span><\/span><\/span>\u00a0, os <\/b><\/span><\/span>acr\u00e9scimos de press\u00e3o<\/b><\/span><\/span><\/span> sofridos por um ponto de um l\u00edquido <\/b><\/span><\/span>s\u00e3o transmitidos integralmente (em todas as dire\u00e7\u00f5es e sentidos) a todos os pontos do l\u00edquido e das paredes do recipiente onde este est\u00e1 contido.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta \"\" <\/b><\/span><\/span><\/span>veja (I).<\/b><\/span><\/span><\/span><\/p>\n

III. Falsa \"\" <\/b><\/span><\/span><\/span>veja (I).<\/b><\/span><\/span><\/span><\/p>\n

IV. Correta<\/b><\/span><\/span><\/span> \"\" veja (I)<\/b><\/span><\/span><\/p>\n

V. Falsa<\/b><\/span><\/span><\/span> \"\" veja (I)<\/b><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Para passar de <\/b><\/span><\/span>\u00e1gua a 25<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a \u00e1gua a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C <\/b><\/span><\/span><\/span>voc\u00ea deve <\/b><\/span><\/span>retirar<\/b><\/span><\/span><\/span> Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub> =mc(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) = 200.1.(0 \u2013 25) = <\/b><\/span><\/span>– 5000 cal<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

Para passar de <\/b><\/span><\/span>\u00e1gua a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a gelo a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C voc\u00ea deve <\/b><\/span><\/span>retirar<\/b><\/span><\/span><\/span> Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub> =mL = – 200.80 = <\/b><\/span><\/span>– 16000cal.<\/b><\/span><\/span><\/span><\/p>\n

Para passar de \u00e1gua a 25<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a gelo a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C voc\u00ea deve retirar Q\u2019 = -5000 \u2013 16000 = – 21000 cal.<\/b><\/span><\/span><\/span><\/p>\n

Para passar de <\/b><\/span><\/span>gelo a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>C a gelo a 0C <\/b><\/span><\/span><\/span>voc\u00ea deve <\/b><\/span><\/span>ceder<\/b><\/span><\/span><\/span> Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub> =mc(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) = 400.0,5.(0 \u2013 (-25)) = <\/b><\/span><\/span> 5000 cal<\/b><\/span><\/span><\/span><\/p>\n

Para passar de <\/b><\/span><\/span>gelo a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a \u00e1gua a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>voc\u00ea deve <\/b><\/span><\/span>ceder<\/b><\/span><\/span><\/span> Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub> =mL = 400.80 = <\/b><\/span><\/span>32000 cal.<\/b><\/span><\/span><\/span><\/p>\n

Para passar de gelo a – 25<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C a \u00e1gua a 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C voc\u00ea deve retirar Q\u2019 = 5000 + 32000 = 32500 cal.<\/b><\/span><\/span><\/span><\/p>\n

Observe que <\/b><\/span><\/span>todo calor retirado da \u00e1gua (- 21000cal)<\/b><\/span><\/span><\/span> <\/b>n\u00e3o <\/b><\/span><\/span><\/span>\u00e9 suficiente para <\/b><\/span><\/span>derreter todo gelo que precisa de 32000 cal, e no final voc\u00ea ter\u00e1 mais gelo que \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>Como o <\/b><\/span><\/span>alum\u00ednio se dilata mais que o vidro<\/b><\/span><\/span><\/span>, pois tem maior coeficiente de dilata\u00e7\u00e3o, se voc\u00ea <\/b><\/span><\/span>aquecer<\/b><\/span><\/span><\/span> <\/b>ambos, o externo, de alum\u00ednio, se dilatar\u00e1 mais e se soltar\u00e1.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/span><\/span><\/span>O <\/b><\/span><\/span>maior valor para a tra\u00e7\u00e3o<\/b><\/span><\/span><\/span> no fio ocorre quando <\/b><\/span><\/span>q est\u00e1 passando pelo extremo inferior do fio:<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/sub><\/span> <\/b>=mg=10.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10 = 10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup> <\/b>= 0,1 N<\/b><\/span><\/span><\/span><\/p>\n

Sendo as duas cargas <\/b><\/span><\/span>positivas<\/b><\/span><\/span><\/span> a for\u00e7a el\u00e9trica entre elas \u00e9 de <\/b><\/span><\/span>repuls\u00e3o<\/b><\/span><\/span><\/span> e ter\u00e1 <\/b><\/span><\/span>intensidade<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> kQq\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> = 9.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.0,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/(10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> = 9.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup> \"\" <\/b><\/span><\/span>F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = 9.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>= 0,9 N.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo a for\u00e7a de tra\u00e7\u00e3o no fio de <\/b><\/span><\/span>T = 11 N,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span>for\u00e7a resultante centripeta<\/b><\/span><\/span><\/span> (dirigida para o centro da circunfer\u00eancia, para cima) ter\u00e1 <\/b><\/span><\/span>intensidade F<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span> = 11 \u2013 (0,1 + 0,9) <\/b><\/span><\/span>= 10N.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R \"\" 10 = 10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span> \"\" 1 = 10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> =1\/10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> = 10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 100 \"\" V = \u221a100 = 10 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/span><\/span><\/span>No final da d\u00e9cada de 1920, <\/b><\/span><\/span>Heisenberg<\/b><\/span><\/span><\/span> formulou o chamado <\/b><\/span><\/span>princ\u00edpio da incerteza<\/b><\/span><\/span><\/span>. De acordo com esse princ\u00edpio, <\/b><\/span><\/span>n\u00e3o podemos determinar com precis\u00e3o e simultaneamente a posi\u00e7\u00e3o e o momento de uma part\u00edcula.<\/b><\/span><\/span><\/span><\/p>\n

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

09- <\/span><\/span><\/span>I. Falsa<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>\u00a0A figura abaixo mostra o gr\u00e1fico da <\/b><\/span><\/span>energia cin\u00e9tica<\/b><\/span><\/span><\/span> do el\u00e9tron extra\u00eddo em fun\u00e7\u00e3o da <\/b><\/span><\/span>frequ\u00eancia<\/b><\/span><\/span><\/span> da radia\u00e7\u00e3o incidente, para uma mesma placa met\u00e1lica <\/b><\/span><\/span>(c\u00e9lula fotoel\u00e9trica).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 a <\/b><\/span><\/span>freq\u00fc\u00eancia m\u00ednima<\/b><\/span><\/span><\/span> (frequ\u00eancia de corte) necess\u00e1ria para <\/b><\/span><\/span>produzir o efeito fotoel\u00e9trico<\/b><\/span><\/span><\/span>. Se <\/b><\/span><\/span>f=f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0o el\u00e9tron \u00e9 liberado, mas sua energia cin\u00e9tica \u00e9 nula.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

II. Falsa<\/b><\/span><\/span><\/span> \"\" veja (I).<\/b><\/span><\/span><\/p>\n

III. Correta<\/b><\/span><\/span><\/span> \"\" para valores superiores a f<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>, o <\/b><\/span><\/span>n\u00famero de el\u00e9trons<\/b><\/span><\/span><\/span> arrancados \u00e9 diretamente proporcional <\/b><\/span><\/span>\u00e0 intensidade<\/b><\/span><\/span><\/span> da radia\u00e7\u00e3o eletromagn\u00e9tica incidente, ou seja, <\/b><\/span><\/span>aumentando a intensidade<\/b><\/span><\/span><\/span> <\/b>da radia\u00e7\u00e3o<\/b><\/span><\/span><\/span> incidente no metal, aumenta-se o n\u00edvel energ\u00e9tico dos f\u00f3tons incidentes, <\/b><\/span><\/span>aumentando assim n\u00famero de el\u00e9trons arrancados.<\/b><\/span><\/span><\/span><\/p>\n

IV. Correta \"\" veja (I).<\/b><\/span><\/span><\/span><\/p>\n

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

<\/h2>\n

Voltar para os exerc\u00edcios.<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- Colocando a origem do referencial no ponto de partida da part\u00edcula 1 e orientando-a para a direita, vamos deduzir a fun\u00e7\u00e3o hor\u00e1ria de cada part\u00edcula, lembrando que elas est\u00e3o freando ( a e V tem sentidos contr\u00e1rios) com movimento retardado e, para cada uma, a e V devem possuir sinais contr\u00e1rios: Part\u00edcula 1 So1 = 0 Vo1 =<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":843,"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-845","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/845","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=845"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/845\/revisions"}],"predecessor-version":[{"id":846,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/845\/revisions\/846"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/843"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=845"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}