{"id":1734,"date":"2015-12-31T03:20:19","date_gmt":"2015-12-31T03:20:19","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1734"},"modified":"2024-08-19T17:12:44","modified_gmt":"2024-08-19T17:12:44","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-segunda-lei-de-ohm-resistividade","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/segunda-lei-de-ohm-resistividade\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-segunda-lei-de-ohm-resistividade\/","title":{"rendered":"Segunda lei de Ohm (Resistividade) – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Segunda lei de Ohm (Resistividade)<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0Como s\u00e3o de mesmo material, possuem a mesma resistividade el\u00e9trica\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0A resist\u00eancia R \u00e9 inversamente proporcional \u00e0 espessura (S)\u00a0 —\u00a0 maior espessura, menor R, maior corrente i\u00a0 —\u00a0 brilha mais\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0Mesmo fio \u00a0—\u00a0 mesma resistividade\u00a0\u03c1\u00a0 —\u00a0 mesma S\u00a0 —\u00a0 L cai pela metade\u00a0 —\u00a0 R \u00e9 diretamente proporcional a L\u00a0 —\u00a0 R tamb\u00e9m cai pela metade\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R \u00e9 inversamente proporcional \u00e0 \u00e1rea de se\u00e7\u00e3o reta transversal S\u00a0 —\u00a0 S=\u03c0r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a nova \u00e1rea ficar\u00e1 4 vezes maior, pois R est\u00e1 ao quadrado, e a resist\u00eancia R ficar\u00e1 4 vezes menor\u00a0 —\u00a0 R \u00e9 diretamente proporcional ao comprimento L\u00a0 —\u00a0 se L \u00e9 reduzido \u00e0 metade, R tamb\u00e9m ser\u00e1 reduzido \u00e0 metade\u00a0 —\u00a0 R\u2019=R\/4.2=R\/8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Reduzindo o comprimento \u00e0 metade\u00a0 —\u00a0 R fica 2 vezes menor\u00a0 —\u00a0 reduzindo o di\u00e2metro (r\/2) \u00e0 metade\u00a0 —\u00a0 R fica 4 vezes maior\u00a0 —\u00a0 R\u2019=(1\/2 x 4)R\u00a0 —\u00a0 R\u2019=2R\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0(01) Quando a temperatura da maioria dos metais aumenta, h\u00e1 um aumento no movimento vibrat\u00f3rio de seus \u00e1tomos e mol\u00e9culas, que dificultam a passagem da corrente el\u00e9trica diminuindo-a, o que provoca um aumentando de sua resist\u00eancia el\u00e9trica. Como a resistividade \u00e9 diretamente proporcional \u00e0 intensidade da corrente el\u00e9trica, ela tamb\u00e9m aumenta\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0Falsa.<\/b><\/span><\/span><\/p>\n

(02)\u00a0<\/b><\/span><\/span>Correta<\/b><\/span><\/span>\u00a0 —\u00a0 resistor \u00f4hmico, R constante\u00a0 —\u00a0 U \u00e9 diretamente proporcional a i.<\/b><\/span><\/span><\/p>\n

(04)<\/b><\/span><\/span>\u00a0Falsa<\/b><\/span><\/span>\u00a0 —\u00a0 Se a \u00e1rea S dobra, a resist\u00eancia R cai pela metade.<\/b><\/span><\/span><\/p>\n

(08)\u00a0<\/b><\/span><\/span>Falsa<\/b><\/span><\/span>\u00a0 — W=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t=i.U.\u0394t\u00a0 —\u00a0 i \u00e9 a intensidade da corrente que independe de seu sentido e, observe na f\u00f3rmula, depende da tens\u00e3o U.<\/b><\/span><\/span><\/p>\n

(16)\u00a0<\/b><\/span><\/span>Falsa\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 corrente el\u00e9trica \u00e9 um movimento ordenado de el\u00e9trons.<\/b><\/span><\/span><\/p>\n

(32)\u00a0<\/b><\/span><\/span>Verdadeira<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/p>\n

R- (02 + 32)=34<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=R\u00a0 —\u00a0 R \u00e9 inversamente proporcional \u00e0 S\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=R\/2\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/(R\/2)\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R x R\/2U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0Sendo o mesmo material, a resistividade \u03c1 \u00e9 a mesma\u00a0 —\u00a0 quando ele \u00e9 estirado, como mant\u00e9m o mesmo volume, o comprimento<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

passa a 2L e a nova espessura a S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V=S.L\u00a0 —\u00a0 V=S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 S.L=S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.2L\u00a0 —\u00a0 S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=S\/2\u00a0 —\u00a0 R=\u03c1.L\/S (I)\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03c1.L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03c1.2L\/S\/2\u00a0\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4\u03c1.L\/S (II)\u00a0 —\u00a0 comparando (II) com (I)\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0Maci\u00e7o\u00a0 —\u00a0 R<\/b><\/span><\/span>maci\u00e7o<\/b><\/span><\/span><\/sub>=\u03c1.L\/\u03c0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 oco\u00a0 —\u00a0 S<\/b><\/span><\/span>maior<\/b><\/span><\/span><\/sub>=\u03c0(R)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>menor<\/b><\/span><\/span><\/sub>=\u03c0( r )<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=\u03c0.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u2013 \u03c0.r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=\u03c0( R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 a express\u00e3o da resist\u00eancia fica\u00a0 —\u00a0 R<\/b><\/span><\/span>oco<\/b><\/span><\/span><\/sub>=\u03c1.L\/\u03c0(R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 R<\/b><\/span><\/span>maci\u00e7o<\/b><\/span><\/span><\/sub>\/R<\/b><\/span><\/span>oco<\/b><\/span><\/span><\/sub>= \u03c1.L\/\u03c0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0x \u03c0(R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\/\u03c1L\u00a0 —\u00a0 R<\/b><\/span><\/span>maci\u00e7o<\/b><\/span><\/span><\/sub>\/R<\/b><\/span><\/span>oco<\/b><\/span><\/span><\/sub>=(R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= 1 \u2013 r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mesmo material, mesma densidade \u00a0—\u00a0 d=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=4m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.25\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.75\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 75S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4.25.S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/3\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03c1L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

4= \u03c1.25\/S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c1=4.S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/25\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= \u03c1.L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=4.S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/25.75\/(4S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)\/3\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=75.3\/25\u00a0 —\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=9\u2126<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0I. Entre A e C\u00a0 —\u00a0 maior resist\u00eancia\u00a0 —\u00a0 menor corrente\u00a0 —\u00a0 menor pot\u00eancia\u00a0 —\u00a0 aquece menos\u00a0 —\u00a0 correta<\/b><\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 veja I.<\/b><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 quente \u00e9 entre B e C (menor R, maior pot\u00eancia, aquece mais)\u00a0 e n\u00e3o entre A e C.<\/b><\/span><\/span><\/p>\n

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

12-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=\u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.L\/\u03c0(r<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.L\/\u03c0(0,25)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=16.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>L\/\u03c0 —\u00a0 R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.L\/\u03c0(r<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1,2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.L\/\u03c0.(0,25)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=19,2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>L\/\u03c0\u00a0 —\u00a0 R<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=\u03c1<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.L\/\u03c0(r<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=1,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.L\/\u03c0.(0,2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=37,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>L\/\u03c0 \u00a0—\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0a) L=3.(4.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>)=12.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 R=\u03c1.L\/A\u00a0 —\u00a0 R=\u03c1.L\/\u03c0.(r)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 1,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>.12.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>\/3,2.(2.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=19,2.10<\/b><\/span><\/span>-18<\/b><\/span><\/span><\/sup>\/12,8.10<\/b><\/span><\/span>-20<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R=1,5.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u2126<\/b><\/span><\/span><\/p>\n

b)R=U\/i=10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\/8.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=0,12.10<\/b><\/span><\/span>5\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>R=1,2.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u2126<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>A resist\u00eancia por unidade de comprimento=\u03c1\/S\u00a0 —\u00a0 R=\u03c1L\/S=5.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.6.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=30.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u2126\u00a0 —\u00a0 R=Ui\u00a0 —\u00a0 3.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=U\/10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 U=3.10<\/b><\/span><\/span>-3\u00a0<\/b><\/span><\/span><\/sup>\u00a0A\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0a) R=\u03c1L\/S=5,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>.31,4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/\u03c0(4\/2.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=175,84.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>\/3,14.4.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>=175,84.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>\/12,56.10<\/b><\/span><\/span>-10<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R=14\u2126<\/b><\/span><\/span><\/p>\n

b) P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 100=(120)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 R=14.400\/100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R=144\u2126<\/b><\/span><\/span>\u00a0(observe que, com a l\u00e2mpada acesa a temperatura aumenta, aumentando assim o movimento vibrat\u00f3rio dos \u00e1tomos e mol\u00e9culas do tungst\u00eanio.o que dificulta a passagem da corrente el\u00e9trica, aumentando portanto a resist\u00eancia).<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0a) R=\u03c1L\/S=1,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>.10\/4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=1,6.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>\/4.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=0,4.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R=0,04\u2126<\/b><\/span><\/span><\/p>\n

b) Corrente que chega \u00e0 resid\u00eancia\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=i.U\u00a0 —\u00a0 3.300=i.110\u00a0 —\u00a0 i=30 A\u00a0 —\u00a0 da caixa de rel\u00f3gio at\u00e9 a resid\u00eancia existem dois fios\u00a0 —\u00a0 R=2.0,04=8.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u2126\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=8.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.900\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=72W<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0A ponte de Wheatstone \u00e9 um dispositivo que permite determinar o valor de resist\u00eancia el\u00e9trica desconhecida\u00a0 —\u00a0 R<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=\u03c1L\/S\u00a0 —\u00a0 R<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0varia de 10\u2126 at\u00e9 12\u2126\u00a0 —\u00a0 R<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=10\u2126\u00a0 —\u00a0 10=\u03c1<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>L\/S (I)\u00a0 —\u00a0 R<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=12\u2126\u00a0 —\u00a0 12= \u03c1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>L\/S (II)\u00a0 —\u00a0 subtraindo (II) de (I)\u00a0 —\u00a0 2=( \u03c1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0– \u03c1<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>).L\/S\u00a0 —\u00a0 2=\u0394\u03c1.L\/S\u00a0 —\u00a0 2=\u0394\u03c1.10\/0,1.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394\u03c1=2.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u2126m\u00a0 —\u00a0 pelo gr\u00e1fico\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 \u0394\u03c1 (2.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u2126m) \u2013 \u0394t\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>C\u00a0 —\u00a0 (\u03c1 \u2013 \u03c1<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) (0,8.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u2126m) \u2013 89<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 0,8.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>.\u0394t=160.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394t=200<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>=R(6R\/\u00a8)\u00a0 —\u00a0 comprimento 6 vezes menor R<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>=5R\/6\u00a0 —\u00a0 P<\/b><\/span><\/span>oantes<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 P<\/b><\/span><\/span>odepois<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/5R\/6=6U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/5R\u00a0 —\u00a0 P<\/b><\/span><\/span>oantes<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>odepois<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R x5R\/6U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>oantes<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>odepois<\/b><\/span><\/span><\/sub>=5\/6\u00a0 —<\/b><\/span><\/span>\u00a0 R- E<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=2R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=2L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 AS=S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/4\u00a0 —\u00a0 \u03c1=R.S\/L\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=2R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.(S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/4)\/2L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/\u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= 2R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.(S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/4)\/2L<\/b><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub>x L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 \u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/\u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/2L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0x L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0\u00a0\u03c1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/\u03c1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n


\n<\/b><\/span><\/span>20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Para que a corrente seja a mesma em cada linha, as duas devem ter a mesma resist\u00eancia el\u00e9trica\u00a0 —\u00a0 R=\u03c1L\/S\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

S=\u03c0( r )<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0=\u03c0.(d\/2 )<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=\u03c0d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4\u00a0 —\u00a0 R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03c1(72 + 72)\/\u03c0.(d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/4)=4.144\/\u03c0d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u03c1100\/\u03c0(d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/4)=4.100\/ \u03c0d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 — 4.144\/\u03c0d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=4.100\/ \u03c0d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 — (d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=144\/100\u00a0 —\u00a0 d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=12\/10\u00a0 —<\/b><\/span><\/span>\u00a0d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1,2<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=R\u00a0 —\u00a0 R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,8R\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/0,8R\u00a0 —\u00a0 P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R x 0,8R\/U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/0,8\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1,25P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=125%P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0W=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t=(U<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R).\u0394t\u00a0 —\u00a0 maior pot\u00eancia, menor resist\u00eancia, consome mais energia\u00a0 —\u00a0 menor resist\u00eancia \u2013 substituindo os valores de cada alternativa na equa\u00e7\u00e3o R=\u03c1.L\/S, chega-se \u00e0 resposta\u00a0 —<\/b><\/span><\/span>\u00a0\u00a0R- C<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0a) R=\u03c1L\/S=1,7.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>.1\/3,5.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=0,486.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R=4,86.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u2126m<\/b><\/span><\/span><\/p>\n

b) R=\u03c1L\/S\u00a0 —\u00a0 2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=1,7.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>.10\/S\u00a0 —\u00a0 S=1,7.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>\/2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=85.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=8,5mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=\u03c0.r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 8,5=3,14.r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 r=\u221a2,71\u00a0 —\u00a0 r=1,6mm\u00a0 —\u00a0 d=2.r=2.1,6\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=3,2mm\u00a0 —\u00a0 calibre 8<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0I- Correta\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 quanto maior i, maior P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/p>\n

II- Correta\u00a0 —\u00a0 R=\u03c1L\/S\u00a0 —\u00a0 quanto maior \u03c1, maior R e menor i<\/b><\/span><\/span><\/p>\n

III- Falsa\u00a0 —\u00a0 \u00e9 a raz\u00e3o U\/i que \u00e9 constante e n\u00e3o o produto U.i<\/b><\/span><\/span><\/p>\n

IV- Correta\u00a0 —\u00a0 Falsa\u00a0 —\u00a0 depende do tipo de associa\u00e7\u00e3o (s\u00e9rie, paralelo ou mista)<\/b><\/span><\/span><\/span><\/p>\n

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

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>(01) Errada. Resistor \u00e9 um dispositivo el\u00e9trico para CONTROLAR a intensidade da corrente el\u00e9trica.<\/b><\/span><\/span><\/p>\n

(02) Errada. A resistividade de um material s\u00f3 depende dele pr\u00f3prio e de sua temperatura.<\/b><\/span><\/span><\/p>\n

(04) Correta. Quando se acrescentam novas l\u00e2mpadas em s\u00e9rie, aumenta-se a resist\u00eancia equivalente do conjunto, diminuindo a corrente. Como P = Ri<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, reduz-se tamb\u00e9m a pot\u00eancia dissipada em cada uma delas, diminuindo as intensidades luminosas.<\/b><\/span><\/span><\/p>\n

(08) Correta. De acordo com a Segunda Lei de Ohm, a resist\u00eancia depende: do material de que \u00e9 feito o condutor, das dimens\u00f5es do condutor e da temperatura.\u00a0<\/b><\/span><\/span><\/p>\n

R- (04 + 08) = 12<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Sendo \u03c1 a resistividade do material, L o comprimento do condutor e A a \u00e1rea de sua sec\u00e7\u00e3o transversal, a segundo lei da Ohm nos d\u00e1 que a resist\u00eancia (R) desse condutor \u00e9\u00a0 —\u00a0 R=\u03c1L\/A\u00a0 —\u00a0 dobrando o comprimento e reduzindo \u00e0 metade a \u00e1rea de sua sec\u00e7\u00e3o transversal, a nova resist\u00eancia passa a ser\u00a0 —\u00a0 R\u2019=(\u03c12L)\/A\/2=4(\u03c1L\/A)\u00a0 —\u00a0 R\u2019=4R\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Dados\u00a0 —\u00a0 A = 1,6.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0mm2 = 1,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 L = 2 m\u00a0 —\u00a0 do gr\u00e1fico\u00a0 — quando a temperatura \u00e9 T = 3.000 \u00b0C, a resistividade \u00e9 \u03c1 = 8.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>\u00a0\u03a9.m\u00a0 —\u00a0 segunda lei de Ohm\u00a0 —\u00a0 R= \u03c1L\/S=8.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>.2\/1,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R=100\u03a9<\/b><\/span><\/span><\/p>\n

b) Dados\u00a0 — \u00a0\u03bb = 12.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0\u00b0C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 T\u2019 = 20 \u00b0C\u00a0 —\u00a0 T = 3.000 \u00b0C\u00a0 — \u00a03.000 \u00b0C, o volume inicial \u00e9\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= S.L = 1,6.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>. 2 \u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>= 3,2.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 varia\u00e7\u00e3o volum\u00e9trica\u00a0 —\u00a0 \u0394V = V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03bb.(T\u2019 \u2013 T) = 3,2.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>.12.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0(20 \u2013 3.000) \u00a0—\u00a0\u00a0<\/b><\/span><\/span>\u0394V= \u2013 \u00ad1,1.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>\u00a0m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=-1,1mm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

—\u00a0<\/b><\/span><\/span>\u00a0o sinal (\u2013) indica que o material sofreu contra\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 — \u00a0D = 2 cm = 2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0m\u00a0 —\u00a0 L = 2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0m\u00a0 —\u00a0\u00a0 i = 10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0A;\u00a0 —\u00a0 \u00a0\u0394t = 1 h = 3,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0s.<\/b><\/span><\/span><\/p>\n

A resist\u00eancia da linha \u00e9 dada pela 2\u00aa lei de Ohm\u00a0 — \u00a0R=\u03c1L\/S\u00a0 —\u00a0 \u00a0\u00e1rea da sec\u00e7\u00e3o transversal\u00a0 —\u00a0\u00a0 S= \u03c0r<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0=\u03c0(D\/2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 S=\u03c0D<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4\u00a0 —<\/b><\/span><\/span><\/p>\n

R=\u03c1L\/S=4\u03c1L\/(\u03c0D<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4)=4(1,57.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>).(2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)\/3,14.(2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=0,1\u03a9\u00a0\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=\u0394W\/\u0394t\u00a0 —\u00a0 \u0394W=P<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u0394W=R.i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.\u0394t\u00a0 —\u00a0 \u0394W=0,1.(10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.3,6.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394W=3,6.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

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

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 — \u00a0c = 4.200 J\u00a0 — \u00a0\u0394T = (100 \u2013 28) = 72 \u00b0C\u00a0 —\u00a0 \u00a0Q = \u0394W = 3,6.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0J\u00a0 —\u00a0 equa\u00e7\u00e3o do calor sens\u00edvel\u00a0 —\u00a0 Q = m c \u0394T\u00a0 —\u00a0 \u00a0m = Q\/c\u0394t=3,6.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\/4,2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.72\u00a0 —\u00a0 m=1,2.10<\/b><\/span><\/span>3\u00a0<\/b><\/span><\/span><\/sup>kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Baseado nas\u00a0 figuras:<\/b><\/span><\/span><\/p>\n

(1) Mantendo-se a sec\u00e7\u00e3o transversal constante e dobrando-se o comprimento (\u2113) do fi o, a resist\u00eancia (R) dobra\u00a0 —\u00a0 ent\u00e3o, a proporcionalidade entre \u2113 e R \u00e9 direta.<\/b><\/span><\/span><\/p>\n

(2) Mantendo-se o comprimento constante e dobrando-se a \u00e1rea da sec\u00e7\u00e3o transversal (A), a resist\u00eancia (R) fica dividida por dois\u00a0 —\u00a0 ent\u00e3o, a proporcionalidade entre A e R \u00e9 inversa.<\/b><\/span><\/span><\/p>\n

(3) Mantendo-se a resist\u00eancia constante e dobrando-se o comprimento (\u2113) do fio, a \u00e1rea da sec\u00e7\u00e3o transversal (A)<\/b><\/span><\/span><\/p>\n

dobra. Assim, a proporcionalidade entre \u2113 e A \u00e9 direta.<\/b><\/span><\/span><\/p>\n

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

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>As figuras 1 e 2 ilustram a situa\u00e7\u00e3o descrita.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Considere que na Fig 1 a resist\u00eancia el\u00e9trica do fio \u00e9 R e a corrente \u00e9 i\u00a0 —\u00a0 sendo U a ddp fornecida pela bateria, aplicando a 1\u00aa lei de Ohm\u00a0 —\u00a0 R=U\/i\u00a0 —\u00a0 i=U\/R\u00a0 —\u00a0 de acordo com a 2\u00aa lei de Ohm, a resist\u00eancia el\u00e9trica \u00e9 diretamente proporcional ao comprimento Ent\u00e3o, ao se cortar o fio ao meio, a resist\u00eancia el\u00e9trica de cada peda\u00e7o \u00e9 metade da resist\u00eancia do fio inteiro\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=R\/2\u00a0 —\u00a0 colocando-se os dois peda\u00e7os em paralelo como na Fig 2, a resist\u00eancia do circuito \u00e9\u00a0 —\u00a0 R\u2019=(R\/2)\/2\u00a0 —\u00a0 R\u2019=R\/4\u00a0 —\u00a0 corrente i\u2019 no circuito\u00a0 —\u00a0 U\u2019=U\/R\u2019=U\/(R\/4)\u00a0 —\u00a0 U\u2019=4U\/R\u00a0 —\u00a0 U\u2019=4i\u00a0 —\u00a0 correntes nos peda\u00e7os\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=i\u2019\/2=4i\/2\u00a0 —\u00a0 i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=i<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2i\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Segunda lei de Ohm (Resistividade) 01-\u00a0Como s\u00e3o de mesmo material, possuem a mesma resistividade el\u00e9trica\u00a0 —\u00a0\u00a0R- B 02-\u00a0A resist\u00eancia R \u00e9 inversamente proporcional \u00e0 espessura (S)\u00a0 —\u00a0 maior espessura, menor R, maior corrente i\u00a0 —\u00a0 brilha mais\u00a0 —\u00a0\u00a0R- E 03-\u00a0Mesmo fio \u00a0—\u00a0 mesma resistividade\u00a0\u03c1\u00a0 —\u00a0 mesma S\u00a0 —\u00a0 L cai pela metade\u00a0 —\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1730,"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-1734","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1734","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=1734"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1734\/revisions"}],"predecessor-version":[{"id":10557,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1734\/revisions\/10557"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1730"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1734"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}