EDs 3° semestre engenharia unip
1101 palavras
5 páginas
Estudos Diciplinares1-
F1,3= K.(MODULO DE Q1).(MODULO Q2) / D² = 9.10^9.10.10^-6.4.10^-3 / 10²=36N
F2,3=K.(MODULO DE Q2).(MODULO Q3) / D² =9.10^9.10.10^-6.4.10^-3 / 8²=338N
θ= arctg=6/8 θ=36,9º Fx=v(36² +338² +(2.36.338.COS (36,9))=6,62N Alternativa=A
2-
F2,3 = F2,3.COStéta
F2,3x=3,38.cos36,9
F2,3x= 2,7N
F2,3y=F2,3.sentéta
F2,3y=3,38.sen36,9
F2,3y= 2,03N -->
F1,3= F1,3= -36i
F2,3= - F2,3i + F2,3j
-->
F2,3= - 27i +2,63j θ= arctg 2,03/6,3= 17,86º Alternativa= E
3-
F1= 1/4pi.epsílon0 . (MODULO DE QB).(MODULO DE Q)/ (x-0,5d)² --> =9.10^9.1.10^-3.5.10^-4/(4-0,001)²= 281,39N
F2=9.10^9 . 10^-3 . 5X10-4 /(4+0,001)²= 281,1 N
m.a= SF--> a=281,39-281,11/01 = 2,8m/s² Alternativa= A
4- E1= 9.10^9 . 1.10^-3 /(x-0,5d)² = 9.10^9 .1x10^-3 /(4-0,001)² = 562781,35 N/C
E2= 9.10^9. 10^-3 /(4+0,001)² = 562218,85 N/C
E= E1-E2=562781,35-562218,85=562,5 N/C Alternativa= B
5-
E=Q/(4π.epsílon.0).x(x^2+r^2)^(-3/2)
(r^2+x^2)^(-3/2)-3x^2(r^2+x^2)^(-5/2)=0.(4π.epsílon.0)/Q
(r^2+x^2)^(-3/2).(r^2+x^2)^(-5/2)=3x^2
(r^2+x^2)=3x^2
3x^2-x^2=r^2
2x^2=r^2
x^2=8 x=2,8 Alternativa= C 6-
E= K.Q/(R²+X²)^3/2
E= K.Q/(X²)^3/2
E= K.Q/(X)^6/2
E= K.Q/X³ Alternativa= B 7 -
dE= 1/(4π.epsílon.0).(dQ)/(x+a)^2
dE= 1/(4π.epsílon.0).(Q/L).[-1/(x+a]de 0 à L
E= 9.10^9.5.10^-6.0,179
E= 803,57 N/C Alternativa= A 8 -
E= 9.10^9.(5.10^-6)/10.[-0,011111+0,0125]
E= 4500.0,0014
E= 6,25 N/C Alternativa= E
9
r1= (9.10^9.5.10^-6)/200 = 225M r2= (9.10^9.5.10^-6)/400 = 112,5M r3 = (9.10^9.5.10^-6)/600 = 75 M r4= (9.10^9.5.10^-6)/2800 =56,25M