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slavik2212
@slavik2212
June 2022
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помогите решить 1 вариант
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IUV
Verified answer
2)
Б
3)
E=U/d
d=U/E=5000/25000 м = 0,2 м
4)
С34=С3+С4=6+4=10 мкф
С12=С1*С2/(С1+С2)=1*2/(1+2)=2/3 мкф
С=С12*С34/(С12+С34)=2/3*10/(2/3+10)=20/32=5/8 мкф
5)
mg-T*cos(pi/6)=0
k*q^2/(2*L*sin(pi/6))^2-T*sin(pi/6)=0
*************
mg=T*cos(pi/6)
k*q^2/(2*L*sin(pi/6))^2=T*sin(pi/6)
*************
T=mg/cos(pi/6) = k*q^2/(2*L*sin(pi/6))^2/sin(pi/6)
*************
q^2=m*g*4*L^2*sin(pi/6)^3/(k*cos(pi/6)
*************
q= корень(m*g*4*L^2*sin(pi/6)^3/(k*cos(pi/6)) = корень(0,005*10*4*0,4^2*sin(pi/6)^3/(8,99*10^9*cos(pi/6)) Кл =
7,17E-07
Кл
*************
1 votes
Thanks 1
slavik2212
а С34 и С12 это что?
slavik2212
в 4
slavik2212
вроде понял, спасибо
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Answers & Comments
Verified answer
2)Б
3)
E=U/d
d=U/E=5000/25000 м = 0,2 м
4)
С34=С3+С4=6+4=10 мкф
С12=С1*С2/(С1+С2)=1*2/(1+2)=2/3 мкф
С=С12*С34/(С12+С34)=2/3*10/(2/3+10)=20/32=5/8 мкф
5)
mg-T*cos(pi/6)=0
k*q^2/(2*L*sin(pi/6))^2-T*sin(pi/6)=0
*************
mg=T*cos(pi/6)
k*q^2/(2*L*sin(pi/6))^2=T*sin(pi/6)
*************
T=mg/cos(pi/6) = k*q^2/(2*L*sin(pi/6))^2/sin(pi/6)
*************
q^2=m*g*4*L^2*sin(pi/6)^3/(k*cos(pi/6)
*************
q= корень(m*g*4*L^2*sin(pi/6)^3/(k*cos(pi/6)) = корень(0,005*10*4*0,4^2*sin(pi/6)^3/(8,99*10^9*cos(pi/6)) Кл = 7,17E-07Кл
*************