求一道初中物理奥赛题
\u521d\u4e2d\u7269\u7406\u5965\u8d5b\u8bd5\u9898\uff081\uff09\u5148\u4ece\u6700\u4e0a\u9762\u5206\u6790
\u5b9a\u6ed1\u8f6e\u4e0e\u4e0a\u9762\u5899\u7684\u90a3\u6839\u7ef3\uff0c\u5b83\u627f\u8f7d\u4e86\u6240\u6709\u7684\u529b\uff0cF\uff1dG1+G2
\u4ee5\u5b9a\u6ed1\u8f6e\u4e3a\u7814\u7a76\u5bf9\u8c61\uff0c\u5411\u4e0a\u7684\u529b\uff0c\u7b49\u4e8e\u4e24\u8fb9\u7684\u529b\uff0c\u5219\u6bcf\u4e00\u8fb9\u7684\u529b\u7b49\u4e8e1/2(G1+G2)
\u518d\u5230\u52a8\u6ed1\u8f6e\u8fd9\uff0c\u540c\u6837\uff0c\u6bcf\u4e00\u8fb9\u7684\u529b\u90fd\u662f\u4e00\u534a\u7b49\u4e8e1/4(G1+G2)
\u6240\u4ee5\uff0c\u4eba\u75281/4(G1+G2)\u7684\u529b\u62c9\u7ef3
\uff083\uff09\u5bf9\u677f\u7684\u538b\u529b\u4e3aF\uff1dG1\uff0dF\u62c9\uff1d3/4G1-1/4G2
\uff082\uff09\u6839\u636e\u6760\u6746\u5e73\u8861\u539f\u7406
\u4ee5\u5de6\u7aef\u4e3a\u652f\u70b9O\uff0c\u6709F1*L1\uff1dF2*L2
L1\uff1d1/4(G1+G2)*L/(3/4G1-1/4G2)
\u4e5f\u5c31\u662f\u79bb\u5de6\u7aef\u7684\u8ddd\u79bb\u4e3a\u677f\u957f\u7684\uff08G1+G2)/\uff083G1-G2)\u7684\u5730\u65b9
\u56de\u7b54\u5b8c\u6bd5\uff0c\u62ff\u5206\u8d70\u4eba\u3002\u3002\u3002\u3002
\u8bbe\u539f\u6765\u7535\u6d41\u4e3aI\uff0c\u540e\u6765\u7684\u7535\u6d41\u4e3aI'\uff0c\u7535\u6e90\u7535\u538b\u4e3aU\u3002
\u636e\u9898\u610f\u77e5\uff0cU1=(5/4)U2\uff0cU2'=(5/4)U2
\u5217\u51fa\u4e0b\u5217\u65b9\u7a0b\uff1a
I*I*(R1+R2)=I*U1
\u5373\uff1aI^2*R1+P2=I*(5/4)U2------------------(1)
I*U2=8--------------------(2)
I*U=I*U1+Pa
\u5373\uff1aI*U=I*(5/4)U2+(8-P2)--------------------(3)
I'*U2'=5
\u5373\uff1aI'*U2'=5
I'*(5/4)U2=5--------------------(4)
I'*R1+U2'=U
\u5373\uff1aI'*R1+(5/4)U2=U--------------------(5)
\u7531(2)\u3001(4)\u5f97(\u6d88\u53bbU2)\uff1aI=2*I'
\u7531(1)\u3001(5)\u5f97(\u6d88\u53bbR1)\uff1a{I*(5/4)U2-P2}/I^2={U-(5/4)U2}/I'--------------------(6)
\u7531(3)\u3001(6)\u6d88\u53bbU\u5f97\uff1a
(5/4)*U2*I*I'-P2*I'=I^2*{(5/4)U2-(8-P2)/I}-(5/4)U2*I^2
\u6574\u7406(\u4ee3\u5165I=2*I')\uff1a
P2=16-(5/2)*I'*U2=16-(5/2)*4=6W
\u6545\u201cR2\u6d88\u8017\u7684\u529f\u7387\u201d\u4e3a\uff1aP2=6W
A由于带正电,产生电场,使导体的电势大于0,用手摸导体,使得导体的电势变为0,这个过程大地有负电流向导体,导体总电量为负.
答案只能在AC中选.
如果C端带正电,那么C端正电荷的电场线无论怎样都会引出矛盾,因此只能C端不带电,B带负电.
似乎应该选择【 A、B端带负电,C端带正电】?!
A接近导体, 由于静电感应, 使B端作为近端带负电, C端为远端带正电, 当手接触B端, 导体和大地形成一个导体, 导体作为近端成负电, 当手放开后, 导体就带负电, 而A没有移走, 使负电荷聚集在B端, C端便不带电! 所以选c
a接近b后b感应带负电,由电荷守恒得c带正电,手触摸后将枕形导体上电荷导走,此时导体整体显负性,由于a仍处于原处,所以将负电荷聚集在b处,因此枕形体b端显负电 c不带电,
D
B
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