7.D【解析】由图甲根据U=E-I可得电动势E=1.5V,内阻r1.5-1.0。1a≈0.17故A项错误;由图乙可得功率越大则电压越大,与坐示原点连线的斜率越大,故电阻越大,B项错误;由图乙可得当小灯泡两端的电压U=0.5V时,电流Ⅰ=6A,故U0.5电阻R=T-6=122,故C项错误;把电源和小灯泡组成闭合回路,在图中作出电源的U-I图线,两U-I曲线的交点即小灯泡的电压、电流,故U=0.5V,r=6A,所以小灯泡的功率P=UI=3W,故D项正确。2.01.00123456789MA

One possible versionStudents' Voluntary Work in a CommunityDuring this winter vacation, I, together with severalmembers of the Students' Union, took part in a socialpracticeWe reached the community at 8: 00 in the morningThen we were organized to watch a short video about thepossible harm resulting from lighting fireworks. After thatwe helped to hand out booklets to citizens to make kinownthe ban on fireworksThough it was tiring, I found it meaningful. Not onlydid I learn to communicate with strangers, I also picked upsome extracurricular knowledge

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