2019年青海省中考物理試卷(含答案與解析)

經(jīng)典word整理文檔，僅參考，雙擊此處可刪除頁眉頁腳。本資料屬于網(wǎng)絡(luò)整理，如有侵權(quán)，請聯(lián)系刪除，謝謝！6()物理在此g取10N/kg。________________號21((()))2V時，通過導體甲的電流為0.2A.將兩導體并聯(lián)后接在3V0.6A卷7()生考__2_____上_________8UUVU____________________名__()答318dm姓___200g__℃1m/s_____________校學業(yè)畢題LLLL12124(())9()無4.74N4.14N4.11N0.6N()5610m53效1.0510kg/m236.910kg/m33第1第2300N0.1m/s20s72NJFN，W()()時RS112.0V2.8VS1R2()，與。S1R1R2428711m166V3Wm9VW。1t172.18J[c10)]水。第3第4在移________________號此4題3題4題6題7卷生考____甲乙丙___________126.2g；上__cm3；___________________名__甲乙丙g；答姓__℃。gkg/m。___。。_____________校2.5V學業(yè)畢無16cm。甲乙P效第5第6176R①②；；R042W。A/AIP/W1234010minR025300。282916000kg400cm220，40s400m0.025kg6.410J/kg7220V1100W第7第8ABCD1B8AABCD2CA不BCDUUUVLVDL1A1燈0B不LLU21CLL122DL9CABCD3DAgB不C約為DDFGF4.74N4.14N0.6NA4CABCD浮水F浮gV可得，金屬塊的體積液排F0.6Ng110kg/m10N/kgVV=610mB53水336mVV53FGF'4.74N4.11N0.63N浮液5AgV排F浮液F0.63NABCD==1.0510kg/m，C正確，金屬塊的質(zhì)量33610mgVG10N/kg53排4.74Nm0.474kg，則金屬的密度g10N/kgm0.474kg金V610mD。7.910kg/m3353DD6B3V0.6A，，0.3AII，I甲I乙甲乙＜ARR甲乙B2VAC3VBII甲0.6A0.3A0.9ADI乙A7C第9第BCD2.110JQ82.18Jt50℃放cm4.210J/kgC10kg33水DA測AR112R1表AP1R1AC2Dss有用fsW72N2m144J；滑輪組的機械效率為則，則總功為144J60%W有用=240J3n總240J6mWs's32m6mF40N總2s240J20sW12WP?？?mt022m5:2串615SS和RRR11212VUII0.05A、S1R211R140U2(6V)2R額12即電源電壓UV；所以和串聯(lián)時，兩端電壓R2R1R23WLP0.8V額，燈泡正常發(fā)光時電路中的電流UUUUVVVR2216；R與RUUU9V6VI0.05A1221RL6VU3VII=U==0.5AR6額R120.5A額RI405。162QI2RtRLI2RtR111PUI3VW。QRR222m1t103kg第第1O1123210360cm3V和剩余鹽水的質(zhì)量為ggggg；⑤鹽水的質(zhì)量mgggm72g，則鹽水的密度V60cm31.2g/cm1.210kg/m。333SNm=V第第U220VI1A總220PUI220V1A220W最小071P23）由題圖甲可知，電流表選用0~AAAVAUI2.5V0.28A0.7W。額額P額4PUI）FGmg(16000kg2000kg)10N/kg1.810N5F1.810N5pS4004.510Pa610m42）Ff0.01G0.011.810N1.810N53WFs1.810N400m7.210J35W7.210J5P1.810W6t40s）Qmq0.025kg6.410J/kg1.610J76W7.210J5