文档介绍：该【2021-2022学年(上)1月厦门市初三质量检测物理卷参考答案 】是由【1781111****】上传分享，文档一共【4】页，该文档可以免费在线阅读，需要了解更多关于【2021-2022学年(上)1月厦门市初三质量检测物理卷参考答案 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。14小题,每小题2分,共28分。题号1234567891011121314答案ADBCBCBCACDBDD二、填空题:本题共6小题,每空1分,共12分。(沸腾)(吸收)、作图题:本题2小题,每小题2分,共4分。、简答题:本题共1小题,共4分。:钻孔时,钻头克服摩擦力做功,将机械能转化为内能,钻头内能增大,温度升高,造成钻头过热。(2分)浇水时,水温低于钻头,发生热传递;水的比热容大,吸收热量多;水汽化会吸收热量,使钻头迅速降温。(2分,答出一点即可得2分)五、实验题:本题共5小题,共30分。第26(3)、28(3)小题每空2分,其它每空1分,答案合理即给分。24.(1)酒精灯外焰高度(2)98(~)不变(3)温度测量误差(各小组杯内水面上方气压大小不同;各小组杯内水含有杂质不同)(4)沸腾需要吸热(试管内水的温度与杯中水的温度相等,无吸热)法25.(1)质量实际功率(功率、型号、规格)(2)加热时间(3)不同种液体吸热能力不同(相同质量的甲、乙液体升高相同的温度,吸收的热量不同;甲、乙液体比热容不同;甲液体比热容小于乙液体)(4)乙26.(1)断开(2)保护电路(电路安全)(3)导体电阻一定,通过导体的电流与导体两端的电压成正比(电阻一定,电流与电压成正比)(4)电池使用一段时间后,电压减小(电池组电压低于3V;电路中其它元件有电阻)(5)520127(1)如右图(2)D学生电源+-(3)(4)增大P(5)①电流表的示数为额③I2(R-R)额2128.(1)减小(2)0~(~)(3)如下图,采用电阻串联方式连接且电压表连接正确得1分,电阻数量合理得1分风速测量通道风速测量通道VRv外二个及以上电阻串联RR其它答案合理即得分风速测风速测量通道量通道VRRVvvRRRRR风速测量通道外三个及以上风速测量通道内二个及以上电阻串联电阻并联六、计题题:本题共3小题,共22分。29.(6分)(1)W=Pt=110W×200s=×104J·························································2分(2)热风档时,电动机与加热电阻并联由P=UI,得P110WI===···································································2分1U220VU220VI===4A········································································1分2R55Ω=+····························································分II1I2=+4A=(7分)m()由,得1Vm=ρV=×103kg/m3×20×10-3m3=20kg···································1分水水水Q=cmΔt=×103J/(kg?℃)×20kg×(90℃-20℃)=×106J吸水水水································2分(2)Q=W=12kW?h=×107J···························································1分放由=Q,得VQ=×107JV=放=······················································×107J/m3气(3)W′=Pt=2000W×3600s=×106J················································×106J?=吸×100%=×100%≈%·············································1分W'×106J31.(9分)()由图象可得,当时,1I==16VU16VR=2==80Ω·····································································()、串联2R1R2由U=U+U=IR+U,得总1212当I==16V,U=+16··············································1分2总1当I==8V,U=+8·················································1分2总1联立方程组,解得:························································分R1=40Ω1U=24V总U220VI=额==·································································1分额R140Ω()==(-)-2····················································分3P2U2I2440II=24I40I1由公式可得-图象如右图,由图象可知:滑动变P2P2I阻器功率先增大后减小,最小功率在电流最小P2I或最大处IOUU24V当R=0Ω时,I=总=总==+R40Ω+<==24I-40I2=(24×-40×)=2W····························1分2maxmaxUU=24V当R′=120Ω时,I=总=总='R+R'40Ω+120Ω总12则P'=24I-40I2=(24×-40×)W=··························1分2minmin<′P2P2根据以上结果可知,滑动变阻器最小功率为2W其它方法合理即可得分ΔU16V-8V(2)由图象斜率求出R===40Ω·····································3分1?-=额==····································································1分额R140ΩU2242(3)P=UI=I2R=(总)R=()R·····································1分222R+R240+R2122由公式可得-图象如右图,由图象可知:滑动P/WP2R22变阻器功率先增大后减小P2考虑电流表和滑动变阻器允许最大电流Imax==总===R-R=48Ω-40Ω=8Ω2min总min1242242P=()R=()×8=2W·····································1分240+R2min40+82min滑动变阻器最大值R2max=120Ω242242P'=()R=()×120=·····························1分22max40+R2max40+120<′P2P2根据以上结果可知,滑动变阻器最小功率为2W4