福建省福州市2020年初中毕业班质量检测物理试卷-

九年级物理参考答案-1-（共3页）2019-2020学年度福州市九年级质量检测物理参考答案及评分标准一、选择题：本题共16小题，每小题2分，共32分。在每小题给出的四个选项中，只有一项是符合题目要求的。二、填空题：本题共6小题，每空1分，共12分。17.无规则运动热传递18.核裂变1.26×101019.150083.320.N缩短21.电阻RL短路22.＜3∶7三、作图题：本题共2小题。每小题2分，共4分。23.如答图124.如答图2四、简答题：本题共1小题。共4分。25.答：（1）大功率用电器工作时，根据I=PU，家庭电路电压U=220V一定，电功率P较大，因此电路中的电流I也较大。（2分）（2）在其他条件相同时，粗导线的横截面积大，电阻小。根据P=I2R，在电路电流一定的情况下，粗导线的电阻小，导线的发热功率P就小，所以大功率用电器工作时，连接的粗导线不怎么热。（2分）12345678910111213141516BAABDACABDCCDBCD答图1答图2九年级物理参考答案-2-（共3页）五、实验题：本题共5小题。共28分。26.（5分）（1）同一高度（2）重力势能（3）惯性（4）远（5）匀速直线27.（5分）（1）温度计（2）93小于（3）还未（4）继续吸热28.（5分）（1）10.0cm（2）照相机左放大（3）D29.（5分）（1）3248（2）④将金属环放入装满水的瓶子中，并盖上盖子，用天平测出此时烧杯的总质量m4⑤(m1m1+m3-m4)ρ水30.（8分）（1）断开（2）如答图3（3）35（4）电压表示数为U0（5）610～40（6）通过导体的电流与电阻I与1R成正比，则可知I与R成反比六、计算题：本题3小题。共20分。31.（5分）（1）320s（2）105N解：（1）t=st=1.6×104m50m/s=320s·······························································（2分）（2）W=Pt=5×106W×320s=1.6×109JF牵=Ws=1.6×109J1.6×104m=105N·························································（3分）32.（8分）（1）10Ω（2）0.3W（3）R不均匀解：（1）当没有载物时，只有电阻R0接入电路。根据欧姆定律得R0=U总I0=3V0.3A=10Ω·································································（2分）（2）最大称量时，R=km=20Ω/kg×1kg=20Ω····················································（1分）此时电路电流：I=U总R0+R=3V10Ω+20Ω=0.1A电子秤的总功率P=U总I=3V×0.1A=0.3W·····································（2分）（3）R·························································································································（1分）电压表并联到电阻R两端后，U=IR=U总RR0+R=U总kmR0+km解得m=UR0U总k－Uk=U×10Ω3V×20(Ω/kg)－U×20Ω/kg=(U6V－2U)kg···········（1分）此函数不是一次函数，所以电压表改装后的电子秤刻度分布不均匀。····（1分）答图3九年级物理参考答案-3-（共3页）33.（7分）（1）2000Pa（2）10N（3）39cm解：（1）由图可知，当20cm≤h≤35cm时F浮不变则A恰能上浮时容器中的水面高度h1=20cm=0.2mp=ρ水gh1=1.0×103kg/m3×10N/kg×0.2m=2000Pa········································（2分）（2）图中F浮不变的阶段，A处于漂浮状态∵漂浮∴GA=F浮1=10N·············································································（2分）（3）A漂浮时，V排=F浮1ρ水g=10N1.0×103kg/m3×10N/kg=1×10-3m3此时A浸入水中部分的高度l=h1=20cmSA=V排l=1×10-3m30.2m=0.005m2······························································································（1分）注水结束时A受到的浮力F浮2=GA+F故从A接触到B至对B施加2N压力时，ΔF浮=F=2NA的排水体积增加了ΔV排=ΔF浮ρ水g=2N1.0×103kg/m3×10N/kg=2×10-4m3水面升高Δh=Δl=ΔV排SA=2×10-4m30.005m2=0.04m=4cm·····································（1分）由图可知A接触到B时水面高度h2=35cm则注水结束时水面高度h0=h2+Δh=35cm+4cm=39cm··································（1分）