第第頁高2024級高二上期第一學(xué)月檢測物理參考答案題號12345678910答案BCCADCCCDBCBC11．(1)微小量放大法（2分）(2)控制變量法（2分）(3)力F和A、C間距離r的平方成反比（2分）12．(1)1（1分）負（1分）B（1分）(2)電荷量（1分）(3)0.47（1分）不變（1分）(4)減小（1分）向下運動（1分）升高（1分）13．解：(1)平行板電容器電容··································（4分）(2)電場強度·······························（4分）(3)靜電力做功·························（4分）14．解：（1）在A點，小球的受力情況如圖所示················································（2分）由平衡條件有·················································（3分）解得·····················································（4分）（2）結(jié)合小問（1）受力分析圖，由平衡條件有·················（6分）解得··········································（7分）（3）根據(jù)小球受力情況可知小球帶正電，把小球從A點移到B點，電場力做負功，電勢能增加，AB在水平方向的距離··········································（8分）解得··························································（9分）電場力做的功·········································（10分）解得·························································（11分）因···························································（12分）解得··········································································（13分）即電勢能增加量為2.1J?！ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぃ?4分）15．解：（1）設(shè)帶電體恰好能通過C點時的速度為，根據(jù)牛頓第二定律有············（2分）解得···········································································（3分）設(shè)帶電體通過B點時的速度為，設(shè)軌道對帶電體的支持力大小為，帶電體在B點時，根據(jù)牛頓第二定律有·········································································（5分）帶電體從B運動到C的過程中，根據(jù)動能定理有························（7分）聯(lián)立解得·······································································（8分）根據(jù)牛頓第三定律帶電體對軌道的壓力·············································（9分）（2）設(shè)帶電體從最高點C落至水平軌道上的D點經(jīng)歷的時間為t根據(jù)運動的分解有，·················································（11分）聯(lián)立解得0·········································································（12分）由P到B帶電體做加速運動，故最大速度一定出現(xiàn)在從B經(jīng)C到D的過程中，在此過程中只有重力和電場力做功，這兩個力大小相等，其合力與重力方向成45°夾角斜向右下方，故最大速度必出現(xiàn)在B點右側(cè)對應(yīng)圓心角為45°處。設(shè)帶電體的最大動能為，根據(jù)動能定理有·············