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高二物理答案试题第1页(共2页)2024学年第一学期丽水五校高中发展共同体10月联考高二年级物理学科参考答案命题：遂昌中学王宇阳审稿：遂昌中学郑卿武一．选择题二．非选择题16-Ⅰ（2）D（3）1.421.33下落

过程中有阻力做功16-Ⅱ（1）BDF（2）乙（3）2.98~3.021.5~1.716-Ⅲ（1）1.193~1.198（2）B（3）280Ω（4）������2���1���4������2−���117.（1）根据匀变速直线运动公式，可得v2

－v02=2gh··········································（1分）解得v=15m/s···················································································

······（1分）（2）根据匀变速直线运动公式，可得v2－v02=2axa=45m/s2····························（1分）根据牛二定律，F浮－mg=ma····································

································（1分）解得F浮=2200N·········································································

··········（1分）（3）由匀变速直线运动公式可得：t1=∆������=2s·······································································

·················（1分）t2=∆������=13s························································································（1

分）t总=t1+t2=73s·························································································（1分）18.（1）静电力公式F1=������1���2�

��2·········································································（1分）对小物块进行受力分析，可得F1=������������������···············

··········································（1分）联立解得Q2=4.8×10－5C·················································································（1分）

（2）对小球进行受力分析，可得F1=���2���tan���················································（1分）由（1）中方程可解得F1=4.8N······································

·································（1分）故可解���2=0.64kg····················································································

····（1分）（3）由动能定理m2gL（1－cos���）=12���2���2·····················································（2分）对小球最低点进行受力分析FN

－m2g=���2���2���·························································（2分）解得FN=8.96N··············

···············································································（1分）123456789101112131415BBABADBABBBDCADBC高二物理答案试题第2页

(共2页)19.（1）刚好过C点，即重力等于向心力���������2���=mg，解得vc=2m/s························（1分）根据动能定理，可列出12���������2－12���������2=2mgR········

······································（1分）解得vc=25m/s······················································································（1分）（2）对

AB段的摩擦生热：Q1=mgµL1=0.3J······················································（1分）长木板上的位移：v2－v02=2ax，其中a=gµ=

5m/s2········································（1分）解得x=0.4m<L2，············································

······································（1分）故长木板上的摩擦生热Q2=mgµx=0.2J，Q总=Q1+Q2=0.5J································（1分）（3）不脱

离轨道ABC且不会滑上木板的最大弹性势能：Epmax=mgR+µmgL1=0.7J······（1分）刚好能滑上木板的弹性势能：Epmin=12���������2+µmgL1=1.3J·················

·················（1分）刚好滑到木板的最右端的弹性势能：Epmax=2������R+µmg（L1+L2）=1.5J··············（1分）综上，0≤Ep≤0.7J或1.3J≤Ep≤1.5J··········

····················································（1分）20.（1）由动能定理可得Ue=12������2·······························

···································（1分）解得v=2���������··················································································

··（1分）（2）设在偏转电场中时间为t1，出偏转电场后的运动时间设为t2，则可列出���1=���1������，���2=���2������·····································································

·············（1分）偏转电场中运动的高度为h1=12������12·························································（1分）

出偏转电场后运动的高度为ℎ2=���������2且h=h1+ℎ2········································（1分）根据牛顿第二定律可知，������=������2，其中���=���������··················

·······················（1分）联立可解得E=4���ℎ2���1���2+���12··································································

········（1分）（3）vy=at1=���������∙���1������····························································

·························（1分）将vx=2���������，E=4���ℎ2���1���2+���12代入，可解得vy=2ℎ2���2+���12���������································（2分）故���=�

�����2+������2=1+4ℎ22���2+���12∙2���������·····················································（1分）