2025.4 物理一模答案（第二稿）

2025年初中学业水平模拟考试（一）物理试题答案第Ⅰ卷选择题（共40分)1-8题：共24分．每小题给出的四个选项中，只有一个是正确的，选对的每小题得3分．9-12题：共16分．每小题给出的四个选项中，至少有两个是正确的，选对的每小题得4分，选对但不全的得2分，选错或不选的得0分.题号123456789101112答案BACDDCCDADBDADAB第Ⅱ卷非选择题（共60分）作图题（6分）（3分）①找到M的对称点，据此绘制出水面处的反射光线、入射光线（1分）（注：若学生将A点做对称也可）②N点位于M对称点的竖直下方（1分）③辅助线、光线的绘制符合规范（1分）14.（3分）①S（1分）②F磁正确（1分）③力臂正确（1分）四、实验探究题（本大题共3个小题，共25分）15.(第1小题第2、3空每空1分，其余每空2分，共6分)（1）30.0倒立缩小；（2）BC16.(每空2分，共8分)（1）便于测量摩擦力的大小（意思相近即可）（2）同一地板砖，同一运动鞋（3）A（4）不必匀速拉动物体（意思相近即可）17.（除第3小问第一个空1分，第四小问共2分外，其余每空2分，共11分）（1）（2）小灯泡短路（3）2.55（4）S、S2S1（5）（I-0.5A）·R0/0.5A18.（共9分）解：（1）桩锤重力G=m桩锤·g=100kg×10N/kg=1000N·····································································1分提升装置对桩锤做的功W有=G·h=1000N×2.4m×50=1.2×105J············································2分（2）消耗的汽油所产生的热量Q=m汽油·q=0.015kg×4.0×107J/kg=6×105J·····································2分（3）内燃机产生的机械功W总=Q·η内燃机=6×105J×25%=1.5×105J·················································1分提升装置的机械效率η机械=W有/W总=1.2×105J/1.5×105J=80%·················································2分整体代数规范···················································································································1分19.（共10分）解：（1）其中一台挂烫机中水吸收的热量Q吸=cm△t=4.2×103J/（kg·℃）×0.2kg×80℃=6.72×104J·············2分（2）不计热损失，电流做的功W=Q吸···············································································1分因此其中一台挂烫机的电功率P1=W/t=6.72×104J/60s=1120W··············································2分（3）R1=R2=U2/P1=（220V）2/1120W=605/14Ω······································································1分R总=R1+R2=605/7Ω··································································································1分此时的电功率P2=U2/R总=（220V）2/605/7Ω=560W···························································2分整体代数规范···················································································································1分20.（共10分）解：（1）浮体完全浸入水中时，杠杆对浮体施加的压力F2=F1·OB/OA=360N/5=72N················2分（2）当F1=360N时，R1=80Ω，此时I1=0.1A，此时R总1=U/I1=12V/0.1A=120Ω···················1分此时R0=120Ω-80Ω=40Ω··························································································2分（3）当I=0.03A时，R总2=U/I2=12V/0.03A=400Ω，R2=400Ω-40Ω=360Ω···································1分此时对应压力F3=120N，则有对A的压力F4=F3·OB/OA=120N/5=24N·································1分则有：24N+G=3/5F浮①72N+G=F浮②···················································································1分 解得：F浮=120N，G=48N浮体质量m=G/g=48N/10N/kg=4.8kg浮体体积V物=V排=F浮/ρ水g=120N/（1.0×103kg/m