2025年初中學業水平模擬考試（一）物理試題答案第Ⅰ卷選擇題（共40分)1-8題：共24分．每小題給出的四個選項中，只有一個是正確的，選對的每小題得3分．9-12題：共16分．每小題給出的四個選項中，至少有兩個是正確的，選對的每小題得4分，選對但不全的得2分，選錯或不選的得0分.題號123456789101112答案BACDDCCDADBDADAB第Ⅱ卷非選擇題（共60分）作圖題（6分）（3分）①找到M的對稱點，據此繪制出水面處的反射光線、入射光線（1分）（注：若學生將A點做對稱也可）②N點位于M對稱點的豎直下方（1分）③輔助線、光線的繪制符合規范（1分）14.（3分）①S（1分）②F磁正確（1分）③力臂正確（1分）四、實驗探究題（本大題共3個小題，共25分）15.(第1小題第2、3空每空1分，其余每空2分，共6分)（1）30.0倒立縮小；（2）BC16.(每空2分，共8分)（1）便于測量摩擦力的大小（意思相近即可）（2）同一地板磚，同一運動鞋（3）A（4）不必勻速拉動物體（意思相近即可）17.（除第3小問第一個空1分，第四小問共2分外，其余每空2分，共11分）（1）（2）小燈泡短路（3）2.55（4）S、S2S1（5）（I-0.5A）·R0/0.5A18.（共9分）解：（1）樁錘重力G=m樁錘·g=100kg×10N/kg=1000N·····································································1分提升裝置對樁錘做的功W有=G·h=1000N×2.4m×50=1.2×105J············································2分（2）消耗的汽油所產生的熱量Q=m汽油·q=0.015kg×4.0×107J/kg=6×105J·····································2分（3）內燃機產生的機械功W總=Q·η內燃機=6×105J×25%=1.5×105J·················································1分提升裝置的機械效率η機械=W有/W總=1.2×105J/1.5×105J=80%·················································2分整體代數規范···················································································································1分19.（共10分）解：（1）其中一臺掛燙機中水吸收的熱量Q吸=cm△t=4.2×103J/（kg·℃）×0.2kg×80℃=6.72×104J·············2分（2）不計熱損失，電流做的功W=Q吸···············································································1分因此其中一臺掛燙機的電功率P1=W/t=6.72×104J/60s=1120W··············································2分（3）R1=R2=U2/P1=（220V）2/1120W=605/14Ω······································································1分R總=R1+R2=605/7Ω··································································································1分此時的電功率P2=U2/R總=（220V）2/605/7Ω=560W···························································2分整體代數規范···················································································································1分20.（共10分）解：（1）浮體完全浸入水中時，杠桿對浮體施加的壓力F2=F1·OB/OA=360N/5=72N················2分（2）當F1=360N時，R1=80Ω，此時I1=0.1A，此時R總1=U/I1=12V/0.1A=120Ω···················1分此時R0=120Ω-80Ω=40Ω··························································································2分（3）當I=0.03A時，R總2=U/I2=12V/0.03A=400Ω，R2=400Ω-40Ω=360Ω···································1分此時對應壓力F3=120N，則有對A的壓力F4=F3·OB/OA=120N/5=24N·································1分則有：24N+G=3/5F浮①72N+G=F浮②···················································································1分 解得：F浮=120N，G=48N浮體質量m=G/g=48N/10N/kg=4.8kg浮體體積V物=V排=F浮/ρ水g=120N/（1.0×103kg/m