2008年大连市物理中考试卷答案
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推荐回答(1个)
2008年大连市初中毕业升学统一考试
物理参考答案与评分标准
一、填空题(本题共12小题,每小题2分,共24分)
1.运动;游艇2.正;中子3.音调;音色4.不可:裂5.增大;压力6.化学;电磁波7.大;长度8.漫;黑9.N;减小10.如图1(折射光线1分;光线位置1分)11.如图2(两灯并联1分;开关在干路上1分)12.如图3(力臂1分;拉力示意图1分)
二、选择题(本题共12小题,每小题2分,共24分)
题号 13 14 15 16 17 18 19 20 21 22 23 24
选项 A B A B D C D C D ACD AB AD
三、计算题(本题共3小题,共20分)
25.(6分)解:
(1)Q=cm(t-t0)………………………………………………………………•(1分)
=4.2×103J/kg•℃×50 kg×(58℃一20℃)=7.98×106J……………(1分)
(2)因为P=W/t…………………………………………………………………(1分)
所以W=Pt=2000W×70×60s=8.4×106J……………………………… (1分)
(3)η=Q/W×100%……………………………………………………………(1分)
=7.98×106J/8.4×106J×100%=95%………………………………………(1分)
答:(1)这50kg水吸收的热量是7.98×106J。
(2)热水器消耗的电能是8.4×lO6J。
(3)热水器的效率是95%。
26.(7分)解:
(1)因为ρ=m/V………………………………………………………………… (1分)
所以m1=ρV=0.8×103kg/m3×6m3=4.8×103kg………………………………(1分)
(2)G=(m1+m2)g=(4.8×103kg+7.2×103kg)×10N/kg=1.2×105N……………(1分)
p=F/S=G/S…………………………………………………………………… (1分)
=1.2×105N/0.2m2=6×105Pa"…………………………………………(1分)
(3)因为W=Fs……………………………………………………………………(1分)
所以F=W/s=1.8×107J/9000m=2×103N ………………………………(1分)
答:(1)油罐中最多能装4.8×103kg的煤油。
(2)装满煤油的油罐车对地面的压强是6×105Pa。
(3)油罐车在水平路面上匀速行驶时,发动机的牵引力是2×103N。
27.(7分)解:
(1)p′=UI′……………………………………………………………………(1分)
=220V×O.1A=22W……………………………………………………(1分)
(2)I1=U/R1………………………………………………………………………(1分)
=220V/110Ω=2A………………………………………………………(1分)
(3)I=P/U=11OOW/220V=5A……………………………………………… (1分)
I2=I-I1-I′=5A一2A一0.1A=2.9A…………………………………(1分)
R2=U/I2=220V/2.9A=75.86Ω…………………………………………… (1分)
答:(1)吹冷风时,电动机的功率是22W。
(2)吹温风时,通过尼的电流是2A。
(3)吹热风时,屉的电阻约为75.86Ω。
四、简答题(本题共2小题,每小题3分,共6分)
28.从冰箱冷冻室取出的冰块温度低于O℃,放到水中后水放热降温,冰吸热升温(1分);
当冰块附近的水的温度降到O℃时,冰的温度还低于O℃(1分);冰块继续吸热,冰块
附近O'C的水继续放热,凝固成冰(1分)。所以冰块会“粘”到一起。
29.运动员从A点运动到B点过程中,重力不变,受到向上的弹力越来越火(1分);弹力
小于重力时,运动员所受合力方向向下,且合力方向与速度方向相同,运动员速度越来
越大(1分);当弹力增大到大于重力时,运动员所受合力方向向上,合力方向与速度
方向相反,运动员速度越来越小(1分)。所以运动员的速度先变大后变小。
五、综合题(本题共5小题,共26分)
30.(5分)
(1)见下图(电压表接大量程1分;电流表接小量程1分)
(2)见下表(1分)
实验
次数 电压
U/V 电流
I/A 电阻
R/Ω 待测电阻
R′/Ω
l
2
3
(3)12.5(1分)
不正确。因为灯泡的电阻随温度变化而变化,取平均值没有物理意义(1分)。
31.(5分)
(1)相同(1分);吸收的热量相同(1分);B(1分)
(2)C(1分);C(1分)
32.(5分)
(1)左(1分)
(2)不能得出探究结论(1分);
因为一次实验获得的数据具有偶然性,不能反映普遍规律(1分)
(3)杠杆在水平位置的实验方案好,便于测量力臂(1分)
(4)右(1分)
33.(5分)‘
(1)见下图(1分)
(2)液体密度一定时,浮力大小,与物体排开液体的体积V成正比(1分)
(3)0.75(1分)
(4)a.量程为lN的测力计(1分)
b.0.02N(1分)
34.(6分)
(1)缩小(1分);
光屏上的像清晰(1分)
(2)便于比较像与物的大小(1分)
(3)<(1分)
(4)因为看到的像是倒立的,所以是实像,成像点在凸透镜的右侧(1分); 发光物体
发出的光线经凸透镜折射后是会聚的,经过成像点后是发散的。所以眼睛必须在成
像点以外的光线发散区域才能看到这个像(1分)。
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