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福州市初中毕业班质量检测物理试卷参考答案及评分标准

2018 年福州市初中毕业班质量检测 物理试卷参考答案及评分标准

一、选择题:本题共 16 小题,每小题 2 分,共 32 分。在每小题给出的四个选项中,只有一 项是符合题目要求的。

1.D 2.B 3.A 4.D 5.B 6.A 7.C 8.A

9.B 10.C 11.C 12.D 13.B 14.D 15.C 16.B

二、填空题:本题共 6 小题,每空 1 分,共 12 分。

17. 音色 空气 18. 无规则 反射 19. 电 裂变

20. 75% 不变 21. 不变

不变 22. 1∶3 1∶2

三、作图题:本题共 2 小题。每小题 2 分,共 4 分。

23.如答图 1 所示

24.如答图 2 所示

四、简答题:本题共 1 小题。共 4 分。

25. (1)家庭电路中电流过大的原因:①短路;②用电器总功率过大(也称为负载过大);

(2)根据焦耳定律 Q=I2Rt,在通电时间 t 和电阻 R 不变情况下,电流 I 越大,电线产生的热

量 Q 就越多,所以容易引起火灾。

五、实验题:本题共 5 小题。共 28 分。

26.(5 分)⑴99

⑵热传递

⑶液化 ⑷99 小于

27.(8 分)⑴如答图 3 ⑵断开 B

⑶灯断路

⑷A 2.5 0.3 0.75

28.(5 分)⑴①10 ②缩小 ③放大镜

⑵远离

⑶30

29.(5 分)⑴右 ⑵2 ⑶变大

⑷②OC 的长度为 l3

l2ρ水 ③l2-l3

30.(5 分)⑴3.5 ⑵如答图 4(2 分) ⑶1.5

⑷弹簧片的长度(厚度、宽度、材料等)

六、计算题:本题共 3 小题。共 20 分。

答图 4

31.(5 分)解:⑴Q 放=Vq=Vnq 天然气 =1m3×164×4.2×107J/m3 =6.888×109J·················································································2 分

⑵由 Q 吸=cmΔt 得:

Q吸 Q放 m= cΔt = cΔt
6.888×109J = 4.2×103J/(kg·℃)×(70℃-20℃)
=3.28×104kg ··················································································3 分

32.(7 分)解:⑴v=72km/h=20m/s,t=10min=600s

s 由 v=t得: s=vt=20m/s×600s=1.2×104m··································································2 分

⑵W=UIt=300V×100A×600s =1.8×107J·······················································································2 分

⑶水平匀速直线运动

∴F=f=1350N

W 有用=Fs=1350N×1.2×104m=1.62×107J

W有用

W有用

1.62×107J

η= W总 ×100%= W ×100%= 1.8×107J ×100%=90%····································3 分

33.(8 分)解:R1、R2、R3 串联,电压表 V1 测 U2+U3,电压表 V2 测 U3

⑴当滑片在 a 端时,R2=0,电流达到最大值

由图像可知,此时 I=0.3A·····································································1 分

U3 1.5V ⑵R3= I = 0.3A=5Ω·············································································3 分

⑶由图像得:

U=4.5V+0.1A·R1 解得 U=1.5V+0.3A·R1 滑片在 b 端时,I′=0.1A

R1=15Ω U=6V·························································2 分

P=I′2R1=(0.1A)2×15Ω=0.15W······························································2 分



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