福建省福州市2011-2012学年度第一学期期末高二模块考试 物理试卷_图文

福州市 2011—2012 学年第一学期期末高二模块测试

物理(3-1)试卷参考答案
一、选择题(本大题共 10 小题,每小题 4 分,共 40 分。每小题只有一个选项正确) 题号 答案 1 B 2 D 3 C 4 D 5 A 6 A 7 B 8 B 9 D 10 D

二、实验题(本大题共 2 小题,每空 2 分,共 16 分。请按题目的要求作答或画图) 11. A、C、E (错一个扣一分,共 2 分,不倒扣) ; 1.46(1.46~1.47 均可) ;0.67(0.65~0.72 均可) 12.(1)24.12(24.11~24.14 均可) ;0.618(0.617~0.620 均可) (2)偏小;

U ?d 2 4 IL

(3) 如右图所示 (连在其他等势位置同样得分, 每条线 1 分) 。 三、计算题(本大题共 4 小题,共 44 分。在答题卷上解答应写出必要的文字说明、方程式和重要 演算步骤,只写最后答案不得分。有数值计算的题,答案应明确写出数值和单位) 13.(10 分)解: (1)根据电场强度公式:

E?

F 9.0 ?10?5 ··········· ·········· ·········· ·········· ? N/C? 9.0 ?105 N / C ·····················(3 分) q 1.0 ?10?10
Qq 得: ······················· (2 分) ··········· ·········· ·· ·········· ··········· ·· r2

(2)根据库仑定律: F ? k 点电荷电量 Q ?

Fr2 9.0 ?10?5 ? (0.1)2 ·········· ·········· ? C ? 1.0 ?10?6 C ·········· (2 分) 9 ?10 kq 9.0 ?10 ? 1.0 ?10

(3) U ?

W 6 ?10?10 ··········· ·········· ···· ·········· ··········· ··· ? V ? 6V ·························(3 分) q 1.0 ?10?10

14.(11 分)解: (1)通过灯泡 L 的电流 I ?

P 12 ··················· (3 ·········· ········· ? A ? 2A ···················· 分) U 6

(2)电动机两端的电压 UM ? U ? U L ? 18V ? 6V ? 12V ············(2 分) ··········· · ·········· · 电动机的输入电功率 P ? U M I ? 12 ? 2W ? 24W ················ (2 分) ··········· ····· ·········· ······ M

(3)电动机的发热功率 P ? I 2 RM ? 22 ? 0.5W ? 2W ···············(2 分) ··········· ···· ·········· ···· r 电动机输出的机械功率 P? ? P ? P ? 24W ? 2W ? 22W ············(2 分) ··········· · ·········· · M r 15.(10 分)解: (1)金属棒 MN 所受安培力 F ? BIL ? 0.5 ? 5 ? 0.2N ? 0.5N ·········· 分) ········· (3 ········· 安培力方向竖直向下 ·································(2 分) ··········· ·········· ··········· · ·········· ··········· ··········· (2)金属棒 MN 的重力 G ? mg ? 0.05 ? 10 N ? 0.5N ·············· (1 分) ··········· ··· ·········· ···· 根据金属棒 MN 稳定后的左视受力图知 稳定后导线与竖直方向的夹角 tan? ?

F 0.5 ···· ··· ? ? 1 ····(1 分) G 0.5

解得 θ=450 ·························· ·························· 分) ························· (1 导线对金属棒的拉力 T ? ·········· ·········· 2G ? 0.7N ·········· (1 分)

每根导线对金属棒的拉力 T ? ? 16.(13 分)解:

T ? 0.35N …………………………………………(1 分) 2

(1)微粒在加速电场中做匀加速直线运动,由动能定理得:qU1=

2 mv0 ① ···· 分) ··· (2 ··· 2

解得 v0 = 1.0× 4m/s ·································(1 分) 10 ··········· ·········· ··········· · ·········· ··········· ··········· (2)微粒在偏转电场中做类平抛运动,有:a=

aL qU 2 ,vy=at= , ·······(2 分) ······· ······ v0 md
? U2L 1 ? 2U 1 d 3
② ······ (1 分) ······ ······

飞出电场时,速度偏转角的正切为: tan ? ?

vy vx

· · · · · · · ·· · · · · · · · ·· · · · · · · ·· · · · · · · ·· · · · · · · ·· · · · · · · · · · · · · ·· · · · · · · ·· · · · · · · ·· · · · · · · · ·· · · · · · · ·· · · · · 解得 θ = 30o ························································································(1 分)

(3)进入磁场时微粒的速度是:v=

v0 cos?

③ ··················(1 分) ··········· ······· ·········· ·······

由轨迹图的几何关系有:D=r+rsinθ ④ ······················· 分) ······················ (1 ·········· ··········· ·

mv 2 洛伦兹力提供向心力:Bqv= ⑤ ························(2 分) ··········· ·········· ··· ·········· ··········· ·· r

由③~⑤联立得: D ?

mv0 ?1 ? sin ? ? ·······················(1 分) ··········· ·········· ·· ·········· ··········· · qB cos?

代入数据解得:D=10cm ······························· 分) ······························ (1 ·········· ··········· ·········


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