高三物理答案 第 1页（共 3页）
2019－2020 学年度第一学期第二学段模块检测
高三物理答案及评分标准
一、单项选择题：本大题共 8 小题，每小题 3 分，共 24 分。
1——5：A、D、D、C、B；6——8：C、B、D
二、多项选择题：本大题共 4 小题，每小题 4 分，共 16 分，选不全得 2 分，有选错得 0 分。
9．ABCD、10．AC、11．BD、12．BC
三、非选择题。
13．（1）BC （3 分）说明：选不全得 1 分；（2）2.0 （3 分）。
14.（1）I-t 图象与坐标轴围成的面积就是总的电荷量 （2 分）、2.4×10－3～2.6×10－3C (2 分)、
（2）3.0×10－4～3.3×10－4F（2 分）；（3）不变（2 分）。
15.（8 分）
(1) 根据理想气体状态方程的分列式，得
p0V＋p0nV′＝4p0V···························································································································（2 分）
其中 V＝5.7×10－3m3－4.2×10－3m3＝1.5×10－3m3····················································· （1 分）
V′＝0.25×10－3m3····························································································· （1 分）
代入数值，解得 n＝18····················································································· （1 分）
(2)当空气完全充满储液桶后，如果空气压强仍然大于标准大气压，则药液可以全部喷出．
由于温度不变，根据玻意耳定律 p1V1＝p2V2，得
0
3
4
5.7 10
p Vp ······························································································· （2 分）
解得 p＝1.053p0>p0···········································································································································（1 分）
所以药液可以全部喷出。
16．(11 分)
(1) (1)由题图乙可知：mgsin 37°＝12 N·································································（2 分）
解得 m＝2 kg··································································································（1 分）
(2)题图乙中图线与横轴所围成的面积表示力 F 所做的功：
WF＝ 1 1-12 0.06 + 20 16-6 =0.642 2
（ ） ······························································（2 分）高三物理答案 第 2页（共 3页）
从 A 点 B 点的过程中由能量守恒可得：
0sin37 2.56JP FE W mgx ············································································ （2 分）
(3)撤去力 F，设物体返回至 A 点时速度大小为 v0，
从 A 出发两次返回 A 处的过程应用动能定理：W＝ 2
0
1
2 mv ········································（2 分）
解得：v0＝0.8 m/s····························································································（2 分）
17. （13 分）
（1）电荷的运动轨迹如右图所示，设电荷在磁场中的轨迹半径为 R：
由几何关系可知： 0sin30R R L ······························（1 分）
解得： 3
LR ……………………………………………………(1 分)
洛伦兹力提供向心力，由牛顿第二定律得：
2
0
0
mvBqv R
···································································································（2 分）
解得磁感应强度： 03mvB qL
············································································· （1 分）
（2）若电子能进入电场后， 从 C 点射入电场的电子做类似平抛运动，设穿出电场时速度的偏
转角为θ，则： 22 1=3 2
qEL tm
g ··············································································（2 分）
0
tan =2
qE tm
v
······························································································· （2 分）
由于电场中偏转位移的夹角α的正切值
1tan = tan =2
OD
OC
l
l
········································（1 分）
解得：lOD= 2
3 L ······························································································· （1 分）
之后电荷做匀速直线运动达到荧光屏上的 Q 点
2 2( )tan3 3PQl L L L ·················································································· （2 分）
18．（14 分）
（1）ab 边产生电动势：E＝BLv0·············································································································· （2 分）高三物理答案 第 3页（共 3页）
因此 0
3
4abU Blv ···························································································· （1 分）
(2) abF BIl ································································································· （1 分）
0BlvI R
······································································································· （1 分）
对火箭主体受力分析可得： Fab－mg＝ma··············································································· （2 分）
解得：
2 2
0B l va gmR
······················································································ （1 分）
(3)设下落 t 时间内火箭下落的高度为 h，对火箭主体由动量定理：
mgt－F
－
abt＝0－mv0··········································································································································· （2 分）
即 mgt－
2 2B L h
R ＝0－mv0
化简得 h＝ 0
2 2
( )mR v gt
B L
···················································································· （1 分）
根据能量守恒定律，产生的电能为：
E＝ 2
0
1
2mgh mv ····························································································· （2 分）
代入数据可得：
2
20
02 2
( ) 1
2
m gR v gtE mv
B L
···························································（1 分）
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