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2026年山西省太原市高三模拟考试(一)物理答案_12.2026年4月全国各地高考模拟卷_2026届山西省太原市高三模拟考试(一)全科_04物理_山西省太原市高三模拟考试(一)

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《2026年山西省太原市高三模拟考试(一)物理答案_12.2026年4月全国各地高考模拟卷_2026届山西省太原市高三模拟考试(一)全科_04物理_山西省太原市高三模拟考试(一)》第1页预览
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太原市 2026 年高三年级模拟考试(一)
物理参考答案与评分建议
一、选择题:本题共7小题,每小题4分,共28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。
题目 1 2 3 4 5 6 7
答案 D D C A D C B
二、多项选择题:本题包含3小题,每小题6分,共18分。在每小题给出的四个选项中,
至少有两个选项正确,全部选对的得6分,选对但不全的得3分,有选错的得0分。
题目 8 9 10
答案 AC AD BD
公众号:高考试卷君
三、实验题:本题包含2小题,共16分。
11.(8分)
(1)0.650(3分)
d
(2) (2分)
t
1
(M 5m)d2
(3) (3分)
2mkL
12.(8分)
L
(1)  (2分)
S
1
(2) (2分)
I
b
(2分)
aS
b
R (2分)
S 0四、计算题:本题包含3小题,共38分。解答应写出必要的文字说明、方程式和重要演算
步骤,只写出最后答案的不能得分。有数值计算的题,答案中必须明确写出数值和单位。
13.(9分)
(1)瓶内气体发生等容变化
p p
1  2 ···············································································(3分)
T T
1 2
T t 273K 400K
1 1
T t 273K 300K
2 2
p 1.5atm··········································································(1分)
2
(2)瓶内原有气体,据理想气体状态方程
pV pV
1  3 ··········································································(2分)
T T
1 2
m V-V
漏  ·········································································(2分)
m V
原 公众号:高考试卷君
m 1
漏  ···············································································(1分)
m 3
原
14.(12分)
(1)S拨向1,电容器充电
q
C  ·················································································(2分)
E
q CE ················································································(1分)
(2)S拨向2,电容器放电,线框加速运动并切割磁感线产生感应电动势,
直到电容器两极间电压与线框的感应电动势相等时,电容器放电结束,线
框获得的速度最大,此时电容器的电压为U,电容器的电荷量为q
1
U  BLv··············································································(1分)
q
C  1
U
对于线框,由动量定理2iLBt mv0································································(1分)
q qq 2it·····························································(1分)
1
CEBL
v ······································································(1分)
CB2L2 m
1
(3)根据图像可得电容器储能公式为 CU2 ·······························(1分)
2
根据能量守恒定律
1 1 1
CE2  CU2  mv2 Q ················································· (2分)
总
2 2 2
PQ边在加速过程中发出的热量
1
Q  Q ·············································································(1分)
总
2
mCE2
Q  ·································································(1分)
4(mCB2L2)
15.(17分)
(1)离子带负电·····公·······众········号·······:·······高·······考········试·······卷·······君·····(3分)
(2)离子从ab边界进入磁场到N点做匀速圆周运动,根据几何条件,
离子在磁场中的轨迹半径为r
1
r  R··················································································(1分)
1
v2
qv B m 1 ··········································································(1分)
1 r
1
离子从N点到P点做类平抛运动
qE ma ·············································································(1分)
1 1R vt ················································································(1分)
11
1
R  at2 ·············································································(1分)
2 11
1
v  BkR,t 
1 1 Bk
E  E 2kRB2 ·····································································(1分)
1
(3)离子从距M点0.8R处射入磁场,做匀速圆周运动,半径为r ,
2
Rsinr 0.8R·································································(1分)
2
Rcosr  R·····································································(1分)
2
sin2cos21
r 0.2R
2
公众号:高考试卷君
离子做匀速圆周运动周期为T ,运动时间为t
磁1 磁1
v2
qv B m 2
2 r
2
v 0.2BkR
2
2πr
T  2
磁1
v
2π
t  ············································································(1分)
磁1
2Bk
离子从C点进入电场到Q点做类平抛运动,运动时间为t ,OC与OM的
2
夹角θ=37°
RsinR v t ···································································(1分)
2 2
1
0.8R  kE t2 ······································································(1分)
2 2 2
从Q点离开电场的速度大小为v ,方向与OQ夹角为α
3
v
v  2
3 cos
8
t  ,45,v 0.2 2BkR
2 Bk 3
离子从Q点离开电场进入QON直线上方磁场做匀速圆周运动,运动半径为r ,
3
周期为T ,运动时间为t ,根据几何条件
磁2 磁2
r  2R··············································································(1分)
3
2πr
T  3
磁2
v
3 公众号:高考试卷君
3
t  T
磁2
4
磁2
15π
t  ············································································(1分)
磁2
2Bk
离子从N点返回电场到D点做类斜抛运动,运动时间为t ,从D点垂直电场
3
离开,据运动的对称性
8
t t 
3 2 Bk
离子第二次离开电场,从D点到ab边界做匀速圆周运动,用时t ,据运动
磁3
的对称性
π
t t 
磁3 磁1
2Bk
t t t t t t
总 磁1 2 磁2 3 磁3
17π32
t  ·······································································(1分)
总
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