2020年宝鸡市高考模拟检测理科综合物理参考答案

222020宝鸡市高考拟检测二）物理参答案第部（择）14A．16C．18B19BD20．．BD第卷非择）22分①BC组成的系统（分）②

(mm)dCt2

2

()gh

（每空2分23分）（1）P（每空分）（2100mA（空）（3如图所示（2分（4（2分（5相同分）24分）解）子射穿木块后子弹速度为v，块速度为v，者地时间为t，平抛运动可12得：

h

12

gt

························································································）2分代入数据可得：

0.5s

···································································（2）1

S1s2tt

m

·············································（）子弹射穿木块过程中由动量守恒定律可得：

0

2

··················（）2分代入数据可得：

mkg

·································································（）2分（2子弹射穿木块的过程中，对木块的冲击力是木块受到的合外力。对于木块，由动量定理可得：

I

F

2

·······················································································（6代入数据可得：

I

F

，方向水平向右···········································（7）25分）解：（1由微粒运动到c点做一次完整的圆周运动可知：1

2222qE2

·····················································································（1）2分解得：

q

2

kg

·······························································（2（2分析可得微粒从bc从cd均做匀速直线运动，设速度为，则由受力分析及平衡条件得：

qvBqE2

····························（3

E1解得：

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

0

合由题意分析可知，微粒从a到做加速直线运动，力一定由指向b受力分析如图甲所示，所以可知：Fmg·································（）2分合1由动能定理得：FS·····································（）2分合ab

mg

图甲解得：

S

AB

m

···························································（7（3微粒在正交场中做匀速圆周运动，可得：

qvB

v2

········）1分解得轨道半径：

qB

·········································（）

图乙微粒做匀速圆周运动的周期：

2v

0.628

································（）由于和

T

均恒定不变正交场的宽度L好等于R时交变电场

E

2

的周期最小图乙所示。································································································（11）1分微粒从图中动到c所用的时间

t

s0.1sv2.0

························（12）1分故电场E变化周期T的最小：

min

s

·························（13）33分）（1分）BCE（210分）解：①由题意可知，集热器内气体在温度升高的过程中，体积不变，所以有：112

·························································································（1其中：

KK

························································（）2分T221.2T11

·················································································（3）2

即：压强升为原来的1.2倍②由等温过程可得：

PVV22

3

·························································（）2分所以有：

52332

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

················································································（）2分即：放出气体的质量与集热器内原有质量的比值为

1634分）（1分ADE（210分）解：若波由AB传，假设波长f，由题意得：k

14

(2,3,4,5)

···························································

（1）又

······················································································

（2）由（1）得：

f(40k10)Hz(k

·························

（3）由（）式可知，当kf最大，且f=210Hz····································（）若波由B向A传，假设波长、率，则由题意得：3k4

(4,5)

·························································

（5）又

f

·····················································································

（6）由（5）得：

f

(