高二物理期末考试答案

参考答案

二．非选择题 14．（1） 3 （数对但没单位的得1分） 15．（1）D C F （2）5.0 （3）15

16．解：由闭合电路的欧姆定律可知，电动机两端的电压为U=E-Ir=120V， ········· （2分） 故电动机的输入功率为P=UI=600W，·········································································· （2分） 由题意知，电动机的输出功率为P1=mgv=500W， ······················································· （2分） 故电动机的热功率为P2=I2R=P-P1=100W， ·································································· （2分） 解得R=4Ω。 ··················································································································· （2分） 17．解：（1）输电电流I PU 5006A 2503A， ························································ （2分）

输电线损耗功率PE4800

损 损t 24kW 200kW， ·

····················································· （2分） 故输电线电阻R

P损 105 28.8 ； ·

···························································· （3分） I

2

2 250

2

3

（2）当采用U′=2.5×104 V的电压输电时，输电电流I

P

U

20A， 输电线损耗功率P

2损 I R 11.52kW， ····································································· （2分）

占总功率百分比

P损

P 11.52

500

100% 2.304%，超过了2%，没有达到节能要求。

（1分）18．解：ab杆在位移L到3L的过程中，由动能定理

F(3L－L)＝12

m(v22－v2

1)， ································································································· （3分） ab杆在磁场中移动L过程中，由功能关系

FL＝12

2

1＋Q1， ············································································································· （3分）

解得电阻R上产生的电热Qmv2－3v2

1＝4

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

ab杆在离开磁场前瞬间， 感应电动势为

E BLv1，·

······················································································································ （1分） 感应电流为I

EBLv1

R

R

， ·························································································· （1分） 水平方向上受安培力FB2L2v1

安 ILB R

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

由牛顿第二定律得ab杆加速度a＝F F22

B2安v2－vL2vm

＝1

4LmR。 ······························ （2分）

19．（1）解：以a表示粒子在电场作用下的加速度，有qE=ma， ····························· ①（1分） 加速度沿y轴负方向。

设粒子从A点进入电场时的初速度为v0，由A点运动到C点经历的时间为t，则有

h

12

2

at， ·

···················································································································· ② 2h=v0t， ···································································③

y

由②③式得v0 2ah， ········································④（1分） A 粒子经过C点时，速度沿y轴的分量

D O αv··············································x y 2ah， ·············⑤（1分） β R

由①④⑤得v 2

qEh

m

， ·······································⑥（1分） P

设粒子经过C点时，速度方向与x轴的夹角为α，则有

tan

vyv 1， ·

·····················································⑦ 0

故 45 ； ····························································（1分）

（2）粒子经过C点进入磁场后，在磁场中做速率为v的圆周运动。若圆周运动的半径为R，则有

qvB mv2

R

， ················································································································· ⑧（1分） 设圆心为P，则PC必与v垂直，且有PC=PA=R。用β表示PA与y轴的夹角，由几何关系得

Rsin 2h Rcos45 ， ····························································································· ⑨（2分） Rcos h Rsin45 ， ·

······························································································ ⑩（2分） 由⑨⑩解得R

5

22h， ·

······························································································ （1分） 由⑧ 解得B 2qEmh5qh

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