2020年宝鸡市高考模拟检测（二）物理参考答案
第Ⅰ部分（选择题）
14．A 15．D 16．C 17．C 18．B 19．BD 20．BC 21．BD
第Ⅱ卷（非选择题）
22．（6分）
① B、C 组成的系统 （2分）
(mB?mC)d2② (mC?mB)gh（每空2分） 22t23．（10分）
（1）P b（每空1分） （2）100mA R1（每空1分） （3）如图所示（2分） （4）28（2分） （5）相同（2分） 24．（13分）
解：（1）设子弹射穿木块后子弹速度为v1，木块速度为v2，二者落地时间为t，由平抛运动可得：h?12························································································ （1）2分 gt 2代入数据可得：t?0.5s ···································································· （2）2分
v1?S1S?80ms v2?2?42ms ············································· （3）2分 tt子弹射穿木块过程中由动量守恒定律可得：mv0?mv1?Mv2 ··················· （4）2分 代入数据可得：m?0.1kg ································································· （5）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：IF?Mv2 ······················································································· （6）2分
代入数据可得：IF?42Ns，方向水平向右 ··········································· （7）1分 25．（18分）
解：（1）由微粒运动到c点后做一次完整的圆周运动可知：
1
qE2?mg ······················································································ （1）2分
解得：q······························································· （2）1分 ?5.0?102Ckg ·mE1q
（2）分析可得微粒从b到c和从c到d均做匀速直线运动，设速度为v，则由受力分析及平衡条件得：qvB?qE2?mg ····························· （3）2分
解得：v?2.0ms ····························································· （4）1分 由题意分析可知，微粒从a到b做匀加速直线运动，合力一定由a指向b，
受力分析如图甲所示，所以可知：F合?mg ································· （5）2分 mg图甲
由动能定理得：F合Sab?450 F合
12····································· （6）2分 mv?0 ·
2解得：SAB?0.2m ···························································· （7）1分
v2（3）微粒在正交场中做匀速圆周运动，可得：qvB?m ········ （8）1分
R解得轨道半径：R?mv?0.2m ·········· ····························· ·· （9）1分 qB图乙
微粒做匀速圆周运动的周期：T??2?R································ （10）1分 ?0.628s ·
v由于R和T?均恒定不变，故正交场的宽度L恰好等于2R时交变电场E2的周期T最小，如图乙所示。 ································································································ （11）1分
微粒从图中b运动到c所用的时间 t?R0.2························ （12）1分 ?s?0.1s ·
v2.0故电场E2变化周期T的最小值：Tmin?t?T??0.728s ························· （13）2分 33．（15分） （1）（5分）BCE （2）（10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：
P1P2? ························································································· （1）2分 T1T2其中：T1?300K T2?360K ························································ （2）2分
P2T2??1.2 ················································································· （3）1分 P1T1
2
即：压强升为原来的1.2倍。
②由等温过程可得：P2V2?P3V3 ························································· （4）2分 所以有：
V2P35?? ····································································· （5）1分 V3P26?m?V1?? ················································································ （6）2分 mV36即：放出气体的质量与集热器内原有质量的比值为34．（15分） （1）（5分）ADE （2）（10分）
解：若波由A向B传播，假设波长为λ、频率f，则由题意得：
1 61k????1.24(k?2,3,4,5)（1）1分
···························································
又 v?f? ······················································································ （2）1分 由（1）（2）解得： f?(40k?10)Hz (k?2,3,4,5) ························· （3）1分 由（3）式可知，当k=5时f最大，且fmax=210Hz ···································· （4）1分 若波由B向A传播，假设波长为λ′、频率f ′，则由题意得：
3k??????1.24(k?2,3,4,5)（5）1分
·························································
又 v?f???····················································································· （6）1分 由（5）（6）解得： f??(40k?30)Hz (n?2,3,4,5) ························ （7）1分 由（7）式可知，当k=5时f ′最大，且 f ′max=230Hz ·································· （8）1分 比较（4）（9）可知，该简谐横波波的最大频率为230Hz ··························· （9）2分
3
2020年宝鸡市高考模拟检测(二)理科综合物理参考答案
