广东省卷C英语周报2021-2022七年级答案

读后续写Possible version lI stared at the paper and then looked around. I was cold. Although I could not see my sistersnor the new house, which was hidden from my sight by a grey wall, I got the feeling that I was notalone. Walking along the shore was a tall familiar figure who looked lost in his thoughts. I lookedgain at the paper in my hand, closed my eyes and wishedAs /opened my eyes, I found my father was standing right beside me. Happy birthday, sonhe said, patting my shoulder gently. I smiled up at him and took his hand. Looking back, life fromthat day on marked the start of a close bond and a strong friendship with my father. What was mybirthday surprise? Wishes do come truePossible version 2I stared at the paper and then looked around. On the backside, there was a map. I noticed thatit was clearly leading me in a direction, but who could have sent this? I stood there in silence andstudied it. I decided to follow the map and it led me down the beach until I reached a staircase. Iooked back and could no longer see my siblings. I continued to follow the point on the map, untilI unexpectedly lost my footing and fell zksq down the stepsAs I opened my eyes, I found my father was standing right beside me. Happy birthday, sonhe said, patting my shoulder gently. Then he asked how I had come upon this place? I explained tohim that I had found the bottle in the ocean and when I opened it had the words on it. Shocked bythe message, my father realized what I had found I threw this bottle into the ocean many years agowhen I was alone. he said. I used to come to this exact place to get away from everyone else. NowI get to share this experience with you. What a birthday surprise

34.(1)ACE解析:由于该简谐横波向右传播,则质点b的振动方向沿y轴的负方向,如果振源位于原点O,则振源的起振方向沿y轴的负方向,选项C正确;由质点的振动方向与波的传播方向的关系可知,t=0时刻质点a的振动方向沿y轴的负方向,质点a经1个周期第二次出现在波峰所以1T=0.7s,解得T=0.4s,则该波的频率为2.5Hz,选项A正确;质点只是在平衡位置附近往复振动,不会随波迁移,选项B错误;产生稳定干涉现象的条件:只有两列光波的频率相同,才能产生稳定的干涉现象,选项D错误;产生明显衍射现象的条件是障碍物或孔的尺寸小于或等于光的波长,选项E正确(2)解:①作出光在透明介质中的光路图,如图所示根据光的折射定律有n=51(1分)由题意可知i=45°解得(1分)假设光在该透明介质中全反射的临界角为C,则有n=sinC(1分)解得C=45°(1分)根据几何关系可知,光线由介质射向空气时入射角为a=15°<45°因此光线在右侧界面上不会发生全反射,光线能从右侧界面射出透明介质(1分)②假设AB=x,则由正弦定理得 in r sin p"又P46(4分)整理得x=5n4d=2d(1分)光在透明介质中的传播速度为v=(1分)则光在透明介质中的传播时间为t=(1分)整理得c=24(1分)

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