英语周报2018到2022初三第九期答案

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

13.(1)3(3)【解析】(1)由导体棒与导轨间的动摩擦因数=tan0可知,对导体棒1和2分别有mgsin 0=umgcos e2mgsin 8= 2umgcos 8(1分)根据右手定则和左手定则可判定,两根导体棒受到的安培力始终大小相等、方向相反,经过足够长的时间,二者以相同的加速度做匀加速直线运动,结合以上分析对导体棒1、2整体有F=3ma(2分解得a=F(1分)(2)经过足够长的时间,两导体棒以共同的加速度a做匀加速直线运动,设安培力大小为FA,对导体棒2根据牛顿第二定律有FA=2ma(1分)设最终回路中产生的感应电动势为E,电流为I,则有FA=BIL(1分)E(1分)联立解得I3BL,E=2FRB(1分)(3)设经过足够长的时间后,导体棒1和2速度的差值为△v则电动势E=BL△v(1分)经足够长的时间后,两根导体棒加速度相等,速度的差值恒定,回路中电流恒定,导体棒1和2之间的距离每增加s所用的时间t=(1分△v回路中产生的焦耳热为Q=12×3Rt(1分联立解得Q=2F(1分

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