英语周报20182022七年级新目标第答案

22.(1)BD(2分)(2)0.88(1分)(3)—(2分)(4)偏大(1分)(1)滑块A受到的拉力可由力传感器读出,故砂和砂桶的质量可以不用测出;也不需要保证砂和砂桶的质量m远小于小车的质量M,故AC错误;调整滑轮的位置,使绳与桌面平行,从而保证实验过程中拉力恒定,故B正确;滑块靠近打点计时器,先接通电源,再释放小车,故D正确;所以需要进行的操作是BD.(2)依题意,可得相邻计数点时间间隔为T=5×0.02s=0.1s利用逐差法可得:滑块A的加速度大小为a=(8.64+7.75-6.87-6.00×10-24×0.12m/s2=0.88m/s2(3)因为力传感器示数F为横坐标,加速度a为纵坐标,对滑块A,由牛顿第二定律2F-mMg=Ma得a=MF=mg由图丙可得—b=-g,解得Pbg(4)由于实验过程中,由于打点计时器与纸带之间存在摩擦力,故测量值比真实值偏大,

书面表达One possible version:NOTICEA sports meeting will be held in the playground of our school from next Thursday to FridayAs you know, the pressure of study is very heavy now, especially for those senior 3. So the purpose of the sports meeting is to letevery student get relaxed, as a result of which we students can live happily and heal thilyEveryone is welcome to take part in it. Those who perform excellently at the sports meeting will get prizes. But don' t take theresults so serously because taking part is more important than the result. Good luck to everyone!

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