020-2022英语周报高三课标HZ第14期答案

书面表达One possible versionDear PeterI'm writing to invite you to see the Chinese Painting Exhibition to be held in our city.The exhibition will start at 8: 30 am on Saturday and last till 5: 00 pm on Sunday in the city museum It'ssaid that a large number of Chinese paintings, some of which are original works by famous painters like ZhangDaqian and Qi Baishi, will be on display. Besides, there will be various souvenirs to be sold. I know you re veryinterested in traditional Chinese culture, so I can t wait to tell you the news and hope we can go together.Looking forward to your reply.You

25解(1)由匀变速直线运动公式x=0+1a得一+2m(1分)结合图线可知图线与纵轴的交点表示滑块A的初速度、斜率k=2a则v=12滑块A的加速度大小为a1=4m/s2(1分)0~2s的时间内,对滑块A由牛顿第二定律得Fcos a-H(mgcos a+ Fsin a)-mgsin a mat (155)整理得F= mgcoS o+mgg+m(1分)代入数据解得F=56N(1分)(2)撒走外力的瞬间,滑块A的速度为v=t+a1t1=(12+4×2)m/s=20m/s0-2s内滑块A的位移大小为x1=0+2m42=(2×2+×4x×2)m=32m(1分)撒走外力后滑块A沿斜面体向上做匀减速直线运动,则其加速度大小为ay-Agcos at gsin a=10 m/s2 (1 4+)撒走外力后滑块A沿斜面向上运动的位移为x2=2=20m滑块A向上运动的时间为t2=卫=2s(1分)滑块A沿斜面体向上滑动的距离为x=x1+x2=52m(1分)物体由最高点返回的过程中,做匀加速直线运动,其加速度大小为a3= gsin a- ucos a=2m/s2(1分)滑块A由最高点返回到斜面体底端所需的时间为=√a-√。=72所以滑块A从施加外力到滑块A再次回到斜面体的底端所用的总时间为t=+t2+t=1.2s(1分)(3)由(2)可知滑块A返回到斜面底端的速度为=a2b3=2×7.2m/s=14.4m/s(1分)滑块A滑上长木板的上表面后,对滑块A由牛顿第二定律得pmg=ma解得a4=gk=3.6m/s2(1分)对滑块A与长木板B组成的系统由动量守恒定律得mUA=2mA解得vA=7.2m/s(1分)则滑块A在长木板上滑动的时间为t4=二vA=1.4=7.2s=2s则长木板的长度为sUA+UAtat4=14.4m(1分)对滑块A与滑块C组成的系统,由动量守恒定律得mvA=(m+Mw(1分)碰后滑块由最低点到最高点的过程,由机械能守恒定律得(m+Mw2=(m+Mg·2R+y(m+Mv2(1分)滑块在最高点时,由牛顿第二定律得(m+MDg=R整理解得M=4kg(1分)

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