英语周报七年级新目标第37期4开6版2020~2021学年答案

25.解:(1)小滑块a从光滑斜面顶端滑到平台上,根据机械能守恒定律得mgH=m(2分)解得小滑块a刚滑上c时速度的大小v=3m/s(1分)(2)a滑上c后水平方向受向右的滑动摩擦力,由牛顿第二定律得:pmg=ma.(1分)代入数据解得an=1m/s2(1分)若b、c相对静止,则b、c整体受a对它的向左的滑动摩擦力由牛顿第二定律得pmg=2ma(1分)代入数据解得a=0.5m/s2(1分)由题意可知b与c间的动摩擦因数为0.1,故b在c表面的最大加速度为ahm=pg=1m/s2(1分)故b、c可以保持相对静止,即aa=0.5m/s2(1分)则当a、b将要相碰时满足va-2a.2-2a1=L(1分)解得t=1s(另一值舍掉)(1分)此时a的速度v1=v-a=2m/s(1分b和c的速度v2=aat=0.5m/s(1分)a、b碰后粘连在一起运动,设共同速度为v,则由动量守恒mn+mv=2mv(1分)解得小滑块a、b碰后瞬间速度的大小v=1.25m/s(1分)(3)设三者共速时速度为v,则由动量守恒定律m+2mv2=3mv(1分)由能量关系2pmg△x=2×2m2+2m-2×3m2(2分)解得△x=3m(1分)当三者共速时a、b距离c右端的距离d=L+△x=2m≈2.3m(1分)

第二节Dear JackI'm writing to ask for your help by lending me a bookRecently I,ve been busy preparing an essay. One of the books that my history teacherbelieves would be helpful to complete my essay is called The California Gold Rush, whose authoris Mel Friedman. Unfortunately, I have been unable to find it at our school library or in the citybookstores. It was not until yesterday that I heard you happened to have this book. May I borrowit? I promise to take good care of it and will return it to you within a week.Thank you in advance. I'm looking forward to your early reply.YouLi hua

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