Thinking of thal, my face felt hot. To our amazement, Mrs. Ferguson replied, " You are welcometo plug it in here. When saying that, she pointed to the mirror in her room.That' s really nice ofyou."Alex and I said and walked into her room. When we were fixing our hair Mrs. Ferguson chattedwith us about our tryout for cheerleaders. We had a pleasant chat, during which I found MrsFerguson genuinely cared about her studentsParagraph 2:When we were to leave, Mrs. Ferguson said,"Can I watch your tryout? " Of course, MrsFerguson, we'd be honored, I replied, feeling amazed again. During the tryout, both Alex and Iperformed well and passed the test. Mrs. Ferguson sincerely congratulated us. Meanwhile, sheinded us to study hard when we tried to develop our personality by being cheerleaders. ThrouglFerguson, we'd be honored, " I replied, feeling amazed again. Duringperformed well and passed the test. Mrs. Ferguson sincerely congratulated us. Meanwhile, shereminded us to study hard when we tried to develop our personality by being cheerleaders. Througlthese and many other nice gestures, Mrs. Ferguson gradually gained love and support from all of usHow blessed we were to have another great teacher

14.解:(1)A位于a处时,细绳无张力且物块A静止,故弹簧处于压缩状态对物块B由平衡条件可得kx1= nisin0(1分)当C恰好离开挡板P时,C的加速度为0,故弹簧处于拉伸状态对物块C由平衡条件可得kx2= mosin0(1分)由几何关系可知R=x+x(1分)解得=30°(1分)(2)物块A在a处与在b处时,弹簧的形变量相同,弹性势能相同,故A在a处与b在处时,A、B组成的系统的机械能相等,有MgR(1-cos60°)= meRsin30°+Ma2+mx2(1分)根据A、B的关联速度有v=wcos30(1分)代入数据解得=4m/s(1分)在b处,对A由牛顿第二定律有FN-Mg=M爷(1分)代入数据解得FN=144N由牛顿第三定律,A对圆轨道的压力大小为144N。(1分)(3)物块不脱离圆形轨道有两种情况:①第一种情况,A不超过圆轨道上与圆心的等高点由动能定理A恰能进入圆轨道时需要满足-pMgn=0-yMon2(1分)A恰能到國心等高处时需满足条件-Mgr-pMgx2=0-Mo2(1分)代入数据解得3m≤r≤4m(1分)②第二种情况,A过圆轨道最高点A恰能过最高点时Mg=M(1分)由动能定理有-M12-Mgn=M2-M2(分)代入数据解得0≤n≤1.5m(1分)综上所述3m≤r≤4m或0≤n≤1.5m,(1分)

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