题目

如图甲所示，A、B两个滑块靠在一起放在光滑的水平面上，其质量分别为2m和m，从t=0时刻起，水平力F1、F2同时分别作用在滑块A和B上，已知F1、F2随时间变化的关系如图乙所示，两力作用在同一直线上．则滑块从静止开始滑动到A、B两个滑块发生分离的时间是（ ）A. 2s B. 3s C. 4s D. 6s 答案：【答案】A【解析】先根据数学知识写出水平力F1、F2与时间的关系式．再根据牛顿第二定律对整体和其中一个物体列式，求出A、B两个滑块发生分离的时间．由乙图可得：F1=(60-10t)N，F2=10t N,当A、B两个滑块刚要发生分离时相互间的作用力为零，对整体，由牛顿第二定律得：F1+F2=(2m+m)a对B有：F2=ma联立得：t=2 阅读短文，完成问题。       Hello! My name is Bai Ling. I'm seven years old. I go to school at eight in the morning (上午8点上学). I learn English, too. I can speak (说) English and sing English songs. I can act like a monkey or a cat, jump like a squirrel, fly like a bird, walk like an elephant.       I have crayons. I can draw a teddy-bear: colour the face yellow, the nose green, the eyes blue, the mouth red, the ears purple, the hands and the feet white, the arms and the legs black.       I have some milk and some bread for breakfast (早饭); have some chicken, French fries and some juice forlunch (午饭); have some hamburgers, hot dogs and some Coke for supper (晚饭). I am happy. Let's make friends (做朋友).1. 下面是Bai Ling画的teddy bear。请你按照文中描述给它涂色。2. 填表，Bai Ling三餐的食物名称。 breakfast (早饭)                                                               lunch (午饭) supper (晚饭)