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课时跟踪检测(十四) 牛顿第一定律

课时跟踪检测(十四)  牛顿第一定律
课时跟踪检测(十四)  牛顿第一定律

课时跟踪检测(十四)牛顿第一定律

1.在物理学史上,正确认识运动与力的关系并且推翻“力是维持运动的原因”的物理学家和建立惯性定律的物理学家分别是()

A.亚里士多德、伽利略B.亚里士多德、牛顿

C.伽利略、牛顿D.伽利略、爱因斯坦

解析:选C在物理学史上,正确认识运动与力的关系并且推翻“力是维持运动的原因”的物理学家和建立惯性定律的物理学家分别是伽利略和牛顿。故选C。

2.下列现象中,不属于利用惯性现象的是()

A.用手拍打衣服上的灰尘

B.锤头松了,将锤柄在地上撞击几下

C.运动员采用助跑跳远

D.骑自行车时为了减速捏刹车闸

解析:选D用手拍打衣服上的灰尘,衣服运动时,由于惯性,灰尘要保持静止状态,利用了惯性现象,所以A错误;锤头松了,将锤柄在地上撞击几下,锤柄撞击地面马上静止,锤头由于惯性要保持运动状态,利用了惯性,B错误;利用助跑跳远,由于惯性,人要保持运动的状态,所以跳得更远,故C错误;捏刹车闸,是通过力改变自行车的运动状态,不是利用了惯性,D正确。

3.关于物体的惯性,下列说法中正确的是()

A.射出枪膛的子弹在运动一段距离后连一件棉衣也穿不透,这表明它的惯性小了

B.战斗机在空战时,甩掉副油箱是为了减小惯性,提高飞行的灵活性

C.用力推一个物体,没有推动,原因是物体的惯性太大了

D.高速公路上要限速,是因为车速越大惯性越大

解析:选B惯性是物体本身的一种属性,它不随物体速度的大小变化而变化,所以子弹运动一段距离后,速度变小了,高速公路上的限速等都不是因为物体的惯性发生了变化,所以A、D错误;而用力推一个物体时,没有推动,说明地面对物体有摩擦力的作用,而不是因为惯性太大了,所以C错误;战斗机甩掉副油箱减小了质量,而惯性与质量有关,质量越小惯性越小,改变它的状态就越容易,所以可以提高战斗机的灵活性,B正确。

4.公共汽车进入转弯路口向右转时,车内乘客会()

A.向前倾斜B.向后倾斜

C.向左侧倾斜D.向右侧倾斜

解析:选C根据牛顿第一定律可知,公共汽车进入转弯路口向右转时,车内乘客会保持原来的运动状态,故向左侧倾斜,选项C正确。

5.下列关于惯性的说法正确的是()

A.战斗机战斗前抛弃副油箱,是为了增大战斗机的惯性

B.物体的质量越大,其惯性就越大

C.火箭升空时,火箭的惯性随其速度的增大而增大

D.做自由落体运动的物体没有惯性

解析:选B质量是物体惯性大小的唯一量度,物体的质量越大则惯性越大,故战斗机抛弃副油箱,减小了惯性,增大了战斗机的灵活性。故A错误,B正确。火箭的速度大小与火箭的惯性大小无关,故C错误。在物体自由下落的过程中质量不变,惯性大小不变,故D错误。

6.下列关于惯性的说法,正确的是()

A.高速行驶的公共汽车紧急刹车时,乘客都要向前倾倒,说明乘客都具有惯性

B.短跑运动员最后冲刺时,速度很大,很难停下来,说明速度越大,惯性越大

C.《道路交通安全法》中规定:各种小型车辆前排乘坐的人(包括司机)必须系好安全带,这是因为系好安全带可以防止因车的惯性而造成的伤害

D.把手中的球由静止释放后,球能竖直下落,是由于球具有惯性的缘故

解析:选A高速行驶的公共汽车紧急刹车时,乘客都要向前倾倒,说明乘客都具有保持原来运动状态的性质,即惯性,选项A正确;物体的惯性只与质量有关,与速度大小无关,选项B错误;系安全带的目的是防止由于人本身的惯性造成的伤害,选项C错误;把手中的球由静止释放后,球能竖直下落,是由于球受到竖直向下的重力的缘故,选项D 错误。

7.以下说法不符合物理学史实的是()

A.亚里士多德认为,必须有力作用在物体上,物体才能运动;没有力的作用,物体就要静止在某一个地方

B.伽利略通过理想实验得出结论:力是维持物体运动的原因

C.笛卡尔指出:如果运动中的物体没有受到力的作用,它将以同一速度沿同一直线运动,既不停下来也不偏离原来的方向

D.牛顿第一定律是逻辑思维对事实进行分析的产物,不可能直接用实验验证

解析:选B亚里士多德认为力的作用是使物体运动的原因,A的说法符合物理学史;伽利略通过理想实验得出运动不需要力来维持,故B项说法不符合物理学史;笛卡尔指出如果运动中的物体没有受到力的作用,它将以同一速度沿同一直线运动,既不停下来也不偏离原来的方向,C的说法符合物理学史;牛顿第一定律是逻辑思维对事实进行分析的产物,不可能直接用实验验证,符合物理学史,故选B。

8.火车在长直水平轨道上匀速行驶,门窗紧闭的车厢内有一人向上跳起,发现仍落回

到火车上原处。这是因为()

A.人跳起后,车厢内空气给他一向前的力,带着他随同火车一起向前运动

B.人跳起的瞬间,车厢的地板给他一个向前的力,推动他随同火车一起向前运动

C.人跳起后,车在继续向前运动,所以人落下后必定偏后一些,只是由于时间很短,偏后距离太小,不明显而已

D.人跳起后直到落地,在水平方向上和车始终具有相同的速度

解析:选D人跳起后,由于在水平方向不受外力的作用,根据惯性和惯性定律可知,人跳起后直到落地,在水平方向上和车始终具有相同的速度。

9.如图1所示,在一辆表面光滑足够长的小车上,有质量为m1和m2的两个小球(m1>m2),两小球原来随车一起运动。当车突然停止时,如不考虑其他阻力,则两个小球()

图1

A.一定相碰B.一定不相碰

C.不一定相碰D.无法确定

解析:选B因小车表面光滑,因此球在水平方向上没有受到外力作用。原来两球与小车有相同速度,当车突然停止时,由于惯性,两小球的速度不变,所以不会相碰。

10. (多选)如图2所示,重球系于细绳DC下端,重球下再系一根同样的绳BA,下列说法正确的是()

图2

A.在绳的A端缓慢增加拉力,结果CD绳先断

B.在绳的A端缓慢增加拉力,结果AB绳先断

C.在绳的A端突然猛一拉,结果AB绳先断

D.在绳的A端突然猛一拉,结果CD绳先断

解析:选AC受力如图所示,在绳的A端缓慢增加拉力,使得重球在足够

的时间发生了微小的位移,这个过程进行缓慢,可以认为重球始终处于平衡状态,

即F T2=F T1+mg,随着F T1增大,F T2也增大且F T2总是大于F T1,所以CD绳先

被拉断,A项对,B项错。若在A端猛拉,由于重球质量很大,力的作用时间极

短,由于惯性,故重球向下的位移极小(可以看成运动状态未来得及改变)以致上段绳的拉力

几乎未增加,F T1已达到极限程度,故AB绳先断,C项正确,D项错误。

11.如图3所示,一木块和小车一起做匀速直线运动,当小车遇到一障碍物的瞬间,则:(设小车不反弹)

图3

(1)如果小车上表面粗糙,木块将如何运动?

(2)如果小车上表面光滑,木块将如何运动?

解析:当小车遇到障碍物时,小车将停下。

(1)如果小车上表面粗糙,则木块上部由于惯性将继续向右运动;木块下部受到一个向左的摩擦力,运动状态发生改变,很快停止,故此时木块将向右倾倒。

(2)如果小车上表面光滑,则木块下部不受摩擦力,此时整个木块都将由于惯性而保持向右的匀速直线运动状态。

答案:(1)向右倾倒(2)向右匀速直线运动

12.如图4所示,在瓶内装满水,将乒乓球用细线拴住并按入水中,线的另一端固定在瓶盖上。盖上瓶盖并将瓶子翻转,乒乓球将浮在水中。用手托着瓶子水平向右做加速直线运动,乒乓球在瓶中的位置会如何变化?解释你所观察到的现象。

图4

解析:因为水的质量大于乒乓球的质量,所以水的惯性大于乒乓球的惯性。当用手托着瓶子水平向右做加速直线运动时,水相对瓶子向左移动时,将挤压乒乓球,使乒乓球相对瓶子向右移动。

答案:见解析

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根据上表可得回归方程y ^=b ^x +a ^中的b ^ 为9.4,据此模型预报广告费用为6万元时销售额为( ) A .63.6万元 B .65.5万元 C .67.7万元 D .72.0万元 解析:选B 样本点的中心是(3.5,42), 则a ^=y -b ^ x =42-9.4×3.5=9.1, 所以回归直线方程是y ^ =9.4x +9.1, 把x =6代入得y ^ =65.5. 5.(福建高考)已知x 与y 之间的几组数据如下表: 假设根据上表数据所得线性回归直线方程为y =b x +a ,若某同学根据上表中的前两组数据(1,0)和(2,2)求得的直线方程为y =b ′x +a ′,则以下结论正确的是( ) A.b ^>b ′,a ^>a ′ B.b ^>b ′,a ^ a ′ D.b ^a ′. 二、填空题 6.在一组样本数据(x 1,y 1),(x 2,y 2),…,(x n ,y n )(n ≥2,x 1,x 2,…,x n 不全相等)的散点图中,若所有样本点(x i ,y i )(i =1,2,…,n )都在直线y =1 2x +1上,则这组样本数据 的样本相关系数为________.

2020版高考地理课时跟踪检测二十八交通运输方式和布局含解析

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