大小球判断 球判断提款快吗?

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友情链接:解析 (1)悬线拉力在经过最低点时最大,t=0.2 s时,F有正向最大值,故A选项正确,t=1.1 s时,F有最小值,不在最低点,周期应为T=1.0
s,因振幅减小,故机械能减小,C选项正确.
(2)振幅可从题图上看出甲摆振幅大,故B对.且两摆周期相等,则摆长相等,因质量关系不明确,无法比较机械能.t=0.5
s时乙摆球在负的最大位移处,故有正向最大加速度,所以正确答案为A、B、D.
答案 (1)AC (2)ABD
请选择年级高一高二高三请输入相应的习题集名称(选填):
科目:高中物理
(2006?青浦区模拟)将一个电动传感器接到计算机上,就可以测量快速变化的力,用这种方法测得某单摆摆动时悬线上拉力的大小随时间变化的曲线如图所示,根据此图提供的信息作出以下判断(  )A.摆球的摆动周期T=0.6&sB.t=0.2&s摆球正好经过最低点C.t=1.1&s摆球正好经过最低点D.摆球摆动过程中机械能守恒
科目:高中物理
来源:学年浙江省台州中学高二下学期期中考试物理试卷(带解析)
题型:单选题
将一个电动传感器接在计算机上,就可以测量快速变化的力,用这种方法测得的某单摆摆动时悬线上拉力的大小随时间变化的曲线如图所示。某同学由此图线提供的信息做出了下列判断(&&&&&)A.t=0.2s时摆球正经过最低点B.t=1.1s时摆球正经过最低点C.摆球摆动过程中机械能守恒D.摆球摆动的周期是T=1.2s
科目:高中物理
来源:2014届浙江省高二下期中考试物理试卷(解析版)
题型:选择题
将一个电动传感器接在计算机上,就可以测量快速变化的力,用这种方法测得的某单摆摆动时悬线上拉力的大小随时间变化的曲线如图所示。某同学由此图线提供的信息做出了下列判断(&&&&& )
A.t=0.2s 时摆球正经过最低点
B.t=1.1s 时摆球正经过最低点
C.摆球摆动过程中机械能守恒
D.摆球摆动的周期是T=1.2s。
科目:高中物理
来源:2014届浙江省高二下学期期中考试物理试卷(解析版)
题型:选择题
将一个电动传感器接在计算机上,就可以测量快速变化的力,用这种方法测得的某单摆摆动时悬线上拉力的大小随时间变化的曲线如图所示。某同学由此图线提供的信息做出了下列判断(&&&&&)
A.t=0.2s时摆球正经过最低点
B.t=1.1s时摆球正经过最低点
C.摆球摆动过程中机械能守恒
D.摆球摆动的周期是T=1.2s
吴老师30日19点直播线段的垂直平分线的性质
余老师30日20点直播unit5第二课时 Section A

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