高三英语听说训练软件:河南省新县高中2009答案

河南省新县高中2009-2010学年度高中毕业班第一次月考物理试卷（参考答案）命题：河南省信阳市新县高级中学　吴国富#TRS_AUTOADD_1258618543592 {MARGIN-TOP: 0px; MARGIN-BOTTOM: 0px}#TRS_AUTOADD_1258618543592 P {MARGIN-TOP: 0px; MARGIN-BOTTOM: 0px}#TRS_AUTOADD_1258618543592 TD {MARGIN-TOP: 0px; MARGIN-BOTTOM: 0px}#TRS_AUTOADD_1258618543592 DIV {MARGIN-TOP: 0px; MARGIN-BOTTOM: 0px}#TRS_AUTOADD_1258618543592 LI {MARGIN-TOP: 0px; MARGIN-BOTTOM: 0px}/**---JSON--{"":{"margin-top":"0","margin-bottom":"0"},"p":{"margin-top":"0","margin-bottom":"0"},"td":{"margin-top":"0","margin-bottom":"0"},"div":{"margin-top":"0","margin-bottom":"0"},"li":{"margin-top":"0","margin-bottom":"0"}}--**/DIV.MyFav_1258618157614 P.MsoNormal{TEXT-JUSTIFY: inter-ideograph; FONT-SIZE: 10.5pt; MARGIN: 0cm 0cm 0pt; FONT-FAMILY: "Times New Roman"; TEXT-ALIGN: justify}DIV.MyFav_1258618157614 LI.MsoNormal{TEXT-JUSTIFY: inter-ideograph; FONT-SIZE: 10.5pt; MARGIN: 0cm 0cm 0pt; FONT-FAMILY: "Times New Roman"; TEXT-ALIGN: justify}DIV.MyFav_1258618157614 DIV.MsoNormal{TEXT-JUSTIFY: inter-ideograph; 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　　1．C　2．ACD　3．AB　4．B　5．CD　6．D　7．A　8．D　9．BD　10．B　11．B　12．B　13．ABD　14．AC 

 

　　15．答案：s₁应为  4  cm，a = 1  m/s²。

 

　　16．答案：（1）①bcd；②更换不同的小重物；

 

　　17．(1)A  (2)减小误差

 

　　18．解析:对B球，受力分析如图所示。

 

　　Tcos30°N_Asin30°

 

　　∴T=2mg

 

　　对A球，受力分析如图所示。在水平方向

 

　　Tcos30⁰=N_Asin30⁰

 

　　在竖直方向

 

　　N_Acos30⁰=m_Ag+Tsin30⁰

 

　　由以上方程解得：m_A=2m

 

　　19．(1)设展伞高度为h，速度为v₀，落地速度v_t=5m/s，h₀=224m，a=－12．5m/s²有

 

　　又   解得  h = 99m

 

　　(2)上述运动方式在空中时间最短

 

　　由     得自由落体时间   t₁ = 5s

　　展伞后匀减速运动  由  得展伞后运动的时间t₂=3．6s

 

　　因此运动员在空中的最短时间为t_min=t₁+t₂= 8．6s

 

　　20．解：由题图中左右对称。如图1考虑右半侧有：

 

　　    △与△FDE全等，故虚线FD两侧的图形对称，

 

　　，

 

　　由，

 

　　解得：3／5

 

　　如图2线段CB和CD的拉力等大，设为T₁，线段DE的拉力设为T₂．由C点受力平衡有

 

　　

 

　　代入②得：

 

　　由D点受力平衡有：

 

　　

 

　　把3／5，4／5代入得： 

 

　　解出

 

　　故

 

　　即

 

　　21．解析：(1)儿童下落时间为t  H=gt²

 

　　要使他能接住儿童，他奔跑的时间要小于3s

 

　　X=Vt  得他的平均速度至少为6m/s

 

　　(2)设加速度为由于要求没有水平方向的冲击则V_t=0

 

　　时间上t₁+t₂+t₃=3s位移上   s₁+s₂+s₃=18m

 

　　由上可得

 

　　则加速度应满足