问题 问答题

一人从甲地出发做变速直线运动,以6m/s的速度用10min行至乙地,然后休息5min,又以5m/s的速度用10min到达丙地,此人从甲地到丙地的过程中的平均速度为多少?

答案

前10min内的位移x1=v1t1=6×600m=3600m.

后10min内的位移x2=v2t2=5×600m=3000m.

所以整个过程中的平均速度

.
v
=
x1+x2
t1+t2+t0
=
6600
1500
m/s=4.4m/s.

故人从甲地到丙地的过程中的平均速度为4.4m/s.

综合

读甲、乙两国略图,完成下列要求。

(1)甲图所示国家产量和出口量均居世界首位的农产品是______      _____,乙图所示国家最大的城市是

    ____________,该国分布有世界上最大的____________平原。

(2)甲图中A运河的重要战略意义是:                                                                                                 

                                                                                                                                                           

(3)亚马孙热带雨林为什么被称为“地球之肺”?

                                                                                                                                                           

(4)根据两图显示的信息,试分析人口的分布与自然环境之间的关系?

                                                                                                                                                            

                                                                                                                                                            

                                                                                                                                                           。 

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