Reading Comprehension

试题详情

文章:

    Prior to 1965 geologists assumed that the two giant rock plates meeting at the San Andreas Fault generate heat through friction as they grind past each other, but in 1965 Henyey found that temperatures in drill holes near the fault were not as elevated as had been expected. Some geologists wondered whether the absence of friction-generated heat could be explained by the kinds of rock composing the fault. Geologists' pre-1965 assumptions concerning heat generated in the fault were based on calculations about common varieties of rocks, such as limestone and granite; but "weaker" materials, such as clays,had already been identified in samples retrieved from the fault zone. Under normal conditions, rocks composed of clay produce far less friction than do other rock types.

    In 1992 Byerlee tested whether these materials would produce friction 10 to 15 kilometers below the Earth's surface. Byerlee found that when clay samples were subjected to the thousands of atmospheres of pressure they would encounter deep inside the Earth, they produced as much friction as was produced by other rock types. The harder rocks push against each other, the hotter they become; in other words, pressure itself, not only the rocks' properties, affects frictional heating. Geologists therefore wondered whether the friction between the plates was being reduced by pockets of pressurized water within the fault that push the plates away from each other.

题目:

The passage is primarily concerned with

选项:

A、evaluating a method used to test a particular scientific hypothesis
B、discussing explanations for an unexpected scientific finding
C、examining the assumptions underlying a particular experiment
D、questioning the validity of a scientific finding
E、presenting evidence to support a recent scientific hypothesis

答案:

B

提问:

我是张慧雯老师,本解析专为参与《GMAT阅读能力提升营》的同学录制,点击即可听详细解析喽~
评分: 0
浏览: 1599

提问:

N=a+b+c=XYZ, abc连续,xyz连续,问N除以5的余数 (1)a除以5余1 (2)x除以5余1 思路:根据连续整数的性质, 任何三个连续的整数中一定有一个是3的倍数,且这三个连续整数之积是6的倍数。因此可知a+b+c=3b=XYZ, 则xyz肯定是3的倍数, 条件1, a除以5余1时,a的尾数可能为1或者6,这时b的尾数为2或者7,c的尾数是3或者8,则xyz的尾数为6或者1,则xyz除以5的余数为1,充分 条件2, x除以5余1,则x可能为1或者6,此时可以算出xyz为123或者678,此时xyz除以5的余数为3,充分 ---------------------------------------------------------- 我觉得以上解法没有利用到XYZ连续这个条件。老师是否可以讲解下,谢谢。
评分: 0
浏览: 1470

提问:

老师能不能带着读一遍,回顾一下读的方法。感觉前面有很多观点句,不知道是不是要读的地方。
评分: 1
浏览: 1567
点我领取
免费专项课程
在线咨询