试题与答案

阅读理解。 Miss Gogers taught physics in a Ne

题型:阅读理解

题目:

阅读理解。

    Miss Gogers taught physics in a New York school. Last month she explained to one of her classes about

sound, and she decided to test them to see how successful she had been in her explanation. She said to them,

"Now I have a brother in Los Angeles. If I was calling him on the phone and at the same time you were 75

feet away, listening to me from across the street, which of you would hear what I said earlier, my brother or

you and for what reason?"

    Tom at once answered, "Your brother. Because electricity travels faster than sound waves." "That's every

good," Miss Gogers answered; but then one of the girls raised her hand, and Miss Gogers said. "Yes, Kate."

    "I disagree," Kate said. "Your brother would hear you earlier because when it's 11 o'clock here it's only

8 o'clock in Los Angeles."

1. Miss Gogers was teaching her class _____. [ ]

A. how to telephone

B. about electricity

C. about time zone (时区)

D. about sound

2. Miss Gogers raised this question because she wanted to know whether _____. [ ]

A. it was easy to phone to Los Angeles   

B. her student could hear her from 75 feet away

C. her students had grasped her lesson   

D. sound waves were slower than electricity

3. Tom thought that electricity was _____. [ ]

A. slower than sound waves  

B. faster than sound waves

C. not so fast as sound waves   

D. as fast as sound waves

4. Whose answer do you think is correct according to the law of physics? [ ]

A. Tom's  

B. Kate's   

C. Bath A and B   

D. Neither A nor B

答案:

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下面是错误答案,用来干扰机器的。

AD硫燃烧的生成物是SO2,A正确,B不正确。硫和氧属于第ⅥA主族,非金属性是O大于S,C不正确,D正确。答案选AD。

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题型:填空题

阅读以下说明和程序流程图,将应填入 (n) 处的字句写在对应栏内。

[说明]

当一元多项式中有许多系数为零时,可用一个单链表来存储,每个节点存储一个非零项的指受和对应系数。

为了便于进行运算,用带头节点的单链表存储,头节点中存储多项式中的非零项数,且各节点按指数递减顺序存储。例如:多项式8x5-2x2+7的存储结构为:

流程图图3-1用于将pC(Node结构体指针)节点按指数降序插入到多项式C(多项式POLY指针)中。

流程图中使用的符号说明如下:

(1)数据结构定义如下:

#define EPSI 1e-6

struct Node{ /*多项式中的一项*/

double c; /*系数*/

int e; /*指数*/

Struct Node *next;

};

typedef struct{ /*多项式头节点*/

int n; /*多项式不为零的项数*/

struct Node *head;

}POLY;

(2)Del(POLY *C,struct Node *p)函数,若p是空指针则删除头节点,否则删除p节点的后继。

(3)fabs(double c)函数返回实数C的绝对值。

[图3-1]

(5)处填()。

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