4. What did the explorers find?
A. The coastlines of South America and Africa could fit together.
B. The coastlines of North America and Africa could fit together.
C. The east coastlines of North America and the west coast of Europe could fit together.
D.The coastlines of North America and India could fit together.
(二)
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3. The last word of the third paragraph “tectonics” mean “________”.
A. study of construction
B. study of architecture
C. earth surface
D. structural geology
2. Which of the following is true according to the passage?
A. We didn’t see the Earth from far away until we saw the picture taken in the space.
B. Our ancient thought that the surface of the earth is still.
C. Alfred Wegener was not the first person to investigate the idea that the continents move.
D. The coastline of India and Africa fit together.
1. What does the writer mainly tell us in the passage?
A. The first breath-taking pictures of the Earth taken from space.
B. Human’s recognition of the earth’s surface.
C. The German scientist Alfred Wegener.
D. The early explorers’ discovery.
如圖是一種建立在空間的太陽能電站向地球上固定區(qū)域供電的示意圖。在太陽能收集板上鋪設(shè)太陽能電池,通過光電轉(zhuǎn)換把太陽能轉(zhuǎn)換成電能,再經(jīng)微波轉(zhuǎn)換器將電流轉(zhuǎn)換成微波,并通過天線將電能以微波的形式向地面發(fā)送。地面接收站通過天線把微波能還原成電能。
![]()
(1)如果太陽能電池的硅片能提供的電能為12kW/m2,巨大的太陽能收集板的總面積為5×106m2,則太陽能電站的發(fā)電能力是多少kW?
(2)利用微波傳輸電能,實現(xiàn)了“無纜輸電”。輸電效率可達80%,若微波輸電采用的頻率是2450MHz。求微波的波長及地面接收站接收到的電能是多少kW?
解析:從題目和問題中可提取有用的信息并推導出公式,然后利用公式進行計算。
如圖(甲)所示,光I從空氣斜射如玻璃中,發(fā)生了折射現(xiàn)象,從圖中可以看出折射角r小于入射角i。那么,一般情況下,i與r有什么定量關(guān)系呢?1621年,荷蘭學者斯涅爾通過實驗終于找到了i與r之間的規(guī)律,如圖是光從空氣進入玻璃中時,利用實驗數(shù)據(jù)得到的sinr與sini的關(guān)系圖象。
![]()
(1)在圖(甲)中,畫出光Ⅱ進入玻璃中傳播的大致方向(如圖)
(2)分析圖象可得到的結(jié)論: 。
解析:從圖中可以看出,橫軸為sini,縱軸為sinr;通過圖像可以看出二者成正比。
答案:(2)入射角的正弦和折射角的正弦成正比。
方法歸納:對于圖像類題目,首先要看清圖像的橫軸和縱軸表示的是哪個物理量,然后再根據(jù)圖像的形狀來判斷物理量之間的關(guān)系,從而確定問題的答案。
2.下表是某同學研究“電阻消耗的電功率與該電阻阻值之間的關(guān)系”時記錄的實驗數(shù)據(jù),請你對表格中的數(shù)據(jù)進行分析,歸納出電功率與電阻的關(guān)系:_____________________。
|
R/Ω |
10 |
15 |
20 |
30 |
50 |
60 |
|
U/V |
6 |
6 |
6 |
6 |
6 |
6 |
|
I/A |
0.6 |
0.4 |
0.3 |
0.2 |
0.12 |
0.1 |
|
P/W |
3.6 |
2.4 |
1.8 |
1.2 |
0.72 |
0.6 |
解析:解答此題要找準表中各量的變化規(guī)律:電阻──逐漸變大,電壓──不變,電流──逐漸變小,電功率──逐漸變;且電功率和電阻成反比。
答案:在電壓一定時,電阻消耗的電功率與該電阻阻值成反比。
方法歸納:認真分析表中各物理量的變化情況,特別是要找準變化量和不變量,并從中分析出各物理量之間的關(guān)系。
1.某實驗小組在探究光的折射規(guī)律時,將光從空氣分別射入水和玻璃,測得數(shù)據(jù)如下表:
|
空氣中的入射角i |
0° |
30° |
45° |
60° |
|
水中的折射角r |
0° |
22° |
32° |
40° |
|
玻璃中的折射角 |
0° |
17° |
24° |
30° |
分析表格中的數(shù)據(jù),你肯定能得出一些規(guī)律。請寫出一條:_____________________。
解析:本題考查學生對數(shù)據(jù)的綜合分析能力。解題時,可橫向分析,也可縱向分析。
答案:(1)入射角為0°,折射角也為0°。(2)入射角增大,折射角也隨著增大。(3)光從空氣斜射入不同介質(zhì)中的折射角不同。(4)光從空氣斜射入其他介質(zhì)時,折射角小于入射角。(5)光以相同入射角斜射入水和玻璃時,水中的折射角大于在玻璃中的折射角。
17.在
中,
,
.
(Ⅰ)求
的值;
(Ⅱ)設(shè)
的面積
,求
的長.
(Ⅰ)由
,得
,
由
,得
.
所以
.
(Ⅱ)由
得
,
由(Ⅰ)知
,
故
,又
,
故
,
.所以
.
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