2018年IAO理论低年组第1题-水银镜面 - 版本历史

2019年9月20日 (五) 05:38 Heavens above

2019-09-20T05:38:23Z

2019年9月13日 (五) 07:45 Dg1200

2019-09-13T07:45:15Z

2019年9月13日 (五) 07:41 Dg1200

2019-09-13T07:41:56Z

2019年9月13日 (五) 04:08 Dg1200

2019-09-13T04:08:06Z

Dg1200：/* 中文翻译 */

2019-09-13T04:06:54Z

‎中文翻译

2019年9月13日 (五) 04:06 Dg1200

2019-09-13T04:06:12Z

2019年9月13日 (五) 03:39 Dg1200

2019-09-13T03:39:06Z

2019年9月13日 (五) 03:38 Dg1200

2019-09-13T03:38:53Z

2019年9月13日 (五) 00:30 Dg1200

2019-09-13T00:30:36Z

Dg1200：/* 英文解答 */

2019-09-12T15:12:48Z

‎英文解答

@@ 第34行： / 第34行： @@
 Thus, the correct answer is "'''NO'''", the idea of PolarBear is impracticable.
-'''中文翻译'''
+==中文翻译==
 '''水银镜面''' 想象一个水银镜面。在每一个点上，垂直于镜面的直线与竖直方向都有一个夹角γ=ω<sup>2</sup>r/g，其中ω是水银镜面的角速度，r是水银镜面的半径，g是重力加速度，如果竖直方向（矢量g）一定，那么夹角γ就与镜面的曲率半径R有关，γ=r/R，通过这两个方程可以得到：

@@ 第1行： / 第1行： @@
-==原文题目==
+==英文题目==
 '''Mercury mirror.''' As you know, if one rotates a vessel with mercury, its surface will take a parabolic form. A kiquid mirror obtained in this wan can be used as the main mirror of a telescope. The Polar Bear-astronomer living in the Arctic decided to use this idea and build such a telescope exactly at the Norht Pole, and use the daile revolution of the Earth as the rotation mechanism so that the main mirror would have a long focal length.
@@ 第14行： / 第14行： @@
 '''1.2.''' 请画一个正在进行这项工作的北极熊。
-==英文解答==
+==解答==
 '''Mercury mirror.''' Imagine rotating mercury. At each point, the normal to its surface will be directed at an angle γ =ω<sup>2</sup>r/g to the vertical, where ω is the angular velocity of rotation of mercury, r is the distance from the axis of rotation, g is the gravity acceleraton, If the direction of the vertical(vector g) is fixed, then this angle γ is related to the radius of curvature R of the liquid mirror as γ = r/R. From these two equations we get:
@@ 第31行： / 第34行： @@
 Thus, the correct answer is "'''NO'''", the idea of PolarBear is impracticable.
-==中文翻译==
+'''中文翻译'''
 '''水银镜面''' 想象一个水银镜面。在每一个点上，垂直于镜面的直线与竖直方向都有一个夹角γ=ω<sup>2</sup>r/g，其中ω是水银镜面的角速度，r是水银镜面的半径，g是重力加速度，如果竖直方向（矢量g）一定，那么夹角γ就与镜面的曲率半径R有关，γ=r/R，通过这两个方程可以得到：

←上一版本		2019年9月13日 (五) 07:41的版本
第29行：		第29行：
	But ... The result would be true for a flat Earth. The condition "if the direction of the vertical is the Pole, the direction of the vertical changes 300 times faster (the Earth only reduces the curvature of the liquid surface at the pole by 1/300. The mirror designed by the Bear will be convex, its curvature will be exactly the same as the curvature of the Arctic Ocean.		But ... The result would be true for a flat Earth. The condition "if the direction of the vertical is the Pole, the direction of the vertical changes 300 times faster (the Earth only reduces the curvature of the liquid surface at the pole by 1/300. The mirror designed by the Bear will be convex, its curvature will be exactly the same as the curvature of the Arctic Ocean.

−	Thus, the correct answer is "'''NO'''", the idea of PolarBear is impracticable	+	Thus, the correct answer is "'''NO'''", the idea of PolarBear is impracticable.
		+
		+	==中文翻译==
		+
		+	'''水银镜面''' 想象一个水银镜面。在每一个点上，垂直于镜面的直线与竖直方向都有一个夹角γ=ω<sup>2</sup>r/g，其中ω是水银镜面的角速度，r是水银镜面的半径，g是重力加速度，如果竖直方向（矢量g）一定，那么夹角γ就与镜面的曲率半径R有关，γ=r/R，通过这两个方程可以得到：
		+
		+	$$R = g/{\omega}^2 = gT^2/4{\pi}^2.$$
		+
		+	代入g = 9.8 m/s<sup>2</sup> 和 T = 86164s ，得到：
		+
		+	$$R \approx 1.84{\cdot}10^9m = 1.84\ million\ km.$$
		+
		+	望远镜的焦距是：
		+
		+	$$F = R/2 = R/2 = gT^2/8{\pi}^2 \approx 0.92{\cdot}10^9 m = 920\ thousand\ km.$$
		+
		+	但是……结果只适用于平坦的地球。如果垂直方向是极点，那么垂直方向的变化要快300倍（地球只会使极点处的液面曲率减少至1/300）。这只北极熊的镜面会变成凸面，其曲率与北冰洋曲率完全相同。
		+
		+	所以正确的答案是'''No'''，北极熊的想法是行不通的。

@@ 第1行： / 第1行： @@
-===原文题目===
+==原文题目==
 '''Mercury mirror.''' As you know, if one rotates a vessel with mercury, its surface will take a parabolic form. A kiquid mirror obtained in this wan can be used as the main mirror of a telescope. The Polar Bear-astronomer living in the Arctic decided to use this idea and build such a telescope exactly at the Norht Pole, and use the daile revolution of the Earth as the rotation mechanism so that the main mirror would have a long focal length.
@@ 第6行： / 第6行： @@
 '''1.2.''' Include an artistic picture with an image of the Bear-astronomer, engaged in the implementation of his project.
-===中文翻译===
+==中文翻译==
 '''水银镜面''' 如你所知， 如果转动一个装有水银的容器，水银表面就会形成一个抛物面。可以把这种液体镜面当作一台望远镜的主镜。一位北极熊天文学家想要用这种办法在北极点上安装一台望远镜，这台望远镜完全依靠地球的自转，因此它的焦距会很长。
@@ 第14行： / 第14行： @@
 '''1.2.''' 请画一个正在进行这项工作的北极熊。
-===英文解答===
+==英文解答==
 '''Mercury mirror.''' Imagine rotating mercury. At each point, the normal to its surface will be directed at an angle γ =ω<sup>2</sup>r/g to the vertical, where ω is the angular velocity of rotation of mercury, r is the distance from the axis of rotation, g is the gravity acceleraton, If the direction of the vertical(vector g) is fixed, then this angle γ is related to the radius of curvature R of the liquid mirror as γ = r/R. From these two equations we get:

@@ 第10行： / 第10行： @@
 '''水银镜面''' 如你所知， 如果转动一个装有水银的容器，水银表面就会形成一个抛物面。可以把这种液体镜面当作一台望远镜的主镜。一位北极熊天文学家想要用这种办法在北极点上安装一台望远镜，这台望远镜完全依靠地球的自转，因此它的焦距会很长。
-'''1.1.''' 北极熊的想法可行吗？（用“'''Yes'''”或“'''No'''”回答）如果可行，请计算出这台望远镜的焦距，如果不可行，请给出必要的计算和示意图。
+'''1.1.''' 北极熊的想法可行吗？（用“'''Yes'''”或“'''No'''”回答）如果可行，请计算出这台望远镜的焦距，如果不可行，请写出必要的计算过程并画出示意图。
 '''1.2.''' 请画一个正在进行这项工作的北极熊。

@@ 第2行： / 第2行： @@
 '''Mercury mirror.''' As you know, if one rotates a vessel with mercury, its surface will take a parabolic form. A kiquid mirror obtained in this wan can be used as the main mirror of a telescope. The Polar Bear-astronomer living in the Arctic decided to use this idea and build such a telescope exactly at the Norht Pole, and use the daile revolution of the Earth as the rotation mechanism so that the main mirror would have a long focal length.
-'''1.1.''' Is tge Polar Bear idea feasible? (Write "'''Yes'''" or "'''No'''"in English.)If yes, then find the focal length of the resulting liquid mirror, and if not, prove your conclusion by doing the necessary calculations and drawings.
+'''1.1.''' Is tge Polar Bear idea feasible? (Write "'''Yes'''" or "'''No'''" in English.)If yes, then find the focal length of the resulting liquid mirror, and if not, prove your conclusion by doing the necessary calculations and drawings.
 '''1.2.''' Include an artistic picture with an image of the Bear-astronomer, engaged in the implementation of his project.

←上一版本		2019年9月13日 (五) 00:30的版本
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−	===~~英文题目~~===	+	===原文题目===
	'''Mercury mirror.''' As you know. If one rotates a vessel with mercury, its surface will take a parabolic form. A kiquid mirror obtained in this wan can be used as the main mirror of a telescope. The Polar Bear-astronomer living in the Arctic decided to use this idea and build such a telescope exactly at the Norht Pole, and use the daile revolution of the Earth as the rotation mechanism so that the main mirror would have a long focal length.		'''Mercury mirror.''' As you know. If one rotates a vessel with mercury, its surface will take a parabolic form. A kiquid mirror obtained in this wan can be used as the main mirror of a telescope. The Polar Bear-astronomer living in the Arctic decided to use this idea and build such a telescope exactly at the Norht Pole, and use the daile revolution of the Earth as the rotation mechanism so that the main mirror would have a long focal length.

@@ 第7行： / 第7行： @@
 ===英文解答===
-'''Mercury mirror.''' Lmagint rotating mercury. At each point, the normal to its surface will be directed at an angle γ =ω<sup>2</sup>r/g to the vertical, where ω is the angular velocity of rotation of mercury, r is the distance from the axis of rotation, g is the gravity acceleraton, If the direction of the vertical(vector g) is fixed, then this angle γ is related to the radius of curvature R of the liquid mirror as γ = r/R. From these two equations we get:
+'''Mercury mirror.''' Imagine rotating mercury. At each point, the normal to its surface will be directed at an angle γ =ω<sup>2</sup>r/g to the vertical, where ω is the angular velocity of rotation of mercury, r is the distance from the axis of rotation, g is the gravity acceleraton, If the direction of the vertical(vector g) is fixed, then this angle γ is related to the radius of curvature R of the liquid mirror as γ = r/R. From these two equations we get:
 $$R = g/{\omega}^2 = gT^2/4{\pi}^2.$$