Sophomore geography questions about time zones and solar altitude 10

Day length = 18:30-4:30 = 14h, sunrise time = 12 - half of the day length = 12-7 = 5:

00, the sunrise at 5:00 local time in A, the sunrise at 4:30 Beijing time, the time difference is half an hour, and the longitude is 1 hour apart every 15 ° time, so the longitude is different, and the time is earlier (large) in the place where the relative position on the earth is easterly, A 5:

00> 4:30 Beijing time, so A is in the east, for.

There are a lot of conditions that don't work, and this should be one of the questions in a group.

Day length: 18:30-4:

30=14 and because 12 o'clock in a day is equally divided into the morning and afternoon, so the place of sunrise in Jiacheng should be 12-14 2=5 o'clock, then the place of sunrise in Jiacheng and Beijing time are 5 and 4:30 respectivelyIt is now known to be 120:4 in local time

30. When finding the place, the meridian of 5 points can be obtained to the city of Jia, so the answer is B

This question lacks a known quantity, such as the time of sunrise (Beijing time?). ), the northern and southern hemispheres.

Beijing is 120 degrees east longitude for four and a half o'clock, because the sunrise is always six o'clock, so the time of that place is 6 o'clock, a difference of one and a half hours, that is, the difference of degrees The earth turns from west to east, so that place is east of Beijing, and the answer is b

The 21st of the month is the vernal equinox. The direct point of the sun is at the equator, and the global day and night are equinoxed, with the length of the day and the length of the night being 12 hours. The length of the day is 24 hours, which is the polar day, which does not match the title, and cannot be the day of the vernal equinox, so it is excluded. All meet the conditions of the vernal equinox.

The height of the sun at noon on the Tropic of Cancer is.

The sun is lower in the regions north of the Tropic of Capricorn. As you can see from the figure, and the horizontal line in the middle represents, which is greater than 66°34, so it is not selected. Only less than 66°34, so choose B,

At noon on March 21, when the sun shines directly on the equator, then the global equinox, which is the length of the day, is 12 hours, excluding . The equatorial solar altitude angle is 90 degrees. The Tropic of Cancer is 66 degrees 34 minutes.

Exclude and . Because and the noon solar altitude angle is higher than 66 degrees 34 minutes. It's 90 degrees, it's 45+45 divided by 2.

The answer is b, the day of the vernal equinox, the sun is directly on the equator (point), the global equinox of day and night can be excluded; On that day, the solar altitude of the Tropic of Cancer was 90°-23°26 26=66°34 (point), and the solar height in the north area was smaller than it, so b③④

It's not the same. The height of the sun is the angle between the sun and the ground at any time of the day, the height of the sun at noon is the maximum height of the sun in a day, the height of the sun is infinite in a day, and the height of the sun at noon is only one in a day.

The calculation of the height of the sun has no practical significance, it is enough to know three sentences, the height of the sun on the morning and dusk line is 0, the height of the sun in the day hemisphere is greater than 0, and the height of the sun in the night hemisphere is less than 0

As for the calculation of the height of the sun at noon, two latitudes are needed, that is, the latitude of the place where you require the height of the sun at noon, and the latitude of the point where the sun shines directly.

It is then calculated based on the fact that the height of the sun at noon is equal to 90 degrees minus two latitude differences.

The key is not to get it wrong in the calculation of the latitude difference, if both latitudes are in the same hemisphere, then the latitude difference is the subtraction (large decrease) of the two latitudes, if the two latitudes are a southern hemisphere and a northern hemisphere, then add the two latitudes.

Another: The second empty in the second question in the figure should be 43°08, the third empty is 0°, and the last empty is 0°

The observation position of Figure A is over the North Pole, Figure B is seen from a long distance above the South Pole, and the polar night is within the Arctic Circle, indicating that the sun is directly shining on the Tropic of Capricorn. The solar altitude at point d should be 0 degrees from its geographic latitude minus its geographic latitude from the tangent point between the dawn and dusk lines and north latitude (this method is simpler), and the solar height at point A should be 0 degrees because it is on the morning and dusk coil. (Sorry, the figure is too small).

It's too late to reply Agree with the answer of the smoker!

First of all, you need to be clear about the height of the sun, not the height of the sun at noon. Point A is located on the morning and dusk line, so the altitude of the Sun is 0°. And the solar altitude at point e is 12 o'clock at this time, and its solar altitude is the noon solar height, so the solar height at point e =

a is located on the morning and dusk line, when the sun height is 0; e is located on the equator, and the altitude of the sun at noon on the summer solstice is 66 5°.

There is no picture, but I guessed what you meant, and combined with the title text I imagined the figure.

If the height of the south building is l, then the shadow length of the south building at noon h=l divided by tanh (h is the height of the sun at noon in the title).

The spacing between the buildings is S

When h>s, the north building will be blocked by the south building at noon, of course, there is no need to be early at night.

This way you'll get a range of δ values (e.g. if you find out that it's hypothetical, and you don't have specific data.) δ is h>s from south latitude to 0 degrees, then the absolute value of the range of δ is the distance. Divide by two degrees, which is divided by 47, and the proportion obtained is the proportion of the time of the year when the North Building is present.

My hypothesis calculated to be one-half, then six months were blocked. ）

Divide the range of values by 47 and bring it yourself.

The idea will be the same, and the bottom is the same.

From the equator to the poles, the higher the latitude, the greater the maximum difference in day length, the equator is 0, the north and south poles are Let the shortest day length be x, list the equation: x+(x+5h42min)=24h, find x= international time, that is, 0 degrees longitude is 10h24min, the maximum height of the sun in 1 place is 12h, and the longitude of 1 place is 24 degrees east longitude. At this time, select the bc term, and then you can know that the maximum difference in the height of the sun at noon at 23 degrees 26 minutes north latitude is 46 degrees 52 minutes, and the elimination method, choose c.

If you don't understand, you can ask, I hope it will be useful to you.

This is the first year of high school geography. Completely incomprehensible.

If the house just can't see any lead sunlight when the sun height angle is 45 ° (the distance between the buildings is 30 meters, the height of the building 3 times 20 equals 60 meters, he lives on the 10th floor is equivalent to 30 meters, then the sun to make the sun shine in is 45 ° of the sun height angle, you can understand the specific you draw the oak year on the paper, I can't draw here), when the solar height is 45 °, the direct sun point is 19 °s (the sun altitude angle = 90 ° - (26 ° + x) = 45 °, the solution is x = 19 °The sun does regression movement for 12 months a year, and a total of 23°26 times 2=46°52, so the average monthly walk is 4°, and 19°s is about 4° away from the Tropic of Capricorn, so the house has 11 months of sunshine per year.