In order to make his measurements for determining the Earth-Sun distance, Aristarchus waited for the Moon's phase to be exactly half full while the Sun was still visible in the sky. For this reason, he chose the time of a half (quarter) moon.
<h3 /><h3>How did Aristarchus calculate the distance to the Sun?</h3>
It was now possible for another Greek astronomer, Aristarchus, to attempt to determine the Earth's distance from the Sun after learning the distance to the Moon. Aristarchus discovered that the Moon, the Earth, and the Sun formed a right triangle when they were all equally illuminated. Now that he was aware of the distance between the Earth and the Moon, all he needed to know to calculate the Sun's distance was the current angle between the Moon and the Sun. It was a wonderful argument that was weakened by scant evidence. Aristarchus calculated this angle to be 87 degrees using only his eyes, which was not far off from the actual number of 89.83 degrees. But when there are significant distances involved, even slight inaccuracies might suddenly become significant. His outcome was more than a thousand times off.
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A) It is very dry at the equator because it is so hot.
False. It depends on what else is around.
The Sahara Desert is near the Equator.
But so is the Amazon rain forest.
B) Higher places are usually warmer because they are closer to the Sun.
False. Higher places are usually colder, because the air is thinner.
C) Land masses change the direction of currents.
True
D)The sun provides energy needed for the evaporation process. Gravity allows water droplets to fall as precipitation.
True
Answer:
The answer to your question is B) Au, Ag and Sn
Explanation:
This is a brief explanation of the origin of the symbols
Gold: the symbol (Au) comes from the Latin Aurum
Silver: the symbol (Ag) comes from the Latin Argentum
Tin: the symbol (Sn) comes from the Latin Stannum
Titanium has the chemical symbol Ti-
Answer:
"Apparent weight during the "plan's turn" is 519.4 N
Explanation:
The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is 
Given that,
v = 420 m/s
R = 11000 m
Substitute the values in the above equation,
It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vector
In magnitude,
Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector
Which is quite heavier than his/her true weigh of 519.4 N
