PLEASE ANSWER ASAP!!!

1. why did aristarchus choose the time of a half (quarter) moon to make his measurements for calculating the earth-sun distance?

Stells [14]

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.

To know more about how Aristarchus calculate the distance to the Sun, visit:

brainly.com/question/26241069

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7 0

2 years ago

Is the arctic ocean larger than the indian ocean?

Alik [6]

The Arctic Ocean is the smallest ocean. It is more than five times smaller than Indian and Atlantic Oceans.

5 0

4 years ago

If the pH of a solution decrease's, is the solution getting more acidic or lees acidic

enyata [817]

Answer:

The solution becomes more acidic.

Explanation:

As the pH of a solution decreases, the concentration of hydrogen ions [H+] increases. Acidic solutions have a higher concentration of hydrogen ions and a lower pH.

I hope this helped. :)

6 0

3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius: $0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

$E=U=\frac{GMm}{R}$

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$