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Nutka1998 [239]
3 years ago
12

In the ptolemaic model of the universe, _____.

Physics
2 answers:
bija089 [108]3 years ago
7 0
In the Ptolemaic model of the universe, the sun was the center of the solar system

There are two models that is most commonly used as the model of the universe. One is Geocentric model (which describes that earth is the center of all of the existing orbits) while the other one is Ptolemaic which describes sun is the center of all existing orbits.
Neporo4naja [7]3 years ago
7 0

Explanation:

Ptolemy was an Alexandrian Astronomer and mathematician. In 150AD, he presented his model of the universe. It was a modified version of the then accepted Geocentric model of the Universe. Some of the highlights of the model were:

1. Earth is at the center of the universe and stationary.

2. Every celestial object is revolving around the Earth.

3. In order to explain the motion of the Sun, he introduced eccentric motion of the Sun. As per this model the point around which the Sun revolved was displaced from the Earth. Thus the Sun will move faster when at perigee and slower when at  apogee.

4. To explain the motion of planet, he introduced the concept of epicycles. According to him the planets revolved around a point in a circular motion and that point revolved around Earth. This also accounted for the retrograde motion of the planets.

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Which of the following results when a crest and trough meet?
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D. Destructive interference. An easy way to think about it is the waves are opposite each other, so they essentially cancel each other out, or make an effort to.
6 0
3 years ago
One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit
nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

\lambda = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

\theta=\dfrac{1.22\lambda}{D}

The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

6 0
3 years ago
A 300-n force acts on a 25-kg object. the acceleration of the object is?
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To solve the answer use the equation: a = fnet / m

a = 300 N / 25 kg

300 N / 25 kg = 12m/s

The acceleration of the object is 12m/s

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