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Slav-nsk [51]
3 years ago
13

Linea de tiempo sobre la aportaciones griegas a la astronomia y astrologia

Physics
1 answer:
rusak2 [61]3 years ago
7 0

Answer:

1. 100 CE

Menelaus of Alexandria lived. a Greek mathematician and astronomer

2. 190 BCE - 120 BCE

Hipparchus of Nicea, an Hellenic language mathematician, astronomer and geographer, regarded by many historians as a scientist of the most effective quality and one amongst the most effective astronomical genius amongst ancient Greeks.

3. 276 BCE - 195 BCE

Eratosthenes, an Hellenic language Alexandrian scholar, who was a native of Cyrene and one amongst the most effective geographers in antiquity.

4. c. 310 BCE - c. 230 BCE

Aristarchus of Samos. A Greek astronomer and mathematician

5. 384 BCE - 322 BCE

Aristotle Era.

6. c. 571 BCE - c. 497 BCE

Pythagoras of Samos lived during this era.

7. 585 BCE

Media and Lydia went into battle and broke off immediately as a result an entire eclipse of the sun which occurred causing the two armies to create peace. The eclipse was already predicted by Thales of Miletus.

8. 585 BCE

Thales of Miletus lived during now.

Explanation:

Ancient Greeks were some of the first people known to study the sky and understand what astronomy really entails. They discovered the Earth was spherical in shape and went ahead to devise a means to measure its size. They also were the ones who created the idea of a geocentric solar system, which was incorrect, But assisted us in understanding the universe for over hundreds of years.

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IrinaK [193]

Answer : F = 3.5\times10^{3}\ N

Explanation :

Given that

Radius of sphere r = 5.90\times 10^{-15}\ m

The distance between the centers of the two spheres is

r = 2\times 5.90\times 10^{-15}\ m

The charge of the sphere q = 46\times1.6\times10^{-19} C

The magnitude of the repulsive force between the charges pushing them a part is

Using coulomb law

F = \dfrac {kq_{1}q_{2}}{r^{2}}

F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}

F = 3501.3\ N

F = 3.5\times10^{3}\ N

Hence, this is the required solution.









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3 years ago
Without effective assessment methods, teachers have no documented proof of childrens specific ____ and _____.
schepotkina [342]

Answer:

D strengths and weakneses

Explanation:

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3 years ago
A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?
Nostrana [21]

Answer:

The work done by friction was -4.5\times10^{5}\ J

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

W=\Delta KE

W=K.E_{f}-K.E_{i}

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2

Put the value of m and v

W=0-\dfrac{1}{2}\times1000\times(30)^2

W=-450000
\ J

W=-4.5\times10^{5}\ J

Hence, The work done by friction was -4.5\times10^{5}\ J

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3 years ago
You are discussing a material that has a strong resistance to the flow of electrons. This is a description of a(n)
cupoosta [38]
C. Insulator

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3 years ago
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An owl weighing 40N is sitting in a tree waiting to dive down to catch a mouse. If the owl's potential energy is 800 J with resp
Natali5045456 [20]

Answer:

<em>h = 20 m</em>

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.

The weight of an object of mass m is:

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U=W.h

Solving for h:

\displaystyle h=\frac{U}{W}

The weight of the owl is W=40 N and its GPE is U=800 J.

\displaystyle h=\frac{800}{40}=20

h = 20 m

3 0
3 years ago
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