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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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THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!!
poizon [28]

Answer:

391.67Hz

Explanation:

The fundamental frequency formula in string is expressed as;

Fo = V/2L

V is the velocity of the wave = 329m/s

L is the length of the string = 42cm = 0.42m

Substitute

Fo = 329/2(0.42)

Fo = 329/0.84

Fo = 391.67Hertz

Hence the  fundamental frequency of a mandolin string is 391.67Hz

4 0
3 years ago
What unit can be used to measure length or distance
strojnjashka [21]

Answer:

The most common units that we use to measure length in the metric system are the millimeter, centimeter, meter, and kilometer.

Explanation:

7 0
3 years ago
Erica throws a tennis ball against a wall, and it bounces back. Which force is responsible for sending the ball back to Erica?
Anna71 [15]

OPTIONS :

A.) the force that the ball exerts on the wall

B.) the frictional force between the wall and the ball

C.) the acceleration of the ball as it approaches the wall

D.) the normal force that the wall exerts on the ball

Answer: D.) the normal force that the wall exerts on the ball

Explanation: The normal force acting on an object can be explained as a force experienced by an object when it comes in contact with a flat surface. The normal force acts perpendicular to the surface of contact.

In the scenario described above, Erica's tennis ball experiences an opposite reaction after hitting the wall.This is in relation to Newton's 3rd law of motion, which states that, For every action, there is an equal and opposite reaction.

The reaction force in this case is the normal force exerted on the ball by the wall perpendicular to the surface of contact.

8 0
3 years ago
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
How can the motion of an object that is NOT moving change?​
Semenov [28]
Only if a force acts upon it, it can move.
6 0
3 years ago
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