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3 years ago

How did ancient civilizations perform astronomical measurements

Physics

1 answer:

BlackZzzverrR [31]3 years ago

4 0

Answer:

The early observations in Astronomy were used to determine the length of the year and to determine the seasons, which is an important factor in knowing when plants will grow. In the early days, astronomy consisted only of observations and projections of objects that could be seen by ordinary eyes.

Explanation:

Astronomy is the oldest of the modern natural sciences. It probably came with the need for a calendar. In ancient times, astronomers were monks, shamans, druids, and priests. At that time, astronomy was considered astrological. In some places, early civilizations have created great spaces for some astronomical purposes. In addition to rituals, these early observations were used to determine the length of the year and to determine the seasons, which is an important factor in knowing when plants will grow. In the early days, astronomy consisted only of observations and projections of objects that could be seen by ordinary eyes. According to some estimates, Stonheng, the Ziggurats in Mesopotamia, and the Pyramids of the Stones and the Chinese are the oldest observatory. Moon calendars are still considered very accurate today. Similarly, Chinese star catalogs are a valuable source for studying ancient astronomical events. Greek scholars made a special contribution to the development of astronomy. Aristotle, Ptolemy, Hipparchus, Fales, Aratos were engaged in astronomy along with other scientific discoveries. Ptolemy has developed the Geocentric Theory to explain the planetary movement of planets. According to this theory, the earth remains stationary in the center of the world, and all the celestial bodies revolve around it. Ptolemy commented regularly on Almagest. The training continued until the 16th century. Despite religious resistance, this training has been replaced by training in the world's heliocentric system. It is thought that the 6,500-year-old megaliths (Nabta Playa, Stonehenge), planted by ancient people, were used for astronomical observation. In the Rig Veda, there are 27 constellations and 13 separate constellations associated with the movement of the sun. The most important contribution made by ancient Greeks to astrology is that they have tried to classify stars according to their size.

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3 years ago

A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

Grace [21]

Answer:

The final velocity of the cart is $v_c = 7.02 \ m/s$

Explanation:

From the question we are told that

The mass of the girl is $m_g = 35.4 \ kg$

The mass of the cart is $m_c = 15.23 \ kg$

The speed of the cart and kid(girl) is $v = 4.25 \ m/s$

The final velocity of the girl is $v_g = 3.06 \ m/s$

Let assume that velocity eastward is positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

$p__{T1}} = (m_g + m_c) * v$

substituting values

$p__{T1}} = (35.4 + 15.23) * 4.25$

$p__{T1}} =215.17 \ kg m /s$

The total momentum after she steps off the back of the cart is mathematically evaluated as

$p__{T2}} = (m_g * v_g ) +( m_c * v_c )$

Where $v_c$ is the final velocity of the cart

substituting values

$p__{T2}} = (35.4 * 3.06 ) +( 15.23 * v_c )$

$p__{T2}} = 108. 324 + 15.23 v_c$

Now according to the law of conservation of momentum

$p__{T1}} =p__{T2}}$

$215.17 \ kg m /s = 108. 324 + 15.23 v_c$

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3 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago