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2 years ago

The Ptolemaic model of the universe:

Physics

2 answers:

harkovskaia [24]2 years ago

8 0

Answer:

The Ptolemaic model of the universe A) explained and predicted the motions of the planets with deferents and epicycles.

Explanation:

Ptolemy of Alexandria built an explanation of the observed movements of the planets that remained in force for thirteen centuries. Ptolemy proposed a model of the Universe with the Earth in the center. In the model, the Earth remains stationary while the planets, the Moon and the Sun describe complicated orbits around it. In other words, Ptolemy devised a system in which he used epicycles, deferential and eccentric, and it was necessary to introduce an equating point to reproduce planetary movements. He proposed that:

a) Each planet revolves with constant velocity around a circle called an epicycle.

b) The center of the epicycle is located and moves with constant velocity around another circle called deferential.

c) The center of the deferent is located at a moving point, which travels with constant speed describing another circumference called eccentric.

d) The center of the eccentric coincides with the center of the Universe.

e) Since the Earth is not located in the center of the Universe, but very close to it, it was necessary to introduce an equating point, which is not on Earth, and from which you can see the planet move with constant speed.

However, Ptolemy put forward this geometric theory to explain mathematically the movements and failed to adjust any system of cycles, epicycles and eccentrics that accurately represented the observed movements of the planets.

Finally, The Ptolemaic model of the universe A) explained and predicted the motions of the planets with deferents and epicycles.

Tcecarenko [31]2 years ago

6 0

Answer:

Earth was the center of the universe.

Explanation:

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Answer:

Explanation:

No, a machine cannot be 100% efficient. This is due to the movement of the moving parts siding against each other and causing friction. This friction is the one that creates heat and causes wear and tear between moving ports f the machine hence making the machine to decrease in efficiency with time

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2 years ago

The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between

erma4kov [3.2K]

Answer:

Considering first question

Generally the coefficient of performance of the air condition is mathematically represented as

$COP = \frac{T_i}{T_o - T_i}$

Here $T_i$ is the inside temperature

while $T_o$ is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes $\frac{T_i}{T_o - T_i}$ heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

$Q \ \alpha \ (T_o - T_i)$

=> $Q= k (T_o - T_i)$

Here k is the constant of proportionality

since 1 unit of electricity removes $\frac{T_i}{T_o - T_i}$ amount of heat

E unit of electricity will remove $Q= k (T_o - T_i)$

$E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }$

=> $E = \frac{k}{T_i} (T_o - T_i)^2$

given that $\frac{k}{T_i}$ is constant

=> $E \ \alpha \ (T_o - T_i)^2$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

Considering the second question

Assuming that $T_i = 30 ^oC$

and $T_o = 40 ^oC$

Hence

$E = K (T_o - T_i)^2$

Here K stand for a constant

$E = K (40 - 30)^2$

=> $E = 100K$

Now if the $T_i = 20 ^oC$

Then

$E = K (40 - 20)^2$

=> $E = 400 \ K$

So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher

Considering the third question

Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

$Q = k (T_o - T_i )^{\frac{1}{2} }$

$E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }$

=> $E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }$

Assuming $\frac{k}{T_i}$ is a constant

Then

$E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.

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