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3 years ago

In the ptolemaic model of the universe, _____.

2 answers:

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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Thermal energy depends on an object’s

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The correct answer is D. I alread took this test.

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3 years ago

Read 2 more answers

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

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2 years ago

Which element could provide one atom to make an ionic bond with nitrogen?

aev [14]

Answer:

boron

aluminum

gallium

indium

thallium

Explanation:

Any of these could work. Nitrogen has 5 valence electrons so you just needed to pick an element that has 3 valence electrons that nitrogen could borrow. This periodic table shows valence electron counts:

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2 years ago

Physics {deceleration}

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