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3 years ago

What phases would Venus show if the geocentric model were correct?

Physics

1 answer:

valina [46]3 years ago

4 0

Answer:

It would show - New, waxing crescent, waxing half, waxing gibbous, full, waning gibbous, waning half, waning crescent phases just like moon.

Explanation: Geocentric model also known as geocentrism, gives the description of universe with Earth at the center. It is often exemplified by the Ptolemaic system. Under this system all the planets, stars, moon and the sun orbited the Earth. Ptolemaic order of spheres from Earth outward is Moon, Mercury, Venus, Sun, Mars. Jupiter, Saturn, Fixed stars, Primum Mobile. It was replaced by the heliocentric model where sun is at the center.

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Which refers to any disturbance that carries energy from one place to another through matter and space?

Black_prince [1.1K]

I think frequency it sounds like the correct answer but I am not completely sure if I am correct

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2 years ago

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What is the theory of light?

Lostsunrise [7]

Answer:

Corpuscular theory of light

Explanation:

In optics, the corpuscular theory of light, arguably set forward by Descartes in 1637, states that light is made up of small discrete particles called "corpuscles" which travel in a straight line with a finite velocity and possess impetus. This was based on an alternate description of atomism of the time period.

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3 years ago

A magnetically soft material is placed in a strong magnetic field. What is the most likely outcome?

Svet_ta [14]

D-It will become a temporary magnet because the domains will easily realign.

9 0

2 years ago

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A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

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2 years ago

Penny puts Fluffy in one wagon and pushes the wagon, once, as hard as she can. Penny puts Butch in the other wagon and pushes it

loris [4]

D; the wagons will both move in the same way

4 0

3 years ago

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