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Gemiola [76]
2 years ago
13

Consider again the objects you ranked by distance. Suppose each object emitted a burst of light right now. Rank the objects from

left to right based on the amount of time it would take this light to reach Earth, from longest time to shortest time.a. star on far side of Andromeda Galaxyb. star on near side of Andromeda Galaxyc. star on far side of Milky Way Galaxyd. star near center of Milky Way Galaxy1. Orion Nebula2. Alpha Centauri3. Pluto4. The Sun
Physics
1 answer:
Vika [28.1K]2 years ago
5 0

Answer:

Following are the solution to this question:

Explanation:

That light takes a very long time to hit the planet, and the object is far off the earth. The light of such an item near to the planet takes less time to enter it. The star is 2,5 million light-years from the Planet on the far side of the Andromeda Galaxy. But on the other hand, the moon is 15 crore miles from the earth, so sunlight is quickly reached on the ground as the other thing.  

That milky way away from the earth is 66,500 light-years far, that distance between Earth and Orion nebula is 1,344 light-years, with such a distance of 4,367 light-years. The earth is 5.2261 trillion km apart from Pluto.

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Explanation:

Transparent objects do not form shadows. The light passes completely from the transparent objects thus these objects will not form shadow. ... In such objects, the light gets refracted thus, such objects forms shadow. The refraction is also the reason why we can see such objects.

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In a science fiction story, a microscopic black hole is given an enormous positive charge by firing an un-neutralized ion drive
Olegator [25]

Answer: distance d = 4.73e10m

Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.

Using electric potential V formula:

V = kq / d

Where K = 9.05×10^9Nm^2/C

And e = 1.6×10^-19C

But you don't need to substitute it.

1090 V = 8.99e9N·m²/C² * 5740C /d

Make d the subject of formula

d = 4.73e10 m

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3 years ago
The atmosphere protects us from all of the following except Meteorites The sun's rays Pollution Wind
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3 years ago
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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