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2 years ago

In the context of studying major bodies of our solar system, what category of object does our moon best fit?.

Physics

1 answer:

Mumz [18]2 years ago

4 0

The simplest, obvious category is " Satellites " .

But there's a lot of good reason to call it "Component of Binary Planet" .

As a fraction of the size and mass of its planet, our moon is the largest in the solar system.

Several writers have pointed out that an alien astronomer studying our solar system might describe the 3rd body from the Sun as a "Binary planet".

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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:

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

5 0

3 years ago

Which of the Materials are insulators? (select all that apply)

DENIUS [597]

I believe all the insulators would be glass, wood, plastic, and yarn.

But I’m not entirely sure if mechanical pencil lead is an insulator or conductor.

Hope this helps.

3 0

3 years ago

d is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?

Anna71 [15]

D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
Answer: The sled's potential energy is 882 Joules

5 0

4 years ago

Read 2 more answers

2. Below what depth would a submarine have to submerge so that it would not be swayed by surface waves with a wavelength of 24 m

Travka [436]

Answer: Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

Explanation:

To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.

This wave base is equal to half of the wavelength. The equation becomes:

Wave base = $\frac{\text{Wavelength}}{2}$

We are given:

Wavelength = 24 m

Putting values in above equation, we get:

Wave base = $\frac{24m}{2}=12m$

Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

4 0

3 years ago

A potential energy function is given by u(x)=(3.00n)xâ(1.00n/m2)x3. at what position or positions is the force equal to zero?

qwelly [4]

I believe the correct form of the energy function is:

u (x) = (3.00 N) x + (1.00 N / m^2) x^3

or in simpler terms without the units:

u (x) = 3 x + x^3

Since the highest degree is power of 3, therefore there are two roots or solutions of the equation.

Since we are to find for the positions x in which the force equal to zero, u (x) = 0, therefore:

3 x + x^3 = u (x)

3 x + x^3 = 0

Taking out x:

x (3 + x^2) = 0

So one of the factors is x = 0.

Finding for the other two factors, we divide the two sides by x and giving us:

x^2 + 3 = 0

x^2 = - 3

x = sqrt (- 3)

x = - 1.732 i, 1.732 i

The other two roots are imaginary therefore the force is only equal to zero when the position is also zero.

Answer:

x = 0

5 0

3 years ago