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Sonja [21]
3 years ago
11

Pluto's atmosphere. As recently observed by the New Horizons mission, the surface pressure of Pluto is about 11 microbar. The su

rface temperature is about 37 K.
(a) What is the number density (in units of number per cubic centimeter) of molecules at Pluto's surface (Hint: use ideal gas law)? The radius of Pluto is about 1187 km and the surface gravity is about 0.62 m s. What is the total mass of the atmosphere in terms of Kg?

(b) Calculate the saturation vapor pressure (in units of Pa) of ethane at Pluto's surface. The saturated vapor pressure of ethane can be assumed as: log1o(P)-10.01-1085.0/(T-0.561). T is temperature in K and the vapor pressure (P) in units of millimeters of Hg (~133.32 Pa).

(c) If the volume mixing ratio of ethane on Pluto is about 1%, what is mass mixing ratio of ethane (assume the mean molecular weight is 28 g mol')? What is the partial pressure of ethane at the surface? (Hint: should you use volume mixing ratio or mass mixing ratio to calculate the partial pressure? Think about the physical meaning of gas pressure.) Finally, is ethane condensable at Pluto's surface)

Physics
1 answer:
Komok [63]3 years ago
4 0

Answer:

a) The number density is 3.623 × 10⁻³ \frac{mol}{m^{3} }

The mass of the atmosphere is 1.3 × 10²²Kg

b) The pressure is 10⁻²⁰ Millimeter of mercury

c) The mass mixing ratio is 0.0107

The partial pressure of ethane is 0.01114 Pa

Yes it is condensable because it boiling point is -88.5 C  which is equivalent to 184.5 K i.e is adding 273 to -88.5C and the temperature of the atmosphere  is 37 K.

Explanation:

The explanation is on the first and second uploaded image

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A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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3 years ago
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Answer:

1486.5\frac{Btu}{s}

Explanation:

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The mass flow rates is expressed as:

\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}

The energy balance for the system can the be expresses in the rate form as:

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What are forces that two objects apply on each other
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Answer:

It is sensible heat- the amount of heat absorbed by 1 kg of water when heated at a constant pressure from freezing point 0 degree Celsius to the temperature of formation of steam i.e. saturation temperature

So it is given as - mass× specific heat × rise in temperature

i.e. 4.2 × T

4.2 × (100–0)

So it is 420kj

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So it will be

Q= 1×336 + 1× 4.18 ×100 + 1× 2257

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2 years ago
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