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Shtirlitz [24]
2 years ago
5

A planet's moon travels in an approximately circular orbit of radius 8.6 107 m with a period of 6 h 25 min. Calculate the mass o

f the planet from this information.
Physics
1 answer:
DerKrebs [107]2 years ago
8 0

Answer:

m v^2 / R = G M m / R^2       gravitational attraction = centripetal force

M = v^2 R / G       solving for M

period = 6 h 25 min = (6 * 3600 + 25 * 60) sec = 23,100 sec = T

v = 2 pi R / T

M = 4 pi^2 R^3 / (G T^2)

M = 39.5 * (8.6E7)^3 / (6.67E-11 * 2.31E4^2)

M = 39.5 * 636 / (6.67 * 5.34) * 10^24

M = 7.05 * 10^26 kg

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

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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:

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A. What is the RMS speed of Helium atoms when the temperature of the Helium gas is 343.0 K? (Possibly useful constants: the atom
kkurt [141]

Answer:

(a) 1462.38 m/s

(b) 2068.13 m/s

Explanation:

(a)

The Kinetic energy of the atom can be given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

K.E = Kinetic Energy of atoms = 343 K

T = absolute temperature of atoms

The K.E is also given as:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

v² = 3KT/m

v = √[3KT/m]

where,

m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg

v = RMS Speed of Helium Atoms = ?

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 1462.38 m/s</u>

(b)

For double temperature:

T = 2 x 343 K = 686 K

all other data remains same:

v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 2068.13 m/s</u>

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3 years ago
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