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Irina18 [472]
2 years ago
11

Air that is not saturated will cool or heat at a rate of __________________ as it rises or descends, respectively

Physics
1 answer:
Kay [80]2 years ago
7 0

Air that is not saturated will cool or heat at a rate of 10 degrees C/1000 meters as it rises or descends, respectively

The rate of the change in temperature which is observed while moving upward in the earth's atmosphere with elevation. It can be positive, negative and zero when the temperature decreases, increases or is constant with the elevation respectively.

For the atmosphere, the drop in temperature of rising air that is  unsaturated air is about 10 degrees C/1000 meters (5.5 °F per 1000 feet) altitude.

That means if a there occur a rise of 1000 m , then the temperature of that thing will decrease to 10 degrees. Every 10°C of temperature from the given temperature will decrease at every rise of 1000 m .Air that is not saturated will cool or heat at a rate of 10 degrees C/1000 meters as it rises or descends, respectively

To learn more about unsaturated air here

brainly.com/question/13434971?referrer=searchResults

#SPJ4

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A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.
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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
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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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