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3 years ago

Does air pressure increase or decrease with an increase in altitude?

Physics

2 answers:

adell [148]3 years ago

7 0

Answer:

decrease.

Explanation:

The air pressure is given by

P = h x d x g

where, h is the height of air column above the surface and g be the acceleration due to gravity and d be the density of air.

As we go up, the height of air column decreases, density of air decreases and acceleration due to gravity also decreases, so the value of pressure decreases at altitudes.

Yakvenalex [24]3 years ago

5 0

Answer:

increase

Explanation:

think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing

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3 years ago

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A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats cP /cV = γ moves wi

Vika [28.1K]

Answer:

∆T = Mv^2Y/2Cp

Explanation:

Formula for Kinetic energy of the vessel = 1/2mv^2

Increase in internal energy Δu = nCVΔT

where n is the number of moles of the gas in vessel.

When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas

We say

1/2mv^2 = ∆u

1/2mv^2 = nCv∆T

Since n = m/M

1/2mv^2 = mCv∆T/M

Making ∆T subject of the formula we have

∆T = Mv^2/2Cv

Multiple the RHS by Cp/Cp

∆T = Mv^2/2Cv *Cp/Cp

Since Y = Cp/CV

∆T = Mv^2Y/2Cp k

Since CV = R/Y - 1

We could also have

∆T = Mv^2(Y - 1)/2R k

6 0

4 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$