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belka [17]
3 years ago
10

The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas

ured from the planet's surface). What type of relationship exists between the altitude and the atmospheric density, and what would the atmospheric density be at an altitude of 1,291 kilometers?
Physics
2 answers:
algol133 years ago
7 0
<span> Atmospheric pressure decreases with altitude is because of  gravity . The more  you are away from the surface, the less atmosphere you have above you. The  relationship between the atmospheric pressure and altitude depends on the mass of the planet.
hope this helps </span>
mylen [45]3 years ago
6 0

Answer:

Atmospheric pressure decreases with altitude is because of  gravity . The more  you are away from the surface, the less atmosphere you have above you.

Explanation:

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A block with mass 0.470 kg sits at rest on a light but not long vertical spring that has spring constant 85.0 N/m and one end on
a_sh-v [17]

Answer: elastic potential energy = 20.27 J

Explanation:

Given that the

Mass M = 0.470 kg

Height h = 4.40 m

Spring constant K = 85 N/m

The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.

But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.

That is

K .E = P.E = mgh

Where g = 9.8m/s^2

Substitutes all the parameters into the formula

K.E = 0.470 × 9.8 × 4.4

K.E = 20.27 J

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4 0
3 years ago
A certain substance has a heat of vaporization of 37.51 kJ / mol. At what Kelvin temperature will the vapor pressure be 3.50 tim
stellarik [79]

Answer:

T2=336K

Explanation:

Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:

where:

In(P2/P1) = ΔvapH/R(1/T1 - 1/T2)

p1 and p2 are the vapour pressures at temperatures 

T1 and T2

ΔvapH = the enthalpy  of vaporization of the liquid

R = the Universal Gas Constant

p1=p1, T1=307K

p2=3.50p1; T2=?

ΔvapH=37.51kJ/mol=37510J/mol

R=8.314J.K^-1moL^-1

In(3.50P1/P1)= (37510J/mol)/(8.314J.K^-1)*(1/307 - 1/T2)

P1 and P1 cancelled out:

In(3.50)=4511.667(T2 - 307/307T2)

1.253=14.696(T2 - 307/T2)

1.253=(14.696T2) - (14.696*307)/T2

1.253T2=14.696T2 - 4511.672

Therefore,

4511.672=14.696T2 - 1.253T2

4511.672=13.443T2

So therefore, T2=4511.672/13.443=335.61

Approximately, T2=336K

6 0
3 years ago
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