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bulgar [2K]
3 years ago
14

The atmosphere of Jupiter is essentially made up of hydrogen, H2. For H2, the specific gas constant is 4157 J/(kg K). The accele

ration of gravity of Jupiter is 24.9 m/s2 . Assuming an isothermal atmosphere with a temperature of 150 K and assuming that Jupiter has a definable surface, calculate the altitude above that surface where the pressure is one-half the surface pressure
Physics
1 answer:
Alenkinab [10]3 years ago
3 0

Answer:

h=17357.9m

Explanation:

The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.

To calculate this, you need to use the barometric formula:

P=P_0e^{-\frac{Mg}{RT}h}

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.

Furthermore, the specific gas constant is defined by:

R_{H_2}=\frac{R}{M}

Therefore yo can write the barometric formula as:

P=P_0e^{-\frac{g}{R_{H_2}T}h}

at the surface of the planet (h =0) the pressure is P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}

applying to the previuos equation:

\frac{P_0}{2} =P_0e^{-\frac{g}{R_{H_2}T}h}

solving for h:

h=17357.9m

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A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant bef
inessss [21]

Answer:

29.396988 m/s

Explanation:

Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.

I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

The acceleration, g, is found from

g = GM/r²

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

M = 5.9724e+24 kg

r = 6.378e+6 m

g = 9.799086 m sec⁻²

An approximate answer is found from an equation from constant acceleration kinematics:

v = gt

t = 3.0 sec

v = 29.397259 m/s

Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Here,

a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where

h = the altitude from which the ball is dropped

We can (using some math) develop the following equation:

t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

t − t₀ = 3 sec

r = 6378000 meters

d = r + h

Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,

h = 44.0954 meters

So,

d = 6378044.09538 meters

a = d/2 = 3189022.04769 meters

Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

5 0
3 years ago
Which body systems help these cells get the energy they need?
PSYCHO15rus [73]

I believe there should be some sort of table attached. Unfortunately I cannot answer this question. Sorry!

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3 years ago
The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the
Inga [223]

Answer:

27.5 days

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Explanation:

r = radius of the orbit of moon around the earth = 3.85\times10^{8} m

M = Mass of earth = 5.98\times10^{24} m

T = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

T^{2} = \frac{4\pi ^{2} r^{3}  }{GM}

inserting the values, we get

T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3}  }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days

1 month = 30 days

T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month

6 0
3 years ago
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