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SCORPION-xisa [38]
3 years ago
8

Imagine a particular exoplanet covered in an ocean of liquid methane. At the surface of the ocean, the acceleration of gravity i

s 6.20 m/s2, and atmospheric pressure is 7.00 ✕ 104 Pa. The atmospheric temperature and pressure on this planet causes the density of the liquid methane ocean to be 415 kg/m3. (a) What force (in N) is exerted by the atmosphere on a disk-shaped region 2.00 m in radius at the surface of the ocean? N (b) What is the weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m? (Enter your answer in N.) N (c) What is the pressure (in Pa) at a depth of 10.0 m in the methane ocean?
Physics
1 answer:
AlladinOne [14]3 years ago
8 0

Answer:

Explanation:

Atmospheric pressure = 7 x 10⁴ Pa

force on  a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere  = pressure x area

= 7 x 10⁴ x 3.14 x 2 x 2

= 87.92 x 10⁴ N

b )

weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m

Pressure x area

height x density x acceleration of gravity x π r²

= 10 x 415 x 6.2 x 3.14 x 2 x 2

=323168.8 N

c ) Pressure at a depth of 10 m

atmospheric pressure + pressure due to liquid column

= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)

= 7 x 10⁴ + 10 x 415 x 6.2

(7 + 2.57 )x 10⁴ Pa

9.57 x 10⁴ Pa

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7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

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Explanation:

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4 years ago
If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37 ∘C, what is the final state of the
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Answer:

Vapors

Explanation:

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Q_{water} = 50 \times \times 4180 \times (100-37)\\= 1.132 \times 10^7 \ J

Q_{change\ to\ steam} = 50 \times 2.256 \times 106 \\= 1.128 \times 10^8 \ J

Q_{total} = 1.132 \times 10^7 + 1.128 \times 10^8\\= 1.126 \times 10^8 \ J

From the lightning we received 10^{10} \ J of energy, out of which 1.126 \times 10^8 has been used to convert the water into steam.

Energy left = 10^{10} - 1.126 \times 10^8 = 9.88 \times 10^9 \ J

We use this energy to convert steam into vapors.

Q = \Delta E

Q = \Delta E = mc (T_{f} - T{i})\\T_{f} = \frac {\Delta E}{mc} + T_{i}\\ \\T_{f} = 100 + \frac{9.88 \times 10^{10}}{50 \times 1970}\\T_{f} = 100 + 10^8\\T_{f} = 10 ^{ \ 8} \ {^ \circ } C

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3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

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\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

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Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

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PART B) The new amplitude A' that is half ot the wavelength of the wave is

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P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

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