6. A 2000 kg spacecraft is located 9.2x10 m from the center of the earth. The mass of the earth
1 answer:
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We will make the comparison between each of the sizes against the known wavelengths.
In the case of the <em>hydrogen atom</em>, we know that this is equivalent to m on average, which corresponds to the wavelength corresponding to X-rays.
In the case of the <em>Virus</em> we know that it is oscillating in a size of 30nm to 200 nm, so the size of the virus is equivalent to the range of the wavelength of an ultraviolet ray.
In the case of <em>height</em>, it fluctuates in a person around to
m, which falls to the wavelength of a radio wave.
Answer:
More extreme weather.
Explanation:
The Conveyor Belt of tides functions on a local and global level to spread out the cold and hot temperature differences on the planet. It is a delicate but important process that is easily disrupted, which causes it to slow down. And when it slows down, all those temperature differences will become more concentrated, causing colder places to be colder and hotter places to be hotter, ultimately leading to more extreme weather events as these cold and hot spots collide more violently than before.
Here's a picture I found on it:
Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v + xv
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v + x₂v
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ