This would be gas, due to it not essentially having a definite volume.
To solve this problem it is necessary to apply the concept related to wavelength, specifically when the wavelength is observed from a source that is in motion to the observer.
By definition the wavelength is given defined by,
![\lambda_{obs} = \lambda_s \sqrt{\frac{1+u/c}{1-u/c}}](https://tex.z-dn.net/?f=%5Clambda_%7Bobs%7D%20%3D%20%5Clambda_s%20%5Csqrt%7B%5Cfrac%7B1%2Bu%2Fc%7D%7B1-u%2Fc%7D%7D)
Where
= Observed wavelength
= Wavelength of the source
c = Speed of light in vacuum
u = Relative velocity of the source to the observer
According to our data we have that the wavelength emitted from the galaxy is 1875nm which is equal to the wavelength from the source, while the wavelength from the observer is ![\lambda_{obs}=1945nm](https://tex.z-dn.net/?f=%5Clambda_%7Bobs%7D%3D1945nm)
Therefore replacing in the previous equation we have,
![1945 = 1875 \sqrt{\frac{1+\frac{u}{c} }{1-\frac{u}{c} }}](https://tex.z-dn.net/?f=1945%20%3D%201875%20%5Csqrt%7B%5Cfrac%7B1%2B%5Cfrac%7Bu%7D%7Bc%7D%20%7D%7B1-%5Cfrac%7Bu%7D%7Bc%7D%20%7D%7D)
![\sqrt{\frac{1+u/c}{1-u/c}} = 1.03733](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B1%2Bu%2Fc%7D%7B1-u%2Fc%7D%7D%20%3D%201.03733)
![\frac{1+\frac{u}{c} }{1-\frac{u}{c} }=1.03733^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%2B%5Cfrac%7Bu%7D%7Bc%7D%20%7D%7B1-%5Cfrac%7Bu%7D%7Bc%7D%20%7D%3D1.03733%5E2)
![1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )](https://tex.z-dn.net/?f=1%2B%5Cfrac%7Bu%7D%7Bc%7D%20%3D1.03733%5E2%2A%281-%5Cfrac%7Bu%7D%7Bc%7D%20%29)
Solving for u,
![1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )](https://tex.z-dn.net/?f=1%2B%5Cfrac%7Bu%7D%7Bc%7D%20%3D1.03733%5E2%2A%281-%5Cfrac%7Bu%7D%7Bc%7D%20%29)
![1+\frac{u}{c} =1.03733^2-1.03733^2(\frac{u}{c} )](https://tex.z-dn.net/?f=1%2B%5Cfrac%7Bu%7D%7Bc%7D%20%3D1.03733%5E2-1.03733%5E2%28%5Cfrac%7Bu%7D%7Bc%7D%20%29)
![\frac{u}{c} +1.03733^2\frac{u}{c} =1.03733^2-1](https://tex.z-dn.net/?f=%5Cfrac%7Bu%7D%7Bc%7D%20%2B1.03733%5E2%5Cfrac%7Bu%7D%7Bc%7D%20%3D1.03733%5E2-1)
![2.88595\frac{u}{c}=1.03733^2-1](https://tex.z-dn.net/?f=2.88595%5Cfrac%7Bu%7D%7Bc%7D%3D1.03733%5E2-1)
![\frac{u}{c} = \frac{1.03733^2-1}{2.88595}](https://tex.z-dn.net/?f=%5Cfrac%7Bu%7D%7Bc%7D%20%3D%20%5Cfrac%7B1.03733%5E2-1%7D%7B2.88595%7D)
![u = \frac{1.03733^2-1}{2.88595}*c](https://tex.z-dn.net/?f=u%20%3D%20%5Cfrac%7B1.03733%5E2-1%7D%7B2.88595%7D%2Ac)
![u = 0.02635c](https://tex.z-dn.net/?f=u%20%3D%200.02635c)
Therefore the speed of the gas relative to earth is 0.02635 times the speed of light.
Answer: Comets are basically floating chunks of ice, dust, and frozen gases. So, when comets are in space they cannot melt, So I'll say no :)
Answer - No, comets doesn't have a liquid center.
Explanation:
I think it's three days. I read it in assignment potion before but it's kinds fuzzy but I believe it's three days. Hopefully thats correct.
Information that is given:
a = -5.4m/s^2
v0 = 25 m/s
---------------------
S = ?
Calculate the S(distance car traveled) with the formula for velocity of decelerated motion:
v^2 = v0^2 - 2aS
The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
S = (625 m^2/s^2)/(10.8 m/s^2)
S = 57.87 m