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,
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 
Therefore replacing in the previous equation we have,
Solving for u,
Therefore the speed of the gas relative to earth is 0.02635 times the speed of light.
Answer:1.81
(a) Explanation:the turn ratio= input voltage÷output voltage.
400÷220=1.81.
Don't know how to solve b part...
Answer:
Explanation:
The process during which pressure remains constant is called an isobaric process.
We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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