Answer:
A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm
, this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.
we have a fluctuation of the intensity emitted by the stars. Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.
Explanation:
The radiation of a black body is characterized by its temperature, with Wien's law of displacement we can find the maximum wavelength emitted by each star.
λ T = 2,898 10⁻³
therefore the emission the star of A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm
The emission of the premiere is in the ultraviolet light range, as this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.
The burst with A = 4300K has a bad emission maximum = 6.7395 10⁻⁷ m = 673.95 nm, which corresponds to an emission in the visible in the orange range, giving a blackbody spectrum of this range, but since the emission is formed by two stars, we see that when the two are placed one in front of the other the intensity of the emission must increase significantly and when they are placed next to each other the intensity reaches its minimum, consequently we have a fluctuation of the intensity emitted by the stars.
Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.
Answer:
What displacement must the physics professor give the car
= 12.91 METERS
Explanation:
Check the attached file for explanation
Answer:
True
Explanation:
Magnitude is the "value" the greater the value the greater the force is and vice versa
Answer:
An Object is positively charged if it has more Positive Electrons in that object
