Answer:
hii there
The correct answer is option ( C ) reduced air pressure
Explanation:
hope it helps
have a nice day :)
Answer:
Increasing
Explanation:
It’s increasing because your are adding more weight
A. Layer C is the right answer out of all the choices
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.
Search Results<span>Use BFS to determine the length of the shortest v-w-path. Then use DFS to find thenumber of the v-w-shortest paths such that two nodes are connected and the length of path equals to the output of BFS. But the running time of this plan is O(m+n)+O(m+n). Also I've tried to modify the Dijkstra algorithm.</span>
