Answer:
no its not like the undertow in the ocean
Explanation:
In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is

where D=5.00 m is the distance of the screen from the slits, and

is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:

And from the relationship between frequency and wavelength,

, we can find the frequency of the light:
In short, and in general:
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Accordingly Newton's findings, astronomy and physics have industrialized
hugely over the period. Scientists now recognize that every object in the
world has a force that draws each other and the power of the force hinge on the
mass of the object. Also, Newton's Laws of Motion offer individuals a
better understanding of what is likely concerning movement. This is very helpful,
particularly in mechanics and space travel. Generally, Newton had a
huge and permanent impact on science.
Answer:

Explanation:
M = Mass of each star
T = Time period = 15.5 days
v = Orbital velocity = 230 km/s
G = Gravitational constant = 
Radius of orbit is given by

We have the relation

The mass of each star is 