Answer:
the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Explanation:
Given the data in the question;
wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m
Index of refraction; n = 1.35
Now, the thinnest thickness of the soap film can be determined from the following expression;
= ( λ / 4n )
so we simply substitute in our given values;
= ( 463 × 10⁻⁹ m ) / 4(1.35)
= ( 463 × 10⁻⁹ m ) / 5.4
= ( 463 × 10⁻⁹ m ) / 4(1.35)
= 8.574 × 10⁻⁸ m
= 85.74 × 10⁻⁹ m
= 85.74 nm
Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Answer:
Explanation:
It is given that,
Radius of the circle, 
The area of the circle is given by :
or
As there is no uncertainty given in the radius of the circle. So, the area of the circle is 
. Hence, this is the required solution.
Answer:
yesssssssssssssssssssssssssssssssssssss
The portion of the object submerged in water is determined as 0.25.
<h3>Fraction of the object submerged in water</h3>
The fraction of the object submerged in water is calculated as follows;
S.G = (density of object) / (density of water)
where;
S.G = (250 kgm⁻³) / (1000 kgm⁻³)
S.G = 0.25 = ¹/₄
Thus, the portion of the object submerged in water is determined as 0.25.
Learn more about density here: brainly.com/question/1354972
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Sorry this is late...
The lowest point of a transverse wave is called the trough, while the highest is called the crest.
