Answer:
Explanation:
This problem relates to interference of light in thin films .
The condition of bright fringe in thin films which is sandwitched by two layers of medium having lesser refractive index is as follows.
2nt = (2n+1) λ / 2 , n is refractive index of thin layer , t is its thickness , λ is wavelength of light .
2 x 1.5 t = λ / 2 , if n = 0 for minimum thickness.
2 x 1.5 t = 600 / 2 nm
t = 100 nm .
That particular group of elements is reffered to as the "Noble Gasses"--a title that comes from the fact that these gases are very "secure" and don't mix well with other elements.
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>