Answer: 117.8 nm
Explanation:
Given,
Nonreflective coating refractive index : n = 1.21
Index of refraction:
= 1.52
Wave length of light = λ = 570 nm = 


Hence, the minimum thickness of the coating that will accomplish= 117.8 nm
Answer:
The correct option is A
Explanation:
From the question we are told that
The mass number is 
Generally the mean radius is mathematically evaluated as

Here
is a constant with a value 
So



(1.9 yr) x (365.24 day/yr) x (86,400 sec/day) x (10⁹ nsec/sec)
= (1.9 x 365.24 x 86,400 x 10⁹) nanosec
= 6.00 x 10¹⁶ nanoseconds
Answer:
Yes, yes it would since we need light
Explanation:
<span>92.96 million mi..........</span>