Answer:
The correct option is;
The index of refraction of the second medium is lower
Explanation:
The index of refraction of a material indicates the magnitude of the optical density of a material. The index of refraction or the refractive index, n, are indices (ratio) of the speed of light through an optically dense medium relative to the speed of light through a vacuum.
The definition of the refractive index is the number of times light travelling through a medium would be slower than light travelling through vacuum
Therefore, the index of refraction of a second medium that is less optically dense than a first medium from which light originates and travels through it would be lower than the index of refraction of the first medium
Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
Yes it is possible. Spectrum of emitted light depends upon the chemical composition of the source. and the way of its excitation. a clear example to us is that of sun.
11m if you add 6+5 you get 11 but of course you need the “m” in the mix so 11m but correct me if I’m wrong.
Answer:
v = 2.928 10³ m / s
Explanation:
For this exercise we use Newton's second law where the force is the gravitational pull force
F = ma
a = F / m
Acceleration is
a = dv / dt
a = dv / dr dr / dt
a = dv / dr v
v dv = a dr
We substitute
v dv = a dr
∫ v dv = 1 / m G m M ∫ 1 / r² dr
We integrate
½ v² = G M (-1 / r)
We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon
v² = 2G M (- 1 / R +2.73 10³+ 1 / R)
We calculate
v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61 - 10⁻³ /(5.61 + 2.73))
v² = 14.6828 10⁷ (0.1783 -0.1199)
v = √8.5748 10⁶
v = 2.928 10³ m / s
