They both make a thing go faster and slower but the relationship is force.
In this question force is measured in g cm/s2 so we know that to get the answer we times g by cm/s2
50 × 20 = 1000
•THAT THE PROPAGATION OF SOUND WAVES NEED MEDIUM TO TRAVEL
•THE MEDIUM SHOULD POSSES ELASTICITY
•FOR THE FASTER PROPAGATION OF SOUND THE PARTICLES SHOULD BE VERY CLOSE TO EACH OTHER
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
Answer:
a)θ=71.89°
b)NO
Explanation:
Given that
For glass n= 1.38
We know that for air n'=1
The angle for total internal reflection θc given as
sin θc=n'/n
By putting the values
sin θc=n'/n
sin θc=1/1.38
θc=46.43°
n'sinθ = n sinθref
sinθref = cosθc
n'sinθ = n cosθc
1 x sinθ =1.38 x cos 46.43°
θ=71.89°
b)
NO
