Answer:
v_f = 24.3 m / s
Explanation:
A) In this exercise there is no friction so energy is conserved.
Starting point. On the roof of the building
Em₀ = K + U = ½ m v₀² + m g y₀
Final point. On the floor
Em_f = K = ½ m v_f²
Emo = Em_g
½ m v₀² + m g y₀ = ½ m v_f²
v_f² = v₀² + 2 g y₀
let's calculate
v_f = √(10² + 2 9.8 25)
v_f = 24.3 m / s
The kinetic energy would be 53,775J:)
1,4,6
a bow is drawn back
a gun is loaded w/ a dart
a bungee cord is stretched
Answer:
The film thickness is 4.32 * 10^-6 m
Explanation:
Here in this question, we are interested in calculating the thickness of the film.
Mathematically;
The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;
ΔN = (2L/λ) (n-1)
where λ is the wavelength of the light used
Let’s make L the subject of the formula
(λ * ΔN)/2(n-1) = L
From the question ΔN = 8 , λ = 540 nm, n = 1.5
Plugging these values, we have
L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m
