Answer:
Explanation:
The path length difference = extra distance traveled
The destructive interference condition is:

where m =0,1, 2,3........
So, ←
![\Delta d = (m+1/2)\lamb da9/tex]so [tex]\Delta d = \frac{\lambda}{2}](https://tex.z-dn.net/?f=%5CDelta%20d%20%3D%20%28m%2B1%2F2%29%5Clamb%20da9%2Ftex%5D%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3Eso%20%3C%2Fstrong%3E%5Btex%5D%5CDelta%20d%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%7D)
⇒ λ = 2Δd = 2×10 = 20
Answer:

Explanation:
The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

Here, r is the radius and v is the tangential speed, which is given by:

Here
is the angular velocity, we replace (2) in (1):

Recall that
.
Solving for
:

Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º