Since it's a transverse wave, a particle on the string moves left and right as the wave passes by, but the particle doesn't travel forward or backward at all.
So the little red dot moves 'A' to the left, then 'A' back to the center, then 'A' to the right, then 'A' back to the center again.
All together, the red dot moves a total distance of <em>4A . (choice 'a')</em>
3.1 the only reason i know this is cause i got it wrong
Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :



v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.
Answer:
the wavelength λ of the light when it is traveling in air = 560 nm
the smallest thickness t of the air film = 140 nm
Explanation:
From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)
Now for constructive interference;
Δx= 
replacing ;
Δx = 2t ; we have:
2t = 
Given that thickness t = 700 nm
Then
2× 700 =
--- equation (1)
For thickness t = 980 nm that is next to constructive interference
2× 980 =
----- equation (2)
Equating the difference of equation (2) and equation (1); we have:'
λ = (2 × 980) - ( 2× 700 )
λ = 1960 - 1400
λ = 560 nm
Thus; the wavelength λ of the light when it is traveling in air = 560 nm
b)
For the smallest thickness 
∴ 



Thus, the smallest thickness t of the air film = 140 nm