The problem is about magnitude of the displacement vector of the lady bug and its directions. so the magnitude of the displacement of the lady bug is 16 in / 2 because it started from the center so the magnitude is 8 in. and the direction is the rotation of the turn table which is 60 degrees
Answer:
15.7 m/s
Explanation:
The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:

where
u = 0 is the initial velocity
g = 9.8 m/s^2 is the acceleration
t is the time
If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:

Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m
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