I would say that it would be A.) Been insulted. This is because you are angered by someone else. It does not necessarily mean that something has de-escelated, you've lost control of your car, or you've nothing to lose. Hopefully that helps. :)
The relationship between the frequency and wavelength of a wave is given by the equation:
v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency.
If we divide the equation by f we get:
λ=v/f
From here we see that the wavelength and frequency are inversely proportional. So as the frequency increases the wavelength decreases.
So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.
A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

and substituting t = 75 seconds, we find

In degrees, it is

So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by

where we have

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find

Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
<span> Allied Forces. they became the allies.</span>