From the solution that I have done, the wavelength in the question that we have is 31.88 cm
<h3>How to solve for the wavelength</h3>
The frequency in the question is given as 40/30 = 1.33 hz
Next we have to solve for V
= 425/10
= 42.5 cm/s
v = frequency * wavelength
we have to put in the values in the formula. This would be
42.5 = 1.33 x wavelength
we have to divide through by 1.33 to get the wavelength. This would be
42.5/1.333 = wavelength
31.88 cm = wavelength
Hence we can say that the wavelength in the question that we have here is 31.88 cm
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Answer:
t = 1.77 s
Explanation:
The equation of a traveling wave is
y = A sin [2π (x /λ -t /T)]
where A is the oscillation amplitude, λ the wavelength and T the period
the speed of the wave is constant and is given by
v = λ f
Where the frequency and period are related
f = 1 / T
we substitute
v = λ / T
let's develop the initial equation
y = A sin [(2π / λ) x - (2π / T) t +Ф]
where Ф is a phase constant given by the initial conditions
the equation given in the problem is
y = 5.26 sin (1.65 x - 4.64 t + 1.33)
if we compare the terms of the two equations
2π /λ = 1.65
λ = 2π / 1.65
λ = 3.81 m
2π / T = 4.64
T = 2π / 4.64
T = 1.35 s
we seek the speed of the wave
v = 3.81 / 1.35
v = 2.82 m / s
Since this speed is constant, we use the uniformly moving ratios
v = d / t
t = d / v
t = 5 / 2.82
t = 1.77 s
Answer:
d = 3.5*10^4 m
Explanation:
In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:
(1)
where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

hence, the displacement of the airplane is 3.45*10^4 m
Answer:
0.2 m
Explanation:
PHASE 1
First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

where u = initial velocity = 0 m/s
a = acceleration = 
t = time = 0.02 s
Therefore:

PHASE 2
Then, for the next 30 ms (0.03 secs), we use the formula:

This speed is the same as the final velocity of the tongue after the first 20 ms.
This can be obtained by using the formula:

Therefore:
distance = 5 * 0.03 = 0.15 m
Therefore, the total distance moved by the tongue in the 50 ms interval is:
0.05 + 0.15 = 0.2 m