I thinks its He uses proof to show the evidence is relevant. But im not totally positive on it hope this helps
Look at the title of the graph, in small print under it.
Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !
'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy' is true for transverse waves only.
'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy' is true for longitudinal waves only.
'Many wave motions in nature are a combination of longitudinal and transverse motion' is true for both longitudinal and transverse waves.
<u>Explanation:</u>
Longitudinal waves are those where the direction of propagation of particles are parallel to the medium' particles. While transverse waves propagate perpendicular to the medium' particles.
As wave motions are assumed to be of standing waves which comprises of particles moving parallel as well as perpendicular to the medium, most of the wave motions are composed of longitudinal and transverse motion.
So the option stating the medium' particle moves perpendicular to the direction of the energy flow is true for transverse waves. Similarly, the option stating the medium' particle moves parallel to the direction of flow of energy is true for longitudinal waves only.
And the option stating that wave motions comprises of combination of longitudinal and transverse motion is true for both of them.
Because the waves in the water with the fan like system.
Answer:
(a) Wavelength is 0.436 m
(b) Length is 0.872 m
(c) 11.518 m/s
Solution:
As per the question:
The eqn of the displacement is given by:
![y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]](https://tex.z-dn.net/?f=y%28x%2C%20t%29%20%3D%20%281.22%20cm%29sin%5B14.4%20m%5E%7B-%201%7Dx%5Dcos%5B%28166%5C%20rad%2Fs%29t%5D)
(1)
n = 4
Now,
We know the standard eqn is given by:
(2)
Now, on comparing eqn (1) and (2):
A = 1.22 cm
K = 
where
A = Amplitude
K = Propagation constant
= angular velocity
Now, to calculate the string's wavelength,
(a) 
where
K = propagation vector
(b) The length of the string is given by:
(c) Now, we first find the frequency of the wave:
Now,
Speed of the wave is given by:
