The medium determines the speed of the wave traveling in it, which also can have a number of other effects, including how much the wave bends (refracts), whether it reflects, etc.
Because waves move through space, they must have a velocity. The velocity of a wave is a function of the type of wave, and the medium it travels through. Electromagnetic waves moving through a vacuum, for instance, travel at roughly 3 x
10
8
m/s. This value is so famous and common in physics it is given its own symbol, c.
Answer:
The displacement in t = 0,
y (0) = - 0.18 m
Explanation:
Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N
v = √ T / μ
v = √20.48 N / 0.02 kg /m = 32 m/s
λ = v / f
λ = 32 m/s / 40 Hz = 0.8
K = 2 π / λ
K = 2π / 0.8 = 7.854
φ = X * 360 / λ
φ = 0.5 * 360 / 0.8 = 225 °
Using the model of y' displacement
y (t) = A* sin ( w * t - φ )
When t = 0
y (0) = 0.25 m *sin ( w*(0) - 225 )
y (0) = 0.25 * -0.707
y (0) = - 0.18 m
Complete Question
The compete question is shown on the first uploaded question
Answer:
The speed is
Explanation:
From the question we are told that
The distance of separation is d = 4.00 m
The distance of the listener to the center between the speakers is I = 5.00 m
The change in the distance of the speaker is by 
The frequency of both speakers is 
Generally the distance of the listener to the first speaker is mathematically represented as
![L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}](https://tex.z-dn.net/?f=L_1%20%20%3D%20%20%5Csqrt%7Bl%5E2%20%2B%20%5B%5Cfrac%7Bd%7D%7B2%7D%20%5D%5E2%7D)
![L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}](https://tex.z-dn.net/?f=L_1%20%20%3D%20%20%5Csqrt%7B5%5E2%20%2B%20%5B%5Cfrac%7B4%7D%7B2%7D%20%5D%5E2%7D)

Generally the distance of the listener to second speaker at its new position is
![L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}](https://tex.z-dn.net/?f=L_2%20%20%3D%20%20%5Csqrt%7Bl%5E2%20%2B%20%5B%5Cfrac%7Bd%7D%7B2%7D%20%5D%5E2%20%2B%20k%7D)
![L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}](https://tex.z-dn.net/?f=L_2%20%20%3D%20%20%5Csqrt%7B5%5E2%20%2B%20%5B%5Cfrac%7B4%7D%7B2%7D%20%5D%5E2%20%2B%200.6%7D)
Generally the path difference between the speakers is mathematically represented as

Here
is the wavelength which is mathematically represented as

=> 
=>
=>
Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves
=>
=>
Does this help?
When an object is
immersed in a fluid (in this case water, but may include both liquids and
gases) the fluid exerts an upward force on the object which is called buoyancy
force or <span>up-thrust. Archimedes’ Principle states that the buoyant
force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by
that object. Thus when an object is
placed in water the rise in the water level is dictated by the mass of that
object.</span>
<span>
</span>
<span>So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).</span>