The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.
<h3>What's the distance between consecutive nodes of the displacement of air molecules?</h3>
- Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.
- So, distance between consecutive nodes = wavelength = 2π÷k
= 2π/(4π÷m)
= m/2
<h3>What's the amplitude after 0.56s of the displacement of air molecules?</h3>
Displacement after 0.56 s = 0.008×cos(50π×0.56s)
=1.75×10^(-4) m
Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.
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Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.
Calculate:
I) the distance between 2 consecutive nodes
ii) the amplitude after 0.565s
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Explanation:
To find the answer use the equation speed of light=wavelength multiplied by frequency (c=lambda*f) by substituting the value for the frequency the the speed of light
The frictional force will be 0.22N.
<h3>What is Frictional force?</h3>
Frictional force is the force generated between two surfaces that are in contact and slide against each other.
Given,
Weight=4N
mass =4.9/9.8=0.5kg
Hieght =30m
velocity=24m/s
Acceration , v²-u²=2as
24²/2×30 =a , u is zero
a= 1.5m/s²
By Using conservation of energy ,
30F+1/2mv²=mgh
30F=1150-144
F= 6/30
F=0.2N
The force will be 0.2N
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