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
Learn more about the wavelength here:
brainly.com/question/10750459
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Answer:
5295.3 N
Explanation:
According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force
mv + P = mV
P = mV - mv = m(V - v)
Since the velocity of the ball before and after is in opposite direction, one of them is negative
P = 0.14(44.8 - (-19.5)) = 9 kg m/s
Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is
F = P/t = 9/0.0017 = 5295.3 N
