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:
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The answer for this question is b because it says how far it goes before he begins to take brake
SI will always be in metric, so the answer is D. Meter.
Phonograph. Hope that helps
Answer:
T = 0.017s
Explanation:
period is the time it takes a particle to make one oscillation
An electric current is periodic in nature
The current reaches 3.8A ten times.
So there must have been 10 cycles (10 periods) in 0.17s. let 'T' be the period:
t is the total time interval
n is the number of oscillations
10T = 0.17
T = 0.17/10 = 0.017s
