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.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
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
#SPJ1
The kinetic energy of the tomato is :
K.E = 1/2 mv^2
K.E = 1/2 x 0.18 kg x 11 m/S^2
K.E = 0.99
Hope this helps
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x 
m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x 
s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x x 3 x )/2
=120m
Answer:33
Explanation:
F = frequency
N = Node count
w = wave lenght
v = wave velocity
L = distance wave traveled
First find wave length of laser
w = (2/(N))*(L)
w = (2/(10))*(8)
w = 1.6
then using (w), find velocity
V = (w)(F)
V = (1.6)*(108)
V = 288
Plug in V and the new frequency to solve for new node count
F = NV/2L
(600) = (N)*(288) / 2 * (8)
(N) = 33.33
there are 33 nodes
