Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Answer:
Explanation:
Work is defined as the scalar product of force and distance
W=F•d
Given that
F = 8.5i + -8.5j. +×-=-
F=8.5i-8.5j
d = 2.5i + cj
If the work in the practice is zero, then W=0
therefore,
W=F•ds
0=F•ds
0=(8.5i -8.5j)•(2.5i + cj)
Note that
i.i=j.j=k.k=1
i.j=j.i=k.i=i.k=j.k=k.j=0
So applying this
0=(8.5i -8.5j)•(2.5i + cj)
0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)
0=21.25-8.5c
Therefore,
8.5c=21.25
c=21.25/8.5
c=2.5