This is a two part question: What is the force of friction acting on a 100 kg steel slab at rest on a steel floor?
1 answer:
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Answer:
distance from speaker is 17.87 m
Explanation:
given data
distance from loudspeaker = 2.5 m
distance between loudspeaker = 3.0 m
room temperature = 20c
wavelength f = 686Hz
to find out
what distances from the speaker
solution
we know sound velocity c = 331.5 + 0.6 × 20c = 343.5
so wavelength of sound λ = c / f
wavelength = 343.5 / 686 = 0.5 m
when the difference in distance of speaker destructive interference will be
d = λ/2 × (2n-1)
for n = 1, 2 3 4 ..
d = 0.5/2 × (2n-1)
d = 0.250 , 0.75 , 1.25 , 1.750............ for n = 1, 2 3 .............
so
for d = 0.250
side of triangle by hypotenuse of triangle are
= 0.250
0.5 x1 = 7.6875
x1 = 15.375 m
for d = 0.75
side of triangle by hypotenuse of triangle are
= 0.75
1.5 x2 = 4.6875
x2 = 3.125 m
for d = 1.250
side of triangle by hypotenuse of triangle are
= 1.250
2.5 x2 = 1.1875
x3 = 0.475 m
for d = 1.750
x4 will be negative so we stop here
so the distance from speaker here is given below
distance = 2.5 + x
here x = 0.475 , 3.125 and 15.375 so
distance 1 = 2.5 + 0.475 = 2.975 m
distance 2 = 2.5 + 3.125 = 5.625 m
distance 3 = 2.5 + 15.375 = 17.875 m
final distance from speaker is 17.87 m