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inna [77]
4 years ago
5

You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in

phase. As you begin to walk directly away from the speaker, at what distances from the speaker do you hear a minimum sound intensity? The room temperature is 20c .
Physics
1 answer:
ahrayia [7]4 years ago
8 0

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

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1) = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2) = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3) = 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

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A particle of mass 4.5 × 10-8 kg and charge +5.4 μC is traveling due east. It enters perpendicularly a magnetic field whose magn
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0.00970 s

Explanation:

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Force due to magnetic field = qvB sin θ

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v = velocity of the charge

B = magnetic field strength = 2.7 T

θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1

F = qvB

Centripetal force responsible for circular motion = mv²/r = mvw

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The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

mvw = qvB

mw = qB

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w = 3.24 × 10² rad/s

w = 324 rad/s

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4 0
3 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

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