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2 years ago

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Two separate speakers each produce 524 Hz notes (the C above Middle C)in air on earth. The sounds would be perfectly in phase if

the speakers were placed next to each other. If both speakers are pointing in the same direction and directly at you, then how far apart would they have to be placed in order for you to experience total deconstructive wave interference

1 answer:

Serjik [45]2 years ago

6 0

Answer:

Explanation:

Frequency of sound = 524 Hz

wavelength of sound λ = velocity / frequency

= 340 / 524

λ = .6488 m

When sound from two sources reach a point simultaneously with a phase difference of π or path difference of λ/2 , or its odd multiple they overlap to give rise to destructive interference

so for destructive interference

path difference = ( 2n+1) λ/2

Here λ/2 = .6488 /2

= .3244 m

or 32.44 cm

So the loudspeaker must be placed with a gap of 32.44 cm or its odd multiple

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Answer:

the magnitude of the work done by the two blocks is the same.

Explanation:

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2 years ago

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

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Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
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2 years ago

A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi

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Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

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=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

$δ=1.18cm$

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2 years ago

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Answer:

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2 years ago

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Answer:

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Torque

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