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Zolol [24]
4 years ago
10

Two speakers emit the same sound wave, identical frequency, wavelength, and amplitude. What other quantity would be necessary to

determine if constructive or destructive interference occurs at a particular point some distance from the speakers?

Physics
1 answer:
Komok [63]4 years ago
6 0

Answer:

Phase Difference

Explanation:

When the sound waves have same wavelength, frequency and amplitude we just need the phase difference between them at a particular location to determine whether the waves are in constructive interference or destructive interference.

Interference is a phenomenon in which there is superposition of two coherent waves at a particular location in the medium of propagation.

When the waves are in constructive interference then we get a resultant wave of maximum amplitude and vice-versa in case of destructive interference.

  • For constructive interference the waves must have either no phase difference or a phase difference of nλ, where n is any natural number.
  • For destructive interference the waves must have a phase difference of n×0.5λ, where n is any odd number.

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A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric
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Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N

F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N

W_{F_x} = F_xd = 227.5\times 4.8 = 1092J

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

W_g = -mg\sin(30^\circ)d

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

W_g = -823J

The energy dissipated by the frictional force can be found as follows:

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N

W_F = Fd = 227.5\times 4.8 = 1092J

W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

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3 years ago
At a certain instant, a rotating turbine wheel of radius RR has angular speed ωω (measured in rad/srad/s). What must be the magn
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Answer:

Explanation:

The magnitude of the acceleration makes an angle of 30° with the tangential velocity.

Resolving the acceleration to tangential and radial acceleration

at = aCos30 = √3a/2

ar = aSin30 = ½a

a = 2•ar

Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:

at = Rα

Then, α = at/R

since at = √3a/2

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α = √3 at/2R

α = √3√(a²-ar²)/2R

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Also, a = 2•ar = 2w²R

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α = 3w²R / 2R

α = 3w²/2

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3 years ago
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