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Ulleksa [173]
3 years ago
14

Two identical speakers are emitting a constant tone that has a wavelength of 0.50 m. Speaker A is located to the left of speaker

B. At which of the following locations would complete destructive interference occur? a. 2.15 m from speaker A and 3.00 m from speaker B b. 2.50 m from speaker A and 1.00 m from speaker B c. 3.75 m from speaker A and 2.50 m from speaker B d. 1.35 m from speaker A and 3.75 m from speaker B e. 200 m from speaker A and 3.00 m from speaker B
Physics
1 answer:
Marianna [84]3 years ago
5 0

Answer:

a) and c).

Explanation:

For a complete destructive interference occur, it must be met the following condition relating the wavelength, and the difference in the paths taken by the sound emitted by the sources until arriving to the listening point:

d = |dA- dB| = (2n-1)*(λ/2)

For n= 1,  d = λ/2 = 0.25 m, it doesn't meet any of the cases.

For n=2, d= 3*(λ/2) = 0.75 m

In the case a) we have dA = 2.15 m and dB = 3.00 m, so dB-dA = 0.75 m, which means that in the location stated by case a) a complete destructive interference would occur.

For n=3, d= 5*(λ/2) = 5*0.25 m = 1.25 m.

This is just the case c) because we have dA = 3.75 m and dB = 2.50 m, so dA-dB = 1.25 m, which means that in the location stated by case c) a complete destructive interference would occur also.

The remaining cases don't meet the condition stated above, so the statements found to be true are a) and c),

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Answer:

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Explanation:

A

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Given that  u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

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=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

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