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Bond [772]
2 years ago
9

Consider a cylindrical pipe with both ends open and fundamental resonance frequency 150 Hz. If this particular pipe has a length

of L, what the wavelengths of three lowest modes produced by this pipe? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a. L,2L, 3L b. 2L, L, 2L/3 с. 2L/3, L, 2L d. L, L/2, L/3 e. 4L, 24, 4L/3 f. 4L, 4L/3, 4L/5 g. Neither of above.
Physics
1 answer:
Naya [18.7K]2 years ago
8 0

Answer:

answers, the correct one is b

Explanation:

For this problem we use the fact that since the ends are open at these points we have a maximum,

fundamental L =λ/2 λ = 2L

second harmonic L = λ λ = 2L / 2

third harmonic L = 3 λ/2 λ = 2L / 3

n harmonic lam = 2L / n

 with n an integer

therefore the length of the first three harmonics are: 2L, L, 2L / 3

when examining the different answers, the correct one is b

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In a parallel connection, the equivalent resistance is the summation of the inverse of each individual resistances. It is mathematically expressed as 1/ Req = 1/10 +1/20 + 1/25 = 5.263 ohms. Also, the voltage across each resistor is equal to the input voltage, therefore I = 100 / 10 = 10 Amps. I hope this helped you.
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ivolga24 [154]

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