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3 years ago

An B flat is played on a french horn, whose frequency is 466Hz.

Physics

2 answers:

Snowcat [4.5K]3 years ago

6 0

Answer:

Explanation:

1. Period = 1/frequency = 1/466 = 0.002146s

2. Wavelength = Speed/frequency = 343/466 = 0.7361m

scoray [572]3 years ago

3 0

Answer:

Period of one vibration = 0.00215 second (Approx.)

Wavelength {Is speed of sound is 343 m/s] = 0.736 m (Approx.)

Explanation:

Given:

Frequency of wave = 466 Hz

Find:

Period of one vibration

Wavelength {Is speed of sound is 343 m/s]

Computation:

Period of one vibration = 1/F

Period of one vibration = 1 / 466

Period of one vibration = 0.00215 second (Approx.)

Wavelength = Velocity / Frequency

Wavelength {Is speed of sound is 343 m/s] = 343 / 466

Wavelength {Is speed of sound is 343 m/s] = 0.736 m (Approx.)

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A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.

charle [14.2K]

Answer: $0.00024\ m/s^2$

Explanation:

Given

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Annette [7]

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3 years ago

A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic

vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 $m^{2}$ </h3>

Explanation:

Given :

No. of turns in the first coil $N_{1} = 41$

No. of turns in the second coil $N_{2} = 123$

Area of first coil $A_{1} = 0.074 m^{2}$

According to the law of electromagnetic induction,

Induced emf = $-N \frac{d \phi}{dt}$

Where $\phi =$ magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

$-\frac{N_{1} d\phi _{1} }{dt_{1} } = -\frac{N_{2} d\phi _{2} }{dt_{2} }$

$\frac{N_{1} A_{1} dB_{1} }{dt_{1} } = \frac{N_{2} A_{2} dB_{2} }{dt_{2} }$

$A_{2} = \frac{N_{1} A_{1} }{N _{2} }$

$A_{2} = \frac{41 \times 0.074 }{123 }$

$A_{2} = 0.0246 = 0.025 m^{2}$

Therefore, the area of second coil is ≅ 0.025 $m^{2}$

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3 years ago