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kipiarov [429]
3 years ago
13

a female vocalist with a soprano voice can sing as high as 1000 hz. given that the speed of sound is 345 m/s what is the wavelen

gth if the sound waves
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0

0.345 m.

<h3>Explanation</h3>

The wavelength is the distance that the wave travels in each cycle. The wave travels 345 meters in each second. Let the wavelength of this wave be \lambda. That's the distance the wave travels in one cycle.

The frequency of the sound wave is 1 000 Hz, meaning that there are 1 000 cycles in each second. The wave travels a distance of 1 000 wavelengths in one second. That would be a distance of 1,000\;\lambda.

From the speed of the wave, the wave travels 345 meters in one second. In other words,

1,000\;\lambda = 345.

\lambda = \dfrac{345}{1,000} = 0.345\;\text{m}.

To generalize:

\lambda = \dfrac{v}{f},

where

  • \lambda wavelength of the wave,
  • v the speed of the wave, and
  • f the frequency of the wave.
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3 years ago
What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds?
slava [35]

Answer:

The tangential velocity of a rotating object is:

v = r*w

where r is the radius, and w is the angular velocity.

w = 2*pi*f

where f is the frequency.

We know that the record plater does 11 revolutions in 20 seconds, then it does:

11 rev/20s = 0.55 rev/s = f

then we have:

w = 2*pi*0.55 s^-1 = 2*3.14*0.55 s^-1 = 3.454 s^-1

The radius of a record player is really variable, it is around 10 inches, so i will use r = 10in, which is the rotating part of the record player.

then the tangential velocity is:

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7 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
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Answer:

v = 7.95 m/s

Explanation:

Given that,

Wavelength of a wave, \lambda=53\ cm=0.53\ m

Frequency of a wave, f = 15 Hz

We need to find the speed of the wave. The speed of a wave is given by :

v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s

So, the wave move with a speed of 7.95 m/s.

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What does the law of conservation of energy state?
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The answer is B. good luck :)
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