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Anuta_ua [19.1K]
3 years ago
5

A tuning fork is used to produce sound waves with a frequency of 514 hertz. The waves travel through the air at 343 m/s. What is

the wavelength of the sound waves?
Physics
1 answer:
Vadim26 [7]3 years ago
5 0

Answer:

0.667 m

Explanation:

Applying,

v = λf................ Equation 1

Where v = velocity of sound, λ = wave length, f = frequency

make λ the subject of the equation

λ = v/f.............. Equation 2

From the question,

Given: v = 343 m/s, f = 514 hertz

Substitute these values into equation 2

λ = 343/514

λ = 0.667 m

Hence the wavelength of the sound wave is 0.667 m

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<h3><u>Answer;</u></h3>

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<h3><u>Explanation</u>;</h3>
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A 75.0 kg person is on a Ferris Wheel that has a radius R = 16 m. The tangential velocity of the Ferris Wheel is 8.25 m/s. Calcu
Dovator [93]

Answer:

The period of rotation is

T=8.025s

Explanation:

The person is undergoing simple harmonic motion on the wheel

Given data

mass of the person =75kg

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Velocity =8.25m/s

The oscillating period of simple harmonic motion is given as

T=(2*pi)/2=2*pi √r/g

Assuming that g=9.81m/s

Substituting our data into the expression we have

T=2*3.142 √ 16/9.81

T=6.284*1.277

T=8.025s

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How does Kepler's first law refine the Copernican model?
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Three charges are on a line. A positive charge is on the far left labeled q Subscript 1 baseline positive 6 Coulombs. The second
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Answer:

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A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista
slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

C' = \frac{1}{2}C

The potential difference across the capacitor is given by

V= \frac{Q}{C}

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

V' = 2 V

Now we can analyze the electric field between the plates of the capacitor, which is given by

E=\frac{V}{d}

we said that:

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- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

E'=\frac{2V}{2d}=\frac{V}{d}=E

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0
3 years ago
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