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ohaa [14]
2 years ago
13

Two identical loudspeakers are driven in phase by a common oscillator at 750 Hz and face each other at a distance of 1.24 m. Loc

ate the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected. (Take the speed of sound in air to be 343 m/s. Choose one speaker as the origin and give your answers in order of increasing distance from this speaker. Enter 'none' in all unused answer boxes.)
Physics
1 answer:
Juli2301 [7.4K]2 years ago
6 0

Answer:

0.2286  m,  0.686 m and  1,143 m

therefore we see that there is respect even where the intensity is minimal

Explanation:

Destructive interference to the two speakers is described by the expression

       Δr = (2n +1) λ/2

where r is the distance,  λ the wavelength and n an integer indicating the order of the interference

let's locate the origin on the left speaker

let's find the wavelength with the equation

        v =  λ f

         λ = v / f

we substitute

      Δr = (2n + 1) v / 2f

let's calculate for difference values ​​of n

     Δr = (2n +1) 343/(2 750)

     Δr = (2n + 1) 0.2286

we locate the different values ​​for a minimum of interim

    n     Δr (m)

     0    0.2286

     1     0.686

     2    1,143

therefore we see that there is respect even where the intensity is minimal

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Given that;

Frequency; f = 30Hz

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using the expression for the relations between wavelength, frequency and speed of wave:

\lambda = \frac{v}{f}

Where \lambda is wavelength, f is frequency and v is speed.

We substitute our given values into the equation

\lambda = \frac{60m/s}{30Hz}\\\\\lambda = \frac{60m/s}{30s^{-1}}\\\\\lambda = 2m

The wavelength is 2m.

Hence, Option c) 2m is the correct answer.

To learn more about wavelength, click here: brainly.com/question/1347107

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3 0
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In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

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3 years ago
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