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sesenic [268]
3 years ago
7

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

ed by each speaker is 172 Hz. You are 8.00 m from speaker A. Take the speed of sound in air to be 344 m/s.What is the closest you can be to speaker B and be at a point of perfectly destructive interference?
Physics
2 answers:
Tamiku [17]3 years ago
8 0

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of  A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am  1 m away from B , the path difference will be

8 - 1 = 7 m  which is  odd multiple of 1 or λ /2

So path difference becomes odd multiple of  λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

Zolol [24]3 years ago
3 0

Answer:

1m.

Explanation:

Frequency = f = 172 Hz

Speed = V = 344 m/s

For the destructive interference,

Wavelength = λ = V/f = 344/172 = 2m

Path difference = d2 – d1 = n λ/2

Where “n” is an odd integer.

If you are 1m away from the loudspeaker B then the path difference will be,

Path difference = d2 - d1 = 8 – 1 = 7m  

Which is odd multiple of λ/2 and also condition for destructive interference. Hence, to hear the perfect destructive interference you have to stay 1m close to loudspeaker B.  

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For what, exactly? XD
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3 years ago
What cloud extending to the gravitational limits of the solar system would comets come from?.
ser-zykov [4K]

The Oort cloud extends to the gravitational limits of the solar system would comets come from.

<h3>What is oort cloud?</h3>

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4 0
2 years ago
slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa
xenn [34]

Answer:

Maximum height reached by the rocket is

y_{max} = 308 m

total time of the motion of rocket is given as

T = 16.44 s

Explanation:

Initial speed of the rocket is given as

v_i = 50 m/s

acceleration of the rocket is given as

a = 2 m/s^2

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

v_f^2 - v_i^2 = 2 a d

v_f^2 - 50^2 = 2(2)(150)

v_f = 55.68 m/s

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 55.68^2 = 2(-9.81)(y - 150)

y_{max} = 308 m

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

v_f - v_i = at

55.68 - 50 = 2 t

t_1 = 2.84 s

2) time to reach ground from this height

\Delta y = v_y t + \frac{1}{2}gt^2

-150 = 55.68 t - \frac{1}{2}(9.81) t^2

t_2 = 13.6 s

so total time of the motion of rocket is given as

T = 13.6 + 2.84 = 16.44 s

3 0
3 years ago
People often live on barrier island even though there not permanent why aren't they permanent
Arlecino [84]
They are dynamic, with winds and waves constantly reworking and moving the barrier island sand.

Changes in sea level also affect these islands.

Most scientists agree that sea level has been gradually rising over the last thousand years, and this rise could be accelerating today due to global warming.

Rising sea level causes existing islands to migrate shoreward.
8 0
3 years ago
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge
Wittaler [7]

Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

        F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

          r₁₂ = y₁ -y₂

          r₁₂ = 0.30 + 0.30

          r₁₂ = 0.60 m

let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

          Fx = -1.357 10-2 N

          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

6 0
3 years ago
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