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natka813 [3]
2 years ago
7

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

eaker 2 is at the origin and speaker 1 is at x = 0.540 m . If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching another maximum when speaker 1 is at x = 0.870 m .
What is the frequency of the sound? Assume velocity of sound is 340m/s.
What is the phase difference between the speakers?
Physics
1 answer:
grandymaker [24]2 years ago
5 0

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

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avanturin [10]

Answer:

\dfrac{dh}{dt} =5\ ft/s

Explanation:

Let

h = height of balloon (in feet).

θ = angle made with line of sight and ground (in radians).

h = 300  tanθ

\dfrac{dh}{d\theta } = 300 sec^2\theta

now  \dfrac{dh}{dt} can be written as

\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}

\dfrac{d\theta }{dt} = \dfrac{1}{120}\at \ \theta =\dfrac{\pi}{4}

When θ = π/4,

\dfrac{dh}{d\theta } = 300 sec^2\theta

\dfrac{dh}{d\theta } = 600

\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}

\dfrac{dh}{dt} =600\times \dfrac{1}{120}

\dfrac{dh}{dt} =5\ ft/s

5 0
2 years ago
A ball with a mass of 0.5 kilograms is lifted to a height of 2.0 meters and dropped. It bounces back to a height of 1.8 meters.
Degger [83]
Hi, thank you for posting your question here at Brainly.

To compute for the change in potential energy, the equation would be:

delta PE =  mg*delta h
delta PE = 0.5*9.81*(2-1.8)
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The potential energy is converted to kinetic energy.
3 0
2 years ago
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A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
A girl pushes an 18.15 kg wagon with a force of 3.63 N. what is the acceleration?
Anton [14]

Divide the force given by mass and you will find the acceleration of the object :-

F = m × a

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3.63 = 18.15a

a = 3.63/18.15

a = 0.2 m/s^2

hope it helps!

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earnstyle [38]

Answer: The spring  of the spring is 25 N/m.

Explanation:

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2 years ago
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