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poizon [28]
2 years ago
11

A source emits sound uniformly in all directions. There are no reflections of the sound. At a distance of 12 m from the source,

the intensity of the sound is 5.0 × 10−3 W/m2. What is the total sound power P emitted by the source?
Physics
1 answer:
Allushta [10]2 years ago
3 0

To solve this problem we will apply the concept related to Intensity, that is, the acoustic power transferred by a sound wave per unit of normal area to the direction of propagation:

I = \frac{P}{A} \rightarrow P= AI

Where,

P = Power

A = Acoustic Area

Our values are given as,

r= 12 m

I = 4.3*10^{-3} W/m^2

The Area then would be

A = 4\pi r^2

A = 4\pi (12)^2

A = 1809.55m^2

Replacing the values we have that

P = (4.3*10^{-3} W/m 2)(1809.55m^2 )

P = 7.781 W

The total sound power P emitted by the source is 7.781W

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2 years ago
wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg
Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg (M_J)

Speed of Jeremy is 3 m/s (V_J)

Speed of Jeremy after collision is (V_{JA}) -2.5 m/s

Mass of Hans is 140 kg (M_H)

Speed of Hans is -2 m/s (V_H)

Speed of Hans after collision is (V_{HA})

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is  

= =\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is  

= =\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}

= 120 × (-2.5) + 140 × V_{HA}

= -300 + 140 × V_{HA}

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × V_{HA}

80 + 300 = 140 × V_{HA}

380 = 140 × V_{HA}

380/140= V_{HA}

V_{HA} = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

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2 years ago
How much displacement will a spring with a constant of 120N / m achieve if it is stretched by a force of 60N?
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Answer:

Explanation:

There's a formula for this:

F = k*displacement

F being force, k being the spring constant, and displacement being the change in x

We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters

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