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luda_lava [24]
3 years ago
9

You are going to an outdoor concert, and you'll be standing near a speaker that emits 60 W of acoustic power as a spherical wave

. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?
Physics
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:

r=44m

Explanation:

β is calculated as:

\beta =(10dB)log_{10}(I/I_{o} )\\ I=I_{o}10^{\frac{\beta }{10dB} }\\ I=(1.0*10^{-12}W/m^{2}  )10^{\frac{\(94dB }{10dB} }\\I=2.51mW/m^{2}

The distance r is defined as the radius of spherical wave.solve for r

We have

I=\frac{P_{source} }{4\pi r^{2} }\\ r=\sqrt{\frac{P_{source}}{4\pi I} }\\ r=\sqrt{\frac{60W}{4\pi (2.51mW/m^{2} )} }\\r=44m

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May you help me answer this​
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1) See three Kepler laws below

2a) Acceleration is 2.2 m/s^2

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

  1. 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
  2. 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
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2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

M g sin \theta - T = Ma (1)

where

Mg sin \theta is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

g=9.8 m/s^2 (acceleration of gravity)

\theta=35^{\circ}

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

T-mg sin \theta = ma (2)

where

m = 3.5 kg (mass of the block)

\theta=35^{\circ}

From (2) we get

T=mg sin \theta + ma

Substituting into (1),

M g sin \theta - mg sin \theta - ma = Ma

Solving for a,

a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2

2b)

The tension in the string can be calculated using the equation

T=mg sin \theta + ma

where

m = 3.5 kg (mass of lighter block)

g=9.8 m/s^2

\theta=35^{\circ}

a=2.2 m/s^2 (acceleration found in part 2)

Substituting,

T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

U=mgh

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m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

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m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

K=\frac{1}{2}(0.5)(3)^2=2.25 J

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