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andrey2020 [161]
3 years ago
11

To understand the concept of intensity; the relationship between the power of the source and the intensity of the wave; and the

dependence of intensity on distance.Since waves transfer energy from one point to another, one can define the power of a wave as the rate at which the wave transports energy. The intensity of a wave, in contrast, is the power relative to a certain surface. Consider a wave traveling across a surface perpendicular to the direction of propagation. The intensity Iof the wave is defined as the ratio of the power P of the wave to the area A of that surface:I=PA.Note that the surface may be real (physical, like an eardrum or a windowpane) or mathematical. Quite frequently, we will be interested in the intensity produced by a relatively small source at a relatively large distance. If the source emits waves uniformly in all possible directions (produces spherical waves), the formula given here makes it possible to find the intensity at a distance r from the source:I=P/4?r^2.Note that, in all parts of this problem, assume that the source generates spherical waves, so that this intensity formula is applicable.Intensity is measured in watts per square meter (W/m^2). All the information presented here is pertinent to any kind of wave. In this problem, we will be focusing on sound waves.A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produce sound at their maximum power.A:Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.0 m.Express your answer numerically in watts per square meter. Use two significant figures.B:Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.5 m.Express your answer numerically in watts per square meter. Use two significant figures.C:Find the intensity I of the sound waves produced by four 60-Wspeakers as heard by the driver. Assume that the driver is located 1.0 m from each of the two front speakers and 1.5 m from each of the two rear speakers.Express your answer numerically in watts per meter squared.The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10^?12W/m^2. Find the distance d from the car at which the sound from the stereo can still be discerned. Assume that the windows are rolled down and that each speaker actually produces 0.06 W of sound, as suggested in the last follow-up comment.
Physics
1 answer:
frez [133]3 years ago
4 0

Answer:

a.  <em>4.77  </em>W/m^{2}

b. 2.122 W/m^{2}

c. <em>13.78</em> W/m^{2}

d. <em>69 km</em>

Explanation:

a.

power P of speaker = 60 W

distance d of speaker = 1.0 m

intensity I of speaker = P/4πd^{2}

I = 60/(4 x 3.142 x 1^{2}) = 60/12.568

Intensity = <em>4.77  </em>W/m^{2}

b.

at a distance of 1.5 m, intensity will be

I = P/4πd^{2}

I = 60/(4 x 3.142 x 1.5^{2}) = 60/28.278

Intensity = 2.122 W/m^{2}

c.

combined intensity of the two front speakers will be 2 x 4.77 = 9.54 W/m^{2}

combined intensity of the two back speaker will be 2 x 2.122 = 4.244 W/m^{2}

<em>total sound intensity perceived by the driver will be the superimposition of these four speakers.</em>

I = 9.54 + 4.244 =<em> 13.78</em> W/m^{2}

d.

Minimum discernible intensity of sound = 10^{-12} W/m^{2}

each speaker produces 0.06 W of power.

<em>We assume that each speaker spreads outward evenly.</em>

from<em> I = P/4π</em>d^{2}<em>,</em>

d = \sqrt{\frac{P}{4*\pi*I } }

d = \sqrt{\frac{0.06} {4*\pi*10^{-12 } } = 69094.35 m ≅ <em>69 km</em>

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Answer:

An acceleration of 5m/s^2 means that the velocity of a body is increasing by 5m/s per second in a certain direction

Explanation:

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As an aid in working this problem, consult Interactive Solution 3.41. A soccer player kicks the ball toward a goal that is 20.0
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Answer:

V=14.9 m/s

Explanation:

In order to solve this problem, we are going to use the formulas of parabolic motion.

The velocity X-component of the ball is given by:

Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s

The motion on the X axis is a constant velocity motion so:

t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s

The whole trajectory of the ball takes 1.48 seconds

We know that:

Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s

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6 0
3 years ago
650 N boy and a 419 N girl sit on a 150 N porch swing that is 2.0 m long. If the swing is supported by a chain at each end, what
Sergio039 [100]

The tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.

<h3>Principle of moments</h3>

The Principle of Moments states that when a body is in equip, the sum of clockwise moment about a point is equal to the sum of anticlockwise moment about the same point.

The formula for calculating moment is given below:

  • Moment = Force × perpendicular distance from the pivot

<h3>Calculating the tension in the chains</h3>

From the principle of moments:

Let tension in chain 1 be T1 and tension in chain 2 be T2.

T1 + T2 = 150 + 650 + 419

T1 + T2 =1219

Taking all distances from chain 1,

Sum of Moments = 0

419 × 0.5 + 150 × 0.85 + 650 × 0.9 = T2 × 1.7

T2 = 922/17

T2 = 542.35 N

Then, T1 = 1219 - 542.35

T1 = 676.65 N

Therefore, the tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.

Learn more about tension and moments at: brainly.com/question/187404

brainly.com/question/14303536

7 0
2 years ago
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0
2 years ago
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