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andrew11 [14]
3 years ago
7

At a maximum level of loudness, the power output of a 75 piece orchestra radiated as sound is 70.0 w. What is the intensity of t

hese sound waves to a listener who is sitting 25.0 m from the orchestra?

Physics
1 answer:
Bogdan [553]3 years ago
6 0

Answer:

0.0089 W/m^2

Explanation:

The intensity of a sound wave is given by

I=\frac{P}{A}

where

P is the power of the wave at the source

A is the area over which the sound wave propagates

In this problem, we have

P = 70.0 W is the power

We want to know the intensity at a distance of r = 25.0 m from the orchestra, so since the sound waves propagates radially, we must consider a spherical surface of radius r, so the area is

A=4\pi r^2 = 4\pi (25.0 m)^2=7850 m^2

Therefore, the intensity of the wave is

I=\frac{70 W}{7850 m^2}=0.0089 W/m^2

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Answer:

\mu_s = 0.62

\mu_k = 0.415

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Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

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F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62

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