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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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Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed
ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

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we know that

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distance = 0.624c × 1.59 × 10⁻⁷ s

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3 0
2 years ago
A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i
snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle=45^{\circ}

Acceleration due to gravity on earth is 9.8 m/s^2

Acceleration due to gravity on moon is \frac{9.8}{6}=1.63 m/s^2

Range of projectile is given by

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R_{earth}=\frac{u^2\sin 2\theta }{g}=5----1

R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}-----2

Divide 1 & 2

\frac{5}{R_{moon}}=\frac{1}{6}

R_{moon}=30 m

4 0
3 years ago
1 point
Alecsey [184]

Answer:

Explanation:

ℎ

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