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3 years ago

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Residents of Hawaii are warned of the approach of a tsunami by sirens mounted on the tops of towers. Suppose a siren produces a

sound that has an intensity level of 120 dB at a distance of 2 m. Treating the siren as a point source of sound, and ignoring reflections and absorption,
a. What is the sound intensity level heard by an observer at a distance of 12 m

b. What is the sound intensity level heard by an observer at a distance of 21 m?

c. How far away can the siren be heard? (I = threshold of hearing)

1 answer:

svet-max [94.6K]3 years ago

5 0

Answer:

a) I₂ = 2.78 10⁻² W / m² , b) I₃ = 9 10⁻² W/m² , c) R₄ = 2 10⁶ m

Explanation:

The intensity of a wave is defined as the emission power per unit area

I = P / A

The unit of decibels is defined by

β = 10 log (I / Io)

With Io the hearing threshold 10⁻¹² W / m²

a) the intensity at r = 12m

Let's use the first equation

P = IA

I₁ A₁ = I₂ A₂

Let's look for intensity (I₁)

β / 10 = log I₁ / I₀

I / I₀ = ${10}^{\beta /10}$

I = I₀ ${10}^{\beta /10}$

I = 1 10⁻¹² ${10}^{120/10}$

I = 1 W / m²

The area of a spherical surface is

A = 4π R²

I₁ r₁² = I₂ r₂²

I₂ = I₁ R₁² / r₂²

I₂ = I₁ 2²/12²

I₂ = 1 2.78 10⁻²

I₂ = 2.78 10⁻² W / m²

b) r = 21 m

I₃ = I₁ r₁² / r₃²

I₃ = 1 2²/21²

I₃ = 9 10⁻² W/m²

c) in this case I₄ = I₀

I₁ r₁² = I₄ r₄²

R₄² = I₁ / I₄ r₁²

R₄² = 1 / 10⁻¹² 2²

R₄² = 4 10¹²

R₄ = 2 10⁶ m

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If something is 50% efficient, how many joules of wasted energy will there be if 750J of energy is put in?

evablogger [386]

Answer:

375J of wasted energy

Explanation:

Efficiency=work output/work input×100%

50=w.o/750×100

750×50=100w.o

w.o=750×50/100

w.o=75×5

w.o=375J

Wasted energy=750J-375J

=375J

Aliter

If the machine is 50% efficient then it is also 50% inefficient

50%×750J=375J

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3 years ago

Question 9 Unsaved A 680-Hz sound wave travels at 340 m/s in air, with a wavelength of 0.5 m. 5 m. 50 m. 500 m. none of the abov

ale4655 [162]

The correct option is 0.5 M

Calculation

Wavelength is defined as the ratio of velocity of a wave to its frequency. It is measure in meters. Mathematically, wavelength is given by the following formula:

Wavelength = wave velocity / frequency

From the details given in the question,

Wavelength =?

Velocity = 340 m/s

Frequency = 680 HZ

Wavelength = 340 /680 = 0.5

Therefore, wavelength = 0.5 M

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3 years ago

Read 2 more answers

An electron has a velocity of 1.50 km/s (in the positive x direction) and an acceleration of 2.00 ✕ 1012 m/s2 (in the positive z

olga nikolaevna [1]

Answer:

see explanation

Explanation:

Given that,

velocity of 1.50 km/s = 1.50 × 10³m/s

acceleration of 2.00 ✕ 1012 m/s2

electric field has a magnitude of strength of 18.0 N/C

$\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]$

$9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)]$ $42.2 \times 10^-^1^9 \hat k = -2.4 \times 10^1^6B_y \hat k + 2.4 \times 10 ^1^6 \hat j B_z\\$

$B_x = undetermined$

$B_y = \frac{42.2 \times 10^-^1^9}{-2.4 \times 10^-^1^6} \\\\= - 0.0176 T$

$B_z = 0T$

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4 years ago

A motorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. Calculate (a) the accelera

8090 [49]

Answer:

acceleration = -0.042 m/s²

velocity at beginning = 14.167 m/s

velocity at end = 5.7183 m/s

Explanation:

given data

distance d1 = 1 km

distance d2 = 2 km

time t1 = 80 s

time t2 = 120 s + 80s = 200 s

to find out

acceleration and velocity at beginning and end

solution

we apply here law of motion that is

d = vt + 1/2×at²

put value

1000 = v(80) + 1/2×a(80)² ........................1

and

2000 = v(200) + 1/2×a(200)² ........................2

so from equation 1 and 2 we get a and v

a = -0.042 m/s² and

v = 14.167 m/s

so by kinematic final velocity will be

V² = v² + 2ad

V² = (14.167)² + 2×(-0.042)×(2000)

V² = 32.70

V = 5.7183 m/s

so

acceleration = -0.042 m/s²

velocity at beginning = 14.167 m/s

velocity at end = 5.7183 m/s

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3 years ago

A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler

Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

$W=K_f - K_i$

Since the cart was initially at rest, $K_i = 0$ , so

$W=K_f = \frac{1}{2}mv^2$ (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, $F_g = 99.5 N$ :

$m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg$

So solving eq.(1) for v, we find the final speed of the cart:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s$

2) $2.51\cdot 10^7 J$

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

$F=4.28 \cdot 10^5 N$

$d=586 m$

So the work done is

$W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J$

3) $2.51\cdot 10^7 J$

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

$W=\Delta K = K_f - K_i$

where

W is the work done

$\Delta K$ is the change in kinetic energy

Therefore, the change in kinetic energy is

$\Delta K = W = 2.51\cdot 10^7 J$

4) 37.2 m/s

According to the work-energy theorem,

$W=\Delta K = K_f - K_i$

where

$K_f$ is the final kinetic energy of the train

$K_i = 0$ is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

$W=K_f = \frac{1}{2}mv^2$

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s$

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4 years ago