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RoseWind [281]
2 years ago
15

Three noise sources produce volume (loudness) levels of 70, 73, and 80 dB when acting separately. When the sources act together,

the resultant intensity is the sum of the individual intensities. Find the sound volume level in decibels when the three sources act at the same time.
Physics
1 answer:
Mashutka [201]2 years ago
8 0

Sound at 70 dB is 70 dB louder than the human reference level.  That's 10⁷ times as much as the reference sound power.

Sound at 73 dB is 73 dB louder than the human reference level.  That's 10⁷.³  or  2 x 10⁷  times as much as the reference sound power.

Sound at 80 dB is 80 dB louder than the human reference level.  That's 10⁸  or 10 x 10⁷ times as much as the reference sound power.

Now we can adumup:

Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)

Intensity = (13 x 10⁷) times the sound power reference intensity.

Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)

Intensity = 70 + 10 log(13)

Intensity = 70 + 10 (1.114)

Intensity = 70 + 11.14

Intensity = <em>81.14 dB</em>

<em>______________________________________</em>

Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct.  For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.

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Match the correct center and scale factor to each figure.
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2 years ago
A snowball that will be used to build a snowman is at the top of a only hill. If the
Darina [25.2K]

Answer:

h = 24.11 m

Explanation:

Given that,

The potential energy of the snowball is 520 J

The mass of the snowball is 2.2 kg

We need to find the height of the hill. The potential energy of an object is given by the formula as follows :

E=mgh

g is acceleration due to gravity

h is height of the hill

h=\dfrac{E}{mg}\\\\h=\dfrac{520}{2.2\times 9.8}\\\\h=24.11\ m

So, the height of the hill is 24.11 m.

5 0
3 years ago
Give two examples of energy being transformed from one type to another
GaryK [48]

Answer:

Explanation:

For example, an ice cube has heat energy and so does a glass of lemonade. If you put the ice in the lemonade, the lemonade (which is warmer) will transfer some of its heat energy to the ice.

8 0
3 years ago
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wave
Firlakuza [10]

Answer:

thinnest soap film is  206.76 nm

Explanation:

Given data

wavelength = 550 nm

index of refraction n  = 1.33

to find out

What is the thinnest soap film

solution

we have wavelength  λ = 550 nm

that is  λ = 550 × 10^{-9} m

and n = 1.3

we will find the thickness of soap film as given by formula that is

thickness = λ/2n

thickness = 550 × 10^{-9}  / 2(1.33)

thickness = 206.76 × 10^{-9}  m

thinnest soap film is  206.76 nm

7 0
2 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
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