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Luba_88 [7]
3 years ago
11

Which statement is true about the loudness of sound?

Physics
2 answers:
weqwewe [10]3 years ago
5 0
Just like the gravitational force, the electrostatic force, the perceived
brightness of a light, and the strength of a radio station . . .

The loudness of a sound is inversely proportional to the square of
the distance between the observer and the source.
kogti [31]3 years ago
5 0

Answer : The correct option is, (B)

Explanation :

There is a direct relation between the loudness of sound and the intensity of sound.

L=10\log \frac{I}{I_o}         ..........(1)

where,

L = loudness of sound

I = intensity of sound

I_o = intensity of reference

From this we conclude that the loudness is directly proportional to the intensity of the sound.

There is an indirect relation between the intensity of sound and the distance of the listener from the source.

\frac{I_1}{I_2}=(\frac{d_2}{d_1})^2          ...................(2)

From this we conclude that the intensity of sound is inversely proportional to the square of the distance of the listener from the source.

From equation (1) and (2) we conclude that the loudness of sound is inversely proportional to the square of the distance of the listener from the source.

Hence, the correct option is, (B)

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If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k​
loris [4]

Recall the definition of the cross product with respect to the unit vectors:

i × i = j × j = k × k = 0

i × j = k

j × k = i

k × i = j

and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)

Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have

a × b = (8i + j - 2k) × (5i - 3j + k)

a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)

… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)

… … … … + 8 (i × k) + (j × k) - 2 (k × k)

a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)

a × b = - 5k - 10j - 24k - 6i - 8j + i

a × b = -5i - 18j - 29k

7 0
3 years ago
Can you predict the speed of a car as it moves down the track?
Elan Coil [88]

Answer I will say no

Explanation:

3 0
3 years ago
Read 2 more answers
What is the relative energy expense of galloping rather than trotting at 3.5 m/s?
labwork [276]

Answer: Trotting uses only 75 percent of the energy as galloping

Explanation: Trotting is only 300 J/m, whereas galloping is roughly 400 J/m

7 0
3 years ago
The index of refraction for red light in water is 1.331 and that for blue light is 1.340. A ray of white light enters the water
Alex Ar [27]

Answer:

(a) 47.08°

(b) 47.50°

Explanation:

Angle of incidence  = 78.9°

<u>For blue light : </u>

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for blue light which is 1.340

n₁ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.340}

{sin\theta_2}=0.7323

Angle of refraction for blue light = sin⁻¹ 0.7323 = 47.08°.

<u>For red light : </u>

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for red light which is 1.331

n₁ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.331}

{sin\theta_2}=0.7373

Angle of refraction for red light = sin⁻¹ 0.7373 = 47.50°.

5 0
4 years ago
A 600-nm-wavelength beam of light is incident on a soap film with air on both sides. The film has an index of refraction of 1.33
sergiy2304 [10]

Answer:

the film's minimum thickness to make the reflected light enhanced in brightness is 112.8 nm

Explanation:

Given the data in the question;

wavelength λ = 600 nm  

index of refraction = 1.33

so the film's minimum thickness to make the reflected light enhanced in brightness = ?

condition for maxima for reflected light is;

2nt = [ ( 2m + 1 )λ ] / 2

we solve for t

4nt = [ ( 2m + 1 )λ ]

t =  [ ( 2m + 1 )λ ] / nt

Now, for minimum thickness m = 1

t = λ / 4n

so we substitute in our values;

t = 600 nm / ( 4 × 1.33 )

t = 600 nm / 5.32

t = 112.78195 ≈ 112.8 nm  { one decimal place }

Therefore, the film's minimum thickness to make the reflected light enhanced in brightness is 112.8 nm

4 0
3 years ago
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