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3 years ago

A brand of earplugs reduces the sound intensity level by 27 dB.By what factor do these earplugs reduce the acoustic intensity?

Physics

1 answer:

Fofino [41]3 years ago

3 0

Answer:

Explanation:

27dB = 2.7 B

So I / I₀ = 10⁻²°⁷ , I₀ is intensity of main sound and I is intensity after reduction.

= 1.99 X 10⁻³

So intensity will reduce by 1.99 X 10⁻³ .

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Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

2 years ago

Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?A. Fluorescent lamps are more

mr_godi [17]

Answer:

B. Fluorescent lamps operate at a higher temperature than incandescent

Explanation:

Fluorescent lamps have a number of advantages over incandescent lamps which are given in the options given in A, C and D. The option available in B is a drawback, not an advantage. This is because it can give out and radiate more heat as a result of working at a higher temperature. Hence B option is correct.

8 0

3 years ago

What building materials do you believe would work well to build a home in that area? Explain why you chose these materials.

Luden [163]

Depends on what the area is. If it’s a rural place, Wood is cheep & easy to build. If there’s a lot of corrosion, strong weather/hurricane, bricks.

8 0

3 years ago

9. ¿Por qué no es lo mismo distancia que desplazamiento?

masya89 [10]

Answer:

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

Explanation:

5 0

3 years ago

Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is

kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

$Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}$

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

$Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa$

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

6 0

3 years ago