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Lelechka [254]
3 years ago
5

A small source of sound waves emits uniformly in all directions. The total power output of the source is P. By what factor must

P increase if the sound intensity level at a distance of 20.0 m from the source is to increase 5.00 dB?
Physics
1 answer:
Paladinen [302]3 years ago
4 0

Answer:

3.16

Explanation:

We know that sound level in dB = 10log(I'/I) where I = initial intensity at 20.0m from source = P/A where P = total power and A = area at 20.0 m and I' = New intensity when the sound level increases by 5.00 dB at 20.0 m from source = P'/A where P' = Power at 5.00 dB and A = area at 20.0 m

dB = 5.00 dB. So,

5.00 dB = 10log(P'/A ÷ P/A)

5.00 dB = 10log(P'/P)

5/10 = log(P'/P)

0.5 = log(P'/P)

taking anti-logarithm of both sides,

10⁰°⁵ = P'/P

P'/P = 3.16

P' = 3.16P

So, P must be increased by a factor of 3.16 for the sound intensity level at a distance of 20.0 m from the source is to increase 5.00 dB.

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Complete question:

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.

Answer:

The electric field inside this metal resistor is 3125 V/m

Explanation:

Given;

length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m

diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m

the potential difference due to electric field between the two ends of the resistor, V = 10 V

The electric field inside this metal resistor is given by;

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E = 10 / (3.2 x 10⁻³ )

E = 3125 V/m

Therefore, the electric field inside this metal resistor is 3125 V/m

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