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2 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?

1 answer:

Paladinen [302]2 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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Explanation:

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3 years ago

A water line with an internal radius of 5.29 x 10-3 m is connected to a shower head that has 15 holes. The speed of the water in

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Answer:

(a) 3.44 x 10^-3 m^3/s

(b) 8.4 m/s

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number of holes, N = 15

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3 years ago

An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J.

AnnyKZ [126]

Answer:

a) $I=733.33\ A$

b) $d=52272.7273\ m$

c) $d'=51948.0519\ m$

Explanation:

Given:

voltage of the battery, $V=12\ V$

energy storage capacity of the battery, $E=2.3\times 10^7\ J$

speed of the car, $v=20\ m.s^{-1}$

a)

power drawn by the car, $P=8.8\ kW$

<u>Now the Current delivered to the motor:</u>

we the relation between the power and electrical current,

$P=V.I$

$8800=12\times I$

$I=733.33\ A$

b)

<u>Distance travelled before battery is out of juice:</u>

we first find the time before the battery runs out,

$t=\frac{E}{P}$

$t=\frac{2.3\times 10^7}{8800}$

$t=2613.636\ s$

Now the distance:

$d=v.t$

$d=20\times 2613.636$

$d=52272.7273\ m$

c)

When the head light of 55 W power is kept on while moving then the power consumption of the car is:

$P'=P+55$

$P'=8800+55$

$P'=8855\ W$

<u>Now the time of operation of the car:</u>

$t'=\frac{E}{P'}$

$t'=\frac{2.3\times 10^7}{8855}$

$t'=2597.4026\ s$

<u>Now the distance travelled:</u>

$d'=v.t'$

$d'=20\times 2597.4025$

$d'=51948.0519\ m$

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