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Mila [183]
3 years ago
10

If the amplitude of a sound wave is made 3.0 times greater, by what factor will the intensity increase? Express your answer usin

g two significant figures. I/I0I / I 0 = nothing Request Answer Part B If the amplitude of a sound wave is made 3.0 times greater, by how many dB will the sound level increase? ββ = nothing dB Request Answer Provide Feedback
Physics
1 answer:
vesna_86 [32]3 years ago
4 0

Answer:

Calculating Sound Intensity Levels: Sound Waves

Calculate the sound intensity level in decibels for a sound wave traveling in air at 0ºC

and having a pressure amplitude of 0.656 Pa.

Explanation:

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PLEASE HELP ASAPPPPPPPPP
Svetlanka [38]
60kg would be the answer
8 0
3 years ago
A 2.0-mm-long, 1.0-mmmm-diameter wire has a variable resistivity given by rho(x)=(2.5×10−6)[1+(x1.0m)2]Ωmrho(x)=(2.5×10−6)[1+(x1
Tresset [83]

Answer:

1.144 A

Explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R = \rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as = \frac{\rho dx}{A}

dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}

Taking the integral of both sides;we have:

\int\limits^R_0  dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx

R = 3.185 [x + \frac {x^3}{3}}]^2__0

R = 3.185 [2 + \frac {2^3}{3}}]

R = 14.863 Ω

Since V = IR

I = \frac{V}{R}

I = \frac{17}{14.863}

I = 1.144 A

∴  the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A

8 0
3 years ago
A circuit has a voltage drop of 24.0 V across a 30.0 resistor that carries a
beks73 [17]

(A) 19.2 W

<u>Explanation:</u>

Given-

Voltage drop, V = 24 V

Resistor = 30Ω

Current, I = 0.8 A

Power, P = ?

We know,

P = VI

P = 24 (0.8)

P = 19.2 W

Therefore, the power conducted by the resistor is 19.2 W

6 0
3 years ago
Read 2 more answers
A motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a spe
Firdavs [7]

To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.

It can be described as

f = \frac{c\pm v_r}{c\pm v_s}f_0

c = Propagation speed of waves in the medium

v_r= Speed of the receiver relative to the medium

v_s= Speed of the source relative to the medium

f_0 =Frequency emited by the source

The sign depends on whether the receiver or the source approach or move away from each other.

Our values are given by,

v_s = 32.2m/s \rightarrow Velocity of car

v_r = 14.8 m/s \rightarrow velocity of motor

c = 343m/s \rightarrow Velocity of sound

f_0 = 523Hz \rightarrowFrequency emited by the source

Replacing we have that

f = \frac{c + v_r}{c - v_s}f_0

f = \frac{343 + 14.8}{343 - 32}(523)

f = 601.7Hz

Therefore the frequency that hear the motorcyclist is 601.7Hz

8 0
3 years ago
Your chances of getting into a collision when talking on a cell phone _________: A. Double B. Triple C. Quadruple D. Remain the
Anit [1.1K]

Answer:

C. Quadruple

Explanation:

¨Drivers who are talking on the phone, even on a hands-free device, are up to four times more likely to be involved in a crash.¨

I hope this helps! Have a great day!

3 0
2 years ago
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