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Firlakuza [10]
3 years ago
7

By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 2?

Physics
2 answers:
alina1380 [7]3 years ago
4 0
I believe that the answer to the question asked  above is the following

sound intensity = sound power / (4 pi R2<span>)
</span>
so if you decrease the intensity by a factor of 2 the sound wave will also decrease by a factor of 2.


Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
dolphi86 [110]3 years ago
4 0

The amplitude of a sound wave must be decreased by a factor of √2

\texttt{ }

<h3>Further explanation</h3>

Let's recall the speed of wave and intensity of wave formula as follows:

\large {\boxed {v = \lambda f}}

<em>f = frequency of wave ( Hz )</em>

<em>v = speed of wave ( m/s )</em>

<em>λ = wavelength ( m )</em>

\texttt{ }

\large {\boxed {I = 2 \pi^2 A^2 f^2 \rho v}}

<em>I = intensity of wave ( W/m² )</em>

<em>A = amplitude of wave ( m )</em>

<em>f = frequeny of wave ( Hz )</em>

<em>ρ = density of medium ( kg/m³ )</em>

<em>v = speed of wave ( m/s )</em>

<em>Let's tackle the problem!</em>

\texttt{ }

<u>Given:</u>

Amplitude of First Wave = A

Intensity of First Wave = I

Intensity of Second Wave = ¹/₂ I

<u>Asked:</u>

Amplitude of Second Wave = ?

<u>Solution:</u>

I : I' = 2 \pi^2 A^2 f^2 \rho v : 2 \pi^2 (A')^2 f^2 \rho v

I : I' = A^2 : (A')^2

I : \frac{1}{2}I = A^2 : (A')^2

2 : 1 = A^2 : (A')^2

2(A')^2 = A^2

(A')^2 = \frac{1}{2}A^2

A' = \frac{A}{\sqrt{2}}

\texttt{ }

<h3>Conclusion:</h3>

<em>The amplitude of a sound wave must be decreased by a factor of </em><em>√2</em>

\texttt{ }

<h3>Learn more</h3>
  • Doppler Effect : brainly.com/question/3841958
  • Example of Doppler Effect : brainly.com/question/810552
  • Sound Waves Cannot Travel In Space. : brainly.com/question/546436
  • Frequency of The Beats - Doppler Effect : brainly.com/question/12367463

\texttt{ }

<h3>Answer details</h3>

Grade: College

Subject: Physics

Chapter: Sound Waves

\texttt{ }

Keywords: Sound, Wave , Wavelength , Doppler , Effect , Policeman , Stationary , Frequency , Speed , Beats, Medium, Space

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He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How
Gwar [14]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

               d)  2.25 m

               e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 =  9.8t^2 / 2

hence d = 4.9t^2  ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t  ---- ( 2 )

equating equation 1 and 2

3 = 7.67t   therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec  = 1.92 m/sec

7 0
3 years ago
……………………………………………………………..?
kotegsom [21]

Answer:

C

Explanation:

First find the electrical wattage

W = I^2 * R

R = 12 ohms

I = 2 amps

Wattage = 2^2 * 12

Wattage = 4* 12

Wattage = 48 watts.

Now you need to use the power formula

Work = Power * Time

Work = ?

Power = 48 watts

Time = 3 minutes = 3 * 60 = 180 seconds.

Work = 48 * 180

Work = 8640 J

That's C

3 0
2 years ago
A 100kg crate slides along a floor with a starting velocity of 21ms If the force due to friction is 8N how long will it take for
GaryK [48]

A 100kg crate slides along a floor with a starting velocity of 21 m/s. If the force due to friction is 8N, then, it will take 262.5 s for the box to come to rest.

We'll begin by calculating the declaration of the box. This can be obtained as follow:

Force (F) = –8 N (opposition)

Mass (m) = 100 Kg

<h3>Deceleration (a) =? </h3>

<h3>F = ma</h3>

–8 = 100 × a

Divide both side by 1000

a = \frac{-8}{100}

<h3>a = –0.08 ms¯²</h3>

Therefore, the deceleration of the box is –0.08 ms¯²

Finally, we shall determine the time taken for the box to come to rest. This can be obtained as follow:

Deceleration (a) = –0.08 ms¯²

Initial velocity (u) = 21 ms¯¹

Final velocity (v) = 0 ms¯¹

<h3>Time (t) =.? </h3>

<h3>v = u + at</h3>

0 = 21 + (–0.08×t)

0 = 21 – 0.08t

Collect like terms

0 – 21 = –0.08t

–21 = –0.08t

Divide both side by –0.08

t = \frac{-21}{-0.08}\\\\

<h3>t = 262.5 s</h3>

Therefore, it will take 262.5 s for the box to come to rest.

Learn more: brainly.com/question/14446351

7 0
2 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
3 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
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