1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vilka [71]
2 years ago
10

A child sets off the firecracker at a distance of 100 m from the family house. what is the sound intensity β100 at the house?

Physics
2 answers:
KATRIN_1 [288]2 years ago
6 0

To solve this problem, we use the formula:

I100 / I1 = [P / 4π(100m)^2] / [P / 4π(1m)^2]

I100 / I1 = 1 / 100^2

I100 / I1 = 10^-4

 

Therefore the change in intensity from 1m to 100m in decibels is:

B100 – B1 = 10 log(10^-4) dB = -40 dB

 

So the intensity at 100m is calculated as:

B100 = B1 – 40 dB = 140 dB – 40 dB = 100 dB

 

Answer:

100 dB

Sonbull [250]2 years ago
3 0

The intensity of the sound produced due to the firecracker at the distance of 100\text{ m} is \fbox{\begin\\100\text{ dB}\end{minispace}}.

Further Explanation:

The vibration, produced in the medium, which travels as an wave of pressure or density through a medium is known as sound. The sound is a longitudinal wave and it requires a medium for its propagation.  

Given:

The distance between the place where firecracker is set off by a child and the family house is 100\text{ m}.  

The intensity of the sound produced by the firecracker at the distance 1\text{ m} is140\text{ dB}.

Concept:

The intensity of the sound wave is defined as the power carried by the sound waves in the direction perpendicular to the direction of propagation per unit time.

The intensity of the sound wave is:

\fbox{\begin\\I=\dfrac{P}{4\pi r^2}\end{minispace}}                                                        ...... (1)

Here, I is the intesnity of the sound wave, P is the power carried by the sound wave and r is the distance between source and the listener.

The intensity of the sound in decible at the house is:

\fbox{\begin\\\beta _{100}=\beta _{1}+10\log \dfrac{I_{100}}{I_{1}}\end{minispace}} ...... (2)

Here, \beta _{100} is the intensity of the sound at the distance of 100\text{ m}, \beta _{1} is the intensity of sound produced by the firecracker at the distance of 1\text{ m}, I_{100} is the intensity of sound produced by firecracker at 100\text{ m} and I_{1} is the intensity of sound produced by firecracker at 1\text{ m}.

The ratio of the intensity of sound at distance 100\text{ m} and 1\text{ m} is:

\fbox{\begin\\\dfrac{{{I_{100}}}}{{{I_1}}}=\frac{{r_1^2}}{{r_2^2}}\end{minispace}}                                                                                     …… (3)

Calculation:

Substitute the values in equation (3).

\begin{aligned}\frac{{{I_{100}}}}{{{I_1}}}&=\frac{{{{\left( 1 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\\&={10^{ - 4}}\\\end{aligned}

Substitute the values in equation (2).

\begin{aligned}{\beta _{100}}&=140\,{\text{dB}} - 40\,{\text{dB}} \\&=100\,{\text{dB}}\\\end{aligned}.

Thus, the intensity of the sound produced due to the firecracker at the distance of 100\text{ m} is \fbox{\begin\\100\text{ dB}\end{minispace}}

Learn more:

1.  The motion of a body under friction brainly.com/question/4033012

2.  A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Conservation of energy brainly.com/question/3943029

Answer Details:

Grade: College

Subject: Physics

Chapter: Waves and Oscillation

Keywords:

Intensity of sound, sound waves, vibration, firecracker, power carried by sound waves, 100db, 100 dB, 100dB, 100 m, bursting of firecrackers, sound inside the house, 140 db, 140 dB.

You might be interested in
a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density
Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³  

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

3 0
2 years ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
Calculate the average speed in metres per second from Glasgow to Edinburgh
mariarad [96]
This is the same question as the one previously but with more details, so I will just use my previous answer.

1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.

So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)
5 0
2 years ago
A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener
trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

U=(m)(9.8)(15)=147 m (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

U=K=\frac{1}{2}mv^2

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

147 m = \frac{1}{2}mv^2

And re-arranging for v, we find the velocity:

v=\sqrt{2\cdot 147}=17.1 m/s

5 0
3 years ago
What is the answer to this definition ?
Paul [167]

Newton's 2nd law of motion is:    <em>Force</em> = (mass) x (acceleration) .

Force is the only way to change an object's velocity.


6 0
3 years ago
Other questions:
  • What is another term means balance in physics?
    15·2 answers
  • A football is kicked from ground level with an initial velocity of 23.6 m/s at angle of 57.5° above the horizontal. How long, in
    12·1 answer
  • What is the wavelength of a wave with a frequency of 330 Hz and a speed of
    12·1 answer
  • Two point charges, with charge magnitudes q and ????, are placed a distance r apart. In this arrangement, each point charge expe
    6·1 answer
  • A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0
    11·1 answer
  • 3.
    8·1 answer
  • A glass rod is charged by the process of ________. *
    8·1 answer
  • At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 55 kg, 90 kg, and 42.5 kg respecti
    11·1 answer
  • Volleyball was invented in 1905.
    7·2 answers
  • Using the graph above determine the acceleration for the object represented between 6.0s and 8.0s
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!