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
Vesna [10]
3 years ago
6

The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that

four of your neighbors are all mowing their lawns using identical "Super-Silent" mowers. When they are each 20 m from your open bedroom window, what is the intensity level of the sound in your bedroom? You can neglect any absorption, reflection, or interference of the sound. The lowest detectable intensity is 1.0 × 10-12 W/m2. The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that four of your neighbors are all mowing their lawns using identical "Super-Silent" mowers. When they are each 20 m from your open bedroom window, what is the intensity level of the sound in your bedroom? You can neglect any absorption, reflection, or interference of the sound. The lowest detectable intensity is 1.0 × 10-12 W/m2. 40 dB 400 dB 104 dB 50 dB 80 dB
Physics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

The 80 dB option is correct.

Explanation:

The sound intensity level is converted to sound intensity with the formula

[D] = 10 log (I/I₀)

I₀ = 10⁻¹² W/m²

[D] = 100 dB

100 = 10 log (I/I₀)

Log (I/I₀) = 10

(I/I₀) = 10¹⁰

I = I₀ × 10¹⁰ = 10⁻¹² × 10¹⁰ = 10⁻² W/m²

But sound intensity varies inversely as the square of the distance of the sound.

I ∝ (1/d²)

I = k/d²

At d = 1 m away, I = 10⁻² W/m²

0.01 = k/1

k = 0.01 W

At d = 20 m, we solve for I

I = 0.01/20² = 2.5 × 10⁻⁵ W/m²

4 people mowing their lawns at the same time,

I = 4 × 2.5 × 10⁻⁵ = 10⁻⁴ W/m⁻²

The sound intensity level in decibels is given as

[D] = 10 log (I/I₀)

[D] = 10 log (10⁻⁴/10⁻¹²)

[D] = 10 log (10⁸)

[D] = 10 × 8 = 80dB

You might be interested in
What is a Vector and a scalar?
vfiekz [6]
Scalars are quantities that are fully described by a magnitude (or numerical value) alone.
Vectors are quantities that are fully described by both a magnitude and a direction.
4 0
3 years ago
A wave will go faster through a liquid at ____________ temperatures
cupoosta [38]
<h3><u>Answer;</u></h3>

Higher temperatures

A wave will go faster through a liquid at <em><u>highe</u></em><u>r </u>temperatures

<h3><u>Explanation;</u></h3>
  • <em><u>Mechanical waves are types of waves that require a material medium for transmission.</u></em> An example of mechanical wave is the sound wave whose transmission occurs in medium such as solids, liquids and gases.  
  • <em><u>The transmission of mechanical waves involves vibration of particles through the medium of transmission, thus transfer of energy from one point to another. </u></em>The vibration of particle may be in the form of a longitudinal wave or a transverse wave.  
  • <em><u>Increasing the temperature in a medium increases the kinetic energy of the particles in the medium and thus increasing the speed at which the particles vibrates and thus aiding a faster transmission of a wave.</u></em>
7 0
3 years ago
Read 2 more answers
A 100kg bag of sand has a weight on 100 N. When dropped it's acceleration is what?​
Vesnalui [34]

Answer:

9.8m/s

Explanation:

acceleration due to gravity is independent of mass

7 0
2 years ago
A 10,000-watt radio station transmits at 535 kHz. Determine the number of joules transmitted per second. 10,000 J/s 10 J/s 535 J
REY [17]
1 nanowatt  =  1 nanojoule/sec
1 watt  =  1 joule/sec
10 watts  =  10 joules/sec
100 watts  =  100 joules/sec
742.914 watts  =  742.914 joules/sec
1,000 watts  =  1,000 joules/sec
10,000 watts  =  10,000 joules/sec
100,000 watts  =  100,000 joules/sec
1 megawatt  =  1 megajoule/sec
1 gigawatt  =  1 gigajoule/sec
1 petawatt  =  1 petajoule/sec

We don't care what frequency the transmission is using,
or who their morning DJ is.

5 0
3 years ago
A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang
Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

3) E=mgh

Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

8 0
3 years ago
Other questions:
  • A 10.0 kilogram mass of iron on earth exerts a downward force of 98 Newtons. On the moon this same 10.0 kilograms of iron is wei
    9·2 answers
  • A bottle rocket takes off with a = 34.5 m/s2. It
    13·1 answer
  • What will be the current when the capacitor has acquired 1/4 of its maximum charge?
    9·1 answer
  • Why the sea does not get any fuller?
    8·2 answers
  • A rectangular certificate has a perimeter of 32 inches. Its area is 63 square inches. What are the dimensions of the certificate
    8·1 answer
  • Two objects, X and Y, are held at rest on a horizontal frictionless surface and a spring is compressed between them. The mass of
    8·1 answer
  • Drawing is a form of<br>​
    8·1 answer
  • In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.2. What is the
    9·1 answer
  • Please help me solve this! :)
    10·1 answer
  • How much work is done in holding a 20 N sack of potatoes while waiting in line at the
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!