PLZZ ANSWER THE QUESTION

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

Gala2k [10]

Answer:

The intensity level in the room is 63 dB

Explanation:

To calculate the intensity of sound in the room, we use the equation of definition of decibels

β = 10 log (I / Io) (1)

With “I” the sound intensity and “Io” the threshold intensity 1.0 10⁻⁻¹² W/m²

To calculate the intensity we will use the initial data and remember the power of the emitted sound is constant, in addition that the sound propagates in three-dimensional form or on a spherical surface

I = P/A ⇒ P = I A

The area of a sphere is 4 π r², where I can calculate of 1

β/10 = log (I/Io)

I / Io = ${10}^{\beta /10}$

I = Io ${10}^{\beta /10}$

I = 1 10⁻¹² 10⁽¹⁰⁰/¹⁰⁾ = 1 10⁻¹² 10¹⁰

I = 1.0 10⁻² W

With this we can calculate the intensity for a distance of 20 m

I = 1.0 10⁻² / ( 4π 20²)

I = 2.0 10⁻⁶ W/m²

We have already found the intensity at the point of interest, so we can calculate the intensity in decibels at this point with equation 1

β = 10 log(2.0 10⁻⁶ / 1.0 10⁻¹²)

β = 10 log ( 2 10⁶) = 10 6.3

β = 63 dB

The intensity level in the room is 63 dB

5 0

4 years ago

Two charged objects have a repulsive force of 0.050 N. If the charge of both of the objects is doubled, then what is the new for

grandymaker [24]

Answer: 4 times grease

Explanation: Force F= C · q1·q2/r². C = Coulomb's constant.

If charges double you have 2q1 and 2q2.

7 0

3 years ago

Which of the following is NOT a result of supernova explosions? The neutron core is completely destroyed. Any planets within a f

DanielleElmas [232]

Answer:

The neutron core is completely destroyed

Explanation:

A earth - supernova is an explosion resulting to the death of a star that occurs close enough to the earth but this does not completely destroy a star. Supernovae are the most violent explosions in the universe. But they do not explode like a bomb explodes, blowing away every bit of the original bomb. Rather, when a star explodes into a supernova, its core survives. The reason for this is that the explosion is caused by a gravitational rebound effect and not by a chemical reaction. Stars are so large that the gravitational forces holding them together are strong enough to keep the nuclear reactions from blowing them apart. It is the gravitational rebound that blows apart a star in a supernova.

4 0

3 years ago

What affects a material’s resistance?

Keith_Richards [23]

Many things can affect a material's resistance, The type of material, how the material is being held (If its laying flat, being pulled, etc). What the material is used for, and how much material there is. Hope this helps!

8 0

3 years ago

Read 2 more answers

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

3 years ago

PLZZ ANSWER THE QUESTION ​

PLZZ ANSWER THE QUESTION