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xeze [42]
2 years ago
12

A sound wave from a police siren has an intensity of 100.0 W/m² at a certain point; a second sound wave from a nearby ambulance

has an intensity level that is 10 db greater than the police siren's sound wave at the same point. What is the sound level of the sound wave due to the ambulance?
Physics
1 answer:
Irina18 [472]2 years ago
3 0

The sound level of the sound wave due to the ambulance is 140.

<h3>What do you mean by sound?</h3>

In terms of physics, the sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave. Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Only acoustic waves with frequencies between roughly 20 Hz and 20 kHz, or the audio frequency range, can cause a human to have an auditory sensation. These are sound waves with wavelengths ranging from 17 meters (56 ft) to 1.7 millimeters in the air at atmospheric pressure (0.67 in). Ultrasounds are sound waves with a frequency higher than 20 kHz that are inaudible to humans. Infrasound refers to sound frequencies below 20 Hz. Animals of different species have different hearing ranges.

To learn more about sound, Visit:

brainly.com/question/9349349

#SPJ4

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The force of gravitation between two spherical bodies is Gm1 m2 /r2, where r is separation between their dash
AnnZ [28]

Explanation:

F = Gm1m2/r^2

kya nikalna hai bhai isme

6 0
3 years ago
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A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A
Sphinxa [80]

Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

7 0
3 years ago
One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

calculation:

Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

4 0
2 years ago
An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17
Softa [21]

Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

t_1 = 3.32 sec

t_2 = 5.08 sec

from equation of motion we know that

d_1 = vt_1 + \frac{1}{2} gt_1^2

where d_1 is distance covered in time t1

sod_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2=

d_1 = 110.78 m

d_2 = vt_2 + \frac{1}{2} gt_2^2

where d_2 is distance covered in time t2

d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2

d_2 = 213.31 m

distance cover is  = 213.31 - 110.78 = 102.53 m

3 0
3 years ago
What is the period of a wave that has a frequency of 300 hz?
Butoxors [25]

Answer:

T = 0.003 s

(Period is written as T)

Explanation:

Period = time it takes for one wave to pass (measured in seconds)

frequency = number of cycles that occur in 1 second

(measured in Hz / hertz / 1 second)

Period : T

frequency : f

So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period

T =  \frac{1}{f}    f = \frac{1}{T}

to find the period (T) of this wave, we need to plug in the frequency (f) of 300

T = \frac{1}{300}

T = 0.00333333333

So, the period of a wave that has a frequency of 300 Hz is 0.003 s

[the period/T of this wave is 0.003 s]

3 0
1 year ago
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