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hammer [34]
3 years ago
15

What are infrasound usages​

Physics
2 answers:
sattari [20]3 years ago
5 0

Answer: it is a sound that you cannot hear but you can feel it

Explanation:

jok3333 [9.3K]3 years ago
5 0

Answer:

Usage of Infrasound are discussed below.

Explanation:

  • Infrasound waves are sound waves so low in frequency that even a human ear can not able to hear or detect them.

Usage of Infrasound

  • The uses of Infrasound waves are for Medical purposes.
  • The Infrasound waves are used in monitoring atmospheric activities.
  • The Infra sound waves are used in Forecasting natural disasters.

 

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
3 years ago
How do you know the earth is rotating on its axis
cupoosta [38]

Answer:

Read below!

Explanation:

You can watch the sun wheel across the sky during the day, and the stars at night. Focus a telescope on any star besides the north star--especially southern stars--and you can watch them drift across your field of view.  

An alternative explanation is that all the stars are painted on (or holes in) some canopy that rotates around the earth. This explanation does not account for the motion of the "wanderers," or planets, as the Greeks called them, or for the path of the moon among the stars.  

As we know the stars are massive bodies of significant and varying distance to the earth, the notion they all swing around us in unison seems highly implausible

4 0
3 years ago
Uranium-238 eventually decays into
GalinKa [24]
Uranium-238 decays<span> by alpha emission </span>into<span> thorium-234, which itself </span>decays<span> by beta emission to protactinium-234, which </span>decays<span> by beta emission to </span>uranium<span>-234, and so on. The various </span>decay<span> products, (sometimes referred to as “progeny” or “daughters”) form a series starting at </span>uranium-238<span>.</span>
4 0
3 years ago
Calculate the weight of a new fast-food sandwich that has a mass of 0.1 kg (approximately the mass of a quarter pound). Think of
iragen [17]

Weight = (mass) x (gravity)

If you plan to sell these things on Earth, then the acceleration of gravity in the neighborhood of your drive-throughs will be 9.81 m/s².

Weight of each sandwich = (0.1 kg) x (9.81 m/s²).

Weight of each sandwich = 0.981 Newton.

This is only  1.9%  less than 1 even Newton.

You should start by setting up one restaurant in New York, one in Chicago, one in LA, and maybe one in Miami or Tulsa.  Sell it with a different name in each place, and see which name sells best.

You might want to try calling it

-- Isaac's burger

-- Gravity grub

-- Prism Patty

-- Mass 'o Meat

-- Unit-wich

and see if anything catches on.

I think I'd simply call it a "Newton Unit".

5 0
3 years ago
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari
kolbaska11 [484]

Answer:

Final volumen first process V_{2} = 98,44 cm^{3}

Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

Replacing : T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}

V_{2} = 98,44 cm^{3}

Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

For Temperature the ideal gases formula is:

T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

P_{3} = 1,317 * 10^{10} Pa

7 0
3 years ago
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