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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choli [55]

Formula for orbital speed, v = √(GM/R)

Where G is the universal gravitational constant, M = Central Mass,

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Given. v = 68 m/s, M = ? , R = 410 km = 410000 m., G = 6.674 * 10⁻¹¹ Nm²/kg²

68 = √(GM/R)

68 = √(6.674 * 10⁻¹¹ * M/410000)

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(68² * 410000) / 6.674 * 10⁻¹¹ = M

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3 years ago

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What is the strength of the electric field ep 0.90 mm from a proton?

dem82 [27]

The electric field strength is inversely related to the square of the distance.so the strength of the electric field is given by

$E=\frac{1}{4\pi \epsilon _{0} } \frac{q}{r^{2} } = \frac{k q}{r^{2} }$

Here, $\frac{1}{4\pi \epsilon _{0} } = k$ is constant depend upon medium and its value is $9.0 \times10^{9} \ N m^2/C^2$ and q is charge and r is the distance.

Given $r = 0.90 mm = 9.0 \times 10^{-4} m$ and we know the charge of proton, $q = 1.6\times 10^{-19} \ C$ .

Therefore,

$E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\ E= 1.77 \times 10^{-3} N/C$