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2 years ago

What are infrasound usages

Physics

2 answers:

sattari [20]2 years ago

5 0

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

Explanation:

jok3333 [9.3K]2 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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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

2 years ago

A car is traveling at 70 km/h. It then uniformly decelerates to a complete stop in 12 s. Find its acceleration ( in m/s2 ).

Yakvenalex [24]

Answer:

Acceleration will be $-1.620m/sec^2$

Explanation:

We have given initial speed of the car is 70 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $70km/hr=70\times \frac{1000m}{3600sec}=19.444m/sec$

It is given that car stops in 12 sec

So final speed of the car v = 0 m/sec

Time t = 12 sec

From first equation of motion v = u+at

So $0=19.444+a\times 12$

$a=\frac{-19.444}{12}=-1.620m/sec^2$ ( negative sign indicates that speed of the car will constantly decrease )

5 0

3 years ago

Read 2 more answers

A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for

faltersainse [42]

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

5 0

2 years ago

During a 980. miles cruise, a ship travelled 590 miles in 7 hours first. Then the ship changed its speed for the rest of the tri

vfiekz [6]

Explanation:

The average speed of a modern cruise ship is roughly 20 knots (23 miles per hour), with maximum speeds reaching about 30 knots (34.5 miles per hour).

7 0

2 years ago

Read 2 more answers

An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the

Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m$

(a) Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2$

(b) Magnification, $m=\dfrac{h'}{h}$

h' is image height

$-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m$

Hence, this is the required solution.

4 0

2 years ago

What are infrasound usages​

What are infrasound usages