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Dimas [21]
3 years ago
8

What is the applications of echo​

Physics
2 answers:
Reil [10]3 years ago
8 0

English:

Applications of echoes

Echoes are used by bats, dolphins and fisherman to detect an object / obstruction. They are also used in SONAR (Sound navigation and ranging) and RADAR(Radio detection and ranging) to detect an obstacle.

Tagalog:

Mga aplikasyon ng dayandang

Ang mga dayandang ay ginagamit ng mga paniki, dolphin at mangingisda upang makita ang isang bagay / sagabal. Ginagamit din ang mga ito sa SONAR (Sound navigation and ranging) at RADAR(Radio detection and ranging) upang makita ang isang balakid.

Explanation:

Sana makatulog ❤️

shepuryov [24]3 years ago
5 0

\huge\tt{{\gray{ANSWER:}}}

When a sound wave is produced in air, some of it is absorbed after hitting with a reflector while some of the sound wave is reflected back to the listener. This reflected sound is called echo.

Applications of Echo:

  1. Since bats cannot see from their eyes, so they use the technique of echolocation to locate their ways. Bats can understand from the reflected sound if there is any object before it. It hunts its prey by using this technique.
  2. This technique is also used to find the depth of sea or distance of submarines
  3. It also helps to estimate the distance or hills and mountains.
  4. This technique is also very helpful in the medical field as well. Doctors used this phenomenon in cardiography, sonogram and many other medical diagnosis.
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2 years ago
A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an
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Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

c) 20mV

Explanation:

a) To find the average magnetic flux trough the inner solenoid you the following formula:

\Phi_B=BA=\mu_oNIA

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

M=\mu_o N_1 N_2 \frac{A_2}{l}

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

\epsilon_1=M\frac{dI_2}{dt}

by replacing you obtain:

\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV

the emf is 20mV

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