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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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The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru
adoni [48]

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

(c) The time required to fill the mold is 1.35 s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

<u>V = 1.85 m/s</u>

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(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

<u>Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s</u>

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

<u>t = 1.35 s</u>

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3 years ago
A 50kg boy runs at a speed of 10.0m/s and jumpsonto a cart
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the mass of the cart is 150 kg

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given,

mass of boy(m) = 50 kg

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initial velocity of cart (u) = 0                    

final velocity of cart(V) = 2.5 m/s              

mass of the cart(M) = ?                              

m v + M u = (m + M ) V......................(1)

50× 10 + 0 = (50 + M ) 2.5

M =\dfrac{500}{2.5} - 50

M = 150 Kg                                          

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Heather and Matt take 34 minutes to walk eastward along a straight road to a store 2.0km away. What is their average velocity in
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The displacement is a vector connecting the initial point to the final point of motion: so, being displacement a vector, velocity is a vector as well, having the same direction of displacement.

Here, Heather and Matt walk 2.0 km eastward, so their displacement is

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While the time they took is

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v=\frac{2000}{2040}=0.98 m/s

And the direction is the same as the displacement (east).

Learn more about average velocity:

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