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3 years ago

Echolocation is based on the use of sound

Physics

2 answers:

shutvik [7]3 years ago

6 0

Answer: In echolocation, you use wave sounds to found objects:

A source emanates a wave of sound, and the sound travels in the medium, eventually rebounds on a wall or something, and the wave returns to the source. Then the source is listening the echoes of the sound ( this is the motive of the name)

Now, if some object is between the walls and the source, then when the wave reaches that object, a small part of the wave will rebound first than the rest of it, and the thing using echolocation will know that there is an object and the exact position of the object.

This is something that some animals use to move in dark places, for example the bats use echolocation when they do that little noises in caves.

Anni [7]3 years ago

5 0

Yes echolocation is based on the use of sound and knowing where the sound comes from without having to look for it a lot of soldiers on the battle field need to know how to use echolocation so they can basically not die

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An unknown material has a mass of 0.447 kg, and its temperature increases by 2.87°C when 943 J of heat are added. What is the sp

Sergeeva-Olga [200]

Answer:

735 J/kg/C

Explanation:

Q = mcT

943 = (0.447)( c )(2.87)

1.28289c = 943

c = 735 J/kg/C (3 sf)

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2 years ago

A mixture in which the parts can be seen separately from the other parts is considered to be what type of mixture

serious [3.7K]

Answer: A mixture is a combination of two or more substances (elements or compounds) which is not done chemically. Mixtures are two types Homogeneous Mixtures and Heterogeneous mixtures.

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2 years ago

A spy camera is said to be able to read the numbers on a car's license plate. If the numbers on the plate are 4.30 cm apart, and

Maslowich

Answer:

D = 2.38 m

Explanation:

This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit

a sin θ = m λ

Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle

θ = λ / a

Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant

θ = 1.22 λ / D

Where D is the circular tightness

Let's apply this equation to our case

D = 1.22 λ / θ

To calculate the angles let's use trigonometry

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ (4.30 10⁻² / 140 10³)

θ = tan⁻¹ (3.07 10⁻⁷)

θ = 3.07 10⁻⁷ rad

Let's calculate

D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷

D = 2.38 m

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2 years ago

Read 2 more answers

How does the size of a planet affected the amount and type of gas in its atmosphere

dalvyx [7]

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if the size of a planet is relatively small it will in turn be that of a smaller area which results in the less area to be covered for gas which basically means higher presence
I can go in depth more but I don't think that would be necessary all you need to know is this ...based on the size and gas will in turn be parallel to it's conformity

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2 years ago

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

$F=\frac{GMm}{r^2}$

Where G = 6.67 x 10⁻¹¹ N m²/kg²

M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

The asteroid is 4.11 x 10¹¹ m far from Sun

3 0

3 years ago