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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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To solve this problem it is necessary to apply the equations related to the conservation of momentum.

This definition can be expressed as

$m_1u_1+m_2u_2 = (m_1+m_2)V_f$

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Rearranging the equation to find the final velocity we have,

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$V_f = \frac{(96)(6.3)+(0.9)(27.4)}{(96+0.9)}$

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3 years ago

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3 years ago

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

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Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

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Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

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Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

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