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lina2011 [118]
3 years ago
9

Dolphin echolocation is similar to ultrasound. Reflected sound waves

Physics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

Answer:

Waves with high frequencies have shorter wavelengths that work better  than low frequency waves for successful echolocation.

Explanation:

To understand why high-frequency waves work better  than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.

The relation between frequency and wavelength is given by

λ = c/f

Where λ is wavelength, c is the speed of light and f is the frequency.

Since the speed of light is constant, the wavelength and frequency are inversely related.

So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.

Thus, waves with high frequencies have shorter wavelengths that work better  than low frequency waves for successful echolocation.

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At which temperature does the molecules of an object stop moving?
PilotLPTM [1.2K]

Answer:

at the melting point and boiling point

Explanation:

the heat abdorbed is used to break down the forces holding the particles together to allow it to change its state

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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
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A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

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The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

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Here, r is the radius and v is the tangential speed, which is given by:

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Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

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How does the electric force vary with distance
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