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Helga [31]
3 years ago
9

A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only i

f the sharks are separated by 3.60 meters, one being that much farther away than the other. (a) If the ultrasound has a frequency of 105 kHz, show this ability is not limited by its wavelength m < 3.50 m (b) If this ability is due to the dolphin's ability to detect the arrival times of echoes, what is the minimum time difference (in ms) the dolphin can perceive? ms
Physics
1 answer:
Vlad1618 [11]3 years ago
6 0

Answer:

a) Wavelength of the ultrasound wave = 0.0143 m <<< 3.5m, hence its ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = Period of oscillation = 0.00952 ms

Explanation:

The frequency of the ultrasound wave = 105 KHz = 105000 Hz. The speed of ultrasound waves in water ≈ 1500 m/s. Wavelength = ?

v = fλ

λ = v/f = 1500/105000 = 0.0143 m <<< 3.5m

This value, 0.0143m is way less than the 3.5m presented in the question, hence, this ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = The period of oscillation = 1/f = 1/105000 = 0.00000952s = 0.00952 ms

Hope this helps!

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Using Newton's second law of motion:

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Given:                      Find:                   Formula:                  Solve for m:

F: 2500N                 mass:?                F=ma Eq.1              m=F/a  Eq. 2

a= 200m/s^2  


Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

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3 years ago
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Explanation:

We'll need two equations.

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x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

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a = 0.0135 m/s²

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Answer:

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Explanation:

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V = 18h

(dV/dt) = 18 (dh/dt)

If (dV/dt) = 0.2 ft³/s

0.2 = 18 (dh/dt)

(dh/dt) = (0.2/18)

(dh/dt) = 0.0111 ft/s

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