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zheka24 [161]
2 years ago
13

A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s

peed is the bat gaining on its prey? Take the speed of sound in air to be 340 m/s.
Physics
1 answer:
3241004551 [841]2 years ago
8 0

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer.  The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\  340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

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Answer:

a. Angular velocity = 0.267rad/s.

b. Centripetal acceleration = 56.25m/s.

Explanation:

<u>Given the following data;</u>

Mass, m = 8kg

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a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

Substituting into the equation, we have;

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Centripetal acceleration = 56.25m/s.

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Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, \rho=1.59\times 10^{-8}\ \Omega-m

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We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

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E=4\times 10^6\times 1.59\times 10^{-8}

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So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

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Answer:

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it's c

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