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zheka24 [161]
3 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]3 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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I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

S = 1.23*10^9 W/m^2

What is the rms value of (a) the electric field and

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Answer:

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b) 2.27*10^3 T

Explanation:

To find the RMS value of the electric field, let's use the formula:

E_r_m_s = sqrt*(S / CE_o)

Where

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E_o = 8.85*10^-^1^2 C^2/N.m^2;

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Therefore

E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}

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b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

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