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A jet engine emits a whine of frequency 3000 Hz. When the engine is moving directly away from the observer at half the speed of sound The value of abcd is __ __ __ __ observer can be expressed as a.bctimes 10^dHz (v_(sound)=340m/s) engine directly away 1. An ocean wave has a wavelength of 10.0 m. Two wave crests passes by a point near a dock every 9.005. Determine the/speed of the wave The tip of a helicopter blade was measured to be traveling at 130m/s. Each blade is measured to be 6.5m long from the center. What would the frequency of the blades be?

Question

A jet engine emits a whine of frequency 3000 Hz. When the engine is moving directly away from the observer at half the speed of sound The value of abcd is __ __ __ __ observer can be expressed as a.bctimes 10^dHz (v_(sound)=340m/s) engine directly away 1. An ocean wave has a wavelength of 10.0 m. Two wave crests passes by a point near a dock every 9.005. Determine the/speed of the wave The tip of a helicopter blade was measured to be traveling at 130m/s. Each blade is measured to be 6.5m long from the center. What would the frequency of the blades be?

A jet engine emits a whine of frequency 3000 Hz. When the engine is moving directly away from the observer at half the speed of sound
The value of abcd is
__ __ __ __
observer can be expressed as a.bctimes 10^dHz
(v_(sound)=340m/s)
engine
directly away
1. An ocean wave has a wavelength of 10.0 m. Two wave crests passes by a point near a
dock every 9.005. Determine the/speed of the wave
The tip of a helicopter blade was measured to be traveling at 130m/s. Each blade is measured to be
6.5m long from the center. What would the frequency of the blades be?

Solution

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ShermanMaster · Tutor for 5 years

Answer

Here's the solution:<br /><br />**1. Doppler Effect for the Jet Engine:**<br /><br />* **Given:**<br /> * Source frequency (f<sub>s</sub>) = 3000 Hz<br /> * Source velocity (v<sub>s</sub>) = 0.5 * speed of sound (v) = 0.5 * 340 m/s = 170 m/s (moving away)<br /> * Speed of sound (v) = 340 m/s<br /><br />* **Formula:** When the source is moving away from the observer, the observed frequency (f<sub>o</sub>) is given by:<br /><br /> f<sub>o</sub> = f<sub>s</sub> * (v / (v + v<sub>s</sub>))<br /><br />* **Calculation:**<br /><br /> f<sub>o</sub> = 3000 Hz * (340 m/s / (340 m/s + 170 m/s))<br /> f<sub>o</sub> = 3000 Hz * (340 / 510)<br /> f<sub>o</sub> = 3000 Hz * (2/3)<br /> f<sub>o</sub> = 2000 Hz<br /><br />* **Expressing in the required format:** f<sub>o</sub> = 2.00 x 10³ Hz<br /><br />* **Therefore:** a=2, b=0, c=0, d=3. The value of abcd is 2003.<br /><br />**2. Speed of the Ocean Wave:**<br /><br />* **Given:**<br /> * Wavelength (λ) = 10.0 m<br /> * Period (T) = 9.00 s (time for one full wavelength to pass)<br /><br />* **Formula:** Wave speed (v) = Wavelength (λ) / Period (T)<br /><br />* **Calculation:**<br /><br /> v = 10.0 m / 9.00 s<br /> v ≈ 1.11 m/s<br /><br />* **Answer:** The speed of the wave is approximately 1.11 m/s.<br /><br />**3. Frequency of the Helicopter Blades:**<br /><br />* **Given:**<br /> * Tangential speed of the blade tip (v) = 130 m/s<br /> * Length of the blade (radius, r) = 6.5 m<br /><br />* **Concept:** The tip of the helicopter blade travels in a circular path. The tangential speed is related to the rotational speed (angular frequency, ω) by: v = ωr<br /><br />* **Calculation:**<br /> * First, find the angular frequency (ω): ω = v / r = 130 m/s / 6.5 m ≈ 20 rad/s<br /> * Then, convert angular frequency (ω) to frequency (f): f = ω / 2π ≈ 20 rad/s / (2 * 3.14) ≈ 3.18 Hz<br /><br />* **Answer:** The frequency of the helicopter blades is approximately 3.18 Hz.<br />
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