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Helen [10]
3 years ago
11

An ambulance is traveling east at 61.9 m/s. Behind it there is a car traveling along the same direction at 28.5 m/s. The ambulan

ce driver hears his siren at a frequency of 694 Hz. 28.5 m/s Car 61.9 m/s Ambulance What is the wavelength of the sound of the ambulance’s siren if you are standing at the position of the car? The velocity of sound in air is 343 m/s . Answer in units of m.
Physics
1 answer:
inessss [21]3 years ago
7 0

Answer: 0.4 m

Explanation:

Given

Speed of ambulance, vs = 61.9 m/s

Speed of car = 28.5 m/s

Frequency of ambulance siren, f = 694 Hz

Velocity of sound in air, v = 343 m/s

With speed of ambulance being (61.9 m/s) -> We solve using

fd = f(v + vr) / (v - vs), where vr = 0

fd = 694 * (343 + 0) / (343 - 61.9)

fd = 694 * (343 / 281.1)

fd = 694 * 1.22

fd = 847 Hz

Recall,

λ = v/f

λ = 343/847

λ = 0.4 m

Therefore, the wavelength of the sound of the ambulance’s siren if you are standing at the position of the car is 0.4 m

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Yuliya22 [10]

Answer:

0.78m/s

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We know that

a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v

vdv=ads

\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}

By using formula:\int x^ndx=\frac{x^{n+1}}{n+1}+C

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1

Substitute s=2

\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)

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A jet taxiing down the runway receives word that it must return to the gate. The jet is traveling 37.6 m/s when the pilot receiv
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Answer:

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Explanation:

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V = U - at

Since the plane is going to stop, the final velocity V = zero.

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Substitute all the parameters into the formula

0 = 37.6 - 5.37a

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Make a the subject of formula

a = 37.6 / 5.37

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