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4 years ago

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A motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a spe

ed of 32.2 m > s. The motorcycle has a speed of 14.8 m > s. What frequency does the motorcyclist hear?

1 answer:

Firdavs [7]4 years ago

8 0

To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.

It can be described as

$f = \frac{c\pm v_r}{c\pm v_s}f_0$

c = Propagation speed of waves in the medium

$v_r$ = Speed of the receiver relative to the medium

$v_s$ = Speed of the source relative to the medium

$f_0 =$ Frequency emited by the source

The sign depends on whether the receiver or the source approach or move away from each other.

Our values are given by,

$v_s = 32.2m/s \rightarrow$ Velocity of car

$v_r = 14.8 m/s \rightarrow$ velocity of motor

$c = 343m/s \rightarrow$ Velocity of sound

$f_0 = 523Hz \rightarrow$ Frequency emited by the source

Replacing we have that

$f = \frac{c + v_r}{c - v_s}f_0$

$f = \frac{343 + 14.8}{343 - 32}(523)$

$f = 601.7Hz$

Therefore the frequency that hear the motorcyclist is 601.7Hz

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Answer:

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Answer

given,

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P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₂² v₂

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{2.25^2}{3.75^2} v_2$

$v_1= 0.36 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)$

$32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)$

$v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}$

$v_2=\sqrt{16.084}$

v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)² x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

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