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3 years ago

6

You are walking along a small country road one foggy morning and come to an intersection. While you are crossing, you hear an am

bulance siren. As a paramedic, you know that the siren sounds at a frequency of 7.87 kHz. But, as an amateur musician with perfect pitch, you can tell that the frequency you are hearing is actually 8.61 kHz. Is the ambulance moving toward or away from you? The ambulance is moving toward you! At what speed is the ambulance traveling?_____ mph?

1 answer:

emmainna [20.7K]3 years ago

7 0

Answer:

64.945 miles per hour

Explanation:

Since the frequency of sound heard is higher than actual frequency, the ambulance is moving towards you!

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = 8.61 KHz

f₀ = real frequency of sound = 7.87 KHz

v₀ = velocity at which the sound source is moving towards the reference point = ?

v = velocity of sound waves = 343 m/s

8.61 = 7.87/(1 - (v₀/v))

1 - (v₀/343) = 0.9141

v₀/343 = 1 - 0.9141 = 0.0859

v₀ = 343 × 0.0859 = 29.48 m/s = 64.945 miles per hour

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a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

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$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

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Explanation:

a)

The electric force exerted on a charged particle is given by

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where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

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- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

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c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

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For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

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And by using the right-hand rule:

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- Middle finger: magnetic field (+y axis)

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