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ozzi
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?
Physics
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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3 years ago
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When driving at night, only use your high-beam headlights___
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B

Explanation:

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3 years ago
A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl
aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0
3 years ago
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