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julsineya [31]
3 years ago
5

Two automobiles are equipped with the same single-frequency horn. When one is at rest and the other is moving toward the first a

t 20 m/sm/s , the driver at rest hears a beat frequency of 5.2 HzHz .
Physics
2 answers:
rosijanka [135]3 years ago
7 0

Answer:

The frequency of the horn is 83.87 Hz

Explanation:

Given:

v = speed of the second horn = 20 m/s

vs = speed of the sound = 343 m/s

Δf = beat frequency = 5.2 Hz

The apparent frequency is equal to:

f_{ap} =f\frac{v_{s}-v_{o}  }{v_{s}-v } \\f_{ap}=f\frac{343-0}{343-20} \\f_{ap}=1.062f

The frequency of the horn is equal:

deltaf=f_{ap} -f\\deltaf=1.062f-f\\5.2=0.062f\\f=83.87Hz

prisoha [69]3 years ago
6 0

Explanation:

This is a good example of Doppler effect

Given

Source velocity Vs=20m/s

Apparent frequency Fa= 5.2 Hz

Recall speed of sound V=340m/s

The frequency of the horn Fh=?

But mathematically Fh=(V-Vs/V)*Fa

(340-20)/340*5.2=4.89Hz

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Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless.  I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
                                    Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
                                                                 </em>
That's about  <em>2.634 m/s²</em>  <em>

</em>
(I'm going to keep the fraction form handy, because the acceleration has to be
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-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

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That's the force that's accelerating the little block, so that must be the tension
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7 0
3 years ago
A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th
irina [24]

The angular momentum is defined as,

L=I\omega

Acording to this text we know for conservation of angular momentum that

L_i=L_f

Where L_iis initial momentum

L_f is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

L_i=mvl

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

L_f=\frac{13}{10}ml^2w

m(0.65)l=\frac{13}{10}ml^2 \omega

\omega=\frac{1}{2l}

Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

Replacing the values and solving

l=\frac{13}{0.54g}

Substituting

l=\frac{13}{0.54(9.8)}

l=2.45cm

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