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3 years ago

10

Determine the length of a wind instrument, assuming that it is modeled as a closed tube and that the lowest sound that it can pl

ay has a frequency of 99 Hz. The speed of sound in air is 343m/s.

1 answer:

Alex787 [66]3 years ago

3 0

Answer:

Length, l = 0.866 meters

Explanation:

Given that,

Frequency of sound, f = 99 Hz

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

To find,

The length of a wind instrument.

Solution,

The standing wave will gets formed in wind instrument. For the closed tube, the closed tube the frequency is given by :

$f=\dfrac{v}{4l}$

Where

l is the length of the instrument

$l=\dfrac{v}{4f}$

$l=\dfrac{343\ m/s}{4\times 99\ Hz}$

l = 0.866 meters

So, the length of a wind instrument is 0.866 meters. Hence, this is the required solution.

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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

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$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

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$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

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We can use the following equation.

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Step 1: list known info

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