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icang [17]
2 years ago
14

True or false: When your ear detects sound, there is an actual collision of particles from the sound wave that hit your ear drum

Physics
2 answers:
balandron [24]2 years ago
7 0

Answer:

True...

Explanation:

When these high amplitude vibrations impinge upon the eardrum, they produce a very forceful displacement of the eardrum from its rest position. The ear's ability to do this allows us to perceive the pitch of sounds by detection of the wave's frequencies, the loudness of sound by detection of the wave's ...

When the sound waves hit your eardrum, they cause it to vibrate—the same way that a real drum vibrates when you hit it with a drumstick. The vibrations in your eardrum are then transferred via three tiny bones inside your ear into a fluid-filled chamber called the cochlea (pronounced KOK-lee-uh).

VashaNatasha [74]2 years ago
4 0

Answer:

True: For a sound wave to propagate, air particles must be displaced in the direction of propagation - it is the displacement of these particles that causes the eardrum to vibrate and recognize the sound.

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Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i
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Answer:

The final speed for the mass 2m is v_{2y}=-1,51\ v_{i} and the final speed for the mass 9m is v_{1f} =0,85\ v_{i}.

The angle at which the particle 9m is scattered is \theta = -66,68^{o} with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

\vec{p}=m\vec{v}

p_{xi} =p_{xf}

In the x axis before the collision we have

p_{xi}=9m\ v_{i} - 2m\ v_{i}

and after the collision we have that

p_{xf} =9m\ v_{1x}

In the y axis before the collision p_{yi} =0

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p_{yf} =9m\ v_{1y} - 2m\ v_{2y}

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p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}

then

p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}

<u>Conservation of kinetic energy:</u>

\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}

so

\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}

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v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}

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v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }

v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}

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As a vector the speed of the particle 2m is:

\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}

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