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2 years ago

What component of a longitudinal sound wave is analogous to a trough of a transverse wave?

Physics

1 answer:

anzhelika [568]2 years ago

6 0

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.

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3 years ago

someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee

cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) $Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \ in \ the \ journey}$

$Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average speed = 6.00 km/hour

$Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour$

Peter's average speed = 8.48 km/hour

4) $Average \ velocicty = \dfrac{Displacement }{Time \ taken}$

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

$Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average velocity = 6.00 km/hour

$Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Therefore, Peter's average velocity is also = 6.00 km/hour

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2 years ago

The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc

Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is $1.163 * 10^{-5} m$

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From the uncertainty Principle,

Δ $_{y}$ Δ $_{p}$ $= \frac{h}{2\pi }$

The momentum P $_{y}$ = (mass of electron)(speed of electron)

= $(9.109 * 10^{-31}kg)(995 * 10^{3} m/s)$

= $9.0638 * 10^{-25}kgm/s$

If the uncertainty is reduced to a 0.0010%, then momentum

= $9.068 * 10^{-30}kgm/s$

Thus the uncertainty in the position would be:

Δ $_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }$

Δ $_{y} \geq 1.163 * 10^{-5}m$

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3 years ago