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Ksju [112]
3 years ago
14

A note is being played on a string on a guitar. The string is then pushed to the side while still in contact with the same fret.

What happens to the frequency of the note then played compared to the original note?
Physics
1 answer:
Brut [27]3 years ago
3 0

Answer:

See explanation

Explanation:

- When the string is pressed on a particular fret, the note is the same. That's because the string will sound from bridge to that fret-wire. It's always the same distance, so will always be the same note.

- Frets is that it splits the fingerboard into discrete diatonic parts, and well made fret-boards will mean the same note gets played on the same fret on the same string every time.

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A thin film of cryolite ( nc = 1.34 ) is applied to a camera lens ( ng = 1.58 ). The coating is designed to reflect wavelengths
rosijanka [135]

Answer:

=99.07nm

Explanation:

minimum thickness

2nd = (m  - 1/2)λ

d = (m - 1/2)(λ/2n)

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minimum thickness m = 1

light wavelength λ = 531nm

d = (1 - 1/2) (531 / (2)(1.34)

d = 531/5.36

  =  99.07nm

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Answer:

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Explanation:

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Based on the replacement reaction, what would the products of the reaction be?
cricket20 [7]

Answer:

\rm Be(OH)_2 and \rm (NH_4)_2 SO_4. The missing ion would be \rm OH^{-}.

Explanation:

In a double replacement reaction, two ionic compounds exchange their ions to produce two different ionic compounds.

In this question, the two ionic compounds are:

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In particular,

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In a binary ionic compound, cations (positive ions) can only bond to anions (negative ions.)

  • \rm Be^{2+} is a cation. In \rm BeSO_4, \rm Be^{2+} was bounded \rm {SO_4}^{2-} anions. During the reaction, it bonds with \rm OH^{-} anions to produce \rm Be(OH)_2.
  • \rm {NH_4}^{+} is also a cation. In \rm NH_4 OH, \rm {NH_4}^{+} was bounded to \rm OH^{-} ions. During the reaction, it bonds with \rm {SO_4}^{2-} anions to produce \rm (NH_4)_2 SO_4.

Hence, the two products will be \rm Be(OH)_2 and \rm (NH_4)_2 SO_4.

Note that charges on the ions must balance. For example, a \rm Be^{2+} ion carries twice as much charge as an \rm {NH_4}^{+} ion. As a result, each \rm Be^{2+} ion would bond with twice as many \rm OH^{-} ions as \rm {NH_4}^{+} would in \rm NH_4 OH.

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So when that force increases, it's an increase in pressure.
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