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mote1985 [20]
3 years ago
8

Two piano strings produce beats. Frequencies of the strings are 1560 Hz. and 1563 Hz. Choose the correct frequency of the beats.

Physics
2 answers:
Naya [18.7K]3 years ago
4 0

Well first of all, I have to point out that the correct frequency of the beats is not on the list of choices that you provided.

When sources with different frequencies combine, two beats are created. The beat frequencies are the sum and difference of the frequencies that produce them.

If the source frequencies are 1,560 Hz and 1,563 Hz, then the beat frequencies are

3 Hz and 3,123 Hz.

The 3 Hz is perceived as a slight 'vibrato' in the piano's tone, so that it sounds warmer and not like a flat sine-wave generator.

The 3,123 Hz doesn't propagate well through the structure of the piano, and isn't noticeable to anyone except an expert listener (like a piano tuner). Only the top 6 keys [out of 88] on a standard piano keyboard have frequencies of 3,123 Hz or higher.

kakasveta [241]3 years ago
3 0

Beat frequency is given by the difference of two frequencies played together

f_{beat} = |f_1 - f_2|

given that

f_1 = 1560 Hz

f_2 = 1563 Hz

Now

f_{beat} = |1560- 1563| Hz

f_{beat} = 3 Hz

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2 years ago
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

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The time needed for the hammer to reach the surface of the Earth is 3.54 s.

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The time of motion is calculated as follows;

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