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sladkih [1.3K]
3 years ago
6

PLEASE HELP!!

Physics
2 answers:
Norma-Jean [14]3 years ago
5 0
Sorry I need a point
g100num [7]3 years ago
3 0

Answer:

A

Explanation:

YAYAYAYAYAYA

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Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note
Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

f_n = n f_1

where

f_1 is the fundamental frequency

Therefore, the frequency of the third harmonic of the A (f_1 = 440 Hz) is

f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz

while the frequency of the second harmonic of the E (f_1 = 659 Hz) is

f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz

So the frequency difference is

\Delta f = 1320 Hz - 1318 Hz = 2 Hz

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

f_B = |f'-f|

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

f_B = |1320 Hz - 1318 Hz|=2 Hz

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

f_1 \propto \sqrt{T}

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

f_B = 4 Hz

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

f_B = |f_3-f_2|

and f_3 = 1320 Hz, we have two possible solutions for f_2:

f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0
3 years ago
A sound wave has a wavelength of 15 meters with a frequency of 2.5 Hz. What would the velocity be for this situation in
rodikova [14]

Answer:

6m/s

Explanation:

V = frequency * wavelength

15 * 2.5 = 6m/s

7 0
2 years ago
An object, with mass 32 kg and speed 26 m/s relative to an observer, explodes into two pieces, one 5 times as massive as the oth
Brut [27]

Answer:

\Delta K = 2164.053\,J

Explanation:

Let consider the observer as an inertial reference frame. The object is modelled after the Principle of Momentum Conservation:

(32\,kg)\cdot (26\,\frac{m}{s} ) = (5.333\,kg)\cdot (0\,\frac{m}{s} )+(26.665\,kg )\cdot v

The speed of the more massive piece is:

v = 31.202\,\frac{m}{s}

The kinetic energy added to the system is:

\Delta K = \frac{1}{2}\cdot [(5.333\,kg)\cdot (0\,\frac{m}{s} )^{2}+(26.665\,kg )\cdot (31.202\,\frac{m}{s} )^{2}]-\frac{1}{2}\cdot (32\,kg)\cdot (26\,\frac{m}{s} )^{2}

\Delta K = 2164.053\,J

3 0
3 years ago
CAN ANY OF YALL ANSWER DIS PLS!!!!!!!! I AM GIVING 20 POINTS BTW!!!!!!!
sveticcg [70]
PHYSICAL CHANGES :
Melting an ice cube.
Boiling water.
Mixing sand and water.
Breaking a glass.
CHEMICAL CHANGES :
Digesting food.
Cooking an egg.
Heating sugar to form caramel.
4 0
3 years ago
David is driving a steady 30.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady
SpyIntel [72]

Answer:

Explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

Distance travelled by Tina

s = ut + 1/2 a t²

= .5 x 2.10 t²

= 1.05 t²

Distance travelled by David

= 30 t ( because of uniform velocity )

1.05 t² = 30t

t = 28.57 s

Distance travelled by Tina

= 1/2 a t²

= .5 x 2.10 x 28.57²

= 857 m approx.

7 0
3 years ago
Read 2 more answers
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