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ivolga24 [154]
2 years ago
15

15. A pattern of weather and climate over three months is called a

Physics
1 answer:
Anna35 [415]2 years ago
6 0

Answer:

Explanation:

weather pattern

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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 light spring has a force constant of 70 N/m and is used to pull a 10 kg box on a horizontal frictionless surface. If the box h
Sergio039 [100]

Answer:

82.4 cm

Explanation:

∑F = ma

kx cos θ = ma

x = ma / (k cos θ)

x = (10 kg) (5 m/s²) / (70 N/m cos 30.0°)

x = 0.824 m

x = 82.4 cm

8 0
3 years ago
A charge of 0.14 C is moved from a position where the electric potential is 20 V to a position where the electric potential is 5
alukav5142 [94]

The change in electric potential energy is given by:

ΔU = ΔVq

ΔU = change in PE, ΔV = potential difference, q = charge

Given values:

ΔV = 50V - 20V = 30V, q = 0.14C

Plug in and solve for ΔU:

ΔU = 30(0.14)

ΔU = 4.2J

7 0
3 years ago
You want to build a snowman, so you accelerate a 2kg snowball across your yard at a rate of 0.5m/s2. Calculate the amount of for
tamaranim1 [39]

Answer:

4

Units:

Newtons

3 0
2 years ago
Factors that affect pressure in liquids
Alona [7]
Two factors influence the pressure of fluids. They are the depth of the fluid and its density.
7 0
3 years ago
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