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4vir4ik [10]
3 years ago
13

Water ripples are produced by making a cork oscillate up and down at a frequency of 5 Hz. The peaks of the ripples are 0.03 mete

rs apart. What is the speed of the water ripples?
Physics
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

The speed of water ripples, v = 0.15 m/s

Explanation:

It is given that,

Water ripples are produced by making a cork oscillate up and down at a frequency of 5 Hz, f = 5 Hz

The peaks of the ripples are 0.03 meters apart, \lambda=0.03\ m

We need to find the speed of the water ripples. The speed of water ripple is given by :

v=f\times \lambda

v=5\ Hz\times 0.03

v = 0.15 m/s

So, the speed of water ripples is 0.15 m/s. Hence, this is the required solution.

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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
The question is on the picture
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Answer:

<em> think 2 also if not im so sorry  but i think it is :)</em>

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What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?
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The kinetic energy K given to the helium nucleus is equal to its potential energy, which is 
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E=K=85~keV=q \Delta V
and from this we can find the potential difference:
\Delta V =  \frac{K}{q}= \frac{85~keV}{2e}=42.5~kV

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Answer:

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