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Akimi4 [234]
3 years ago
12

If all the strings on the guitar in the picture were plucked and allowed to vibrate freely, the string at the top would produce

the highest pitched sound. What could be concluded about the rate of vibration of the strings? The rates of vibration would vary over time for each string. There is no way to tell which string would be vibrating faster. The bottom string would be vibrating the fastest. The top string would be vibrating the fastest.
Physics
2 answers:
ivanzaharov [21]3 years ago
7 0

The top string would be vibrating the fastest

laiz [17]3 years ago
6 0

Answer: The top string would be vibrating the fastest.

Explanation: Here we do not have the picture, but we know that the string in the top would produce the highest pitched sound.

Now, the pitch of the sound is directly related to the frequency at which the string vibrates.

If the frequency is high, then the pitch is also high if the frequency is low, then the pitch is low.

now, let's analyze one of the sentences that are interesting.

"The rates of vibration would vary over time for each string."

As time passes, the intensity of the sound will decrease, not the frequencies.

Now, knowing that the top string is the highest-pitched one and that the high pitch is related with high frequencies, we can know that the top string is the one vibrating the fastest.

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Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

 gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second ,  change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

=  4.859 x 10⁻² oscillations .

5 0
3 years ago
What is it called when velocity changes over time
natka813 [3]
Well,

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3 0
3 years ago
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Jake calculates that the frequency of a wave is 230 hertz, and its wave is moving at 460 m/s. What is the wavelength of the wave
Nesterboy [21]

Wavelength = (speed) / (frequency) = (460 m/s) / (230/sec) = <em>2 meters</em>


3 0
3 years ago
A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?
Delvig [45]

Answer:

u =30 m/s

v = 35 m/s

t = 5 secs

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8 0
2 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

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b )

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= 1/2 m v²

= .5 x 45 x 5.8²

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c ) If  the average frictional force that acts on the sliding bear be F

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= F x 11 J

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F x 11 = 4851 J -   756.9 J

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F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
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