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Radda [10]
3 years ago
8

Select the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz.

Physics
1 answer:
DedPeter [7]3 years ago
6 0

It is defined that a string that is in its two ends, its fundamental frequency and its subsequent frequencies are the product of the whole depending on the number of the harmonic. In other words, harmonics (higher) define the subsequent frequencies under the functions 2f, 3f, 4f, 5f, etc.

Therefore we have that the higher harmonics would be:

1 x 80 Hz = 80 Hz (1st harmonic and Fundalmental Frequency)

2 x 80Hz = 160Hz (2nd harmonic)

3 x 80Hz = 240Hz (3rd harmonic)

4 x 80Hz = 320Hz (4th harmonic)

5 x 80Hz = 400Hz (5th harmonic)

Hence, the frequencies 160Hz and 240Hz are the two higher harmonics of string with a fundamental frequency of 80Hz.

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A 782.10 kg car is brought from 7.60 m/s to 3.61 m/s over a time period of 4.23 seconds. What is the average force the car exper
alexandr402 [8]

Answer:

–735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 782.10 kg

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Force (F) =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Acceleration (a) =?

a = (v – u) / t

a = (3.61 – 7.60) / 4.23

a = –3.99 / 4.23

a = –0.94 m/s²

Finally, we shall determine the force experienced by the car as shown below:

Mass (m) of car = 782.10 kg

Acceleration (a) = –0.94 m/s²

Force (F) =?

F = ma

F = 782.10 × –0.94

F = –735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

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Answer:

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