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2 years ago

Which tem decorbes the perceived frequency of a sound wave?

Physics

2 answers:

JulsSmile [24]2 years ago

5 0

Pitch

Explanation:

The pitch of a sound wave is the perceived frequency of a sound wave.

The quality of sound that makes it discernible to the ear is the pitch.

Sound wave is a longitudinal wave that is transmitted by series of rarefaction and compression.
Pitch helps to distinguish between the different sound qualities.
Pitch is how high or low sound is perceived.

learn more:

Pitch brainly.com/question/9772227

Sound wave brainly.com/question/2845448

#learnwithBrainly

alexandr1967 [171]2 years ago

4 0

Answer:

Like the other guy said, the answer is Pitch

Explanation:

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3 years ago

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

olga nikolaevna [1]

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Total number of oscillation per unit time is called frequency of oscillation. Here, $f=\dfrac{31}{15}=2.06\ Hz$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{15}{4\pi^2\times 2.06^2}$

m = 0.0895 kg

m = 89 g

(b) The maximum speed of the ball that is given by :

$v_{max}=A\times \omega$

$v_{max}=A\times 2\pi f$

$v_{max}=0.075\times 2\pi \times 2.06$

$v_{max}=0.970\ m/s$

$v_{max}=97\ cm/s$

Hence, this is the required solution.

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2 years ago

An archer pulls her bowstring back 0.396 m by exerting a force that increases uniformly from zero to 237 N. (a) What is the equi

tatyana61 [14]

Answer:

(a) The equivalent spring constant is 598.485 N/m

(b) The work done is 46.926 J

Explanation:

From Hooke's law of elasticity

K (spring constant) = F/e

F is the range of force exerted = 237 - 0 = 237 N

e is the extension of bowstring = 0.396 m

K = F/e = 237/0.396 = 598.485 N/m

Work done = 1/2 Fe = 1/2 × 237 × 0.396 = 46.926 J

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3 years ago