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Anastaziya [24]
3 years ago
6

A person produces two sound waves with a flute one immediately after the other. Both sound waves have the same pitch, but the se

cond one is louder. Which of the following properties is greater for the second sound wave?
A) Frequency
B) Amplitude
C) Wavelength
D) Speed in air

Physics
1 answer:
stealth61 [152]3 years ago
4 0

Answer:

Option B is the right choice.

Explanation:

Given:

Two sounds waves lets say S_1 and S_2 having same pitch but

We have to find the property which from the options and identify which one is greater for S_2 .

Lets take one and one analysis of the terms.

a.

Frequency :

  • It is how fast the sound wave is oscillating.
  • Frequency is f=\frac{1}{T} hertz.
  • The faster the sound wave oscillates the higher pitch it will have.

According to the question the pitch is same so the frequency will be same for both the waves.

b.

Amplitude :

  • The loudness of the sound increases with an increase in the amplitude of sound waves.
  • It is the maximum amount of displacement of a particle on the medium from its rest position.

c.

Wavelength :

  • Distance between two consecutive crest (high) or trough (low) is called wavelength.
  • Shorter wavelength will have higher frequency.

Here the frequency is same so the wavelength for S_1,S_2 will be same.

d.

Speed in air:

  • Speed of sound in a same medium is usually same.
  • Speed of sound in air is 343 m/s.

So,

Amplitude of S_2 > S_1 .

Here the amplitude of the louder sound wave will be greater .

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A stable atom possesses full outside ring of electrons while unstable one does not. So, they are happy also because of stability.

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The weld bead in SMAW is formed by the?
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You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta
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Answer:

1)  3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes:    \frac{Distance}{Speed}= \frac{22}{6} =   3.66

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes:    Speed * Time = 34 * 3.66 = 124.44

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : s = u*t + \frac{1}{2}(a * t^2)

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1  = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.

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