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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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ludmilkaskok [199]

Answer:

No

Explanation:

Not all metals stick to magnets. Like aluminum. if you were to stick a magnet on to an aluminum it would fall off.

6 0
3 years ago
If 36 grams of water is to be heated from 24.0°C to 48°C to make a cup of tea, how much heat must be added? The specific heat of
Vinvika [58]

We will have the following:

\begin{gathered} Q=mc\Delta T\Rightarrow Q=(36)(4.18)(48-24) \\  \\ \Rightarrow Q=3611.52 \end{gathered}

So, the heat to add is 3611.52 Joules.

3 0
1 year ago
calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s​
zheka24 [161]

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

5 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
Two identical charged pith balls are brought together to touch each other. They are then
sergejj [24]

Answer:

-17.5 nC

Explanation:

charge A = -30 nC

charge B = -5 nC

After adding them it would be the average of the two charges because of the getting same voltage difference. so

c = (-30+(-5)) / 2 nC

c= -17.5 nC

answer is -17.5 nC

4 0
3 years ago
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