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Zina [86]
3 years ago
13

Which feature of a soundwave is related to pitch

Physics
2 answers:
elena-s [515]3 years ago
6 0

Answer:

loudness

Explanation:

both loudness and pitch are determined by intensity of a sound wave, though pitch has another factor known as <em><u>f</u></em><em><u>r</u></em><em><u>e</u></em><em><u>q</u></em><em><u>u</u></em><em><u>e</u></em><em><u>n</u></em><em><u>c</u></em><em><u>y</u></em>

almond37 [142]3 years ago
3 0

Answer:

Frequency

Explanation:

As we know that the characteristics of sound is as following

Loudness : loudness of sound depends on the amplitude of sound so  increasing the loudness will cause the amplitude to increase.

Quality of sound : Quality of sound depends on the type of waveform. For different waveform the quality of sound is also different.

Pitch of sound : Pitch depends on the frequency of the sound which means if the frequency of sound is more then its frequency will be more

the correct answer will be

Frequency

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Name the substance in which starch becomes dark blue in colour.<br>​
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Any substance that contains starch turns blue-black in presence of <u>iodine solution.</u>

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2 years ago
A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp
maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance D, which is divided into two segments d=10 miles:

D=d+d=2d (1)

Where d=10mi \frac{1609.34 m}{1 mi}=16093.4 m

On the other hand, we know speed is defined as:

S=\frac{d}{t} (2)

Where t is the time, which can be isolated from (2):

t=\frac{d}{S} (3)

Now, for the first segment d=16093.4 m the car has a speed S_{1}=22m/s, using equation (3):

t_{1}=\frac{d}{S_{1}} (4)

t_{1}=\frac{16093.4 m}{22m/s} (5)

t_{1}=731.518 s (6) This is the time it takes to travel the first segment

For the second segment d=16093.4 m the car has a speed S_{1}=30m/s,  hence:

t_{2}=\frac{d}{S_{2}} (7)

t_{2}=\frac{16093.4 m}{30m/s} (8)

t_{2}=536.44 s (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed S_{ave}:

S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}} (10)

S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s} (11)

Finally:

S_{ave}=25.38 m/s

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