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3 years ago

Why does the output of a microphone increase as the frequency of the sound waves which it receives increases

Physics

1 answer:

Sloan [31]3 years ago

7 0

Answer:

See Explanation

Explanation:

The frequency of sound waves received by the microphone influences the output or pitch of the sound obtained from the microphone.

The higher the frequency of the sound received by the microphone, the higher the output of the microphone and vice versa. This is because, the higher the frequency of sound, the higher the oscillations produced and the greater the output of the microphone.

The rise and fall in the pitch of sound waves as the frequency of sound waves varies is called inflection.

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Lynn rubs a balloon with a piece of wool, which causes the balloon to pick up some of the electric charges from the wool. Lynn t

Hitman42 [59]

Answer:

electrostatics

gravitational forces

Explanation:

rubbing the ballon leads to it acquiring charges that are opposite to that in the wall.the electostatic forces attract the ballon to the wall and tyhe gravitational pull later acts on the ballon causing it to fall down

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3 years ago

Read 2 more answers

A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface

givi [52]

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

v(max)² = 0.018 / 0.5

v(max)² = 0.036

v(max) = √0.036

v(max) = 0.19 m/s

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

K = 1/4 * 0.02

K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J

4 0

2 years ago

Which of the following is quantitative data? a. Color c. Shape b. Odor d. Volume

Papessa [141]

Hello there! Quantitive data has to do with measurements that can be shown with numbers. Examples of this are things like your height and the length of your arms. With that alone, A and B are eliminated, because those answer choices make no sense. They can't be expressed by numbers and you can't measure colors or odors mathematically. Volume is a way to measure something that CAN be written down by numbers. D is the only answer choice that fits the definition of quantitive data. The answer is D: volume.

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3 years ago

Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius

Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

$a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2$

The centripetal acceleration for the second radius; 4.0 m

$a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2$

The centripetal acceleration for the third radius; 6.0 m

$a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2$

The centripetal acceleration for the fourth radius; 8.0 m

$a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2$

The centripetal acceleration for the fifth radius; 10.0 m

$a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2$

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3 years ago

Please help!!!!!!!!!

IgorC [24]

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3 years ago