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4 years ago

The _(height or width) of a sound wave determines its _(loudness or pitch).

Physics

2 answers:

postnew [5]4 years ago

5 0

<h2>Answer: The height of a sound wave determines its loudness </h2>

Sound waves are longitudinal mechanical waves, that is, they depend on a medium to propagate.

Among the characteristics of a sound wave, it is the amplitude, that is the degree of movement of the molecules of the medium in which the wave propagates.

Depending on how high this amplitude is, the sound will be louder.

Vikki [24]4 years ago

3 0

Answer:

1. height

2. loudness

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A 26-cm-long wire with a linear density of 20 g/m passes across the open end of an 86-cm-long open-closed tube of air. If the wi

damaskus [11]

Answer: T = 472.71 N

Explanation: The wire vibrates thus making sound waves in the tube.

The frequency of sound wave on the string equals frequency of sound wave in the tube.

L= Length of wire = 26cm = 0.26m

u=linear density of wire = 20g/m = 0.02kg/m

Length of open close tube = 86cm = 0.86m

Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as

L = 3λ/4

0.86 = 3λ/4

3λ = 4 * 0.86

3λ = 3.44

λ = 3.44/3 = 1.15m.

Speed of sound in the tube = 340 m/s

Hence to get frequency of sound, we use the formulae below.

v = fλ

340 = f * 1.15

f = 340/ 1.15

f = 295.65Hz.

f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.

The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as

L = λ/2

0.26 = λ/2

λ = 0.52m

The speed of sound in string is given as v = fλ

Where λ = 0.52m f = 295.65 Hz

v = 295.65 * 0.52

v = 153.738 m/s.

The velocity of sound in the string is related to tension, linear density and tension is given below as

v = √(T/u)

153.738 = √T/ 0.02

By squaring both sides

153.738² = T / 0.02

T = 153.738² * 0.02

T = 23,635.372 * 0.02

T= 472.71 N

3 0

3 years ago

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

melisa1 [442]

Answer:

$H_2 = 91.55 km$

Explanation:

Gravity on the surface of planet is given as

$g = \frac{GM}{R^2}$

as we know that

$M = 8.93 \times 10^{22} kg$

$R = 1821 km$

now gravity on the planet is

$g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}$

so we have

$g = 1.8 m/s^2$

now we know that

$H_{max} = \frac{v^2}{2g}$

so we will say

$\frac{H_1}{H_2} = \frac{g_2}{g_1}$

$\frac{500}{H_2} = \frac{9.81}{1.8}$

$H_2 = 91.55 km$

5 0

3 years ago

. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in

kupik [55]

Answer:

Explanation:

The position is $r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}$ .

The velocity is the first time-derivative of r(t).

$v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}$

The acceleration is the first time-derivative of the velocity.

$a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}$

Since a(t) does not have the variable t, it is constant. Hence, at any time,

$a = -16\hat{j}$

Its magnitude is 16 m/s².

4 0

3 years ago

A jet aircraft is traveling at 260 m/s in horizontal flight. The engine takes in air at a rate of 53.3 kg/s and burns fuel at a

IrinaVladis [17]

Answer:

The thrust of the jet engine is 4188.81 N.

Explanation:

Given that,

Speed = 260 m/s

Rate in air= 53.3 kg/s

Rate of fuel = 3.63 kg/s

Relative speed = 317 m/s

We need to calculate the rate of mass change in the rocket

Using formula of rate of mass

$\dfrac{dM}{dt}=\dfrac{dM_{a}}{dt}+\dfrac{dM_{f}}{dt}$

Put the value into the formula

$\dfrac{dM}{dt}=53.3+3.63$

$\dfrac{dM}{dt}=56.93\ kg/s$

We need to calculate the thrust of the jet engine

Using formula of thrust

$T=\dfrac{dM}{dt}u-\dfrac{dM_{a}}{dt}v$

Put the value into the formula

$T=56.93\times317-53.3\times260$

$T=4188.81\ N$

Hence, The thrust of the jet engine is 4188.81 N.

7 0

3 years ago

A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance betw

mario62 [17]

Answer:

$0.2$ mm

Explanation:

As we know that

$Y = \frac{m\lambda * D}{d}$

where

m represents the order of minimum

y represents the distance on the screen of the minimum from central axis

λ is the wavelength of the light

D is the distance between screen-to-slit

d represents the width of the slit

For first minima

$y_1 = \frac{1 * 633 * 10^{-9}*1.5}{d}$

For second minima

$y_2 = \frac{2 * 633 * 10^{-9}*1.5}{d}$

$Y_2 - Y_1 = 0.00475m\\\frac{633 * 10^{-9} * 1.5 }{d} = 0.00475\\d = 0.0002 m\\d = 0.2 mm$

3 0

3 years ago

The _____(height or width) of a sound wave determines its _____(loudness or pitch).

The _(height or width) of a sound wave determines its _(loudness or pitch).