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2 years ago

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A band director instructs students to play a pitch louder. How will the sound wave change when the band plays the same pitch lou

der?
The speed of the sound wave will increase.
The frequency of the sound wave will increase.
The wavelength of the sound wave will increase.
The amplitude of the sound wave will increase.

2 answers:

Fofino [41]2 years ago

6 0

Explanation :

Loudness is defined as how our ear perceives the sound. It is basically related to the intensity of a sound wave.

When the same pitch is played louder, its amplitude changes. Amplitude is the maximum displacement covered by a wave.

This is the only difference between the loudness to that of shrillness. The amplitude of the loud sound is more as compared to that of shrill sound.

So, the correct option is (d).

seropon [69]2 years ago

3 0

The amplitude will increse

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hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

For light moving from air to water:

$n_1 \sim 1.00$ is the index of refraction of air

$n_2 = 1.33$ is the index of refraction ofwater

In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

For light moving from air to glass,

$n_2 = 1.51$ is the index of refraction of glass

And so

$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

From eq.(1), we see that the angle of refraction $\theta_2$ is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

$n=\frac{c}{v}$

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

The higher the index of refraction, the slower the light
The lower the index of refraction, the faster the light

26)

From Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, $n_1 > n_2$ , so $\frac{n_1}{n_2}>1$ , and since $sin \theta_2$ cannot be larger than 1, there exists a maximum value of the angle of incidence $\theta_c$ (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

For the interface glass-water,

$n_1 \sim 1.51\\n_2 = 1.33$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}$

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0

2 years ago

A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at

Vinvika [58]

Answer:

Explanation:

One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m

These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.

Applying Gauss's theorem

Flux through spherical surface = charge inside sphere / ε₀

= 0 / ε₀

= 0 Ans .

3 0

2 years ago

A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric

Tcecarenko [31]

Answer:

a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

Fe + Fm = m a

a = (Fe + Fm) / m

the electric force is

Fe = q E k ^

Fe = 4.25 10-4 60 k ^

Fe = 2.55 10-2 k ^

the magnetic force is

Fm = q v x B

Fm = 4.25 10⁻⁴ $\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]$

fm = 4.25 10⁻⁴ (-j ^ 30 4)

fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

aₓ = 0

there is no force

Axis y

ay = -5.10 10²/5 10⁻⁵ j ^

ay = -1.02 107 j ^ / s2

z axis

az = 2.55 10⁻² / 5 10⁻⁵ k ^

az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

x = vo t + ½ aₓ t₂2

x = 30 0.02 + 0

x = 0.6m

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

y = I + go t + ½ ay t²

y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

y = 1 - 2.04 10³

y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

z = zo + v₀ t + ½ az t²

z = 0 + 0 + ½ 5.1 10² 0.02²

z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

since there is no acceleration the speed remains constant

vₓ = 30.0 m / s

Axis y

let's use the equation v = v₀ + $a_{y}$ t

$v_{y}$ = 0 + -1.02 10⁷ 0.02

v_{y} = 2.04 10⁵ m / s

z axis

v_{z} = vo + az t

v_{z} = 0 + 5.1 10² 0.02

v_{z} = 1.02 10⁻¹m / s

8 0

2 years ago

If isomerization requires breaking the pi bond, what minimum energy is required for isomerization in j/mol?

aliina [53]

<span>The minimum energy required for isomerization is 267 000 J/mol
</span>

The isomerization of cis-but-2-ene to trans-but-2-ene requires breaking of the π bond.

The bond energy of a C-C σ bond is 347 kJ/mol.

The bond energy of a C=C double bond (σ + π) is 614 kJ/mol.

So the bond energy of a π bond is (614 – 347) kJ/mol = 267 kJ/mol =
267 000 J/mol.

6 0

2 years ago

A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C

Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 $\frac{J}{kg k}$

The heat of vaporization is = 2256 $\frac{KJ}{kg}$

Mass = 0.030 kg

Energy needed to heat water to boiling point = $m c ( T_{2} - T_{1} )$

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy ( $E_{1}$ ) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

$E_{2}$ = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E = $E_{1}$ + $E_{2}$

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

9 0

2 years ago