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Taya2010 [7]
2 years ago
13

The most soaring vocal melody is in Johann Sebastian Bach's Mass in B Minor. In one section, the basses, tenors, altos, and sopr

anos carry the melody from a low D to a high
A. In concert pitch, these notes are now assigned frequencies of 146.8Hz and 880.0 Hz . Find the wavelengths of(c) the initial note and
Physics
1 answer:
inysia [295]2 years ago
8 0

The wavelength of the initial note is 2.34m

<h3>What do you mean by the term wavelength?</h3>

The distance between two successive crests or troughs of a wave is known as its wavelength. It is measured in the wave's direction. The distance a wave travels between its crests or troughs is known as its wavelength (which may be an electromagnetic wave, a sound wave, or any other wave). The wave's crest is its highest point, while its dip is its lowest. Because a wavelength represents a distance or length, it is expressed in length units like meters, centimeters, millimeters, nanometers, etc. Light's wavelength changes with color, or it differs for each hue. While violet has the shortest wavelength, red has the longest.

wavelength = v/f

Given:

Initial frequency, f_{initial} = 146.8 Hz

Final frequency, f_{final} = 880\;Hz

Sound Level, \beta = 79\;dB

We know that,

Speed of the sound, v=343\;m/s

Density of air, \rho=1.2\;kg/m^{3}

Wavelength, \lambda = \frac{v}{f}

Wavelength of initial note is,

\lambda = \frac{v}{f_{initial} }

Substitute the known values in the above equation,

\lambda = \frac{343}{146.8}

\lambda = 2.34\;m

To learn more about wavelength, Visit:

brainly.com/question/10750459

#SPJ4

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a 1 m length of string is wrapped around a solid disk, of mass .25 kg and a radius of .30m, mounted on a frictionless axle. the
Nezavi [6.7K]

Answer:

60 rad/s

Explanation:

∑τ = Iα

Fr = Iα

For a solid disc, I = ½ mr².

Fr = ½ mr² α

α = 2F / (mr)

α = 2 (20 N) / (0.25 kg × 0.30 m)

α = 533.33 rad/s²

The arc length is 1 m, so the angle is:

s = rθ

1 m = 0.30 m θ

θ = 3.33 rad

Use constant acceleration equation to find ω.

ω² = ω₀² + 2αΔθ

ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)

ω = 59.6 rad/s

Rounding to one significant figure, the angular velocity is 60 rad/s.

8 0
3 years ago
Which material would result in the least amount of energy transfer?
Dovator [93]

Answer:

The answer is A , aka, a reflector that is bright color and smooth

Explanation:

your welcome

3 0
3 years ago
PLEASE HELP 100 POINTS! Please fill in the scale distance from sun and diversion factor, listing off the numbers works
steposvetlana [31]

Solved your another question same like this with scaling to Cm this time we go with metre(m)

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7 0
2 years ago
Which of the following would most likely be considered nonpoint source pollution
fredd [130]
The best answer is b) increased turbidity from erosion.

Nonpoint source pollution generally happens as a result of many systems interacting, and is not directly attributed to one event or pollutant. Generally, natural environmental systems participate in pollution of this kind, regardless of whether or not human activity was a factor. Examples include water runoff, or erosion. 

The other pollutants listed have a direct cause and direct effect, the animal waste goes directly from the animals to the ground they live on, the car shop directly sumps the oil on the ground, and the oil tank leaks directly into the earth. Erosion causing turbidity is a less direct form of pollution, and is due to the synthesis of several natural phenomena<span />
8 0
4 years ago
A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.
AveGali [126]

Answer:

The magnitude of electric force is  7.2\times10^{-3} N

Explanation:

Coulomb's Law:

The force of attraction or repletion is

  • directly proportional to the products of charges i.e F\propto q_1q_2
  • inversely proportional to the square of distance i.e F\propto \frac{1}{r^2}

\therefore F\propto \frac{q_1q_2}{r^2}

\Rightarrow F=k \frac{q_1q_2}{r^2}    [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5

we know,

cos \theta = \frac{base }{hypotenuse}

\Rightarrow cos \theta = \frac{4 }{r}

Total force F_Q = 2.F_1 cos\theta \hat{i}

                         =2k\frac{Qq_1}{r^2} cos\theta \hat i

                         =2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i

                         =8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i     [ r=5]

                         =7.2\times10^{-3}\hat i   N

The magnitude of electric force is  7.2\times10^{-3} N

                         

3 0
3 years ago
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