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steposvetlana [31]
3 years ago
10

which of the following would not be taken into consideration when describing the quality of a sound? a. the number of the overto

nes in the sound b. the source producing the overtones in the sound c. the frequency of the overtones in the sound d. the intensity of the overtones in the sound
Physics
2 answers:
maks197457 [2]3 years ago
8 0
B. the <span>source producing the overtones in the sound</span>
Tpy6a [65]3 years ago
5 0

Answer;

B. the source producing the overtones in the sound

Explanation;

-Sound quality describes the characteristics of the sound which allow the ear to distinguish sounds which have the same pitch and loudness. The quality or timbre of a sound depends on its wave form, which varies with the number of overtones, or harmonics, that are present, their frequencies, and their relative intensities.

-The source producing the overtones in the sound does not affect the timbre or the quality of sound and thus it is not taken into consideration when describing the quality of a sound.

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ra1l [238]

Answer:

yes

Explanation:

5 0
3 years ago
An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move
lions [1.4K]
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
8 0
3 years ago
Read 2 more answers
An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance
marshall27 [118]

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

4 0
3 years ago
. As we increase the quantum number of an electron in a one-dimensional, infinite potential well, what happens to the number of
Natalija [7]

Answer:

It increases.

Explanation:

For the electron to escape the photon needs energy is equal to the difference between initial and its non quantised region energy , then only it will be able to escape finite well.

E ∝ n^2

n= energy state quantum number

so if , n increases maximum point of probability density increases.

8 0
4 years ago
What is the value of the composite constant (Gme,/r2e) to be multiplied by the mass of the object mo, in equation below:
Sedbober [7]

To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.

The equation for the gravitational force between the Earth and the object on the surface of the Earth is

F_g = \frac{Gm_em_o}{r^2_e}

Where,

G = Universal gravitational constant

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m_o= Mass of Object

The equation for the gravitational pulling force on the object due to gravitational acceleration is

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Equation the two expression we have

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This the acceleration due to gravity which is composite constant.

Replacing with our values we have then

g = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5.98*10^{24}kg)}{6378km(\frac{10^3m}{1km})^2}

g = 9.8m/s^2

The value of composite constant is 9.8m/s^2. Here, the composite constant is nothing but the acceleration due to gravity which is constant always.

8 0
3 years ago
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