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3 years ago

When two notes are played simultaneously, creating a discordant sound, it is called _____.

Physics

1 answer:

mart [117]3 years ago

5 0

Answer:

The correct answer is option D.

Explanation:

Acoustic : A branch of physics which study the properties of sound.

Consonance: Combination of notes occurring simultaneously due to relationship between their respective frequencies.

Timbre: A characteristic of a musical note which makes it distinct from another wave which also have same pitch and intensity.

Dissonance :When combination of two notes are played simultaneously with lack of harmony in between them.

Hence, the correct answer is option D.

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how do I solve how much time does it take for a satellite moving 15000 miles per hour travel to travel once around the earth 249

iogann1982 [59]

Answer:

The time taken by the satellite to orbit earth at its surface, t = 1.66 hr

Explanation:

Given data,

The velocity of the satellite, v = 15000 miles/hr

The distance of travel, d = 24901 miles,

It is equal to the circumference of earth,

So the time taken by the satellite to orbit earth at its surface,

v = d/t

t = d/v

Substituting the given values in the above equation,

t = 24901 / 15000

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Hence, the time taken by the satellite to orbit earth at its surface, t = 1.66 hr

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3 years ago

A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is

Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d= $\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m$

Wavelength of light, $\lambda=510 nm=510\times 10^{-9} m$

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

$sin\theta=\frac{m\lambda}{d}$

For first bright fringe, =1

$sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305$

$\theta=sin^{-1}(0.305)=17.76^{\circ}$

The distance between two m=1 fringes

$x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m$

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

$sin\theta=1$

$sin\theta=\frac{m\lambda}{d}$

Substitute the values

$1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}$

$m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3$

Maximum number of bright fringes on the scree= $2m+1=2(3)+1=7$

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Explanation:

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