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xxMikexx [17]
3 years ago
11

How does frequency and pitch affect sound

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
7 0
<span>The word that musicians use for </span>frequency<span> is </span>pitch<span>. The shorter the wavelength, the higher the </span>frequency<span>, and the higher the </span>pitch<span>, of the </span>sound<span>. In other words, short waves </span>sound<span> high; long waves </span>sound<span> low.</span>
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Which language style would be most appropriate for the given situation?
polet [3.4K]

Answer:

informal language

Explanation:

you do not need to be formal! you are not at a business conference. technical is not needed either since you are not discussing the intricacies of your job or some computer language.

5 0
2 years ago
Margaret wants to go for a swim, and decides to jump in using the diving board that measures 3-m long.
otez555 [7]

Answer:

The horizontal component of her velocity is approximately 1.389 m/s

The vertical component of her velocity is approximately 7.878 m/s

Explanation:

The given question parameters are;

The initial velocity with which Margaret leaps, v = 8.0 m/s

The angle to the horizontal with which she jumps, θ = 80° to the horizontal

The horizontal component of her velocity, vₓ = v × cos(θ)

∴ vₓ = 8.0 × cos(80°) ≈ 1.389

The horizontal component of her velocity, vₓ ≈ 1.389 m/s

The vertical component of her velocity, v_y = v × sin(θ)

∴ v_y = 8.0 × sin(80°) ≈ 7.878

The vertical component of her velocity, v_y ≈ 7.878 m/s.

6 0
3 years ago
An electric field of intensity 3.80 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.
lakkis [162]

Answer: 1.55 x 10⁴ Nm²c^-1

Explanation: The electric flux, electric field intensity and area are related by the formulae below.

Φ= EAcosθ,

Where Φ= electric flux (Nm²c^-1)

E =electric field intensity (N/m²)

A = Area (m²)

θ= this is angle between the planar area and the magnetic flux

For our question E=3.80KN/c= 3800 N/c

A= 0.700 x 0.350= 0.245m²

θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).

Hence Φ= 3800 x 0.245 x cos(0)

= 3800 x 0.245 x 1 (value of cos 0° =1)

= 1.55 x 10⁴ Nm²c^-1

Thus the electric field is 1.55 x 10⁴ Nm²c^-1

4 0
3 years ago
A charged particle is located in an electric field where the magnitude of the electric field strength is 2.0x10^3 newtons per co
Alexxandr [17]

Answer:

20.7

Explanation:

:0 because basis  of the daily occured

6 0
2 years ago
Which TWO statements about the substances in the experiment are true?
nataly862011 [7]

Answer:

A and B are correct both are correct

4 0
2 years ago
Read 2 more answers
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