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klio [65]
1 year ago
10

If a key is pressed on a piano, the frequency of the resulting sound will determine the ________, and the amplitude will determi

ne the ________ of the perceived musical note.
Physics
2 answers:
Arada [10]1 year ago
8 0

Answer:

The frequency will determine the pitch

the amplitude will determine the loudness

Explanation:

The frequency of a sound refers to the number of vibrations made by the sound wave produced in a unit of time. This usually affects how high or how low a note is perceived in music. High-frequency sounds have higher pitches, while low-frequency sounds have lower pitches.

The amplitude of a sound wave refers to the height between the wave crests and the equilibrium line in a sound wave. It shows how loud a sound will be. High amplitude sounds are loud while low amplitude sounds are quiet.

MakcuM [25]1 year ago
5 0

Answer:

If a key is pressed on a piano, the frequency of the resulting sound will determine the ___PITCH_____, and the amplitude will determine the _____LOUDNESS___ of the perceived musical note.

Explanation:

The frequency of a vibrating string is primarily based on three factors:

The sounding length (longer is lower, shorter is higher)

The tension on the string (more tension is higher, less is lower)

The mass of the string, normally based on a uniform density per unit length (higher mass is lower, lower mass is higher)

To make a shorter string (such as in an upright piano) sound the same fundamental frequency as a longer string (such as in a 9' grand piano), either the thickness of the string must be increased (which increases the density and the mass) or the tension must be decreased, and usually it's a bit of both.

Thicker strings are often stiffer and that creates more inharmonic partials, and lower tension is associated with other problems, so the best way to make a string sound lower is the make it longer, but it is not practical to make a piano from strings that are all the same density and tension, because the lowest strings would have to be ridiculously long. Nine feet is already a great demand on space for a single musical instrument, and of course those pianos are extremely expensive and difficult to move.

And alsoBesides the pitch of a musical note, perhaps the most noticeable feature in how loud the note is. The loudness of a sound wave is determined from its amplitude. While loudness is only associated with sound waves, all types of waves have an amplitude. Waves on a calm ocean may be less than 1 foot high. Good surfing waves might be 10 feet or more in amplitude. During a storm the amplitude might increase to 40 or 50 feet.

Many things can influence the amplitude.

What is producing the sound?

How far are you from the source of the sound? The farther away the smaller the amplitude.

Intervening material. Sound does not travel through walls as well as air.

Depends on what is detecting the wave sound. Ear vs. microphone.

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From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:
bazaltina [42]

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

T^2\propto a^3\\\\T^2=ka^3

It is mentioned that, an asteroid with an orbital period of 8 years. So,

(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

7 0
1 year ago
A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an
Sonbull [250]

Answer:

(a) k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b)  τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = \frac{M}{3} (a^{2} +b^{2} )   ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k =\frac{1}{2} Iw^{2} --------------------------------2

Putting equation 1 into equation 2, we have;

k = \frac{M}{6} (a^{2} +b^{2} )w^{2}

k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = \frac{Mw}{3} (a^{2} +b^{2} )

But

τ = dτ/dt = \frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}   ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

8 0
2 years ago
A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel
Vikki [24]

acceleration of rocket is given here as

a_y = 2.60* t

now we know that

\frac{dv}{dt} = 2.60t

now integrating both sides

\int dv = \int 2.60t dt

v = 2.60\frac{t^2}{2}

v = 1.30 t^2

here since its given that rocket will accelerate for t = 10 s

so here we have

v = 1.30 * 10^2

v = 130 m/s

so after t = 10 s the speed of rocket will be 130 m/s upwards

5 0
1 year ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Darya [45]

Answer:

y = -8.37 cm

Explanation:

As we know that the equation of SHM is given as

y = A cos(\omega t)

here we know that

\omega = \frac{2\pi}{T}

here we have

T = 0.66 s

now we have

\omega = \frac{2\pi}{0.66}

\omega = 3\pi

now we have

y = (8.8 cm) cos(3\pi t)

now at t = 2.3 s we have

y = (8.8 cm) cos(3\pi \times 2.3)

y = -8.37 cm

6 0
1 year ago
Read 2 more answers
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
1 year ago
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