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IgorC [24]
2 years ago
10

I will mark you brainlist. How can you use a tuning fork to tune a piano?

Physics
1 answer:
Phoenix [80]2 years ago
6 0

A tuning fork's job is to establish a single note that everybody can tune to.

Most tuning forks are made to vibrate at 440 Hz, a tone known to musicians as "concert A." To tune a piano, you would start by playing the piano's "A" key while ringing an "A" tuning fork. If the piano is out of tune, you'll hear a distinct warble between the note you're playing and the note played by the tuning fork; the further apart the warbles, the more out-of-tune the piano. By either tightening or loosening the piano's strings, you reduce the warble until it's in line with the tuning fork. Once the "A" key is in tune, you would then adjust all of the instrument's 87 other keys to match. The method is much the same for most other instruments. Whether you're tuning a clarinet or guitar, simply play a concert A and adjust your instrument accordingly

Explanation:

It can be a bit tricky to hold a tuning fork while manipulating an instrument, which is why some musicians decide to clench the base of a ringing tuning fork in their teeth. This has the unique effect of transmitting sound through your bones, allowing your brain to "hear" the tone through your jaw. According to some urban legends, touching your teeth with a vibrating tuning fork is enough to make them explode. It's a myth, obviously, but if you have a cavity or a chipped tooth, you'll quickly find this method to be unbelievably painful.

Luckily, you can also buy tuning forks that come mounted on top of a resonator, a hollow wooden box designed to amplify a tuning fork's vibrations. In 1860, a pair of German inventors even devised a battery-powered tuning fork that musicians didn't need to ring again and again

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What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?
kipiarov [429]

Answer:

\boxed {\boxed {\sf a= -7 \ m/s^2}}

Explanation:

Acceleration is the change in velocity over time.

a= \frac {v_f-v_i}{t}

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s

Substitute the values into the formula.

a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}

Solve the numerator.

a= \frac { -35 \ m/s}{5 \ s}

Divide.

a= -7 \ m/s/s

a= -7 \ m/s^2

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0
2 years ago
Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo
zubka84 [21]

Responder:

A) ω = 565.56 rad / seg

B) f = 90Hz

C) 0.011111s

Explicación:

Dado que:

Velocidad = 5400 rpm (revolución por minuto)

La velocidad angular (ω) = 2πf

Donde f = frecuencia

ω = 5400 rev / minuto

1 minuto = 60 segundos

2πrad = I revolución

Por lo tanto,

ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)

ω = (5400 * 2πrad) / 60 s

ω = 10800πrad / 60 s

ω = 180πrad / seg

ω = 565.56 rad / seg

SI)

Dado que :

ω = 2πf

donde f = frecuencia, ω = velocidad angular en rad / s

f = ω / 2π

f = 565.56 / 2π

f = 90.011669

f = 90 Hz

C) Periodo (T)

Recordar T = 1 / f

Por lo tanto,

T = 1/90

T = 0.0111111s

3 0
3 years ago
A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir
Pachacha [2.7K]

Answer:

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

\vec{F} = q \vec{E}

11\hat{i} = -5 \times 10^{-6} \vec{E}

\vec{E} =-2.2 \times 10^{6}\hat{i}

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

\vec{F} = q \vec{E}

\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Force experienced by the photon in the same field is equal to \vec{F}= -3.52\times 10^{-13}\hat{i}\ N

3 0
2 years ago
(01.06 MC)
Cerrena [4.2K]

Answer:

c

Explanation:

7 0
3 years ago
How long would this acceleration last
Aleksandr-060686 [28]
It would last as long as the applied force continued, or until the accelerating object hit something.
6 0
3 years ago
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