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julia-pushkina [17]
2 years ago
5

Los músicos de una sinfónica siempre ""calientan"" sus instrumentos de viento soplándolos antes de un concierto. ¿Para qué sirve

esto?
Physics
1 answer:
Snowcat [4.5K]2 years ago
8 0

El calentamiento del instrumento viento antes de un concierto asegura que todos los instrumentos estén afinados y existan un mejor sonido, debido al cambio de frecuencia del instrumentos por los cambios de velocidad del aire debido al cambio de temperatura

Los instrumentos musicales trabajan por procesos de resonancia, ya sean de cuerda o viento; en los instrumentos de viento la frecuencia que emites esta dada por la relación  

                   fₙ = n \ \frac{v_s}{2L}    n = 1, 2, 3, ...

Donde f es la frecuencia emitida, L la longitud del tubo n es una constante entera y v_s es la velocidad del sonido

La velocidad del sonido en el aire depende de la temperatura del aire, según la relacion          

               vs = vo + 0,6 T  

Donde v₀ es la temperatura del aire a 0ºC, v₀ = 331 m/s y T la temperatura en grados centígrados.

De esta dos expresiones podemos ver que la frecuencia que emite el instrumento de viento depende  de la temperatura del aire, además en los instrumentos con boquilla la frecuencia de resonancia de la boquilla también depende de la temperatura de la boquilla que por ser liviana cambia fácilmente.

En conclusión el calentamiento del instrumento antes de un concierto asegura que todos los instrumentos estén afinados y existan un mejor sonido, debido al cambio de frecuencia del instrumentos por los cambios de velocidad del aire debido al cambio de temperatura

aprende mas acerca velocidad sonido aquí:  

brainly.com/question/22160656

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Answer:

B. Tomatos reflect red light

Explanation:

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4 0
3 years ago
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sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N
yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

m=\frac{W}{g} =\frac{455}{9.81}=46.38kg

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2

for point b we use the equations of motion with constant acceleration to find the velocity

Vf=\sqrt{X(2)(a)+Vo^2}

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s

4 0
3 years ago
What is the momentum of a 5 kg object that has a velocity of 1.2 m/s? 3.8 kg • m/s 4.2 kg • m/s 6.0 kg • m/s 6.2 kg • m/s
Gnesinka [82]

Answer:

Your answer will be 6.0kg•m/s

Explanation:

In the given question all the required details d given. Using these information's a person can easily find the momentum of the object. In the question it is already given that the mass of the object is 5 kg and the velocity at which it is traveling is 1.2 m/s.We know the equation of finding momentum asMomentum = mass * velocity                   = 5 * 1.2                    = 6So the momentum of the object is 6 Newton.

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3 years ago
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Q4: Two fixed charges, 1 c and -3 C are
White raven [17]

Answer:

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5 0
2 years ago
A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0
3 years ago
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